3.43.9 \(\int \frac {e^{\frac {2 (-e^6+x+4 e^3 x-4 x^2+(2+e^6-x-4 e^3 x+4 x^2+(e^6 x-x^2-4 e^3 x^2+4 x^3) \log (4)) \log (x))}{-1+(1+x \log (4)) \log (x)}} (-4-2 x-8 e^3 x+16 x^2+(4 x+16 e^3 x-32 x^2+(4 x^2+16 e^3 x^2-32 x^3) \log (4)) \log (x)+(-2 x-8 e^3 x+16 x^2+(-4 x-4 x^2-16 e^3 x^2+32 x^3) \log (4)+(-2 x^3-8 e^3 x^3+16 x^4) \log ^2(4)) \log ^2(x))}{x+(-2 x-2 x^2 \log (4)) \log (x)+(x+2 x^2 \log (4)+x^3 \log ^2(4)) \log ^2(x)} \, dx\)

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

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Rubi [F]  time = 36.82, 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 {\exp \left (\frac {2 \left (-e^6+x+4 e^3 x-4 x^2+\left (2+e^6-x-4 e^3 x+4 x^2+\left (e^6 x-x^2-4 e^3 x^2+4 x^3\right ) \log (4)\right ) \log (x)\right )}{-1+(1+x \log (4)) \log (x)}\right ) \left (-4-2 x-8 e^3 x+16 x^2+\left (4 x+16 e^3 x-32 x^2+\left (4 x^2+16 e^3 x^2-32 x^3\right ) \log (4)\right ) \log (x)+\left (-2 x-8 e^3 x+16 x^2+\left (-4 x-4 x^2-16 e^3 x^2+32 x^3\right ) \log (4)+\left (-2 x^3-8 e^3 x^3+16 x^4\right ) \log ^2(4)\right ) \log ^2(x)\right )}{x+\left (-2 x-2 x^2 \log (4)\right ) \log (x)+\left (x+2 x^2 \log (4)+x^3 \log ^2(4)\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((2*(-E^6 + x + 4*E^3*x - 4*x^2 + (2 + E^6 - x - 4*E^3*x + 4*x^2 + (E^6*x - x^2 - 4*E^3*x^2 + 4*x^3)*Lo
g[4])*Log[x]))/(-1 + (1 + x*Log[4])*Log[x]))*(-4 - 2*x - 8*E^3*x + 16*x^2 + (4*x + 16*E^3*x - 32*x^2 + (4*x^2
+ 16*E^3*x^2 - 32*x^3)*Log[4])*Log[x] + (-2*x - 8*E^3*x + 16*x^2 + (-4*x - 4*x^2 - 16*E^3*x^2 + 32*x^3)*Log[4]
+ (-2*x^3 - 8*E^3*x^3 + 16*x^4)*Log[4]^2)*Log[x]^2))/(x + (-2*x - 2*x^2*Log[4])*Log[x] + (x + 2*x^2*Log[4] +
x^3*Log[4]^2)*Log[x]^2),x]

[Out]

