3.28.8 \(\int \frac {-4 e^{3 x}+8 x+8 x^2+e^{2 x} (16+16 x)+e^x (-16-36 x-16 x^2)+(8 x-4 e^{3 x} x+e^{2 x} (16 x+16 x^2)+e^x (-20 x-28 x^2-16 x^3)) \log (x)+(2 x-e^{3 x} x+e^{2 x} (4 x+4 x^2)+e^x (-5 x-7 x^2-4 x^3)) \log ^2(x)}{(e^{4 x} x+x^3+e^{3 x} (-4 x-4 x^2)+e^x (-4 x^2-4 x^3)+e^{2 x} (4 x+10 x^2+4 x^3)) \log ^2(x)} \, dx\)

Optimal. Leaf size=27 \[ \frac {4+\log (x)}{\left (e^x+\frac {x}{-2+e^x-2 x}\right ) \log (x)} \]

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Rubi [F]  time = 8.99, 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 {-4 e^{3 x}+8 x+8 x^2+e^{2 x} (16+16 x)+e^x \left (-16-36 x-16 x^2\right )+\left (8 x-4 e^{3 x} x+e^{2 x} \left (16 x+16 x^2\right )+e^x \left (-20 x-28 x^2-16 x^3\right )\right ) \log (x)+\left (2 x-e^{3 x} x+e^{2 x} \left (4 x+4 x^2\right )+e^x \left (-5 x-7 x^2-4 x^3\right )\right ) \log ^2(x)}{\left (e^{4 x} x+x^3+e^{3 x} \left (-4 x-4 x^2\right )+e^x \left (-4 x^2-4 x^3\right )+e^{2 x} \left (4 x+10 x^2+4 x^3\right )\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-4*E^(3*x) + 8*x + 8*x^2 + E^(2*x)*(16 + 16*x) + E^x*(-16 - 36*x - 16*x^2) + (8*x - 4*E^(3*x)*x + E^(2*x)
*(16*x + 16*x^2) + E^x*(-20*x - 28*x^2 - 16*x^3))*Log[x] + (2*x - E^(3*x)*x + E^(2*x)*(4*x + 4*x^2) + E^x*(-5*
x - 7*x^2 - 4*x^3))*Log[x]^2)/((E^(4*x)*x + x^3 + E^(3*x)*(-4*x - 4*x^2) + E^x*(-4*x^2 - 4*x^3) + E^(2*x)*(4*x
+ 10*x^2 + 4*x^3))*Log[x]^2),x]

[Out]

