3.84.100 \(\int \frac {(2 e^2 x+2 x^2) \log (3)+(-2 e^2-2 x) \log ^2(3)+(-4 e^2 x^3+4 e^2 x^2 \log (3)) \log (\frac {12 e^2+12 x}{x})+(2 e^2 x^3+2 x^4) \log ^2(\frac {12 e^2+12 x}{x})-4 e^2 x^4 \log ^3(\frac {12 e^2+12 x}{x})+(2 e^2 x^4+2 x^5) \log ^4(\frac {12 e^2+12 x}{x})}{e^2 x^3+x^4} \, dx\)

Optimal. Leaf size=29 \[ \frac {\left (x-\log (3)+x^2 \log ^2\left (\frac {12 \left (e^2+x\right )}{x}\right )\right )^2}{x^2} \]

________________________________________________________________________________________

Rubi [F]  time = 1.86, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {\left (2 e^2 x+2 x^2\right ) \log (3)+\left (-2 e^2-2 x\right ) \log ^2(3)+\left (-4 e^2 x^3+4 e^2 x^2 \log (3)\right ) \log \left (\frac {12 e^2+12 x}{x}\right )+\left (2 e^2 x^3+2 x^4\right ) \log ^2\left (\frac {12 e^2+12 x}{x}\right )-4 e^2 x^4 \log ^3\left (\frac {12 e^2+12 x}{x}\right )+\left (2 e^2 x^4+2 x^5\right ) \log ^4\left (\frac {12 e^2+12 x}{x}\right )}{e^2 x^3+x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((2*E^2*x + 2*x^2)*Log[3] + (-2*E^2 - 2*x)*Log[3]^2 + (-4*E^2*x^3 + 4*E^2*x^2*Log[3])*Log[(12*E^2 + 12*x)/
x] + (2*E^2*x^3 + 2*x^4)*Log[(12*E^2 + 12*x)/x]^2 - 4*E^2*x^4*Log[(12*E^2 + 12*x)/x]^3 + (2*E^2*x^4 + 2*x^5)*L
og[(12*E^2 + 12*x)/x]^4)/(E^2*x^3 + x^4),x]

[Out]

