3.33.99 \(\int \frac {-16 e^x x-16 e^x \log (x)+(e^x (-8-4 x+4 x^2)+e^x (4+4 x) \log (x)) \log (9 x^2)}{(x^3+3 x^2 \log (x)+3 x \log ^2(x)+\log ^3(x)) \log ^3(9 x^2)} \, dx\)

Optimal. Leaf size=22 \[ -2+\frac {4 e^x x}{(x+\log (x))^2 \log ^2\left (9 x^2\right )} \]

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Rubi [F]  time = 2.65, 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 {-16 e^x x-16 e^x \log (x)+\left (e^x \left (-8-4 x+4 x^2\right )+e^x (4+4 x) \log (x)\right ) \log \left (9 x^2\right )}{\left (x^3+3 x^2 \log (x)+3 x \log ^2(x)+\log ^3(x)\right ) \log ^3\left (9 x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-16*E^x*x - 16*E^x*Log[x] + (E^x*(-8 - 4*x + 4*x^2) + E^x*(4 + 4*x)*Log[x])*Log[9*x^2])/((x^3 + 3*x^2*Log
[x] + 3*x*Log[x]^2 + Log[x]^3)*Log[9*x^2]^3),x]

[Out]

-16*Defer[Int][E^x/((x + Log[x])^2*Log[9*x^2]^3), x] - 8*Defer[Int][E^x/((x + Log[x])^3*Log[9*x^2]^2), x] - 4*
Defer[Int][(E^x*x)/((x + Log[x])^3*Log[9*x^2]^2), x] + 4*Defer[Int][(E^x*x^2)/((x + Log[x])^3*Log[9*x^2]^2), x
] + 4*Defer[Int][(E^x*Log[x])/((x + Log[x])^3*Log[9*x^2]^2), x] + 4*Defer[Int][(E^x*x*Log[x])/((x + Log[x])^3*
Log[9*x^2]^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 e^x \left (-4 x+\left (-2-x+x^2\right ) \log \left (9 x^2\right )+\log (x) \left (-4+(1+x) \log \left (9 x^2\right )\right )\right )}{(x+\log (x))^3 \log ^3\left (9 x^2\right )} \, dx\\ &=4 \int \frac {e^x \left (-4 x+\left (-2-x+x^2\right ) \log \left (9 x^2\right )+\log (x) \left (-4+(1+x) \log \left (9 x^2\right )\right )\right )}{(x+\log (x))^3 \log ^3\left (9 x^2\right )} \, dx\\ &=4 \int \left (-\frac {4 e^x}{(x+\log (x))^2 \log ^3\left (9 x^2\right )}+\frac {e^x (1+x) (-2+x+\log (x))}{(x+\log (x))^3 \log ^2\left (9 x^2\right )}\right ) \, dx\\ &=4 \int \frac {e^x (1+x) (-2+x+\log (x))}{(x+\log (x))^3 \log ^2\left (9 x^2\right )} \, dx-16 \int \frac {e^x}{(x+\log (x))^2 \log ^3\left (9 x^2\right )} \, dx\\ &=4 \int \left (\frac {e^x (-2+x+\log (x))}{(x+\log (x))^3 \log ^2\left (9 x^2\right )}+\frac {e^x x (-2+x+\log (x))}{(x+\log (x))^3 \log ^2\left (9 x^2\right )}\right ) \, dx-16 \int \frac {e^x}{(x+\log (x))^2 \log ^3\left (9 x^2\right )} \, dx\\ &=4 \int \frac {e^x (-2+x+\log (x))}{(x+\log (x))^3 \log ^2\left (9 x^2\right )} \, dx+4 \int \frac {e^x x (-2+x+\log (x))}{(x+\log (x))^3 \log ^2\left (9 x^2\right )} \, dx-16 \int \frac {e^x}{(x+\log (x))^2 \log ^3\left (9 x^2\right )} \, dx\\ &=4 \int \left (-\frac {2 e^x}{(x+\log (x))^3 \log ^2\left (9 x^2\right )}+\frac {e^x x}{(x+\log (x))^3 \log ^2\left (9 x^2\right )}+\frac {e^x \log (x)}{(x+\log (x))^3 \log ^2\left (9 x^2\right )}\right ) \, dx+4 \int \left (-\frac {2 e^x x}{(x+\log (x))^3 \log ^2\left (9 x^2\right )}+\frac {e^x x^2}{(x+\log (x))^3 \log ^2\left (9 x^2\right )}+\frac {e^x x \log (x)}{(x+\log (x))^3 \log ^2\left (9 x^2\right )}\right ) \, dx-16 \int \frac {e^x}{(x+\log (x))^2 \log ^3\left (9 x^2\right )} \, dx\\ &=4 \int \frac {e^x x}{(x+\log (x))^3 \log ^2\left (9 x^2\right )} \, dx+4 \int \frac {e^x x^2}{(x+\log (x))^3 \log ^2\left (9 x^2\right )} \, dx+4 \int \frac {e^x \log (x)}{(x+\log (x))^3 \log ^2\left (9 x^2\right )} \, dx+4 \int \frac {e^x x \log (x)}{(x+\log (x))^3 \log ^2\left (9 x^2\right )} \, dx-8 \int \frac {e^x}{(x+\log (x))^3 \log ^2\left (9 x^2\right )} \, dx-8 \int \frac {e^x x}{(x+\log (x))^3 \log ^2\left (9 x^2\right )} \, dx-16 \int \frac {e^x}{(x+\log (x))^2 \log ^3\left (9 x^2\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.48, size = 20, normalized size = 0.91 \begin {gather*} \frac {4 e^x x}{(x+\log (x))^2 \log ^2\left (9 x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-16*E^x*x - 16*E^x*Log[x] + (E^x*(-8 - 4*x + 4*x^2) + E^x*(4 + 4*x)*Log[x])*Log[9*x^2])/((x^3 + 3*x
^2*Log[x] + 3*x*Log[x]^2 + Log[x]^3)*Log[9*x^2]^3),x]

