How do I solve this integral assuming assumption in Maple?

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Question:How do I solve this integral assuming assumption in Maple?

Posted: ahmadi 10 Product: Maple 15

August 17 2016

`assuming`([int((((2/3)*t^3-2*t^2+t)/(t+m*((2/3)*t^3-2*t^2+t)))^2*(t+m*((2/3)*t^3-2*t^2+t)), t = 0 .. 1)+int(((-(2/3)*t^3+2*t^2-t-2/3)/(2-t+m*(-(2/3)*t^3+2*t^2-t-2/3)))^2*(2-t+m*(-(2/3)*t^3+2*t^2-t-2/3)), t = 1 .. 2)], [-1 < m and m < 1])

-(1/6)*(3*3^(1/2)*(2-m)^(1/2)*(-m)^(1/2)*ln(3)+3*3^(1/2)*(2-m)^(1/2)*(-m)^(1/2)*ln(1+m)+18*arctanh(m*3^(1/2)/(-m*(2-m))^(1/2))*m-18*arctanh((1/3)*m*3^(1/2)/(-m*(2-m))^(1/2))*m+2*3^(1/2)*m*(2-m)^(1/2)*(-m)^(1/2)-3*ln(-m+3)*3^(1/2)*(2-m)^(1/2)*(-m)^(1/2))*3^(1/2)/(m^3*(2-m)^(1/2)*(-m)^(1/2))

(1)

(2)

`assuming`([int(-ln(t+m*((2/3)*t^3-2*t^2+t))*(t+m*((2/3)*t^3-2*t^2+t)), t = 0 .. 1)+int(-ln(2-t+m*(-(2/3)*t^3+2*t^2-t-2/3))*(2-t+m*(-(2/3)*t^3+2*t^2-t-2/3)), t = 1 .. 2)], [-1 < m and m < 1])

3^(1/2)*(-m*(-2+m)^2*arctanh((1/3)*m*3^(1/2)/(m*(-2+m))^(1/2))+m*(-2+m)^2*arctanh(m*3^(1/2)/(m*(-2+m))^(1/2))+(1/4)*((-1-m^2+(8/3)*m)*ln(-m+3)+(-4*m+m^2+1)*ln(1+m)+(1+m^2-(8/3)*m)*ln(3)+(14/3)*m-m^2)*3^(1/2)*(m*(-2+m))^(1/2))/((m*(-2+m))^(1/2)*m)

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Question: How do I solve this integral assuming assumption in Maple?

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