(x - Log[3])^2/x^2 + 2*(E^2 + x)*Log[12 + (12*E^2)/x]^2 - 4*E^2*(E^2 + x)*Log[12 + (12*E^2)/x]^3 + 4*E^2*Log[1
2]*Log[x] + 4*Log[3]*(2 + Log[12])*Log[x] - 2*Log[3]*Log[x]^2 - 2*(E^2 + Log[3])*Log[12*(E^2 + x)]^2 + 4*E^2*(
1 + E^2/x)*x*Log[(12*(E^2 + x))/x]^3 + 4*E^4*Log[-(E^2/x)]*Log[(12*(E^2 + x))/x]^3 - E^4*Log[(12*(E^2 + x))/x]
^4 + x^2*Log[(12*(E^2 + x))/x]^4 + 4*Log[3]*Log[x]*(Log[x] + Log[(12*(E^2 + x))/x] - Log[12*E^2 + 12*x]) - 4*(
E^2 + Log[3])*Log[E^2 + x]*(Log[x] + Log[(12*(E^2 + x))/x] - Log[12*E^2 + 12*x]) + 4*(E^2 + Log[3])*Log[x]*Log
[1 + x/E^2] + 4*E^2*PolyLog[2, -(E^2/x)] - 4*Log[3]*PolyLog[2, -(x/E^2)] + 4*(E^2 + Log[3])*PolyLog[2, -(x/E^2
)] + 12*E^4*Log[(12*(E^2 + x))/x]^2*PolyLog[2, (E^2 + x)/x] - 24*E^4*Log[(12*(E^2 + x))/x]*PolyLog[3, (E^2 + x
)/x] + 24*E^4*PolyLog[4, (E^2 + x)/x] + 4*E^4*Defer[Int][Log[12 + (12*E^2)/x]^3/(E^2 + x), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (2 e^2 x+2 x^2\right ) \log (3)+\left (-2 e^2-2 x\right ) \log ^2(3)+\left (-4 e^2 x^3+4 e^2 x^2 \log (3)\right ) \log \left (\frac {12 e^2+12 x}{x}\right )+\left (2 e^2 x^3+2 x^4\right ) \log ^2\left (\frac {12 e^2+12 x}{x}\right )-4 e^2 x^4 \log ^3\left (\frac {12 e^2+12 x}{x}\right )+\left (2 e^2 x^4+2 x^5\right ) \log ^4\left (\frac {12 e^2+12 x}{x}\right )}{x^3 \left (e^2+x\right )} \, dx\\ &=\int \frac {2 \left (x-\log (3)+x^2 \log ^2\left (\frac {12 \left (e^2+x\right )}{x}\right )\right ) \left (\left (e^2+x\right ) \log (3)-2 e^2 x^2 \log \left (\frac {12 \left (e^2+x\right )}{x}\right )+x^2 \left (e^2+x\right ) \log ^2\left (\frac {12 \left (e^2+x\right )}{x}\right )\right )}{x^3 \left (e^2+x\right )} \, dx\\ &=2 \int \frac {\left (x-\log (3)+x^2 \log ^2\left (\frac {12 \left (e^2+x\right )}{x}\right )\right ) \left (\left (e^2+x\right ) \log (3)-2 e^2 x^2 \log \left (\frac {12 \left (e^2+x\right )}{x}\right )+x^2 \left (e^2+x\right ) \log ^2\left (\frac {12 \left (e^2+x\right )}{x}\right )\right )}{x^3 \left (e^2+x\right )} \, dx\\ &=2 \int \left (\frac {(x-\log (3)) \log (3)}{x^3}+\frac {2 e^2 (-x+\log (3)) \log \left (12+\frac {12 e^2}{x}\right )}{x \left (e^2+x\right )}+\log ^2\left (12+\frac {12 e^2}{x}\right )+\frac {2 e^2 x \log ^3\left (12+\frac {12 e^2}{x}\right )}{-e^2-x}+x \log ^4\left (12+\frac {12 e^2}{x}\right )\right ) \, dx\\ &=2 \int \log ^2\left (12+\frac {12 e^2}{x}\right ) \, dx+2 \int x \log ^4\left (12+\frac {12 e^2}{x}\right ) \, dx+\left (4 e^2\right ) \int \frac {(-x+\log (3)) \log \left (12+\frac {12 e^2}{x}\right )}{x \left (e^2+x\right )} \, dx+\left (4 e^2\right ) \int \frac {x \log ^3\left (12+\frac {12 e^2}{x}\right )}{-e^2-x} \, dx+(2 \log (3)) \int \frac {x-\log (3)}{x^3} \, dx\\ &=\frac {(x-\log (3))^2}{x^2}+2 \left (e^2+x\right ) \log ^2\left (12+\frac {12 e^2}{x}\right )-2 \operatorname {Subst}\left (\int \frac {\log ^4\left (12+12 e^2 x\right )}{x^3} \, dx,x,\frac {1}{x}\right )+\left (4 e^2\right ) \int \frac {\log \left (12+\frac {12 e^2}{x}\right )}{x} \, dx+\left (4 e^2\right ) \int \left (-\log ^3\left (12+\frac {12 e^2}{x}\right )+\frac {e^2 \log ^3\left (12+\frac {12 e^2}{x}\right )}{e^2+x}\right ) \, dx+\left (4 e^2\right ) \int \frac {(-x+\log (3)) \log \left (\frac {12 e^2+12 x}{x}\right )}{x \left (e^2+x\right )} \, dx\\ &=\frac {(x-\log (3))^2}{x^2}+2 \left (e^2+x\right ) \log ^2\left (12+\frac {12 e^2}{x}\right )+x^2 \log ^4\left (\frac {12 \left (e^2+x\right )}{x}\right )-\left (4 e^2\right ) \int \log ^3\left (12+\frac {12 e^2}{x}\right ) \, dx-\left (4 e^2\right ) \int \frac {(-x+\log (3)) \log (x)}{x \left (e^2+x\right )} \, dx+\left (4 e^2\right ) \int \frac {(-x+\log (3)) \log \left (12 e^2+12 x\right )}{x \left (e^2+x\right )} \, dx-\left (4 e^2\right ) \operatorname {Subst}\left (\int \frac {\log \left (12+12 e^2 x\right )}{x} \, dx,x,\frac {1}{x}\right )-\left (48 e^2\right ) \operatorname {Subst}\left (\int \frac {\log ^3\left (12+12 e^2 x\right )}{x^2 \left (12+12 e^2 x\right )} \, dx,x,\frac {1}{x}\right )+\left (4 e^4\right ) \int \frac {\log ^3\left (12+\frac {12 e^2}{x}\right )}{e^2+x} \, dx-\left (4 e^2 \left (-\log (x)+\log \left (12 e^2+12 x\right )-\log \left (\frac {12 e^2+12 x}{x}\right )\right )\right ) \int \frac {-x+\log (3)}{x \left (e^2+x\right )} \, dx\\ &=\frac {(x-\log (3))^2}{x^2}+2 \left (e^2+x\right ) \log ^2\left (12+\frac {12 e^2}{x}\right )-4 e^2 \left (e^2+x\right ) \log ^3\left (12+\frac {12 e^2}{x}\right )+4 e^2 \log (12) \log (x)+x^2 \log ^4\left (\frac {12 \left (e^2+x\right )}{x}\right )-4 \operatorname {Subst}\left (\int \frac {\log ^3(x)}{x \left (-\frac {1}{e^2}+\frac {x}{12 e^2}\right )^2} \, dx,x,12+\frac {12 e^2}{x}\right )-\left (4 e^2\right ) \int \left (\frac {\left (-e^2-\log (3)\right ) \log (x)}{e^2 \left (e^2+x\right )}+\frac {\log (3) \log (x)}{e^2 x}\right ) \, dx+\left (4 e^2\right ) \int \left (\frac {\left (-e^2-\log (3)\right ) \log \left (12 e^2+12 x\right )}{e^2 \left (e^2+x\right )}+\frac {\log (3) \log \left (12 e^2+12 x\right )}{e^2 x}\right ) \, dx-\left (4 e^2\right ) \operatorname {Subst}\left (\int \frac {\log \left (1+e^2 x\right )}{x} \, dx,x,\frac {1}{x}\right )+\left (4 e^4\right ) \int \frac {\log ^3\left (12+\frac {12 e^2}{x}\right )}{e^2+x} \, dx-\left (12 e^4\right ) \int \frac {\log ^2\left (12+\frac {12 e^2}{x}\right )}{x} \, dx-\left (4 e^2 \left (-\log (x)+\log \left (12 e^2+12 