-2*(1 + 4*E^3)*Defer[Int][E^((2*(-E^6 + (1 + 4*E^3)*x - 4*x^2 + (2 + E^6 - x - 4*E^3*x + 4*x^2 + (E^6*x - x^2
- 4*E^3*x^2 + 4*x^3)*Log[4])*Log[x]))/(-1 + (1 + x*Log[4])*Log[x])), x] + 16*Defer[Int][E^((2*(-E^6 + (1 + 4*E
^3)*x - 4*x^2 + (2 + E^6 - x - 4*E^3*x + 4*x^2 + (E^6*x - x^2 - 4*E^3*x^2 + 4*x^3)*Log[4])*Log[x]))/(-1 + (1 +
x*Log[4])*Log[x]))*x, x] - 2*Log[16]*Defer[Int][E^((2*(-E^6 + (1 + 4*E^3)*x - 4*x^2 + (2 + E^6 - x - 4*E^3*x
+ 4*x^2 + (E^6*x - x^2 - 4*E^3*x^2 + 4*x^3)*Log[4])*Log[x]))/(-1 + (1 + x*Log[4])*Log[x]))/(1 + x*Log[4])^2, x
] - 4*Defer[Int][E^((2*(-E^6 + (1 + 4*E^3)*x - 4*x^2 + (2 + E^6 - x - 4*E^3*x + 4*x^2 + (E^6*x - x^2 - 4*E^3*x
^2 + 4*x^3)*Log[4])*Log[x]))/(-1 + (1 + x*Log[4])*Log[x]))/(x*(-1 + Log[x] + x*Log[4]*Log[x])^2), x] - 4*Log[4
]*Defer[Int][E^((2*(-E^6 + (1 + 4*E^3)*x - 4*x^2 + (2 + E^6 - x - 4*E^3*x + 4*x^2 + (E^6*x - x^2 - 4*E^3*x^2 +
4*x^3)*Log[4])*Log[x]))/(-1 + (1 + x*Log[4])*Log[x]))/((1 + x*Log[4])^2*(-1 + Log[x] + x*Log[4]*Log[x])^2), x
] - 8*Log[4]*Defer[Int][E^((2*(-E^6 + (1 + 4*E^3)*x - 4*x^2 + (2 + E^6 - x - 4*E^3*x + 4*x^2 + (E^6*x - x^2 -
4*E^3*x^2 + 4*x^3)*Log[4])*Log[x]))/(-1 + (1 + x*Log[4])*Log[x]))/((1 + x*Log[4])^2*(-1 + Log[x] + x*Log[4]*Lo
g[x])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (\frac {2 \left (-e^6+x+4 e^3 x-4 x^2+\left (2+e^6-x-4 e^3 x+4 x^2+\left (e^6 x-x^2-4 e^3 x^2+4 x^3\right ) \log (4)\right ) \log (x)\right )}{-1+(1+x \log (4)) \log (x)}\right ) \left (-4+\left (-2-8 e^3\right ) x+16 x^2+\left (4 x+16 e^3 x-32 x^2+\left (4 x^2+16 e^3 x^2-32 x^3\right ) \log (4)\right ) \log (x)+\left (-2 x-8 e^3 x+16 x^2+\left (-4 x-4 x^2-16 e^3 x^2+32 x^3\right ) \log (4)+\left (-2 x^3-8 e^3 x^3+16 x^4\right ) \log ^2(4)\right ) \log ^2(x)\right )}{x+\left (-2 x-2 x^2 \log (4)\right ) \log (x)+\left (x+2 x^2 \log (4)+x^3 \log ^2(4)\right ) \log ^2(x)} \, dx\\ &=\int \frac {\exp \left (\frac {2 \left (-e^6+\left (1+4 e^3\right ) x-4 x^2+\left (2+e^6-x-4 e^3 x+4 x^2+\left (e^6 x-x^2-4 e^3 x^2+4 x^3\right ) \log (4)\right ) \log (x)\right )}{-1+(1+x \log (4)) \log (x)}\right ) \left (-4+\left (-2-8 e^3\right ) x+16 x^2+\left (4 x+16 e^3 x-32 x^2+\left (4 x^2+16 e^3 x^2-32 x^3\right ) \log (4)\right ) \log (x)+\left (-2 x-8 e^3 x+16 x^2+\left (-4 x-4 x^2-16 e^3 x^2+32 x^3\right ) \log (4)+\left (-2 x^3-8 e^3 x^3+16 x^4\right ) \log ^2(4)\right ) \log ^2(x)\right )}{x (1-\log (x)-x \log (4) \log (x))^2} \, dx\\ &=\int \left (\frac {2 \exp \left (\frac {2 \left (-e^6+\left (1+4 e^3\right ) x-4 x^2+\left (2+e^6-x-4 e^3 x+4 x^2+\left (e^6 x-x^2-4 e^3 x^2+4 x^3\right ) \log (4)\right ) \log (x)\right )}{-1+(1+x \log (4)) \log (x)}\right ) \left (-1-4 e^3+8 x^3 \log ^2(4)+x^2 \log (4) \left (16-\log (4)-4 e^3 \log (4)\right )+x \left (8-8 e^3 \log (4)-\log (16)\right )-\log (16)\right )}{(1+x \log (4))^2}-\frac {4 \exp \left (\frac {2 \left (-e^6+\left (1+4 e^3\right ) x-4 x^2+\left (2+e^6-x-4 e^3 x+4 