[Out]

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

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fricas [B]  time = 0.54, size = 64, normalized size = 2.91 \begin {gather*} \frac {x e^{x}}{x^{2} \log \relax (3)^{2} + 2 \, {\left (x + \log \relax (3)\right )} \log \relax (x)^{3} + \log \relax (x)^{4} + {\left (x^{2} + 4 \, x \log \relax (3) + \log \relax (3)^{2}\right )} \log \relax (x)^{2} + 2 \, {\left (x^{2} \log \relax (3) + x \log \relax (3)^{2}\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x+4)*exp(x)*log(x)+(4*x^2-4*x-8)*exp(x))*log(9*x^2)-16*exp(x)*log(x)-16*exp(x)*x)/(log(x)^3+3*x
*log(x)^2+3*x^2*log(x)+x^3)/log(9*x^2)^3,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.40, size = 78, normalized size = 3.55 \begin {gather*} \frac {x e^{x}}{x^{2} \log \relax (3)^{2} + 2 \, x^{2} \log \relax (3) \log \relax (x) + 2 \, x \log \relax (3)^{2} \log \relax (x) + x^{2} \log \relax (x)^{2} + 4 \, x \log \relax (3) \log \relax (x)^{2} + \log \relax (3)^{2} \log \relax (x)^{2} + 2 \, x \log \relax (x)^{3} + 2 \, \log \relax (3) \log \relax (x)^{3} + \log \relax (x)^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x+4)*exp(x)*log(x)+(4*x^2-4*x-8)*exp(x))*log(9*x^2)-16*exp(x)*log(x)-16*exp(x)*x)/(log(x)^3+3*x
*log(x)^2+3*x^2*log(x)+x^3)/log(9*x^2)^3,x, algorithm="giac")

[Out]

