3.26.80 \(\int \frac {((16 x+e^x (-4 x-4 x^2)) \log (x)+(-8 x^3+2 e^x x^3+(16-4 e^x) \log (x)+(-16+4 e^x) \log ^2(x)) \log (-4 x+e^x x)+(16 x^2+e^x (-4 x^2-4 x^3)+(16 x-4 e^x x+(-16 x+4 e^x x) \log (x)) \log (-4 x+e^x x)) \log (\frac {1}{3} \log (-4 x+e^x x))) \log (\frac {-x^3+\log ^2(x)+2 x \log (x) \log (\frac {1}{3} \log (-4 x+e^x x))+x^2 \log ^2(\frac {1}{3} \log (-4 x+e^x x))}{x^2})}{(4 x^4-e^x x^4+(-4 x+e^x x) \log ^2(x)) \log (-4 x+e^x x)+(-8 x^2+2 e^x x^2) \log (x) \log (-4 x+e^x x) \log (\frac {1}{3} \log (-4 x+e^x x))+(-4 x^3+e^x x^3) \log (-4 x+e^x x) \log ^2(\frac {1}{3} \log (-4 x+e^x x))} \, dx\)

Optimal. Leaf size=33 \[ 5-\log ^2\left (-x+\left (\frac {\log (x)}{x}+\log \left (\frac {1}{3} \log \left (\left (-4+e^x\right ) x\right )\right )\right )^2\right ) \]

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Rubi [F]  time = 39.59, 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 (\left (16 x+e^x \left (-4 x-4 x^2\right )\right ) \log (x)+\left (-8 x^3+2 e^x x^3+\left (16-4 e^x\right ) \log (x)+\left (-16+4 e^x\right ) \log ^2(x)\right ) \log \left (-4 x+e^x x\right )+\left (16 x^2+e^x \left (-4 x^2-4 x^3\right )+\left (16 x-4 e^x x+\left (-16 x+4 e^x x\right ) \log (x)\right ) \log \left (-4 x+e^x x\right )\right ) \log \left (\frac {1}{3} \log \left (-4 x+e^x x\right )\right )\right ) \log \left (\frac {-x^3+\log ^2(x)+2 x \log (x) \log \left (\frac {1}{3} \log \left (-4 x+e^x x\right )\right )+x^2 \log ^2\left (\frac {1}{3} \log \left (-4 x+e^x x\right )\right )}{x^2}\right )}{\left (4 x^4-e^x x^4+\left (-4 x+e^x x\right ) \log ^2(x)\right ) \log \left (-4 x+e^x x\right )+\left (-8 x^2+2 e^x x^2\right ) \log (x) \log \left (-4 x+e^x x\right ) \log \left (\frac {1}{3} \log \left (-4 x+e^x x\right )\right )+\left (-4 x^3+e^x x^3\right ) \log \left (-4 x+e^x x\right ) \log ^2\left (\frac {1}{3} \log \left (-4 x+e^x x\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(((16*x + E^x*(-4*x - 4*x^2))*Log[x] + (-8*x^3 + 2*E^x*x^3 + (16 - 4*E^x)*Log[x] + (-16 + 4*E^x)*Log[x]^2)
*Log[-4*x + E^x*x] + (16*x^2 + E^x*(-4*x^2 - 4*x^3) + (16*x - 4*E^x*x + (-16*x + 4*E^x*x)*Log[x])*Log[-4*x + E
^x*x])*Log[Log[-4*x + E^x*x]/3])*Log[(-x^3 + Log[x]^2 + 2*x*Log[x]*Log[Log[-4*x + E^x*x]/3] + x^2*Log[Log[-4*x
+ E^x*x]/3]^2)/x^2])/((4*x^4 - E^x*x^4 + (-4*x + E^x*x)*Log[x]^2)*Log[-4*x + E^x*x] + (-8*x^2 + 2*E^x*x^2)*Lo
g[x]*Log[-4*x + E^x*x]*Log[Log[-4*x + E^x*x]/3] + (-4*x^3 + E^x*x^3)*Log[-4*x + E^x*x]*Log[Log[-4*x + E^x*x]/3
]^2),x]

[Out]