x^2+\left (e^6 x-x^2-4 e^3 x^2+4 x^3\right ) \log (4)\right ) \log (x)\right )}{-1+(1+x \log (4)) \log (x)}\right ) \left (1+x^2 \log ^2(4)+x \log (64)\right )}{x (1+x \log (4))^2 (-1+\log (x)+x \log (4) \log (x))^2}-\frac {8 \exp \left (\frac {2 \left (-e^6+\left (1+4 e^3\right ) x-4 x^2+\left (2+e^6-x-4 e^3 x+4 x^2+\left (e^6 x-x^2-4 e^3 x^2+4 x^3\right ) \log (4)\right ) \log (x)\right )}{-1+(1+x \log (4)) \log (x)}\right ) \log (4)}{(1+x \log (4))^2 (-1+\log (x)+x \log (4) \log (x))}\right ) \, dx\\ &=2 \int \frac {\exp \left (\frac {2 \left (-e^6+\left (1+4 e^3\right ) x-4 x^2+\left (2+e^6-x-4 e^3 x+4 x^2+\left (e^6 x-x^2-4 e^3 x^2+4 x^3\right ) \log (4)\right ) \log (x)\right )}{-1+(1+x \log (4)) \log (x)}\right ) \left (-1-4 e^3+8 x^3 \log ^2(4)+x^2 \log (4) \left (16-\log (4)-4 e^3 \log (4)\right )+x \left (8-8 e^3 \log (4)-\log (16)\right )-\log (16)\right )}{(1+x \log (4))^2} \, dx-4 \int \frac {\exp \left (\frac {2 \left (-e^6+\left (1+4 e^3\right ) x-4 x^2+\left (2+e^6-x-4 e^3 x+4 x^2+\left (e^6 x-x^2-4 e^3 x^2+4 x^3\right ) \log (4)\right ) \log (x)\right )}{-1+(1+x \log (4)) \log (x)}\right ) \left (1+x^2 \log ^2(4)+x \log (64)\right )}{x (1+x \log (4))^2 (-1+\log (x)+x \log (4) \log (x))^2} \, dx-(8 \log (4)) \int \frac {\exp \left (\frac {2 \left (-e^6+\left (1+4 e^3\right ) x-4 x^2+\left (2+e^6-x-4 e^3 x+4 x^2+\left (e^6 x-x^2-4 e^3 x^2+4 x^3\right ) \log (4)\right ) \log (x)\right )}{-1+(1+x \log (4)) \log (x)}\right )}{(1+x \log (4))^2 (-1+\log (x)+x \log (4) \log (x))} \, dx\\ &=2 \int \left (\exp \left (\frac {2 \left (-e^6+\left (1+4 e^3\right ) x-4 x^2+\left (2+e^6-x-4 e^3 x+4 x^2+\left (e^6 x-x^2-4 e^3 x^2+4 x^3\right ) \log (4)\right ) \log (x)\right )}{-1+(1+x \log (4)) \log (x)}\right ) \left (-1-4 e^3\right )+8 \exp \left (\frac {2 \left (-e^6+\left (1+4 e^3\right ) x-4 x^2+\left (2+e^6-x-4 e^3 x+4 x^2+\left (e^6 x-x^2-4 e^3 x^2+4 x^3\right ) \log (4)\right ) \log (x)\right )}{-1+(1+x \log (4)) \log (x)}\right ) x-\frac {\exp \left (\frac {2 \left (-e^6+\left (1+4 e^3\right ) x-4 x^2+\left (2+e^6-x-4 e^3 x+4 x^2+\left (e^6 x-x^2-4 e^3 x^2+4 x^3\right ) \log (4)\right ) \log (x)\right )}{-1+(1+x \log (4)) \log (x)}\right ) \log (16)}{(1+x \log (4))^2}\right ) \, dx-4 \int \left (\frac {\exp \left (\frac {2 \left (-e^6+\left (1+4 e^3\right ) x-4 x^2+\left (2+e^6-x-4 e^3 x+4 x^2+\left (e^6 x-x^2-4 e^3 x^2+4 x^3\right ) \log (4)\right ) \log (x)\right )}{-1+(1+x \log (4)) \log (x)}\right )}{x (-1+\log (x)+x \log (4) \log (x))^2}+\frac {\exp \left (\frac {2 \left (-e^6+\left (1+4 e^3\right ) x-4 x^2+\left (2+e^6-x-4 e^3 x+4 x^2+\left (e^6 x-x^2-4 e^3 x^2+4 x^3\right ) \log (4)\right ) \log (x)\right )}{-1+(1+x \log (4)) \log (x)}\right ) \log (4)}{(1+x \log (4))^2 (-1+\log (x)+x \log (4) \log (x))^2}\right ) \, dx-(8 \log (4)) \int \frac {\exp \left (\frac {2 \left (-e^6+\left (1+4 e^3\right ) x-4 x^2+\left (2+e^6-x-4 e^3 x+4 x^2+\left (e^6 