3.17.59 $\int \frac{e^{\frac{1}{-x-4\log (3)+\log (x+e^x\log (x^2))}}(e^x(-16-4x)-8x+6x^2+2x^3+2x^4+16x^3\log (3)+32x^2{\log }^2(3)+e^x(2x^3+16x^2\log (3)+32x{\log }^2(3))\log (x^2)+(-4x^3-16x^2\log (3)+e^x(-4x^2-16x\log (3))\log (x^2))\log (x+e^x\log (x^2))+(2x^2+2e^{x}x\log (x^2)){\log }^2(x+e^x\log (x^2)))}{x^4+8x^3\log (3)+16x^2{\log }^2(3)+e^x(x^3+8x^2\log (3)+16x{\log }^2(3))\log (x^2)+(-2x^3-8x^2\log (3)+e^x(-2x^2-8x\log (3))\log (x^2))\log (x+e^x\log (x^2))+(x^2+e^{x}x\log (x^2)){\log }^2(x+e^x\log (x^2))}dx$

Optimal. Leaf size=28 $$2e^{\frac{1}{-x-4\log (3)+\log (x+e^x\log \left(x^2\right))}}(4+x)$$

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Rubi [F]  time = 180.00, 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 = {} $$\begin{array}{c}\text{\$Aborted}\end{array}$$

Verification is not applicable to the result.

[In]

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

[Out]

$Aborted

Rubi steps

Aborted

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Mathematica [F]  time = 0.50, size = 0, normalized size = 0.00 $$\begin{array}{c}\int \frac{e^{\frac{1}{-x-4\log (3)+\log (x+e^x\log \left(x^2\right))}}(e^x(-16-4x)-8x+6x^2+2x^3+2x^4+16x^3\log (3)+32x^2{\log }^2(3)+e^x(2x^3+16x^2\log (3)+32x{\log }^2(3))\log \left(x^2\right)+(-4x^3-16x^2\log (3)+e^x(-4x^2-16x\log (3))\log \left(x^2\right))\log (x+e^x\log \left(x^2\right))+(2x^2+2e^{x}x\log \left(x^2\right)){\log }^2(x+e^x\log \left(x^2\right)))}{x^4+8x^3\log (3)+16x^2{\log }^2(3)+e^x(x^3+8x^2\log (3)+16x{\log }^2(3))\log \left(x^2\right)+(-2x^3-8x^2\log (3)+e^x(-2x^2-8x\log (3))\log \left(x^2\right))\log (x+e^x\log \left(x^2\right))+(x^2+e^{x}x\log \left(x^2\right)){\log }^2(x+e^x\log \left(x^2\right))}dx\end{array}$$