-2*Defer[Int][(x^2*Log[-x + Log[x]^2/x^2 + (2*Log[x]*Log[Log[(-4 + E^x)*x]/3])/x + Log[Log[(-4 + E^x)*x]/3]^2]
)/(x^3 - Log[x]^2 - 2*x*Log[x]*Log[Log[(-4 + E^x)*x]/3] - x^2*Log[Log[(-4 + E^x)*x]/3]^2), x] + 4*Defer[Int][(
Log[x]*Log[-x + Log[x]^2/x^2 + (2*Log[x]*Log[Log[(-4 + E^x)*x]/3])/x + Log[Log[(-4 + E^x)*x]/3]^2])/(x*(x^3 -
Log[x]^2 - 2*x*Log[x]*Log[Log[(-4 + E^x)*x]/3] - x^2*Log[Log[(-4 + E^x)*x]/3]^2)), x] - 4*Defer[Int][(Log[x]^2
*Log[-x + Log[x]^2/x^2 + (2*Log[x]*Log[Log[(-4 + E^x)*x]/3])/x + Log[Log[(-4 + E^x)*x]/3]^2])/(x*(x^3 - Log[x]
^2 - 2*x*Log[x]*Log[Log[(-4 + E^x)*x]/3] - x^2*Log[Log[(-4 + E^x)*x]/3]^2)), x] + 4*Defer[Int][(Log[x]*Log[-x
+ Log[x]^2/x^2 + (2*Log[x]*Log[Log[(-4 + E^x)*x]/3])/x + Log[Log[(-4 + E^x)*x]/3]^2])/(Log[-4*x + E^x*x]*(x^3
- Log[x]^2 - 2*x*Log[x]*Log[Log[(-4 + E^x)*x]/3] - x^2*Log[Log[(-4 + E^x)*x]/3]^2)), x] + 4*Defer[Int][(x*Log[
x]*Log[-x + Log[x]^2/x^2 + (2*Log[x]*Log[Log[(-4 + E^x)*x]/3])/x + Log[Log[(-4 + E^x)*x]/3]^2])/(Log[-4*x + E^
x*x]*(x^3 - Log[x]^2 - 2*x*Log[x]*Log[Log[(-4 + E^x)*x]/3] - x^2*Log[Log[(-4 + E^x)*x]/3]^2)), x] + 16*Defer[I
nt][(x*Log[x]*Log[-x + Log[x]^2/x^2 + (2*Log[x]*Log[Log[(-4 + E^x)*x]/3])/x + Log[Log[(-4 + E^x)*x]/3]^2])/((-
4 + E^x)*Log[-4*x + E^x*x]*(x^3 - Log[x]^2 - 2*x*Log[x]*Log[Log[(-4 + E^x)*x]/3] - x^2*Log[Log[(-4 + E^x)*x]/3
]^2)), x] + 2*Log[9]*Defer[Int][Log[-x + Log[x]^2/x^2 + (2*Log[x]*Log[Log[(-4 + E^x)*x]/3])/x + Log[Log[(-4 +
E^x)*x]/3]^2]/(-x^3 + Log[x]^2 + 2*x*Log[x]*Log[Log[(-4 + E^x)*x]/3] + x^2*Log[Log[(-4 + E^x)*x]/3]^2), x] - 2
*Log[9]*Defer[Int][(Log[x]*Log[-x + Log[x]^2/x^2 + (2*Log[x]*Log[Log[(-4 + E^x)*x]/3])/x + Log[Log[(-4 + E^x)*
x]/3]^2])/(-x^3 + Log[x]^2 + 2*x*Log[x]*Log[Log[(-4 + E^x)*x]/3] + x^2*Log[Log[(-4 + E^x)*x]/3]^2), x] + 4*Def
er[Int][(x*Log[Log[-4*x + E^x*x]/3]*Log[-x + Log[x]^2/x^2 + (2*Log[x]*Log[Log[(-4 + E^x)*x]/3])/x + Log[Log[(-
4 + E^x)*x]/3]^2])/(Log[-4*x + E^x*x]*(x^3 - Log[x]^2 - 2*x*Log[x]*Log[Log[(-4 + E^x)*x]/3] - x^2*Log[Log[(-4
+ E^x)*x]/3]^2)), x] + 4*Defer[Int][(x^2*Log[Log[-4*x + E^x*x]/3]*Log[-x + Log[x]^2/x^2 + (2*Log[x]*Log[Log[(-
4 + E^x)*x]/3])/x + Log[Log[(-4 + E^x)*x]/3]^2])/(Log[-4*x + E^x*x]*(x^3 - Log[x]^2 - 2*x*Log[x]*Log[Log[(-4 +
E^x)*x]/3] - x^2*Log[Log[(-4 + E^x)*x]/3]^2)), x] + 16*Defer[Int][(x^2*Log[Log[-4*x + E^x*x]/3]*Log[-x + Log[
x]^2/x^2 + (2*Log[x]*Log[Log[(-4 + E^x)*x]/3])/x + Log[Log[(-4 + E^x)*x]/3]^2])/((-4 + E^x)*Log[-4*x + E^x*x]*
(x^3 - Log[x]^2 - 2*x*Log[x]*Log[Log[(-4 + E^x)*x]/3] - x^2*Log[Log[(-4 + E^x)*x]/3]^2)), x] + 4*Defer[Int][(L
og[Log[-4*x + E^x*x]]*Log[-x + Log[x]^2/x^2 + (2*Log[x]*Log[Log[(-4 + E^x)*x]/3])/x + Log[Log[(-4 + E^x)*x]/3]
^2])/(x^3 - Log[x]^2 - 2*x*Log[x]*Log[Log[(-4 + E^x)*x]/3] - x^2*Log[Log[(-4 + E^x)*x]/3]^2), x] + 4*Defer[Int
][(Log[x]*Log[Log[-4*x + E^x*x]]*Log[-x + Log[x]^2/x^2 + (2*Log[x]*Log[Log[(-4 + E^x)*x]/3])/x + Log[Log[(-4 +
E^x)*x]/3]^2])/(-x^3 + Log[x]^2 + 2*x*Log[x]*Log[Log[(-4 + E^x)*x]/3] + x^2*Log[Log[(-4 + E^x)*x]/3]^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (\left (16 x+e^x \left (-4 x-4 x^2\right )\right ) \log (x)+\left (-8 x^3+2 e^x x^3+\left (16-4 e^x\right ) \log (x)+\left (-16+4 e^x\right ) \log ^2(x)\right ) \log \left (-4 x+e^x x\right )+\left (16 x^2+e^x \left (-4 x^2-4 x^3\right )+\left (16 x-4 e^x x+\left (-16 x+4 e^x x\right ) \log (x)\right ) \log \left (-4 x+e^x x\right )\right ) \log \left (\frac {1}{3} \log \left (-4 x+e^x x\right )\right )\right ) \log \left (-x+\frac {\log ^2(x)}{x^2}+\frac {2 \log (x) \log \left (\frac {1}{3} \log \left (-4 x+e^x x\right )\right )}{x}+\log ^2\left (\frac {1}{3} \log \left (-4 x+e^x x\right )\right )\right )}{\left (4-e^x\right ) x \log \left (-4 x+e^x x\right ) \left (x^3-\log ^2(x)-2 x \log (x) \log \left (\frac {1}{3} \log \left (\left (-4+e^x\right ) x\right )\right )-x^2 \log ^2\left (\frac {1}{3} \log \left (\left (-4+e^x\right ) x\right )\right )\right )} \, dx\\ &=\int \left (\frac {16 x \left (-\log (x)-x \log \left (\frac {1}{3} \log \left (\left (-4+e^x\right ) x\right )\right )\right ) \log \left (-x+\frac {\log ^2(x)}{x^2}+\frac {2 \log (x) \log \left (\frac {1}{3} \log \left (\left (-4+e^x\right ) x\right )\right )}{x}+\log ^2\left (\frac {1}{3} \log \left (\left (-4+e^x\right ) x\right )\right )\right )}{\left (4-e^x\right ) \log \left (-4 x+e^x x\right ) \left (x^3-\log ^2(x)-2 x \log (x) \log \left (\frac {1}{3} \log \left (\left (-4+e^x\right ) x\right )\right )-x^2 \log ^2\left (\frac {1}{3} \log \left (\left (-4+e^x\right ) x\right )\right )\right )}+\frac {2 \left (2 x \log (x)+2 x^2 \log (x)-x^3 \log \left (\left (-4+e^x\right ) x\right )-x \log (9) \log \left (\left (-4+e^x\right ) x\right )+2 \log (x) \log \left (\left (-4+e^x\right ) x\right )+2 x \log (3) \log (x) \log \left (\left (-4+e^x\right ) x\right )-2 \log ^2(x) \log \left (\left (-4+e^x\right ) x\right )+2 x^2 \log \left (\frac {1}{3} \log \left (\left (-4+e^x\right ) x\right )\right )+2 x^3 \log \left (\frac {1}{3} \log \left (\left (-4+e^x\right ) x\right )\right )+2 x \log \left (\left (-4+e^x\right ) x\right ) \log \left (\log \left (\left (-4+e^x\right ) x\right )\right )-2 x \log (x) \log \left (\left (-4+e^x\right ) x\right ) \log \left (\log \left (\left (-4+e^x\right ) x\right )\right )\right ) \log \left (-x+\frac {\log ^2(x)}{x^2}+\frac {2 \log (x) \log \left (\frac {1}{3} \log \left (\left (-4+e^x\right ) x\right )\right )}{x}+\log ^2\left (\frac {1}{3} \log \left (\left (-4+e^x\right ) x\right )\right )\right )}{x \log \left (-4 x+e^x x\right ) \left (x^3-\log ^2(x)-2 x \log (x) \log \left (\frac {1}{3} \log \left (\left (-4+e^x\right ) x\right )\right )-x^2 \log ^2\left (\frac {1}{3} \log \left (\left (-4+e^x\right ) x\right )\right )\right )}\right ) \, dx\\ &=2 \int \frac {\left (2 x \log (x)+2 x^2 \log (x)-x^3 \log \left (\left (-4+e^x\right ) x\right )-x \log (9) \log \left (\left (-4+e^x\right ) x\right )+2 \log (x) \log \left (\left (-4+e^x\right ) x\right )+2 x \log (3) \log (x) \log \left (\left (-4+e^x\right ) x\right )-2 \log ^2(x) \log \left (\left (-4+e^x\right ) x\right )+2 x^2 \log \left (\frac {1}{3} \log \left (\left (-4+e^x\right ) x\right )\right )+2 x^3 \log \left (\frac {1}{3} \log \left (\left (-4+e^x\right ) x\right )\right )+2 x \log \left (\left (-4+e^x\right ) x\right ) \log \left (\log \left (\left (-4+e^x\right ) x\right )\right )-2 x \log (x) \log \left (\left (-4+e^x\right ) x\right ) \log \left (\log \left (\left (-4+e^x\right ) x\right )\right )\right ) \log \left (-x+\frac {\log ^2(x)}{x^2}+\frac {2 \log (x) \log \left (\frac {1}{3} \log \left (\left (-4+e^x\right ) x\right )\right )}{x}+\log ^2\left (\frac {1}{3} \log \left (\left (-4+e^x\right ) x\right )\right )\right )}{x \log \left (-4 x+e^x x\right ) \left (x^3-\log ^2(x)-2 x \log (x) \log \left (\frac {1}{3} \log \left (\left (-4+e^x\right ) x\right )\right )-x^2 \log ^2\left (\frac {1}{3} \log \left (\left (-4+e^x\right ) x\right )\right )\right )} \, dx+16 \int \frac {x \left (-\log (x)-x \log \left (\frac {1}{3} \log \left (\left (-4+e^x\right ) x\right )\right )\right ) \log \left (-x+\frac {\log ^2(x)}{x^2}+\frac {2 \log (x) \log \left (\frac {1}{3} \log \left (\left (-4+e^x\right ) x\right )\right )}{x}+\log ^2\left (\frac {1}{3} \log \left (\left (-4+e^x\right ) x\right )\right )\right )}{\left (4-e^x\right ) \log \left (-4 x+e^x x\right ) \left (x^3-\log ^2(x)-2 x \log (x) \log \left (\frac {1}{3} \log \left (\left (-4+e^x\right ) x\right )\right )-x^2 \log ^2\left (\frac {1}{3} \log \left (\left (-4+e^x\right ) x\right )\right )\right )} \, dx\\ &=2 \int \frac {\left (2 \log ^2(x) \log \left (\left (-4+e^x\right ) x\right )-x \left (2 x (1+x) \log \left (\frac {1}{3} \log \left (\left (-4+e^x\right ) x\right )\right )-\log \left (\left (-4+e^x\right ) x\right ) \left (x^2+\log (9)-2 \log \left (\log \left (\left (-4+e^x\right ) x\right )\right )\right )\right )-\log (x) \left (2 x (1+x)+\log \left (\left (-4+e^x\right ) x\right ) \left (2+x \log (9)-2 x \log \left (\log \left (\left (-4+e^x\right ) x\right )\right )\right )\right )\right ) \log \left (-x+\frac {\log ^2(x)}{x^2}+\frac {2 \log (x) \log \left (\frac {1}{3} \log \left (\left (-4+e^x\right ) x\right )\right )}{x}+\log ^2\left (\frac {1}{3} \log \left (\left (-4+e^x\right ) x\right )\right )\right )}{x \log \left (-4 x+e^x x\right ) \left (\log ^2(x)+2 x \log (x) \log \left (\frac {1}{3} \log \left (\left (-4+e^x\right ) x\right )\right )+x^2 \left (-x+\log ^2\left (\frac {1}{3} \log \left (\left (-4+e^x\right ) x\right )\right )\right )\right )} \, dx+16 \int \left (\frac {x \log (x) \log \left (-x+\frac {\log ^2(x)}{x^2}+\frac {2 \log (x) \log \left (\frac {1}{3} \log \left (\left (-4+e^x\right ) x\right )\right )}{x}+\log ^2\left (\frac {1}{3} \log \left (\left (-4+e^x\right ) x\right )\right )\right )}{\left (-4+e^x\right ) \log \left (-4 x+e^x x\right ) \left (x^3-\log ^2(x)-2 x \log (x) \log \left (\frac {1}{3} \log \left (\left (-4+e^x\right ) x\right )\right )-x^2 \log ^2\left (\frac {1}{3} \log \left (\left (-4+e^x\right ) x\right )\right )\right )}+\frac {x^2 \log \left (\frac {1}{3} \log \left (-4 x+e^x x\right )\right ) \log \left (-x+\frac {\log ^2(x)}{x^2}+\frac {2 \log (x) \log \left (\frac {1}{3} \log \left (\left (-4+e^x\right ) x\right )\right )}{x}+\log ^2\left (\frac {1}{3} \log \left (\left (-4+e^x\right ) x\right )\right )\right )}{\left (-4+e^x\right ) \log \left (-4 x+e^x x\right ) \left (x^3-\log ^2(x)-2 x \log (x) \log \left (\frac {1}{3} \log \left (\left (-4+e^x\right ) x\right )\right )-x^2 \log ^2\left (\frac {1}{3} \log \left (\left (-4+e^x\right ) x\right )\right )\right )}\right ) \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [F]  time = 1.17, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (\left (16 x+e^x \left (-4 x-4 x^2\right )\right ) \log (x)+\left (-8 x^3+2 e^x x^3+\left (16-4 e^x\right ) \log (x)+\left (-16+4 e^x\right ) \log ^2(x)\right ) \log \left (-4 x+e^x x\right )+\left (16 x^2+e^x \left (-4 x^2-4 x^3\right )+\left (16 x-4 e^x x+\left (-16 x+4 e^x x\right ) \log (x)\right ) \log \left (-4 x+e^x x\right )\right ) \log \left (\frac {1}{3} \log \left (-4 x+e^x x\right )\right )\right ) \log \left (\frac {-x^3+\log ^2(x)+2 x \log (x) \log \left (\frac {1}{3} \log \left (-4 x+e^x x\right )\right )+x^2 \log ^2\left (\frac {1}{3} \log \left (-4 x+e^x x\right )\right )}{x^2}\right )}{\left (4 x^4-e^x x^4+\left (-4 x+e^x x\right ) \log ^2(x)\right ) \log \left (-4 x+e^x x\right )+\left (-8 x^2+2 e^x x^2\right ) \log (x) \log \left (-4 x+e^x x\right ) \log \left (\frac {1}{3} \log \left (-4 x+e^x x\right )\right )+\left (-4 x^3+e^x x^3\right ) \log \left (-4 x+e^x x\right ) \log ^2\left (\frac {1}{3} \log \left (-4 x+e^x x\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(((16*x + E^x*(-4*x - 4*x^2))*Log[x] + (-8*x^3 + 2*E^x*x^3 + (16 - 4*E^x)*Log[x] + (-16 + 4*E^x)*Log
[x]^2)*Log[-4*x + E^x*x] + (16*x^2 + E^x*(-4*x^2 - 4*x^3) + (16*x - 4*E^x*x + (-16*x + 4*E^x*x)*Log[x])*Log[-4
*x + E^x*x])*Log[Log[-4*x + E^x*x]/3])*Log[(-x^3 + Log[x]^2 + 2*x*Log[x]*Log[Log[-4*x + E^x*x]/3] + x^2*Log[Lo
g[-4*x + E^x*x]/3]^2)/x^2])/((4*x^4 - E^x*x^4 + (-4*x + E^x*x)*Log[x]^2)*Log[-4*x + E^x*x] + (-8*x^2 + 2*E^x*x
^2)*Log[x]*Log[-4*x + E^x*x]*Log[Log[-4*x + E^x*x]/3] + (-4*x^3 + E^x*x^3)*Log[-4*x + E^x*x]*Log[Log[-4*x + E^
x*x]/3]^2),x]