x-x^2-4 e^3 x^2+4 x^3\right ) \log (4)\right ) \log (x)\right )}{-1+(1+x \log (4)) \log (x)}\right )}{(1+x \log (4))^2 (-1+\log (x)+x \log (4) \log (x))} \, dx\\ &=-\left (4 \int \frac {\exp \left (\frac {2 \left (-e^6+\left (1+4 e^3\right ) x-4 x^2+\left (2+e^6-x-4 e^3 x+4 x^2+\left (e^6 x-x^2-4 e^3 x^2+4 x^3\right ) \log (4)\right ) \log (x)\right )}{-1+(1+x \log (4)) \log (x)}\right )}{x (-1+\log (x)+x \log (4) \log (x))^2} \, dx\right )+16 \int \exp \left (\frac {2 \left (-e^6+\left (1+4 e^3\right ) x-4 x^2+\left (2+e^6-x-4 e^3 x+4 x^2+\left (e^6 x-x^2-4 e^3 x^2+4 x^3\right ) \log (4)\right ) \log (x)\right )}{-1+(1+x \log (4)) \log (x)}\right ) x \, dx-\left (2 \left (1+4 e^3\right )\right ) \int \exp \left (\frac {2 \left (-e^6+\left (1+4 e^3\right ) x-4 x^2+\left (2+e^6-x-4 e^3 x+4 x^2+\left (e^6 x-x^2-4 e^3 x^2+4 x^3\right ) \log (4)\right ) \log (x)\right )}{-1+(1+x \log (4)) \log (x)}\right ) \, dx-(4 \log (4)) \int \frac {\exp \left (\frac {2 \left (-e^6+\left (1+4 e^3\right ) x-4 x^2+\left (2+e^6-x-4 e^3 x+4 x^2+\left (e^6 x-x^2-4 e^3 x^2+4 x^3\right ) \log (4)\right ) \log (x)\right )}{-1+(1+x \log (4)) \log (x)}\right )}{(1+x \log (4))^2 (-1+\log (x)+x \log (4) \log (x))^2} \, dx-(8 \log (4)) \int \frac {\exp \left (\frac {2 \left (-e^6+\left (1+4 e^3\right ) x-4 x^2+\left (2+e^6-x-4 e^3 x+4 x^2+\left (e^6 x-x^2-4 e^3 x^2+4 x^3\right ) \log (4)\right ) \log (x)\right )}{-1+(1+x \log (4)) \log (x)}\right )}{(1+x \log (4))^2 (-1+\log (x)+x \log (4) \log (x))} \, dx-(2 \log (16)) \int \frac {\exp \left (\frac {2 \left (-e^6+\left (1+4 e^3\right ) x-4 x^2+\left (2+e^6-x-4 e^3 x+4 x^2+\left (e^6 x-x^2-4 e^3 x^2+4 x^3\right ) \log (4)\right ) \log (x)\right )}{-1+(1+x \log (4)) \log (x)}\right )}{(1+x \log (4))^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 134.27, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{\frac {2 \left (-e^6+x+4 e^3 x-4 x^2+\left (2+e^6-x-4 e^3 x+4 x^2+\left (e^6 x-x^2-4 e^3 x^2+4 x^3\right ) \log (4)\right ) \log (x)\right )}{-1+(1+x \log (4)) \log (x)}} \left (-4-2 x-8 e^3 x+16 x^2+\left (4 x+16 e^3 x-32 x^2+\left (4 x^2+16 e^3 x^2-32 x^3\right ) \log (4)\right ) \log (x)+\left (-2 x-8 e^3 x+16 x^2+\left (-4 x-4 x^2-16 e^3 x^2+32 x^3\right ) \log (4)+\left (-2 x^3-8 e^3 x^3+16 x^4\right ) \log ^2(4)\right ) \log ^2(x)\right )}{x+\left (-2 x-2 x^2 \log (4)\right ) \log (x)+\left (x+2 x^2 \log (4)+x^3 \log ^2(4)\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(E^((2*(-E^6 + x + 4*E^3*x - 4*x^2 + (2 + E^6 - x - 4*E^3*x + 4*x^2 + (E^6*x - x^2 - 4*E^3*x^2 + 4*x
^3)*Log[4])*Log[x]))/(-1 + (1 + x*Log[4])*Log[x]))*(-4 - 2*x - 8*E^3*x + 16*x^2 + (4*x + 16*E^3*x - 32*x^2 + (
4*x^2 + 16*E^3*x^2 - 32*x^3)*Log[4])*Log[x] + (-2*x - 8*E^3*x + 16*x^2 + (-4*x - 4*x^2 - 16*E^3*x^2 + 32*x^3)*
Log[4] + (-2*x^3 - 8*E^3*x^3 + 16*x^4)*Log[4]^2)*Log[x]^2))/(x + (-2*x - 2*x^2*Log[4])*Log[x] + (x + 2*x^2*Log
[4] + x^3*Log[4]^2)*Log[x]^2),x]