x\right )-\log \left (\frac {12 e^2+12 x}{x}\right )\right )\right ) \int \left (\frac {-e^2-\log (3)}{e^2 \left (e^2+x\right )}+\frac {\log (3)}{e^2 x}\right ) \, dx\\ &=\frac {(x-\log (3))^2}{x^2}+2 \left (e^2+x\right ) \log ^2\left (12+\frac {12 e^2}{x}\right )-4 e^2 \left (e^2+x\right ) \log ^3\left (12+\frac {12 e^2}{x}\right )+4 e^2 \log (12) \log (x)+x^2 \log ^4\left (\frac {12 \left (e^2+x\right )}{x}\right )+4 \log (3) \log (x) \left (\log (x)+\log \left (\frac {12 \left (e^2+x\right )}{x}\right )-\log \left (12 e^2+12 x\right )\right )-4 \left (e^2+\log (3)\right ) \log \left (e^2+x\right ) \left (\log (x)+\log \left (\frac {12 \left (e^2+x\right )}{x}\right )-\log \left (12 e^2+12 x\right )\right )+4 e^2 \text {Li}_2\left (-\frac {e^2}{x}\right )-\frac {1}{3} \operatorname {Subst}\left (\int \frac {\log ^3(x)}{\left (-\frac {1}{e^2}+\frac {x}{12 e^2}\right )^2} \, dx,x,12+\frac {12 e^2}{x}\right )+\left (4 e^2\right ) \operatorname {Subst}\left (\int \frac {\log ^3(x)}{x \left (-\frac {1}{e^2}+\frac {x}{12 e^2}\right )} \, dx,x,12+\frac {12 e^2}{x}\right )+\left (4 e^4\right ) \int \frac {\log ^3\left (12+\frac {12 e^2}{x}\right )}{e^2+x} \, dx+\left (12 e^4\right ) \operatorname {Subst}\left (\int \frac {\log ^2\left (12+12 e^2 x\right )}{x} \, dx,x,\frac {1}{x}\right )-(4 \log (3)) \int \frac {\log (x)}{x} \, dx+(4 \log (3)) \int \frac {\log \left (12 e^2+12 x\right )}{x} \, dx+\left (4 \left (e^2+\log (3)\right )\right ) \int \frac {\log (x)}{e^2+x} \, dx-\left (4 \left (e^2+\log (3)\right )\right ) \int \frac {\log \left (12 e^2+12 x\right )}{e^2+x} \, dx\\ &=\frac {(x-\log (3))^2}{x^2}+2 \left (e^2+x\right ) \log ^2\left (12+\frac {12 e^2}{x}\right )-4 e^2 \left (e^2+x\right ) \log ^3\left (12+\frac {12 e^2}{x}\right )+4 e^2 \log (12) \log (x)+4 \log (3) (2+\log (12)) \log (x)-2 \log (3) \log ^2(x)+12 e^4 \log \left (-\frac {e^2}{x}\right ) \log ^2\left (\frac {12 \left (e^2+x\right )}{x}\right )+4 e^2 \left (1+\frac {e^2}{x}\right ) x \log ^3\left (\frac {12 \left (e^2+x\right )}{x}\right )+x^2 \log ^4\left (\frac {12 \left (e^2+x\right )}{x}\right )+4 \log (3) \log (x) \left (\log (x)+\log \left (\frac {12 \left (e^2+x\right )}{x}\right )-\log \left (12 e^2+12 x\right )\right )-4 \left (e^2+\log (3)\right ) \log \left (e^2+x\right ) \left (\log (x)+\log \left (\frac {12 \left (e^2+x\right )}{x}\right )-\log \left (12 e^2+12 x\right )\right )+4 \left (e^2+\log (3)\right ) \log (x) \log \left (1+\frac {x}{e^2}\right )+4 e^2 \text {Li}_2\left (-\frac {e^2}{x}\right )+\frac {1}{3} e^2 \operatorname {Subst}\left (\int \frac {\log ^3(x)}{-\frac {1}{e^2}+\frac {x}{12 e^2}} \, dx,x,12+\frac {12 e^2}{x}\right )-e^2 \operatorname {Subst}\left (\int \frac {\log ^2(x)}{-\frac {1}{e^2}+\frac {x}{12 e^2}} \, dx,x,12+\frac {12 e^2}{x}\right )+\left (4 e^4\right ) \int \frac {\log ^3\left (12+\frac {12 e^2}{x}\right )}{e^2+x} \, dx-\left (4 e^4\right ) \operatorname {Subst}\left (\int \frac {\log ^3(x)}{x} \, dx,x,12+\frac {12 e^2}{x}\right )-\left (288 e^6\right ) \operatorname {Subst}\left (\int \frac {\log \left (-e^2 x\right ) \log \left (12+12 e^2 x\right )}{12+12 e^2 x} \, dx,x,\frac {1}{x}\right )+(4 \log (3)) \int \frac {\log \left (1+\frac {x}{e^2}\right )}{x} \, dx-\frac {1}{3} \left (e^2+\log (3)\right ) \operatorname {Subst}\left (\int \frac {12 \log (x)}{x} \, dx,x,12 e^2+12 x\right )-\left (4 \left (e^2+\log (3)\right )\right ) \int \frac {\log \left (1+\frac {x}{e^2}\right )}{x} \, dx\\ &=\frac {(x-\log (3))^2}{x^2}+2 \left (e^2+x\right ) \log ^2\left (12+\frac {12 e^2}{x}\right )-4 e^2 \left (e^2+x\right ) \log ^3\left (12+\frac {12 e^2}{x}\right )+4 e^2 \log (12) \log (x)+4 \log (3) (2+\log (12)) \log (x)-2 \log (3) \log ^2(x)+4 e^2 \left (1+\frac {e^2}{x}\right ) x \log ^3\left (\frac {12 \left (e^2+x\right )}{x}\right )+4 e^4 \log \left (-\frac {e^2}{x}\right ) \log ^3\left (\frac {12 \left (e^2+x\right )}{x}\right )+x^2 \log ^4\left (\frac {12 \left (e^2+x\right )}{x}\right )+4 \log (3) \log (x) \left (\log (x)+\log \left (\frac {12 \left (e^2+x\right )}{x}\right )-\log \left (12 e^2+12 x\right )\right )-4 \left (e^2+\log (3)\right ) \log \left (e^2+x\right ) \left (\log (x)+\log \left (\frac {12 \left (e^2+x\right )}{x}\right )-\log \left (12 e^2+12 x\right )\right )+4 \left (e^2+\log (3)\right ) \log (x) \log \left (1+\frac {x}{e^2}\right )+4 e^2 \text {Li}_2\left (-\frac {e^2}{x}\right )-4 \log (3) \text {Li}_2\left (-\frac {x}{e^2}\right )+4 \left (e^2+\log (3)\right ) \text {Li}_2\left (-\frac {x}{e^2}\right )+\left (4 e^4\right ) \int \frac {\log ^3\left (12+\frac {12 e^2}{x}\right )}{e^2+x} \, dx-\left (4 e^4\right ) \operatorname {Subst}\left (\int x^3 \, dx,x,\log \left (\frac {12 \left (e^2+x\right )}{x}\right )\right )-\left (12 e^4\right ) \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {x}{12}\right ) \log ^2(x)}{x} \, dx,x,12+\frac {12 e^2}{x}\right )+\left (24 e^4\right ) \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {x}{12}\right ) \log (x)}{x} \, dx,x,12+\frac {12 e^2}{x}\right )-\left (24 e^4\right ) \operatorname {Subst}\left (\int \frac {\log (x) \log \left (-e^2 \left (-\frac {1}{e^2}+\frac {x}{12 e^2}\right )\right )}{x} \, dx,x,12+\frac {12 e^2}{x}\right )-\left (4 \left (e^2+\log (3)\right )\right ) \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,12 