2*Defer[Int][(-2*E^x + E^(2*x) + x - 2*E^x*x)^(-2), x] - Defer[Int][E^x/(-2*E^x + E^(2*x) + x - 2*E^x*x)^2, x]
+ 2*Defer[Int][(E^x*x)/(-2*E^x + E^(2*x) + x - 2*E^x*x)^2, x] + 2*Defer[Int][(-2*E^x + E^(2*x) + x - 2*E^x*x)
^(-1), x] - Defer[Int][E^x/(-2*E^x + E^(2*x) + x - 2*E^x*x), x] - 2*Defer[Int][x/(2*E^x - E^(2*x) - x + 2*E^x*
x), x] - 2*Defer[Int][x/(E^(2*x) + x - 2*E^x*(1 + x))^2, x] - 2*Defer[Int][x^2/(E^(2*x) + x - 2*E^x*(1 + x))^2
, x] + 8*Defer[Int][1/((-2*E^x + E^(2*x) + x - 2*E^x*x)*Log[x]^2), x] - 4*Defer[Int][E^x/(x*(-2*E^x + E^(2*x)
+ x - 2*E^x*x)*Log[x]^2), x] - 8*Defer[Int][1/(x*(2*E^x - E^(2*x) - x + 2*E^x*x)*Log[x]^2), x] + 8*Defer[Int][
1/((-2*E^x + E^(2*x) + x - 2*E^x*x)^2*Log[x]), x] - 4*Defer[Int][E^x/((-2*E^x + E^(2*x) + x - 2*E^x*x)^2*Log[x
]), x] + 8*Defer[Int][(E^x*x)/((-2*E^x + E^(2*x) + x - 2*E^x*x)^2*Log[x]), x] + 8*Defer[Int][1/((-2*E^x + E^(2
*x) + x - 2*E^x*x)*Log[x]), x] - 4*Defer[Int][E^x/((-2*E^x + E^(2*x) + x - 2*E^x*x)*Log[x]), x] - 8*Defer[Int]
[x/((2*E^x - E^(2*x) - x + 2*E^x*x)*Log[x]), x] - 8*Defer[Int][x/((E^(2*x) + x - 2*E^x*(1 + x))^2*Log[x]), x]
- 8*Defer[Int][x^2/((E^(2*x) + x - 2*E^x*(1 + x))^2*Log[x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-4 \left (e^{3 x}-4 e^{2 x} (1+x)-2 x (1+x)+e^x \left (4+9 x+4 x^2\right )\right )-4 x \left (-2+e^{3 x}-4 e^{2 x} (1+x)+e^x \left (5+7 x+4 x^2\right )\right ) \log (x)-x \left (-2+e^{3 x}-4 e^{2 x} (1+x)+e^x \left (5+7 x+4 x^2\right )\right ) \log ^2(x)}{x \left (e^{2 x}+x-2 e^x (1+x)\right )^2 \log ^2(x)} \, dx\\ &=\int \left (\frac {\left (2-e^x-2 x+2 e^x x-2 x^2\right ) (4+\log (x))}{\left (-2 e^x+e^{2 x}+x-2 e^x x\right )^2 \log (x)}-\frac {\left (2-e^x+2 x\right ) \left (4+4 x \log (x)+x \log ^2(x)\right )}{x \left (2 e^x-e^{2 x}-x+2 e^x x\right ) \log ^2(x)}\right ) \, dx\\ &=\int \frac {\left (2-e^x-2 x+2 e^x x-2 x^2\right ) (4+\log (x))}{\left (-2 e^x+e^{2 x}+x-2 e^x x\right )^2 \log (x)} \, dx-\int \frac {\left (2-e^x+2 x\right ) \left (4+4 x \log (x)+x \log ^2(x)\right )}{x \left (2 e^x-e^{2 x}-x+2 e^x x\right ) \log ^2(x)} \, dx\\ &=\int \left (\frac {2}{\left (-2 e^x+e^{2 x}+x-2 e^x x\right )^2}-\frac {e^x}{\left (-2 e^x+e^{2 x}+x-2 e^x x\right )^2}+\frac {2 e^x x}{\left (-2 e^x+e^{2 x}+x-2 e^x x\right )^2}-\frac {2 x}{\left (2 e^x-e^{2 x}-x+2 e^x x\right )^2}-\frac {2 x^2}{\left (2 e^x-e^{2 x}-x+2 e^x x\right )^2}+\frac {8}{\left (-2 e^x+e^{2 x}+x-2 e^x x\right )^2 \log (x)}-\frac {4 e^x}{\left (-2 e^x+e^{2 x}+x-2 e^x x\right )^2 \log (x)}+\frac {8 e^x x}{\left (-2 e^x+e^{2 x}+x-2 e^x x\right )^2 \log (x)}-\frac {8 x}{\left (2 e^x-e^{2 x}-x+2 e^x x\right )^2 \log (x)}-\frac {8 x^2}{\left (2 e^x-e^{2 x}-x+2 e^x x\right )^2 \log (x)}\right ) \, dx-\int \left (-\frac {2}{-2 e^x+e^{2 x}+x-2 e^x x}+\frac {e^x}{-2 e^x+e^{2 x}+x-2 e^x x}+\frac {2 x}{2 e^x-e^{2 x}-x+2 e^x x}-\frac {8}{\left (-2 e^x+e^{2 x}+x-2 e^x x\right ) \log ^2(x)}+\frac {4 e^x}{x \left (-2 e^x+e^{2 x}+x-2 e^x x\right ) \log ^2(x)}+\frac {8}{x \left (2 e^x-e^{2 x}-x+2 e^x x\right ) \log ^2(x)}-\frac {8}{\left (-2 e^x+e^{2 x}+x-2 e^x x\right ) \log (x)}+\frac {4 e^x}{\left (-2 e^x+e^{2 x}+x-2 e^x x\right ) \log (x)}+\frac {8 x}{\left (2 e^x-e^{2 x}-x+2 e^x x\right ) \log (x)}\right ) \, dx\\ &=2 \int \frac {1}{\left (-2 e^x+e^{2 x}+x-2 e^x x\right )^2} \, dx+2 \int \frac {e^x x}{\left (-2 e^x+e^{2 x}+x-2 e^x x\right )^2} \, dx+2 \int \frac {1}{-2 e^x+e^{2 x}+x-2 e^x x} \, dx-2 \int \frac {x}{\left (2 e^x-e^{2 x}-x+2 e^x x\right )^2} \, dx-2 \int \frac {x^2}{\left (2 e^x-e^{2 x}-x+2 e^x x\right )^2} \, dx-2 \int \frac {x}{2 e^x-e^{2 x}-x+2 e^x x} \, dx-4 \int \frac {e^x}{x \left (-2 e^x+e^{2 x}+x-2 e^x x\right ) \log ^2(x)} \, dx-4 \int \frac {e^x}{\left (-2 e^x+e^{2 x}+x-2 e^x x\right )^2 \log (x)} \, dx-4 \int \frac {e^x}{\left (-2 e^x+e^{2 x}+x-2 e^x x\right ) \log (x)} \, dx+8 \int \frac {1}{\left (-2 e^x+e^{2 x}+x-2 e^x x\right ) \log ^2(x)} \, dx-8 \int \frac {1}{x \left (2 e^x-e^{2 x}-x+2 e^x x\right ) \log ^2(x)} \, dx+8 \int \frac {1}{\left (-2 e^x+e^{2 x}+x-2 e^x x\right )^2 \log (x)} \, dx+8 \int \frac {e^x x}{\left (-2 e^x+e^{2 x}+x-2 e^x x\right )^2 \log (x)} \, dx+8 \int \frac {1}{\left (-2 e^x+e^{2 x}+x-2 e^x x\right ) \log (x)} \, dx-8 \int \frac {x}{\left (2 e^x-e^{2 x}-x+2 e^x x\right )^2 \log (x)} \, dx-8 \int \frac {x^2}{\left (2 e^x-e^{2 x}-x+2 e^x x\right )^2 \log (x)} \, dx-8 \int \frac {x}{\left (2 e^x-e^{2 x}-x+2 e^x x\right ) \log (x)} \, dx-\int \frac {e^x}{\left (-2 e^x+e^{2 x}+x-2 e^x x\right )^2} \, dx-\int \frac {e^x}{-2 e^x+e^{2 x}+x-2 e^x x} \, dx\\ &=2 \int \frac {1}{\left (-2 e^x+e^{2 x}+x-2 e^x x\right )^2} \, dx+2 \int \frac {e^x x}{\left (-2 e^x+e^{2 x}+x-2 e^x x\right )^2} \, dx+2 \int \frac {1}{-2 e^x+e^{2 x}+x-2 e^x x} \, dx-2 \int \frac {x}{2 e^x-e^{2 x}-x+2 e^x x} \, dx-2 \int \frac {x}{\left (e^{2 x}+x-2 e^x (1+x)\right )^2} \, dx-2 \int \frac {x^2}{\left (e^{2 x}+x-2 e^x (1+x)\right )^2} \, dx-4 \int \frac {e^x}{x \left (-2 e^x+e^{2 x}+x-2 e^x x\right ) \log ^2(x)} \, dx-4 \int \frac {e^x}{\left (-2 e^x+e^{2 x}+x-2 e^x x\right )^2 \log (x)} \, dx-4 \int \frac {e^x}{\left (-2 e^x+e^{2 x}+x-2 e^x x\right ) \log (x)} \, dx+8 \int \frac {1}{\left (-2 e^x+e^{2 x}+x-2 e^x x\right ) \log ^2(x)} \, dx-8 \int \frac {1}{x \left (2 e^x-e^{2 x}-x+2 e^x x\right ) \log ^2(x)} \, dx+8 \int \frac {1}{\left (-2 e^x+e^{2 x}+x-2 e^x x\right )^2 \log (x)} \, dx+8 \int \frac {e^x x}{\left (-2 e^x+e^{2 x}+x-2 e^x x\right )^2 \log (x)} \, dx+8 \int \frac {1}{\left (-2 e^x+e^{2 x}+x-2 e^x x\right ) \log (x)} \, dx-8 \int \frac {x}{\left (2 e^x-e^{2 x}-x+2 e^x x\right ) \log (x)} \, dx-8 \int \frac {x}{\left (e^{2 x}+x-2 e^x (1+x)\right )^2 \log (x)} \, dx-8 \int \frac {x^2}{\left (e^{2 x}+x-2 e^x (1+x)\right )^2 \log (x)} \, dx-\int \frac {e^x}{\left (-2 e^x+e^{2 x}+x-2 e^x x\right )^2} \, dx-\int \frac {e^x}{-2 e^x+e^{2 x}+x-2 e^x x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.16, size = 35, normalized size = 1.30 \begin {gather*} \frac {\left (e^x-2 (1+x)\right ) (4+\log (x))}{\left (e^{2 x}+x-2 e^x (1+x)\right ) \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4*E^(3*x) + 8*x + 8*x^2 + E^(2*x)*(16 + 16*x) + E^x*(-16 - 36*x - 16*x^2) + (8*x - 4*E^(3*x)*x + E
^(2*x)*(16*x + 16*x^2) + E^x*(-20*x - 28*x^2 - 16*x^3))*Log[x] + (2*x - E^(3*x)*x + E^(2*x)*(4*x + 4*x^2) + E^
x*(-5*x - 7*x^2 - 4*x^3))*Log[x]^2)/((E^(4*x)*x + x^3 + E^(3*x)*(-4*x - 4*x^2) + E^x*(-4*x^2 - 4*x^3) + E^(2*x
)*(4*x + 10*x^2 + 4*x^3))*Log[x]^2),x]