[Out]

Integrate[(E^((2*(-E^6 + x + 4*E^3*x - 4*x^2 + (2 + E^6 - x - 4*E^3*x + 4*x^2 + (E^6*x - x^2 - 4*E^3*x^2 + 4*x
^3)*Log[4])*Log[x]))/(-1 + (1 + x*Log[4])*Log[x]))*(-4 - 2*x - 8*E^3*x + 16*x^2 + (4*x + 16*E^3*x - 32*x^2 + (
4*x^2 + 16*E^3*x^2 - 32*x^3)*Log[4])*Log[x] + (-2*x - 8*E^3*x + 16*x^2 + (-4*x - 4*x^2 - 16*E^3*x^2 + 32*x^3)*
Log[4] + (-2*x^3 - 8*E^3*x^3 + 16*x^4)*Log[4]^2)*Log[x]^2))/(x + (-2*x - 2*x^2*Log[4])*Log[x] + (x + 2*x^2*Log
[4] + x^3*Log[4]^2)*Log[x]^2), x]

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fricas [B]  time = 1.05, size = 80, normalized size = 2.42 \begin {gather*} e^{\left (-\frac {2 \, {\left (4 \, x^{2} - 4 \, x e^{3} - {\left (4 \, x^{2} - 4 \, x e^{3} + 2 \, {\left (4 \, x^{3} - 4 \, x^{2} e^{3} - x^{2} + x e^{6}\right )} \log \relax (2) - x + e^{6} + 2\right )} \log \relax (x) - x + e^{6}\right )}}{{\left (2 \, x \log \relax (2) + 1\right )} \log \relax (x) - 1}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*(-8*x^3*exp(3)+16*x^4-2*x^3)*log(2)^2+2*(-16*x^2*exp(3)+32*x^3-4*x^2-4*x)*log(2)-8*x*exp(3)+16*x
^2-2*x)*log(x)^2+(2*(16*x^2*exp(3)-32*x^3+4*x^2)*log(2)+16*x*exp(3)-32*x^2+4*x)*log(x)-8*x*exp(3)+16*x^2-2*x-4
)*exp(((2*(x*exp(3)^2-4*x^2*exp(3)+4*x^3-x^2)*log(2)+exp(3)^2-4*x*exp(3)+4*x^2-x+2)*log(x)-exp(3)^2+4*x*exp(3)
-4*x^2+x)/((2*x*log(2)+1)*log(x)-1))^2/((4*x^3*log(2)^2+4*x^2*log(2)+x)*log(x)^2+(-4*x^2*log(2)-2*x)*log(x)+x)
,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 2.44, size = 255, normalized size = 7.73 \begin {gather*} e^{\left (\frac {16 \, x^{3} \log \relax (2) \log \relax (x)}{2 \, x \log \relax (2) \log \relax (x) + \log \relax (x) - 1} - \frac {16 \, x^{2} e^{3} \log \relax (2) \log \relax (x)}{2 \, x \log \relax (2) \log \relax (x) + \log \relax (x) - 1} - \frac {4 \, x^{2} \log \relax (2) \log \relax (x)}{2 \, x \log \relax (2) \log \relax (x) + \log \relax (x) - 1} + \frac {4 \, x e^{6} \log \relax (2) \log \relax (x)}{2 \, x \log \relax (2) \log \relax (x) + \log \relax (x) - 1} + \frac {8 \, x^{2} \log \relax (x)}{2 \, x \log \relax (2) \log \relax (x) + \log \relax (x) - 1} - \frac {8 \, x e^{3} \log \relax (x)}{2 \, x \log \relax (2) \log \relax (x) + \log \relax (x) - 1} - \frac {8 \, x^{2}}{2 \, x \log \relax (2) \log \relax (x) + \log \relax (x) - 1} + \frac {8 \, x e^{3}}{2 \, x \log \relax (2) \log \relax (x) + \log \relax (x) - 1} - \frac {2 \, x \log \relax (x)}{2 \, x \log \relax (2) \log \relax (x) + \log \relax (x) - 1} + \frac {2 \, e^{6} \log \relax (x)}{2 \, x \log \relax (2) \log \relax (x) + \log \relax (x) - 1} + \frac {2 \, x}{2 \, x \log \relax (2) \log \relax (x) + \log \relax (x) - 1} - \frac {2 \, e^{6}}{2 \, x \log \relax (2) \log \relax (x) + \log \relax (x) - 1} + \frac {4 \, \log \relax (x)}{2 \, x \log \relax (2) \log \relax (x) + \log \relax (x) - 1}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*(-8*x^3*exp(3)+16*x^4-2*x^3)*log(2)^2+2*(-16*x^2*exp(3)+32*x^3-4*x^2-4*x)*log(2)-8*x*exp(3)+16*x
^2-2*x)*log(x)^2+(2*(16*x^2*exp(3)-32*x^3+4*x^2)*log(2)+16*x*exp(3)-32*x^2+4*x)*log(x)-8*x*exp(3)+16*x^2-2*x-4
)*exp(((2*(x*exp(3)^2-4*x^2*exp(3)+4*x^3-x^2)*log(2)+exp(3)^2-4*x*exp(3)+4*x^2-x+2)*log(x)-exp(3)^2+4*x*exp(3)
-4*x^2+x)/((2*x*log(2)+1)*log(x)-1))^2/((4*x^3*log(2)^2+4*x^2*log(2)+x)*log(x)^2+(-4*x^2*log(2)-2*x)*log(x)+x)
,x, algorithm="giac")