e^2+12 x\right )\\ &=\frac {(x-\log (3))^2}{x^2}+2 \left (e^2+x\right ) \log ^2\left (12+\frac {12 e^2}{x}\right )-4 e^2 \left (e^2+x\right ) \log ^3\left (12+\frac {12 e^2}{x}\right )+4 e^2 \log (12) \log (x)+4 \log (3) (2+\log (12)) \log (x)-2 \log (3) \log ^2(x)-2 \left (e^2+\log (3)\right ) \log ^2\left (12 \left (e^2+x\right )\right )+4 e^2 \left (1+\frac {e^2}{x}\right ) x \log ^3\left (\frac {12 \left (e^2+x\right )}{x}\right )+4 e^4 \log \left (-\frac {e^2}{x}\right ) \log ^3\left (\frac {12 \left (e^2+x\right )}{x}\right )-e^4 \log ^4\left (\frac {12 \left (e^2+x\right )}{x}\right )+x^2 \log ^4\left (\frac {12 \left (e^2+x\right )}{x}\right )+4 \log (3) \log (x) \left (\log (x)+\log \left (\frac {12 \left (e^2+x\right )}{x}\right )-\log \left (12 e^2+12 x\right )\right )-4 \left (e^2+\log (3)\right ) \log \left (e^2+x\right ) \left (\log (x)+\log \left (\frac {12 \left (e^2+x\right )}{x}\right )-\log \left (12 e^2+12 x\right )\right )+4 \left (e^2+\log (3)\right ) \log (x) \log \left (1+\frac {x}{e^2}\right )+4 e^2 \text {Li}_2\left (-\frac {e^2}{x}\right )-4 \log (3) \text {Li}_2\left (-\frac {x}{e^2}\right )+4 \left (e^2+\log (3)\right ) \text {Li}_2\left (-\frac {x}{e^2}\right )+12 e^4 \log ^2\left (\frac {12 \left (e^2+x\right )}{x}\right ) \text {Li}_2\left (\frac {e^2+x}{x}\right )+\left (4 e^4\right ) \int \frac {\log ^3\left (12+\frac {12 e^2}{x}\right )}{e^2+x} \, dx-\left (24 e^4\right ) \operatorname {Subst}\left (\int \frac {\log (x) \text {Li}_2\left (\frac {x}{12}\right )}{x} \, dx,x,12+\frac {12 e^2}{x}\right )\\ &=\frac {(x-\log (3))^2}{x^2}+2 \left (e^2+x\right ) \log ^2\left (12+\frac {12 e^2}{x}\right )-4 e^2 \left (e^2+x\right ) \log ^3\left (12+\frac {12 e^2}{x}\right )+4 e^2 \log (12) \log (x)+4 \log (3) (2+\log (12)) \log (x)-2 \log (3) \log ^2(x)-2 \left (e^2+\log (3)\right ) \log ^2\left (12 \left (e^2+x\right )\right )+4 e^2 \left (1+\frac {e^2}{x}\right ) x \log ^3\left (\frac {12 \left (e^2+x\right )}{x}\right )+4 e^4 \log \left (-\frac {e^2}{x}\right ) \log ^3\left (\frac {12 \left (e^2+x\right )}{x}\right )-e^4 \log ^4\left (\frac {12 \left (e^2+x\right )}{x}\right )+x^2 \log ^4\left (\frac {12 \left (e^2+x\right )}{x}\right )+4 \log (3) \log (x) \left (\log (x)+\log \left (\frac {12 \left (e^2+x\right )}{x}\right )-\log \left (12 e^2+12 x\right )\right )-4 \left (e^2+\log (3)\right ) \log \left (e^2+x\right ) \left (\log (x)+\log \left (\frac {12 \left (e^2+x\right )}{x}\right )-\log \left (12 e^2+12 x\right )\right )+4 \left (e^2+\log (3)\right ) \log (x) \log \left (1+\frac {x}{e^2}\right )+4 e^2 \text {Li}_2\left (-\frac {e^2}{x}\right )-4 \log (3) \text {Li}_2\left (-\frac {x}{e^2}\right )+4 \left (e^2+\log (3)\right ) \text {Li}_2\left (-\frac {x}{e^2}\right )+12 e^4 \log ^2\left (\frac {12 \left (e^2+x\right )}{x}\right ) \text {Li}_2\left (\frac {e^2+x}{x}\right )-24 e^4 \log \left (\frac {12 \left (e^2+x\right )}{x}\right ) \text {Li}_3\left (\frac {e^2+x}{x}\right )+\left (4 e^4\right ) \int \frac {\log ^3\left (12+\frac {12 e^2}{x}\right )}{e^2+x} \, dx+\left (24 e^4\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (\frac {x}{12}\right )}{x} \, dx,x,12+\frac {12 e^2}{x}\right )\\ &=\frac {(x-\log (3))^2}{x^2}+2 \left (e^2+x\right ) \log ^2\left (12+\frac {12 e^2}{x}\right )-4 e^2 \left (e^2+x\right ) \log ^3\left (12+\frac {12 e^2}{x}\right )+4 e^2 \log (12) \log (x)+4 \log (3) (2+\log (12)) \log (x)-2 \log (3) \log ^2(x)-2 \left (e^2+\log (3)\right ) \log ^2\left (12 \left (e^2+x\right )\right )+4 e^2 \left (1+\frac {e^2}{x}\right ) x \log ^3\left (\frac {12 \left (e^2+x\right )}{x}\right )+4 e^4 \log \left (-\frac {e^2}{x}\right ) \log ^3\left (\frac {12 \left (e^2+x\right )}{x}\right )-e^4 \log ^4\left (\frac {12 \left (e^2+x\right )}{x}\right )+x^2 \log ^4\left (\frac {12 \left (e^2+x\right )}{x}\right )+4 \log (3) \log (x) \left (\log (x)+\log \left (\frac {12 \left (e^2+x\right )}{x}\right )-\log \left (12 e^2+12 x\right )\right )-4 \left (e^2+\log (3)\right ) \log \left (e^2+x\right ) \left (\log (x)+\log \left (\frac {12 \left (e^2+x\right )}{x}\right )-\log \left (12 e^2+12 x\right )\right )+4 \left (e^2+\log (3)\right ) \log (x) \log \left (1+\frac {x}{e^2}\right )+4 e^2 \text {Li}_2\left (-\frac {e^2}{x}\right )-4 \log (3) \text {Li}_2\left (-\frac {x}{e^2}\right )+4 \left (e^2+\log (3)\right ) \text {Li}_2\left (-\frac {x}{e^2}\right )+12 e^4 \log ^2\left (\frac {12 \left (e^2+x\right )}{x}\right ) \text {Li}_2\left (\frac {e^2+x}{x}\right )-24 e^4 \log \left (\frac {12 \left (e^2+x\right )}{x}\right ) \text {Li}_3\left (\frac {e^2+x}{x}\right )+24 e^4 \text {Li}_4\left (\frac {e^2+x}{x}\right )+\left (4 e^4\right ) \int \frac {\log ^3\left (12+\frac {12 e^2}{x}\right )}{e^2+x} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.30, size = 55, normalized size = 1.90 \begin {gather*} \frac {\log ^2(3)-x \log (9)+2 x^2 (x-\log (3)) \log ^2\left (\frac {12 \left (e^2+x\right )}{x}\right )+x^4 \log ^4\left (\frac {12 \left (e^2+x\right )}{x}\right )}{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((2*E^2*x + 2*x^2)*Log[3] + (-2*E^2 - 2*x)*Log[3]^2 + (-4*E^2*x^3 + 4*E^2*x^2*Log[3])*Log[(12*E^2 +
12*x)/x] + (2*E^2*x^3 + 2*x^4)*Log[(12*E^2 + 12*x)/x]^2 - 4*E^2*x^4*Log[(12*E^2 + 12*x)/x]^3 + (2*E^2*x^4 + 2*
x^5)*Log[(12*E^2 + 12*x)/x]^4)/(E^2*x^3 + x^4),x]