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.62, size = 28, normalized size = 1.00 $$\begin{array}{c}2(x+4)e^{(-\frac{1}{x+4\log (3)-\log (e^x\log \left(x^2\right)+x)})}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(x)*log(x^2)+2*x^2)*log(exp(x)*log(x^2)+x)^2+((-16*x*log(3)-4*x^2)*exp(x)*log(x^2)-16*x^2*l
og(3)-4*x^3)*log(exp(x)*log(x^2)+x)+(32*x*log(3)^2+16*x^2*log(3)+2*x^3)*exp(x)*log(x^2)+(-16-4*x)*exp(x)+32*x^
2*log(3)^2+16*x^3*log(3)+2*x^4+2*x^3+6*x^2-8*x)*exp(1/(log(exp(x)*log(x^2)+x)-4*log(3)-x))/((x*exp(x)*log(x^2)
+x^2)*log(exp(x)*log(x^2)+x)^2+((-8*x*log(3)-2*x^2)*exp(x)*log(x^2)-8*x^2*log(3)-2*x^3)*log(exp(x)*log(x^2)+x)
+(16*x*log(3)^2+8*x^2*log(3)+x^3)*exp(x)*log(x^2)+16*x^2*log(3)^2+8*x^3*log(3)+x^4),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 4.20, size = 108, normalized size = 3.86 $$\begin{array}{c}2x{e}^{(\frac{x-\log (e^x\log \left(x^2\right)+x)}{4(x\log (3)+4\log (3{)}^2-\log (3)\log (e^x\log \left(x^2\right)+x))}-\frac{1}{4\log (3)})}+8e^{(\frac{x-\log (e^x\log \left(x^2\right)+x)}{4(x\log (3)+4\log (3{)}^2-\log (3)\log (e^x\log \left(x^2\right)+x))}-\frac{1}{4\log (3)})}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(x)*log(x^2)+2*x^2)*log(exp(x)*log(x^2)+x)^2+((-16*x*log(3)-4*x^2)*exp(x)*log(x^2)-16*x^2*l
og(3)-4*x^3)*log(exp(x)*log(x^2)+x)+(32*x*log(3)^2+16*x^2*log(3)+2*x^3)*exp(x)*log(x^2)+(-16-4*x)*exp(x)+32*x^
2*log(3)^2+16*x^3*log(3)+2*x^4+2*x^3+6*x^2-8*x)*exp(1/(log(exp(x)*log(x^2)+x)-4*log(3)-x))/((x*exp(x)*log(x^2)
+x^2)*log(exp(x)*log(x^2)+x)^2+((-8*x*log(3)-2*x^2)*exp(x)*log(x^2)-8*x^2*log(3)-2*x^3)*log(exp(x)*log(x^2)+x)
+(16*x*log(3)^2+8*x^2*log(3)+x^3)*exp(x)*log(x^2)+16*x^2*log(3)^2+8*x^3*log(3)+x^4),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.24, size = 59, normalized size = 2.11

method	result	size
risch	$(2x+8){\mathrm{e}}^{-\frac{1}{-\ln ({\mathrm{e}}^x(2\ln (x)-\frac{i\pi \mathrm{csgn}\left(i{x}^2\right){(-\mathrm{csgn}\left(i{x}^2\right)+\mathrm{csgn}\left(ix\right))}^2}{2})+x)+4\ln (3)+x}}$	$59$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x*exp(x)*ln(x^2)+2*x^2)*ln(exp(x)*ln(x^2)+x)^2+((-16*x*ln(3)-4*x^2)*exp(x)*ln(x^2)-16*x^2*ln(3)-4*x^3)
*ln(exp(x)*ln(x^2)+x)+(32*x*ln(3)^2+16*x^2*ln(3)+2*x^3)*exp(x)*ln(x^2)+(-16-4*x)*exp(x)+32*x^2*ln(3)^2+16*x^3*
ln(3)+2*x^4+2*x^3+6*x^2-8*x)*exp(1/(ln(exp(x)*ln(x^2)+x)-4*ln(3)-x))/((x*exp(x)*ln(x^2)+x^2)*ln(exp(x)*ln(x^2)
+x)^2+((-8*x*ln(3)-2*x^2)*exp(x)*ln(x^2)-8*x^2*ln(3)-2*x^3)*ln(exp(x)*ln(x^2)+x)+(16*x*ln(3)^2+8*x^2*ln(3)+x^3
)*exp(x)*ln(x^2)+16*x^2*ln(3)^2+8*x^3*ln(3)+x^4),x,method=_RETURNVERBOSE)

[Out]