[Out]

Integrate[(((16*x + E^x*(-4*x - 4*x^2))*Log[x] + (-8*x^3 + 2*E^x*x^3 + (16 - 4*E^x)*Log[x] + (-16 + 4*E^x)*Log
[x]^2)*Log[-4*x + E^x*x] + (16*x^2 + E^x*(-4*x^2 - 4*x^3) + (16*x - 4*E^x*x + (-16*x + 4*E^x*x)*Log[x])*Log[-4
*x + E^x*x])*Log[Log[-4*x + E^x*x]/3])*Log[(-x^3 + Log[x]^2 + 2*x*Log[x]*Log[Log[-4*x + E^x*x]/3] + x^2*Log[Lo
g[-4*x + E^x*x]/3]^2)/x^2])/((4*x^4 - E^x*x^4 + (-4*x + E^x*x)*Log[x]^2)*Log[-4*x + E^x*x] + (-8*x^2 + 2*E^x*x
^2)*Log[x]*Log[-4*x + E^x*x]*Log[Log[-4*x + E^x*x]/3] + (-4*x^3 + E^x*x^3)*Log[-4*x + E^x*x]*Log[Log[-4*x + E^
x*x]/3]^2), x]

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fricas [A]  time = 0.58, size = 54, normalized size = 1.64 \begin {gather*} -\log \left (\frac {x^{2} \log \left (\frac {1}{3} \, \log \left (x e^{x} - 4 \, x\right )\right )^{2} - x^{3} + 2 \, x \log \relax (x) \log \left (\frac {1}{3} \, \log \left (x e^{x} - 4 \, x\right )\right ) + \log \relax (x)^{2}}{x^{2}}\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((4*exp(x)*x-16*x)*log(x)-4*exp(x)*x+16*x)*log(exp(x)*x-4*x)+(-4*x^3-4*x^2)*exp(x)+16*x^2)*log(1/3
*log(exp(x)*x-4*x))+((4*exp(x)-16)*log(x)^2+(-4*exp(x)+16)*log(x)+2*exp(x)*x^3-8*x^3)*log(exp(x)*x-4*x)+((-4*x
^2-4*x)*exp(x)+16*x)*log(x))*log((x^2*log(1/3*log(exp(x)*x-4*x))^2+2*x*log(x)*log(1/3*log(exp(x)*x-4*x))+log(x
)^2-x^3)/x^2)/((exp(x)*x^3-4*x^3)*log(exp(x)*x-4*x)*log(1/3*log(exp(x)*x-4*x))^2+(2*exp(x)*x^2-8*x^2)*log(x)*l
og(exp(x)*x-4*x)*log(1/3*log(exp(x)*x-4*x))+((exp(x)*x-4*x)*log(x)^2-exp(x)*x^4+4*x^4)*log(exp(x)*x-4*x)),x, a
lgorithm="fricas")