[Out]

e^(16*x^3*log(2)*log(x)/(2*x*log(2)*log(x) + log(x) - 1) - 16*x^2*e^3*log(2)*log(x)/(2*x*log(2)*log(x) + log(x
) - 1) - 4*x^2*log(2)*log(x)/(2*x*log(2)*log(x) + log(x) - 1) + 4*x*e^6*log(2)*log(x)/(2*x*log(2)*log(x) + log
(x) - 1) + 8*x^2*log(x)/(2*x*log(2)*log(x) + log(x) - 1) - 8*x*e^3*log(x)/(2*x*log(2)*log(x) + log(x) - 1) - 8
*x^2/(2*x*log(2)*log(x) + log(x) - 1) + 8*x*e^3/(2*x*log(2)*log(x) + log(x) - 1) - 2*x*log(x)/(2*x*log(2)*log(
x) + log(x) - 1) + 2*e^6*log(x)/(2*x*log(2)*log(x) + log(x) - 1) + 2*x/(2*x*log(2)*log(x) + log(x) - 1) - 2*e^
6/(2*x*log(2)*log(x) + log(x) - 1) + 4*log(x)/(2*x*log(2)*log(x) + log(x) - 1))

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maple [B]  time = 0.09, size = 99, normalized size = 3.00

method	result	size
risch	\({\mathrm e}^{-\frac {2 \left (8 \ln \relax (x ) {\mathrm e}^{3} \ln \relax (2) x^{2}-8 \ln \relax (x ) \ln \relax (2) x^{3}-2 \ln \relax (x ) \ln \relax (2) {\mathrm e}^{6} x +2 x^{2} \ln \relax (2) \ln \relax (x )+4 x \,{\mathrm e}^{3} \ln \relax (x )-4 x^{2} \ln \relax (x )-\ln \relax (x ) {\mathrm e}^{6}+x \ln \relax (x )-4 x \,{\mathrm e}^{3}+4 x^{2}-2 \ln \relax (x )+{\mathrm e}^{6}-x \right )}{2 x \ln \relax (2) \ln \relax (x )+\ln \relax (x )-1}}\)	\(99\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*(-8*x^3*exp(3)+16*x^4-2*x^3)*ln(2)^2+2*(-16*x^2*exp(3)+32*x^3-4*x^2-4*x)*ln(2)-8*x*exp(3)+16*x^2-2*x)*
ln(x)^2+(2*(16*x^2*exp(3)-32*x^3+4*x^2)*ln(2)+16*x*exp(3)-32*x^2+4*x)*ln(x)-8*x*exp(3)+16*x^2-2*x-4)*exp(((2*(
x*exp(3)^2-4*x^2*exp(3)+4*x^3-x^2)*ln(2)+exp(3)^2-4*x*exp(3)+4*x^2-x+2)*ln(x)-exp(3)^2+4*x*exp(3)-4*x^2+x)/((2
*x*ln(2)+1)*ln(x)-1))^2/((4*x^3*ln(2)^2+4*x^2*ln(2)+x)*ln(x)^2+(-4*x^2*ln(2)-2*x)*ln(x)+x),x,method=_RETURNVER
BOSE)

[Out]

exp(-2*(8*ln(x)*exp(3)*ln(2)*x^2-8*ln(x)*ln(2)*x^3-2*ln(x)*ln(2)*exp(6)*x+2*x^2*ln(2)*ln(x)+4*x*exp(3)*ln(x)-4
*x^2*ln(x)-ln(x)*exp(6)+x*ln(x)-4*x*exp(3)+4*x^2-2*ln(x)+exp(6)-x)/(2*x*ln(2)*ln(x)+ln(x)-1))