[Out]

((E^x - 2*(1 + x))*(4 + Log[x]))/((E^(2*x) + x - 2*E^x*(1 + x))*Log[x])

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fricas [A]  time = 0.58, size = 45, normalized size = 1.67 \begin {gather*} \frac {{\left (2 \, x - e^{x} + 2\right )} \log \relax (x) + 8 \, x - 4 \, e^{x} + 8}{{\left (2 \, {\left (x + 1\right )} e^{x} - x - e^{\left (2 \, x\right )}\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*exp(x)^3+(4*x^2+4*x)*exp(x)^2+(-4*x^3-7*x^2-5*x)*exp(x)+2*x)*log(x)^2+(-4*x*exp(x)^3+(16*x^2+16
*x)*exp(x)^2+(-16*x^3-28*x^2-20*x)*exp(x)+8*x)*log(x)-4*exp(x)^3+(16*x+16)*exp(x)^2+(-16*x^2-36*x-16)*exp(x)+8
*x^2+8*x)/(x*exp(x)^4+(-4*x^2-4*x)*exp(x)^3+(4*x^3+10*x^2+4*x)*exp(x)^2+(-4*x^3-4*x^2)*exp(x)+x^3)/log(x)^2,x,
algorithm="fricas")

[Out]

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

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giac [B]  time = 0.44, size = 209, normalized size = 7.74 \begin {gather*} \frac {2 \, {\left (8 \, x^{4} \log \relax (x)^{2} - 4 \, x^{3} e^{x} \log \relax (x)^{2} + 16 \, x^{4} \log \relax (x) - 8 \, x^{3} e^{x} \log \relax (x) + 8 \, x^{3} \log \relax (x)^{2} + 16 \, x^{3} \log \relax (x) - 16 \, x^{3} + 8 \, x^{2} e^{x} - 16 \, x^{2} \log \relax (x) + 8 \, x e^{x} \log \relax (x) - 2 \, x \log \relax (x)^{2} + e^{x} \log \relax (x)^{2} - 24 \, x^{2} + 8 \, x e^{x} - 12 \, x \log \relax (x) + 2 \, e^{x} \log \relax (x) - 2 \, \log \relax (x)^{2} - 12 \, x - 8 \, \log \relax (x) - 4\right )}}{8 \, x^{4} e^{x} \log \relax (x)^{2} - 4 \, x^{4} \log \relax (x)^{2} - 4 \, x^{3} e^{\left (2 \, x\right )} \log \relax (x)^{2} + 8 \, x^{3} e^{x} \log \relax (x)^{2} - 2 \, x e^{x} \log \relax (x)^{2} + x \log \relax (x)^{2} + e^{\left (2 \, x\right )} \log \relax (x)^{2} - 2 \, e^{x} \log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*exp(x)^3+(4*x^2+4*x)*exp(x)^2+(-4*x^3-7*x^2-5*x)*exp(x)+2*x)*log(x)^2+(-4*x*exp(x)^3+(16*x^2+16
*x)*exp(x)^2+(-16*x^3-28*x^2-20*x)*exp(x)+8*x)*log(x)-4*exp(x)^3+(16*x+16)*exp(x)^2+(-16*x^2-36*x-16)*exp(x)+8
*x^2+8*x)/(x*exp(x)^4+(-4*x^2-4*x)*exp(x)^3+(4*x^3+10*x^2+4*x)*exp(x)^2+(-4*x^3-4*x^2)*exp(x)+x^3)/log(x)^2,x,
algorithm="giac")