[Out]

(Log[3]^2 - x*Log[9] + 2*x^2*(x - Log[3])*Log[(12*(E^2 + x))/x]^2 + x^4*Log[(12*(E^2 + x))/x]^4)/x^2

________________________________________________________________________________________

fricas [A]  time = 0.80, size = 55, normalized size = 1.90 \begin {gather*} \frac {x^{4} \log \left (\frac {12 \, {\left (x + e^{2}\right )}}{x}\right )^{4} + 2 \, {\left (x^{3} - x^{2} \log \relax (3)\right )} \log \left (\frac {12 \, {\left (x + e^{2}\right )}}{x}\right )^{2} - 2 \, x \log \relax (3) + \log \relax (3)^{2}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^4*exp(2)+2*x^5)*log((12*exp(2)+12*x)/x)^4-4*x^4*exp(2)*log((12*exp(2)+12*x)/x)^3+(2*x^3*exp(2)
+2*x^4)*log((12*exp(2)+12*x)/x)^2+(4*x^2*exp(2)*log(3)-4*x^3*exp(2))*log((12*exp(2)+12*x)/x)+(-2*exp(2)-2*x)*l
og(3)^2+(2*exp(2)*x+2*x^2)*log(3))/(x^3*exp(2)+x^4),x, algorithm="fricas")