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

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maple [F]  time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (4 x +4\right ) {\mathrm e}^{x} \ln \relax (x )+\left (4 x^{2}-4 x -8\right ) {\mathrm e}^{x}\right ) \ln \left (9 x^{2}\right )-16 \,{\mathrm e}^{x} \ln \relax (x )-16 \,{\mathrm e}^{x} x}{\left (\ln \relax (x )^{3}+3 x \ln \relax (x )^{2}+3 x^{2} \ln \relax (x )+x^{3}\right ) \ln \left (9 x^{2}\right )^{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((4*x+4)*exp(x)*ln(x)+(4*x^2-4*x-8)*exp(x))*ln(9*x^2)-16*exp(x)*ln(x)-16*exp(x)*x)/(ln(x)^3+3*x*ln(x)^2+3
*x^2*ln(x)+x^3)/ln(9*x^2)^3,x)

[Out]

int((((4*x+4)*exp(x)*ln(x)+(4*x^2-4*x-8)*exp(x))*ln(9*x^2)-16*exp(x)*ln(x)-16*exp(x)*x)/(ln(x)^3+3*x*ln(x)^2+3
*x^2*ln(x)+x^3)/ln(9*x^2)^3,x)

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maxima [B]  time = 0.64, size = 64, normalized size = 2.91 \begin {gather*} \frac {x e^{x}}{x^{2} \log \relax (3)^{2} + 2 \, {\left (x + \log \relax (3)\right )} \log \relax (x)^{3} + \log \relax (x)^{4} + {\left (x^{2} + 4 \, x \log \relax (3) + \log \relax (3)^{2}\right )} \log \relax (x)^{2} + 2 \, {\left (x^{2} \log \relax (3) + x \log \relax (3)^{2}\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x+4)*exp(x)*log(x)+(4*x^2-4*x-8)*exp(x))*log(9*x^2)-16*exp(x)*log(x)-16*exp(x)*x)/(log(x)^3+3*x
*log(x)^2+3*x^2*log(x)+x^3)/log(9*x^2)^3,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int -\frac {16\,{\mathrm {e}}^x\,\ln \relax (x)+\ln \left (9\,x^2\right )\,\left ({\mathrm {e}}^x\,\left (-4\,x^2+4\,x+8\right )-{\mathrm {e}}^x\,\ln \relax (x)\,\left (4\,x+4\right )\right )+16\,x\,{\mathrm {e}}^x}{{\ln \left (9\,x^2\right )}^3\,\left (x^3+3\,x^2\,\ln \relax (x)+3\,x\,{\ln \relax (x)}^2+{\ln \relax (x)}^3\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(16*exp(x)*log(x) + log(9*x^2)*(exp(x)*(4*x - 4*x^2 + 8) - exp(x)*log(x)*(4*x + 4)) + 16*x*exp(x))/(log(9
*x^2)^3*(3*x*log(x)^2 + 3*x^2*log(x) + log(x)^3 + x^3)),x)

[Out]

int(-(16*exp(x)*log(x) + log(9*x^2)*(exp(x)*(4*x - 4*x^2 + 8) - exp(x)*log(x)*(4*x + 4)) + 16*x*exp(x))/(log(9
*x^2)^3*(3*x*log(x)^2 + 3*x^2*log(x) + log(x)^3 + x^3)), x)

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sympy [B]  time = 0.70, size = 90, normalized size = 4.09 \begin {gather*} \frac {x e^{x}}{x^{2} \log {\relax (x )}^{2} + 2 x^{2} \log {\relax (3 )} \log {\relax (x )} + x^{2} \log {\relax (3 )}^{2} + 2 x \log {\relax (x )}^{3} + 4 x \log {\relax (3 )} \log {\relax (x )}^{2} + 2 x \log {\relax (3 )}^{2} \log {\relax (x )} + \log {\relax (x )}^{4} + 2 \log {\relax (3 )} \log {\relax (x )}^{3} + \log {\relax (3 )}^{2} \log {\relax (x )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x+4)*exp(x)*ln(x)+(4*x**2-4*x-8)*exp(x))*ln(9*x**2)-16*exp(x)*ln(x)-16*exp(x)*x)/(ln(x)**3+3*x*
ln(x)**2+3*x**2*ln(x)+x**3)/ln(9*x**2)**3,x)

[Out]

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

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