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maxima [A]  time = 1.98, size = 37, normalized size = 1.12 \begin {gather*} e^{\left (8 \, x^{2} - 8 \, x e^{3} - 2 \, x + \frac {4 \, \log \relax (x)}{{\left (2 \, x \log \relax (2) + 1\right )} \log \relax (x) - 1} + 2 \, e^{6}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*(-8*x^3*exp(3)+16*x^4-2*x^3)*log(2)^2+2*(-16*x^2*exp(3)+32*x^3-4*x^2-4*x)*log(2)-8*x*exp(3)+16*x
^2-2*x)*log(x)^2+(2*(16*x^2*exp(3)-32*x^3+4*x^2)*log(2)+16*x*exp(3)-32*x^2+4*x)*log(x)-8*x*exp(3)+16*x^2-2*x-4
)*exp(((2*(x*exp(3)^2-4*x^2*exp(3)+4*x^3-x^2)*log(2)+exp(3)^2-4*x*exp(3)+4*x^2-x+2)*log(x)-exp(3)^2+4*x*exp(3)
-4*x^2+x)/((2*x*log(2)+1)*log(x)-1))^2/((4*x^3*log(2)^2+4*x^2*log(2)+x)*log(x)^2+(-4*x^2*log(2)-2*x)*log(x)+x)
,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.70, size = 138, normalized size = 4.18 \begin {gather*} x^{\frac {2\,\left ({\mathrm {e}}^6-x-4\,x\,{\mathrm {e}}^3-2\,x^2\,\ln \relax (2)+8\,x^3\,\ln \relax (2)+4\,x^2-8\,x^2\,{\mathrm {e}}^3\,\ln \relax (2)+2\,x\,{\mathrm {e}}^6\,\ln \relax (2)+2\right )}{\ln \relax (x)+2\,x\,\ln \relax (2)\,\ln \relax (x)-1}}\,{\mathrm {e}}^{\frac {2\,x}{\ln \relax (x)+2\,x\,\ln \relax (2)\,\ln \relax (x)-1}}\,{\mathrm {e}}^{-\frac {8\,x^2}{\ln \relax (x)+2\,x\,\ln \relax (2)\,\ln \relax (x)-1}}\,{\mathrm {e}}^{\frac {8\,x\,{\mathrm {e}}^3}{\ln \relax (x)+2\,x\,\ln \relax (2)\,\ln \relax (x)-1}}\,{\mathrm {e}}^{-\frac {2\,{\mathrm {e}}^6}{\ln \relax (x)+2\,x\,\ln \relax (2)\,\ln \relax (x)-1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((2*(x - exp(6) + 4*x*exp(3) + log(x)*(exp(6) - x - 4*x*exp(3) + 2*log(2)*(x*exp(6) - 4*x^2*exp(3) -
x^2 + 4*x^3) + 4*x^2 + 2) - 4*x^2))/(log(x)*(2*x*log(2) + 1) - 1))*(2*x - log(x)*(4*x + 2*log(2)*(16*x^2*exp(3
) + 4*x^2 - 32*x^3) + 16*x*exp(3) - 32*x^2) + 8*x*exp(3) - 16*x^2 + log(x)^2*(2*x + 2*log(2)*(4*x + 16*x^2*exp
(3) + 4*x^2 - 32*x^3) + 8*x*exp(3) + 4*log(2)^2*(8*x^3*exp(3) + 2*x^3 - 16*x^4) - 16*x^2) + 4))/(x - log(x)*(2
*x + 4*x^2*log(2)) + log(x)^2*(x + 4*x^3*log(2)^2 + 4*x^2*log(2))),x)