[Out]

(x^4*log(12*(x + e^2)/x)^4 + 2*(x^3 - x^2*log(3))*log(12*(x + e^2)/x)^2 - 2*x*log(3) + log(3)^2)/x^2

________________________________________________________________________________________

giac [B]  time = 0.36, size = 256, normalized size = 8.83 \begin {gather*} \frac {{\left (e^{8} \log \left (\frac {12 \, {\left (x + e^{2}\right )}}{x}\right )^{4} - \frac {2 \, {\left (x + e^{2}\right )}^{2} e^{4} \log \relax (3) \log \left (\frac {12 \, {\left (x + e^{2}\right )}}{x}\right )^{2}}{x^{2}} + \frac {4 \, {\left (x + e^{2}\right )} e^{4} \log \relax (3) \log \left (\frac {12 \, {\left (x + e^{2}\right )}}{x}\right )^{2}}{x} - 2 \, e^{4} \log \relax (3) \log \left (\frac {12 \, {\left (x + e^{2}\right )}}{x}\right )^{2} + \frac {2 \, {\left (x + e^{2}\right )} e^{6} \log \left (\frac {12 \, {\left (x + e^{2}\right )}}{x}\right )^{2}}{x} - 2 \, e^{6} \log \left (\frac {12 \, {\left (x + e^{2}\right )}}{x}\right )^{2} - \frac {2 \, {\left (x + e^{2}\right )}^{3} e^{2} \log \relax (3)}{x^{3}} + \frac {4 \, {\left (x + e^{2}\right )}^{2} e^{2} \log \relax (3)}{x^{2}} - \frac {2 \, {\left (x + e^{2}\right )} e^{2} \log \relax (3)}{x} + \frac {{\left (x + e^{2}\right )}^{4} \log \relax (3)^{2}}{x^{4}} - \frac {4 \, {\left (x + e^{2}\right )}^{3} \log \relax (3)^{2}}{x^{3}} + \frac {5 \, {\left (x + e^{2}\right )}^{2} \log \relax (3)^{2}}{x^{2}} - \frac {2 \, {\left (x + e^{2}\right )} \log \relax (3)^{2}}{x}\right )} e^{\left (-2\right )}}{\frac {{\left (x + e^{2}\right )}^{2} e^{2}}{x^{2}} - \frac {2 \, {\left (x + e^{2}\right )} e^{2}}{x} + e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^4*exp(2)+2*x^5)*log((12*exp(2)+12*x)/x)^4-4*x^4*exp(2)*log((12*exp(2)+12*x)/x)^3+(2*x^3*exp(2)
+2*x^4)*log((12*exp(2)+12*x)/x)^2+(4*x^2*exp(2)*log(3)-4*x^3*exp(2))*log((12*exp(2)+12*x)/x)+(-2*exp(2)-2*x)*l
og(3)^2+(2*exp(2)*x+2*x^2)*log(3))/(x^3*exp(2)+x^4),x, algorithm="giac")

[Out]

(e^8*log(12*(x + e^2)/x)^4 - 2*(x + e^2)^2*e^4*log(3)*log(12*(x + e^2)/x)^2/x^2 + 4*(x + e^2)*e^4*log(3)*log(1
2*(x + e^2)/x)^2/x - 2*e^4*log(3)*log(12*(x + e^2)/x)^2 + 2*(x + e^2)*e^6*log(12*(x + e^2)/x)^2/x - 2*e^6*log(
12*(x + e^2)/x)^2 - 2*(x + e^2)^3*e^2*log(3)/x^3 + 4*(x + e^2)^2*e^2*log(3)/x^2 - 2*(x + e^2)*e^2*log(3)/x + (
x + e^2)^4*log(3)^2/x^4 - 4*(x + e^2)^3*log(3)^2/x^3 + 5*(x + e^2)^2*log(3)^2/x^2 - 2*(x + e^2)*log(3)^2/x)*e^
(-2)/((x + e^2)^2*e^2/x^2 - 2*(x + e^2)*e^2/x + e^2)

________________________________________________________________________________________

maple [F]  time = 0.78, size = 0, normalized size = 0.00 \[\int \frac {\left (2 x^{4} {\mathrm e}^{2}+2 x^{5}\right ) \ln \left (\frac {12 \,{\mathrm e}^{2}+12 x}{x}\right )^{4}-4 x^{4} {\mathrm e}^{2} \ln \left (\frac {12 \,{\mathrm e}^{2}+12 x}{x}\right )^{3}+\left (2 x^{3} {\mathrm e}^{2}+2 x^{4}\right ) \ln \left (\frac {12 \,{\mathrm e}^{2}+12 x}{x}\right )^{2}+\left (4 x^{2} {\mathrm e}^{2} \ln \relax (3)-4 x^{3} {\mathrm e}^{2}\right ) \ln \left (\frac {12 \,{\mathrm e}^{2}+12 x}{x}\right )+\left (-2 \,{\mathrm e}^{2}-2 x \right ) \ln \relax (3)^{2}+\left (2 \,{\mathrm e}^{2} x +2 x^{2}\right ) \ln \relax (3)}{x^{3} {\mathrm e}^{2}+x^{4}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^4*exp(2)+2*x^5)*ln((12*exp(2)+12*x)/x)^4-4*x^4*exp(2)*ln((12*exp(2)+12*x)/x)^3+(2*x^3*exp(2)+2*x^4)*
ln((12*exp(2)+12*x)/x)^2+(4*x^2*exp(2)*ln(3)-4*x^3*exp(2))*ln((12*exp(2)+12*x)/x)+(-2*exp(2)-2*x)*ln(3)^2+(2*e
xp(2)*x+2*x^2)*ln(3))/(x^3*exp(2)+x^4),x)