[Out]

2*(8*x^4*log(x)^2 - 4*x^3*e^x*log(x)^2 + 16*x^4*log(x) - 8*x^3*e^x*log(x) + 8*x^3*log(x)^2 + 16*x^3*log(x) - 1
6*x^3 + 8*x^2*e^x - 16*x^2*log(x) + 8*x*e^x*log(x) - 2*x*log(x)^2 + e^x*log(x)^2 - 24*x^2 + 8*x*e^x - 12*x*log
(x) + 2*e^x*log(x) - 2*log(x)^2 - 12*x - 8*log(x) - 4)/(8*x^4*e^x*log(x)^2 - 4*x^4*log(x)^2 - 4*x^3*e^(2*x)*lo
g(x)^2 + 8*x^3*e^x*log(x)^2 - 2*x*e^x*log(x)^2 + x*log(x)^2 + e^(2*x)*log(x)^2 - 2*e^x*log(x)^2)

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maple [B]  time = 0.05, size = 69, normalized size = 2.56

method	result	size
risch	\(\frac {-{\mathrm e}^{x}+2 x +2}{2 \,{\mathrm e}^{x} x -{\mathrm e}^{2 x}-x +2 \,{\mathrm e}^{x}}+\frac {-4 \,{\mathrm e}^{x}+8 x +8}{\left (2 \,{\mathrm e}^{x} x -{\mathrm e}^{2 x}-x +2 \,{\mathrm e}^{x}\right ) \ln \relax (x )}\)	\(69\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x*exp(x)^3+(4*x^2+4*x)*exp(x)^2+(-4*x^3-7*x^2-5*x)*exp(x)+2*x)*ln(x)^2+(-4*x*exp(x)^3+(16*x^2+16*x)*exp
(x)^2+(-16*x^3-28*x^2-20*x)*exp(x)+8*x)*ln(x)-4*exp(x)^3+(16*x+16)*exp(x)^2+(-16*x^2-36*x-16)*exp(x)+8*x^2+8*x
)/(x*exp(x)^4+(-4*x^2-4*x)*exp(x)^3+(4*x^3+10*x^2+4*x)*exp(x)^2+(-4*x^3-4*x^2)*exp(x)+x^3)/ln(x)^2,x,method=_R
ETURNVERBOSE)

[Out]

(-exp(x)+2*x+2)/(2*exp(x)*x-exp(2*x)-x+2*exp(x))+4*(-exp(x)+2*x+2)/(2*exp(x)*x-exp(2*x)-x+2*exp(x))/ln(x)