[Out]

x^((2*(exp(6) - x - 4*x*exp(3) - 2*x^2*log(2) + 8*x^3*log(2) + 4*x^2 - 8*x^2*exp(3)*log(2) + 2*x*exp(6)*log(2)
+ 2))/(log(x) + 2*x*log(2)*log(x) - 1))*exp((2*x)/(log(x) + 2*x*log(2)*log(x) - 1))*exp(-(8*x^2)/(log(x) + 2*
x*log(2)*log(x) - 1))*exp((8*x*exp(3))/(log(x) + 2*x*log(2)*log(x) - 1))*exp(-(2*exp(6))/(log(x) + 2*x*log(2)*
log(x) - 1))

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sympy [B]  time = 1.84, size = 87, normalized size = 2.64 \begin {gather*} e^{\frac {2 \left (- 4 x^{2} + x + 4 x e^{3} + \left (4 x^{2} - 4 x e^{3} - x + \left (8 x^{3} - 8 x^{2} e^{3} - 2 x^{2} + 2 x e^{6}\right ) \log {\relax (2 )} + 2 + e^{6}\right ) \log {\relax (x )} - e^{6}\right )}{\left (2 x \log {\relax (2 )} + 1\right ) \log {\relax (x )} - 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*(-8*x**3*exp(3)+16*x**4-2*x**3)*ln(2)**2+2*(-16*x**2*exp(3)+32*x**3-4*x**2-4*x)*ln(2)-8*x*exp(3)
+16*x**2-2*x)*ln(x)**2+(2*(16*x**2*exp(3)-32*x**3+4*x**2)*ln(2)+16*x*exp(3)-32*x**2+4*x)*ln(x)-8*x*exp(3)+16*x
**2-2*x-4)*exp(((2*(x*exp(3)**2-4*x**2*exp(3)+4*x**3-x**2)*ln(2)+exp(3)**2-4*x*exp(3)+4*x**2-x+2)*ln(x)-exp(3)
**2+4*x*exp(3)-4*x**2+x)/((2*x*ln(2)+1)*ln(x)-1))**2/((4*x**3*ln(2)**2+4*x**2*ln(2)+x)*ln(x)**2+(-4*x**2*ln(2)
-2*x)*ln(x)+x),x)

[Out]

exp(2*(-4*x**2 + x + 4*x*exp(3) + (4*x**2 - 4*x*exp(3) - x + (8*x**3 - 8*x**2*exp(3) - 2*x**2 + 2*x*exp(6))*lo
g(2) + 2 + exp(6))*log(x) - exp(6))/((2*x*log(2) + 1)*log(x) - 1))

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