[Out]

int(((2*x^4*exp(2)+2*x^5)*ln((12*exp(2)+12*x)/x)^4-4*x^4*exp(2)*ln((12*exp(2)+12*x)/x)^3+(2*x^3*exp(2)+2*x^4)*
ln((12*exp(2)+12*x)/x)^2+(4*x^2*exp(2)*ln(3)-4*x^3*exp(2))*ln((12*exp(2)+12*x)/x)+(-2*exp(2)-2*x)*ln(3)^2+(2*e
xp(2)*x+2*x^2)*ln(3))/(x^3*exp(2)+x^4),x)

________________________________________________________________________________________

maxima [B]  time = 0.53, size = 528, normalized size = 18.21 \begin {gather*} x^{2} \log \left (x + e^{2}\right )^{4} - 4 \, x^{2} {\left (\log \relax (3) + 2 \, \log \relax (2)\right )} \log \relax (x)^{3} + x^{2} \log \relax (x)^{4} + {\left (2 \, e^{\left (-6\right )} \log \left (x + e^{2}\right ) - 2 \, e^{\left (-6\right )} \log \relax (x) - \frac {{\left (2 \, x - e^{2}\right )} e^{\left (-4\right )}}{x^{2}}\right )} e^{2} \log \relax (3)^{2} + 4 \, {\left (x^{2} {\left (\log \relax (3) + 2 \, \log \relax (2)\right )} - x^{2} \log \relax (x)\right )} \log \left (x + e^{2}\right )^{3} - 4 \, {\left (e^{\left (-2\right )} \log \left (x + e^{2}\right ) - e^{\left (-2\right )} \log \relax (x)\right )} e^{2} \log \relax (3) \log \left (\frac {12 \, e^{2}}{x} + 12\right ) + {\left (\log \relax (3)^{4} + 8 \, \log \relax (3)^{3} \log \relax (2) + 24 \, \log \relax (3)^{2} \log \relax (2)^{2} + 32 \, \log \relax (3) \log \relax (2)^{3} + 16 \, \log \relax (2)^{4}\right )} x^{2} + 2 \, {\left (e^{\left (-4\right )} \log \left (x + e^{2}\right ) - e^{\left (-4\right )} \log \relax (x) - \frac {e^{\left (-2\right )}}{x}\right )} e^{2} \log \relax (3) - 2 \, {\left (e^{\left (-4\right )} \log \left (x + e^{2}\right ) - e^{\left (-4\right )} \log \relax (x) - \frac {e^{\left (-2\right )}}{x}\right )} \log \relax (3)^{2} - 2 \, {\left (6 \, x^{2} {\left (\log \relax (3) + 2 \, \log \relax (2)\right )} \log \relax (x) - 3 \, x^{2} \log \relax (x)^{2} - 3 \, {\left (\log \relax (3)^{2} + 4 \, \log \relax (3) \log \relax (2) + 4 \, \log \relax (2)^{2}\right )} x^{2} - x\right )} \log \left (x + e^{2}\right )^{2} + 2 \, {\left (3 \, {\left (\log \relax (3)^{2} + 4 \, \log \relax (3) \log \relax (2) + 4 \, \log \relax (2)^{2}\right )} x^{2} + x\right )} \log \relax (x)^{2} + 2 \, {\left (\log \relax (3)^{2} + 4 \, \log \relax (3) \log \relax (2) + 4 \, \log \relax (2)^{2}\right )} x - 2 \, {\left (e^{\left (-2\right )} \log \left (x + e^{2}\right ) - e^{\left (-2\right )} \log \relax (x)\right )} \log \relax (3) + 2 \, {\left (\log \left (x + e^{2}\right )^{2} - 2 \, \log \left (x + e^{2}\right ) \log \relax (x) + \log \relax (x)^{2}\right )} \log \relax (3) + 4 \, {\left (3 \, x^{2} {\left (\log \relax (3) + 2 \, \log \relax (2)\right )} \log \relax (x)^{2} - x^{2} \log \relax (x)^{3} + {\left (\log \relax (3)^{3} + 6 \, \log \relax (3)^{2} \log \relax (2) + 12 \, \log \relax (3) \log \relax (2)^{2} + 8 \, \log \relax (2)^{3}\right )} x^{2} + x {\left (\log \relax (3) + 2 \, \log \relax (2)\right )} - {\left (3 \, {\left (\log \relax (3)^{2} + 4 \, \log \relax (3) \log \relax (2) + 4 \, \log \relax (2)^{2}\right )} x^{2} + x\right )} \log \relax (x)\right )} \log \left (x + e^{2}\right ) - 4 \, {\left ({\left (\log \relax (3)^{3} + 6 \, \log \relax (3)^{2} \log \relax (2) + 12 \, \log \relax (3) \log \relax (2)^{2} + 8 \, \log \relax (2)^{3}\right )} x^{2} + x {\left (\log \relax (3) + 2 \, \log \relax (2)\right )}\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^4*exp(2)+2*x^5)*log((12*exp(2)+12*x)/x)^4-4*x^4*exp(2)*log((12*exp(2)+12*x)/x)^3+(2*x^3*exp(2)
+2*x^4)*log((12*exp(2)+12*x)/x)^2+(4*x^2*exp(2)*log(3)-4*x^3*exp(2))*log((12*exp(2)+12*x)/x)+(-2*exp(2)-2*x)*l
og(3)^2+(2*exp(2)*x+2*x^2)*log(3))/(x^3*exp(2)+x^4),x, algorithm="maxima")