[Out]

-log((x^2*log(1/3*log(x*e^x - 4*x))^2 - x^3 + 2*x*log(x)*log(1/3*log(x*e^x - 4*x)) + log(x)^2)/x^2)^2

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((4*exp(x)*x-16*x)*log(x)-4*exp(x)*x+16*x)*log(exp(x)*x-4*x)+(-4*x^3-4*x^2)*exp(x)+16*x^2)*log(1/3
*log(exp(x)*x-4*x))+((4*exp(x)-16)*log(x)^2+(-4*exp(x)+16)*log(x)+2*exp(x)*x^3-8*x^3)*log(exp(x)*x-4*x)+((-4*x
^2-4*x)*exp(x)+16*x)*log(x))*log((x^2*log(1/3*log(exp(x)*x-4*x))^2+2*x*log(x)*log(1/3*log(exp(x)*x-4*x))+log(x
)^2-x^3)/x^2)/((exp(x)*x^3-4*x^3)*log(exp(x)*x-4*x)*log(1/3*log(exp(x)*x-4*x))^2+(2*exp(x)*x^2-8*x^2)*log(x)*l
og(exp(x)*x-4*x)*log(1/3*log(exp(x)*x-4*x))+((exp(x)*x-4*x)*log(x)^2-exp(x)*x^4+4*x^4)*log(exp(x)*x-4*x)),x, a
lgorithm="giac")

[Out]