(2*x+8)*exp(-1/(-ln(exp(x)*(2*ln(x)-1/2*I*Pi*csgn(I*x^2)*(-csgn(I*x^2)+csgn(I*x))^2)+x)+4*ln(3)+x))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 $$\begin{array}{c}\frac{2x^4{e}^{(-\frac{1}{x+4\log (3)-\log (2e^x\log (x)+x)})}}{x^2-x-2e^x}+\frac{16x^3{e}^{(-\frac{1}{x+4\log (3)-\log (2e^x\log (x)+x)})}\log (3)}{x^2-x-2e^x}+\frac{32x^2{e}^{(-\frac{1}{x+4\log (3)-\log (2e^x\log (x)+x)})}\log (3{)}^2}{x^2-x-2e^x}+\frac{4x^3{e}^{(x-\frac{1}{x+4\log (3)-\log (2e^x\log (x)+x)})}\log (x)}{x^2-x-2e^x}+\frac{32x^2{e}^{(x-\frac{1}{x+4\log (3)-\log (2e^x\log (x)+x)})}\log (3)\log (x)}{x^2-x-2e^x}+\frac{64x{e}^{(x-\frac{1}{x+4\log (3)-\log (2e^x\log (x)+x)})}\log (3{)}^2\log (x)}{x^2-x-2e^x}+\frac{2x^3{e}^{(-\frac{1}{x+4\log (3)-\log (2e^x\log (x)+x)})}}{x^2-x-2e^x}+\frac{6x^2{e}^{(-\frac{1}{x+4\log (3)-\log (2e^x\log (x)+x)})}}{x^2-x-2e^x}-\frac{4x{e}^{(x-\frac{1}{x+4\log (3)-\log (2e^x\log (x)+x)})}}{x^2-x-2e^x}-\frac{8x{e}^{(-\frac{1}{x+4\log (3)-\log (2e^x\log (x)+x)})}}{x^2-x-2e^x}-\frac{16e^{(x-\frac{1}{x+4\log (3)-\log (2e^x\log (x)+x)})}}{x^2-x-2e^x}+2\int -\frac{(2(x+4\log (3))\log (2e^x\log (x)+x)-\log {(2e^x\log (x)+x)}^2)e^{(-\frac{1}{x+4\log (3)-\log (2e^x\log (x)+x)})}}{x^2+8x\log (3)+16\log (3{)}^2-2(x+4\log (3))\log (2e^x\log (x)+x)+\log {(2e^x\log (x)+x)}^2}dx\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(x)*log(x^2)+2*x^2)*log(exp(x)*log(x^2)+x)^2+((-16*x*log(3)-4*x^2)*exp(x)*log(x^2)-16*x^2*l
og(3)-4*x^3)*log(exp(x)*log(x^2)+x)+(32*x*log(3)^2+16*x^2*log(3)+2*x^3)*exp(x)*log(x^2)+(-16-4*x)*exp(x)+32*x^
2*log(3)^2+16*x^3*log(3)+2*x^4+2*x^3+6*x^2-8*x)*exp(1/(log(exp(x)*log(x^2)+x)-4*log(3)-x))/((x*exp(x)*log(x^2)
+x^2)*log(exp(x)*log(x^2)+x)^2+((-8*x*log(3)-2*x^2)*exp(x)*log(x^2)-8*x^2*log(3)-2*x^3)*log(exp(x)*log(x^2)+x)
+(16*x*log(3)^2+8*x^2*log(3)+x^3)*exp(x)*log(x^2)+16*x^2*log(3)^2+8*x^3*log(3)+x^4),x, algorithm="maxima")

[Out]