[Out]

x^2*log(x + e^2)^4 - 4*x^2*(log(3) + 2*log(2))*log(x)^3 + x^2*log(x)^4 + (2*e^(-6)*log(x + e^2) - 2*e^(-6)*log
(x) - (2*x - e^2)*e^(-4)/x^2)*e^2*log(3)^2 + 4*(x^2*(log(3) + 2*log(2)) - x^2*log(x))*log(x + e^2)^3 - 4*(e^(-
2)*log(x + e^2) - e^(-2)*log(x))*e^2*log(3)*log(12*e^2/x + 12) + (log(3)^4 + 8*log(3)^3*log(2) + 24*log(3)^2*l
og(2)^2 + 32*log(3)*log(2)^3 + 16*log(2)^4)*x^2 + 2*(e^(-4)*log(x + e^2) - e^(-4)*log(x) - e^(-2)/x)*e^2*log(3
) - 2*(e^(-4)*log(x + e^2) - e^(-4)*log(x) - e^(-2)/x)*log(3)^2 - 2*(6*x^2*(log(3) + 2*log(2))*log(x) - 3*x^2*
log(x)^2 - 3*(log(3)^2 + 4*log(3)*log(2) + 4*log(2)^2)*x^2 - x)*log(x + e^2)^2 + 2*(3*(log(3)^2 + 4*log(3)*log
(2) + 4*log(2)^2)*x^2 + x)*log(x)^2 + 2*(log(3)^2 + 4*log(3)*log(2) + 4*log(2)^2)*x - 2*(e^(-2)*log(x + e^2) -
e^(-2)*log(x))*log(3) + 2*(log(x + e^2)^2 - 2*log(x + e^2)*log(x) + log(x)^2)*log(3) + 4*(3*x^2*(log(3) + 2*l
og(2))*log(x)^2 - x^2*log(x)^3 + (log(3)^3 + 6*log(3)^2*log(2) + 12*log(3)*log(2)^2 + 8*log(2)^3)*x^2 + x*(log
(3) + 2*log(2)) - (3*(log(3)^2 + 4*log(3)*log(2) + 4*log(2)^2)*x^2 + x)*log(x))*log(x + e^2) - 4*((log(3)^3 +
6*log(3)^2*log(2) + 12*log(3)*log(2)^2 + 8*log(2)^3)*x^2 + x*(log(3) + 2*log(2)))*log(x)

________________________________________________________________________________________

mupad [B]  time = 5.73, size = 55, normalized size = 1.90 \begin {gather*} -\frac {\left (\ln \relax (3)-x^2\,{\ln \left (\frac {12\,x+12\,{\mathrm {e}}^2}{x}\right )}^2\right )\,\left (x^2\,{\ln \left (\frac {12\,x+12\,{\mathrm {e}}^2}{x}\right )}^2+2\,x-\ln \relax (3)\right )}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log((12*x + 12*exp(2))/x)*(4*x^3*exp(2) - 4*x^2*exp(2)*log(3)) + log(3)^2*(2*x + 2*exp(2)) - log((12*x +
12*exp(2))/x)^2*(2*x^3*exp(2) + 2*x^4) - log((12*x + 12*exp(2))/x)^4*(2*x^4*exp(2) + 2*x^5) - log(3)*(2*x*exp
(2) + 2*x^2) + 4*x^4*exp(2)*log((12*x + 12*exp(2))/x)^3)/(x^3*exp(2) + x^4),x)

[Out]

-((log(3) - x^2*log((12*x + 12*exp(2))/x)^2)*(2*x - log(3) + x^2*log((12*x + 12*exp(2))/x)^2))/x^2

________________________________________________________________________________________

sympy [B]  time = 0.31, size = 53, normalized size = 1.83 \begin {gather*} x^{2} \log {\left (\frac {12 x + 12 e^{2}}{x} \right )}^{4} + \left (2 x - 2 \log {\relax (3 )}\right ) \log {\left (\frac {12 x + 12 e^{2}}{x} \right )}^{2} + \frac {- 2 x \log {\relax (3 )} + \log {\relax (3 )}^{2}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**4*exp(2)+2*x**5)*ln((12*exp(2)+12*x)/x)**4-4*x**4*exp(2)*ln((12*exp(2)+12*x)/x)**3+(2*x**3*ex
p(2)+2*x**4)*ln((12*exp(2)+12*x)/x)**2+(4*x**2*exp(2)*ln(3)-4*x**3*exp(2))*ln((12*exp(2)+12*x)/x)+(-2*exp(2)-2
*x)*ln(3)**2+(2*exp(2)*x+2*x**2)*ln(3))/(x**3*exp(2)+x**4),x)

[Out]

x**2*log((12*x + 12*exp(2))/x)**4 + (2*x - 2*log(3))*log((12*x + 12*exp(2))/x)**2 + (-2*x*log(3) + log(3)**2)/
x**2

________________________________________________________________________________________