Timed out

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maple [F]  time = 0.10, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (\left (\left (4 \,{\mathrm e}^{x} x -16 x \right ) \ln \relax (x )-4 \,{\mathrm e}^{x} x +16 x \right ) \ln \left ({\mathrm e}^{x} x -4 x \right )+\left (-4 x^{3}-4 x^{2}\right ) {\mathrm e}^{x}+16 x^{2}\right ) \ln \left (\frac {\ln \left ({\mathrm e}^{x} x -4 x \right )}{3}\right )+\left (\left (4 \,{\mathrm e}^{x}-16\right ) \ln \relax (x )^{2}+\left (-4 \,{\mathrm e}^{x}+16\right ) \ln \relax (x )+2 \,{\mathrm e}^{x} x^{3}-8 x^{3}\right ) \ln \left ({\mathrm e}^{x} x -4 x \right )+\left (\left (-4 x^{2}-4 x \right ) {\mathrm e}^{x}+16 x \right ) \ln \relax (x )\right ) \ln \left (\frac {x^{2} \ln \left (\frac {\ln \left ({\mathrm e}^{x} x -4 x \right )}{3}\right )^{2}+2 x \ln \relax (x ) \ln \left (\frac {\ln \left ({\mathrm e}^{x} x -4 x \right )}{3}\right )+\ln \relax (x )^{2}-x^{3}}{x^{2}}\right )}{\left ({\mathrm e}^{x} x^{3}-4 x^{3}\right ) \ln \left ({\mathrm e}^{x} x -4 x \right ) \ln \left (\frac {\ln \left ({\mathrm e}^{x} x -4 x \right )}{3}\right )^{2}+\left (2 \,{\mathrm e}^{x} x^{2}-8 x^{2}\right ) \ln \relax (x ) \ln \left ({\mathrm e}^{x} x -4 x \right ) \ln \left (\frac {\ln \left ({\mathrm e}^{x} x -4 x \right )}{3}\right )+\left (\left ({\mathrm e}^{x} x -4 x \right ) \ln \relax (x )^{2}-{\mathrm e}^{x} x^{4}+4 x^{4}\right ) \ln \left ({\mathrm e}^{x} x -4 x \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((((4*exp(x)*x-16*x)*ln(x)-4*exp(x)*x+16*x)*ln(exp(x)*x-4*x)+(-4*x^3-4*x^2)*exp(x)+16*x^2)*ln(1/3*ln(exp(x
)*x-4*x))+((4*exp(x)-16)*ln(x)^2+(-4*exp(x)+16)*ln(x)+2*exp(x)*x^3-8*x^3)*ln(exp(x)*x-4*x)+((-4*x^2-4*x)*exp(x
)+16*x)*ln(x))*ln((x^2*ln(1/3*ln(exp(x)*x-4*x))^2+2*x*ln(x)*ln(1/3*ln(exp(x)*x-4*x))+ln(x)^2-x^3)/x^2)/((exp(x
)*x^3-4*x^3)*ln(exp(x)*x-4*x)*ln(1/3*ln(exp(x)*x-4*x))^2+(2*exp(x)*x^2-8*x^2)*ln(x)*ln(exp(x)*x-4*x)*ln(1/3*ln
(exp(x)*x-4*x))+((exp(x)*x-4*x)*ln(x)^2-exp(x)*x^4+4*x^4)*ln(exp(x)*x-4*x)),x)

[Out]

int(((((4*exp(x)*x-16*x)*ln(x)-4*exp(x)*x+16*x)*ln(exp(x)*x-4*x)+(-4*x^3-4*x^2)*exp(x)+16*x^2)*ln(1/3*ln(exp(x
)*x-4*x))+((4*exp(x)-16)*ln(x)^2+(-4*exp(x)+16)*ln(x)+2*exp(x)*x^3-8*x^3)*ln(exp(x)*x-4*x)+((-4*x^2-4*x)*exp(x
)+16*x)*ln(x))*ln((x^2*ln(1/3*ln(exp(x)*x-4*x))^2+2*x*ln(x)*ln(1/3*ln(exp(x)*x-4*x))+ln(x)^2-x^3)/x^2)/((exp(x
)*x^3-4*x^3)*ln(exp(x)*x-4*x)*ln(1/3*ln(exp(x)*x-4*x))^2+(2*exp(x)*x^2-8*x^2)*ln(x)*ln(exp(x)*x-4*x)*ln(1/3*ln
(exp(x)*x-4*x))+((exp(x)*x-4*x)*ln(x)^2-exp(x)*x^4+4*x^4)*ln(exp(x)*x-4*x)),x)

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maxima [B]  time = 1.59, size = 191, normalized size = 5.79 \begin {gather*} \log \left (\frac {x^{2} \log \relax (3)^{2} + x^{2} \log \left (\log \relax (x) + \log \left (e^{x} - 4\right )\right )^{2} - x^{3} - 2 \, x \log \relax (3) \log \relax (x) + \log \relax (x)^{2} - 2 \, {\left (x^{2} \log \relax (3) - x \log \relax (x)\right )} \log \left (\log \relax (x) + \log \left (e^{x} - 4\right )\right )}{x^{2}}\right )^{2} - 2 \, \log \left (\frac {x^{2} \log \relax (3)^{2} + x^{2} \log \left (\log \relax (x) + \log \left (e^{x} - 4\right )\right )^{2} - x^{3} - 2 \, x \log \relax (3) \log \relax (x) + \log \relax (x)^{2} - 2 \, {\left (x^{2} \log \relax (3) - x \log \relax (x)\right )} \log \left (\log \relax (x) + \log \left (e^{x} - 4\right )\right )}{x^{2}}\right ) \log \left (\frac {x^{2} \log \left (\frac {1}{3} \, \log \left (x e^{x} - 4 \, x\right )\right )^{2} - x^{3} + 2 \, x \log \relax (x) \log \left (\frac {1}{3} \, \log \left (x e^{x} - 4 \, x\right )\right ) + \log \relax (x)^{2}}{x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((4*exp(x)*x-16*x)*log(x)-4*exp(x)*x+16*x)*log(exp(x)*x-4*x)+(-4*x^3-4*x^2)*exp(x)+16*x^2)*log(1/3
*log(exp(x)*x-4*x))+((4*exp(x)-16)*log(x)^2+(-4*exp(x)+16)*log(x)+2*exp(x)*x^3-8*x^3)*log(exp(x)*x-4*x)+((-4*x
^2-4*x)*exp(x)+16*x)*log(x))*log((x^2*log(1/3*log(exp(x)*x-4*x))^2+2*x*log(x)*log(1/3*log(exp(x)*x-4*x))+log(x
)^2-x^3)/x^2)/((exp(x)*x^3-4*x^3)*log(exp(x)*x-4*x)*log(1/3*log(exp(x)*x-4*x))^2+(2*exp(x)*x^2-8*x^2)*log(x)*l
og(exp(x)*x-4*x)*log(1/3*log(exp(x)*x-4*x))+((exp(x)*x-4*x)*log(x)^2-exp(x)*x^4+4*x^4)*log(exp(x)*x-4*x)),x, a
lgorithm="maxima")