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maxima [A]  time = 0.84, size = 46, normalized size = 1.70 \begin {gather*} -\frac {{\left (\log \relax (x) + 4\right )} e^{x} - 2 \, {\left (x + 1\right )} \log \relax (x) - 8 \, x - 8}{2 \, {\left (x + 1\right )} e^{x} \log \relax (x) - x \log \relax (x) - e^{\left (2 \, x\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*exp(x)^3+(4*x^2+4*x)*exp(x)^2+(-4*x^3-7*x^2-5*x)*exp(x)+2*x)*log(x)^2+(-4*x*exp(x)^3+(16*x^2+16
*x)*exp(x)^2+(-16*x^3-28*x^2-20*x)*exp(x)+8*x)*log(x)-4*exp(x)^3+(16*x+16)*exp(x)^2+(-16*x^2-36*x-16)*exp(x)+8
*x^2+8*x)/(x*exp(x)^4+(-4*x^2-4*x)*exp(x)^3+(4*x^3+10*x^2+4*x)*exp(x)^2+(-4*x^3-4*x^2)*exp(x)+x^3)/log(x)^2,x,
algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {8\,x-4\,{\mathrm {e}}^{3\,x}+\ln \relax (x)\,\left (8\,x+{\mathrm {e}}^{2\,x}\,\left (16\,x^2+16\,x\right )-4\,x\,{\mathrm {e}}^{3\,x}-{\mathrm {e}}^x\,\left (16\,x^3+28\,x^2+20\,x\right )\right )-{\mathrm {e}}^x\,\left (16\,x^2+36\,x+16\right )+{\mathrm {e}}^{2\,x}\,\left (16\,x+16\right )+8\,x^2+{\ln \relax (x)}^2\,\left (2\,x+{\mathrm {e}}^{2\,x}\,\left (4\,x^2+4\,x\right )-x\,{\mathrm {e}}^{3\,x}-{\mathrm {e}}^x\,\left (4\,x^3+7\,x^2+5\,x\right )\right )}{{\ln \relax (x)}^2\,\left (x\,{\mathrm {e}}^{4\,x}-{\mathrm {e}}^x\,\left (4\,x^3+4\,x^2\right )-{\mathrm {e}}^{3\,x}\,\left (4\,x^2+4\,x\right )+{\mathrm {e}}^{2\,x}\,\left (4\,x^3+10\,x^2+4\,x\right )+x^3\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x - 4*exp(3*x) + log(x)*(8*x + exp(2*x)*(16*x + 16*x^2) - 4*x*exp(3*x) - exp(x)*(20*x + 28*x^2 + 16*x^3
)) - exp(x)*(36*x + 16*x^2 + 16) + exp(2*x)*(16*x + 16) + 8*x^2 + log(x)^2*(2*x + exp(2*x)*(4*x + 4*x^2) - x*e
xp(3*x) - exp(x)*(5*x + 7*x^2 + 4*x^3)))/(log(x)^2*(x*exp(4*x) - exp(x)*(4*x^2 + 4*x^3) - exp(3*x)*(4*x + 4*x^
2) + exp(2*x)*(4*x + 10*x^2 + 4*x^3) + x^3)),x)

[Out]

int((8*x - 4*exp(3*x) + log(x)*(8*x + exp(2*x)*(16*x + 16*x^2) - 4*x*exp(3*x) - exp(x)*(20*x + 28*x^2 + 16*x^3
)) - exp(x)*(36*x + 16*x^2 + 16) + exp(2*x)*(16*x + 16) + 8*x^2 + log(x)^2*(2*x + exp(2*x)*(4*x + 4*x^2) - x*e
xp(3*x) - exp(x)*(5*x + 7*x^2 + 4*x^3)))/(log(x)^2*(x*exp(4*x) - exp(x)*(4*x^2 + 4*x^3) - exp(3*x)*(4*x + 4*x^
2) + exp(2*x)*(4*x + 10*x^2 + 4*x^3) + x^3)), x)

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sympy [B]  time = 0.43, size = 54, normalized size = 2.00 \begin {gather*} \frac {- 2 x \log {\relax (x )} - 8 x + \left (\log {\relax (x )} + 4\right ) e^{x} - 2 \log {\relax (x )} - 8}{x \log {\relax (x )} + \left (- 2 x \log {\relax (x )} - 2 \log {\relax (x )}\right ) e^{x} + e^{2 x} \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*exp(x)**3+(4*x**2+4*x)*exp(x)**2+(-4*x**3-7*x**2-5*x)*exp(x)+2*x)*ln(x)**2+(-4*x*exp(x)**3+(16*
x**2+16*x)*exp(x)**2+(-16*x**3-28*x**2-20*x)*exp(x)+8*x)*ln(x)-4*exp(x)**3+(16*x+16)*exp(x)**2+(-16*x**2-36*x-
16)*exp(x)+8*x**2+8*x)/(x*exp(x)**4+(-4*x**2-4*x)*exp(x)**3+(4*x**3+10*x**2+4*x)*exp(x)**2+(-4*x**3-4*x**2)*ex
p(x)+x**3)/ln(x)**2,x)

[Out]

(-2*x*log(x) - 8*x + (log(x) + 4)*exp(x) - 2*log(x) - 8)/(x*log(x) + (-2*x*log(x) - 2*log(x))*exp(x) + exp(2*x
)*log(x))

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