2*x^4*e^(-1/(x + 4*log(3) - log(2*e^x*log(x) + x)))/(x^2 - x - 2*e^x) + 16*x^3*e^(-1/(x + 4*log(3) - log(2*e^x
*log(x) + x)))*log(3)/(x^2 - x - 2*e^x) + 32*x^2*e^(-1/(x + 4*log(3) - log(2*e^x*log(x) + x)))*log(3)^2/(x^2 -
x - 2*e^x) + 4*x^3*e^(x - 1/(x + 4*log(3) - log(2*e^x*log(x) + x)))*log(x)/(x^2 - x - 2*e^x) + 32*x^2*e^(x -
1/(x + 4*log(3) - log(2*e^x*log(x) + x)))*log(3)*log(x)/(x^2 - x - 2*e^x) + 64*x*e^(x - 1/(x + 4*log(3) - log(
2*e^x*log(x) + x)))*log(3)^2*log(x)/(x^2 - x - 2*e^x) + 2*x^3*e^(-1/(x + 4*log(3) - log(2*e^x*log(x) + x)))/(x
^2 - x - 2*e^x) + 6*x^2*e^(-1/(x + 4*log(3) - log(2*e^x*log(x) + x)))/(x^2 - x - 2*e^x) - 4*x*e^(x - 1/(x + 4*
log(3) - log(2*e^x*log(x) + x)))/(x^2 - x - 2*e^x) - 8*x*e^(-1/(x + 4*log(3) - log(2*e^x*log(x) + x)))/(x^2 -
x - 2*e^x) - 16*e^(x - 1/(x + 4*log(3) - log(2*e^x*log(x) + x)))/(x^2 - x - 2*e^x) + 2*integrate(-(2*(x + 4*lo
g(3))*log(2*e^x*log(x) + x) - log(2*e^x*log(x) + x)^2)*e^(-1/(x + 4*log(3) - log(2*e^x*log(x) + x)))/(x^2 + 8*
x*log(3) + 16*log(3)^2 - 2*(x + 4*log(3))*log(2*e^x*log(x) + x) + log(2*e^x*log(x) + x)^2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 $$\begin{array}{c}\int \frac{{\mathrm{e}}^{-\frac{1}{x+4\ln (3)-\ln (x+\ln \left(x^2\right){\mathrm{e}}^x)}}(32x^2{\ln (3)}^2-8x+{\ln (x+\ln \left(x^2\right){\mathrm{e}}^x)}^2(2x^2+2x\ln \left(x^2\right){\mathrm{e}}^x)-{\mathrm{e}}^x(4x+16)+16x^3\ln (3)-\ln (x+\ln \left(x^2\right){\mathrm{e}}^x)(16x^2\ln (3)+4x^3+\ln \left(x^2\right){\mathrm{e}}^x(4x^2+16\ln (3)x))+6x^2+2x^3+2x^4+\ln \left(x^2\right){\mathrm{e}}^x(2x^3+16\ln (3)x^2+32{\ln (3)}^{2}x))}{16x^2{\ln (3)}^2+8x^3\ln (3)+{\ln (x+\ln \left(x^2\right){\mathrm{e}}^x)}^2(x^2+x\ln \left(x^2\right){\mathrm{e}}^x)-\ln (x+\ln \left(x^2\right){\mathrm{e}}^x)(8x^2\ln (3)+2x^3+\ln \left(x^2\right){\mathrm{e}}^x(2x^2+8\ln (3)x))+x^4+\ln \left(x^2\right){\mathrm{e}}^x(x^3+8\ln (3)x^2+16{\ln (3)}^{2}x)}dx\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(x)*ln(x**2)+2*x**2)*ln(exp(x)*ln(x**2)+x)**2+((-16*x*ln(3)-4*x**2)*exp(x)*ln(x**2)-16*x**2
*ln(3)-4*x**3)*ln(exp(x)*ln(x**2)+x)+(32*x*ln(3)**2+16*x**2*ln(3)+2*x**3)*exp(x)*ln(x**2)+(-16-4*x)*exp(x)+32*
x**2*ln(3)**2+16*x**3*ln(3)+2*x**4+2*x**3+6*x**2-8*x)*exp(1/(ln(exp(x)*ln(x**2)+x)-4*ln(3)-x))/((x*exp(x)*ln(x
**2)+x**2)*ln(exp(x)*ln(x**2)+x)**2+((-8*x*ln(3)-2*x**2)*exp(x)*ln(x**2)-8*x**2*ln(3)-2*x**3)*ln(exp(x)*ln(x**
2)+x)+(16*x*ln(3)**2+8*x**2*ln(3)+x**3)*exp(x)*ln(x**2)+16*x**2*ln(3)**2+8*x**3*ln(3)+x**4),x)

[Out]

Timed out

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