[Out]

log((x^2*log(3)^2 + x^2*log(log(x) + log(e^x - 4))^2 - x^3 - 2*x*log(3)*log(x) + log(x)^2 - 2*(x^2*log(3) - x*
log(x))*log(log(x) + log(e^x - 4)))/x^2)^2 - 2*log((x^2*log(3)^2 + x^2*log(log(x) + log(e^x - 4))^2 - x^3 - 2*
x*log(3)*log(x) + log(x)^2 - 2*(x^2*log(3) - x*log(x))*log(log(x) + log(e^x - 4)))/x^2)*log((x^2*log(1/3*log(x
*e^x - 4*x))^2 - x^3 + 2*x*log(x)*log(1/3*log(x*e^x - 4*x)) + log(x)^2)/x^2)

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mupad [B]  time = 2.65, size = 54, normalized size = 1.64 \begin {gather*} -{\ln \left (\frac {-x^3+x^2\,{\ln \left (\frac {\ln \left (x\,{\mathrm {e}}^x-4\,x\right )}{3}\right )}^2+2\,x\,\ln \left (\frac {\ln \left (x\,{\mathrm {e}}^x-4\,x\right )}{3}\right )\,\ln \relax (x)+{\ln \relax (x)}^2}{x^2}\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log((log(x)^2 + x^2*log(log(x*exp(x) - 4*x)/3)^2 - x^3 + 2*x*log(log(x*exp(x) - 4*x)/3)*log(x))/x^2)*(log
(x)*(16*x - exp(x)*(4*x + 4*x^2)) + log(x*exp(x) - 4*x)*(2*x^3*exp(x) + log(x)^2*(4*exp(x) - 16) - 8*x^3 - log
(x)*(4*exp(x) - 16)) - log(log(x*exp(x) - 4*x)/3)*(exp(x)*(4*x^2 + 4*x^3) + log(x*exp(x) - 4*x)*(log(x)*(16*x
- 4*x*exp(x)) - 16*x + 4*x*exp(x)) - 16*x^2)))/(log(log(x*exp(x) - 4*x)/3)^2*log(x*exp(x) - 4*x)*(x^3*exp(x) -
4*x^3) - log(x*exp(x) - 4*x)*(x^4*exp(x) + log(x)^2*(4*x - x*exp(x)) - 4*x^4) + log(log(x*exp(x) - 4*x)/3)*lo
g(x*exp(x) - 4*x)*log(x)*(2*x^2*exp(x) - 8*x^2)),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((4*exp(x)*x-16*x)*ln(x)-4*exp(x)*x+16*x)*ln(exp(x)*x-4*x)+(-4*x**3-4*x**2)*exp(x)+16*x**2)*ln(1/3
*ln(exp(x)*x-4*x))+((4*exp(x)-16)*ln(x)**2+(-4*exp(x)+16)*ln(x)+2*exp(x)*x**3-8*x**3)*ln(exp(x)*x-4*x)+((-4*x*
*2-4*x)*exp(x)+16*x)*ln(x))*ln((x**2*ln(1/3*ln(exp(x)*x-4*x))**2+2*x*ln(x)*ln(1/3*ln(exp(x)*x-4*x))+ln(x)**2-x
**3)/x**2)/((exp(x)*x**3-4*x**3)*ln(exp(x)*x-4*x)*ln(1/3*ln(exp(x)*x-4*x))**2+(2*exp(x)*x**2-8*x**2)*ln(x)*ln(
exp(x)*x-4*x)*ln(1/3*ln(exp(x)*x-4*x))+((exp(x)*x-4*x)*ln(x)**2-exp(x)*x**4+4*x**4)*ln(exp(x)*x-4*x)),x)

[Out]

Timed out

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