3.8.17 \(\int \frac {e^x (-1+2 x) \log (3)+e^x (-2+x) \log (3) \log (9 e^{-x})+e^x (2-x) \log (3) \log (2 x)}{x^3+x^3 \log ^2(9 e^{-x})-2 x^3 \log (2 x)+x^3 \log ^2(2 x)+\log (9 e^{-x}) (2 x^3-2 x^3 \log (2 x))} \, dx\)

Optimal. Leaf size=30 \[ \frac {e^x \log (3)}{x \left (x+x \log \left (9 e^{-x}\right )-x \log (2 x)\right )} \]

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Rubi [F]  time = 1.75, 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 {e^x (-1+2 x) \log (3)+e^x (-2+x) \log (3) \log \left (9 e^{-x}\right )+e^x (2-x) \log (3) \log (2 x)}{x^3+x^3 \log ^2\left (9 e^{-x}\right )-2 x^3 \log (2 x)+x^3 \log ^2(2 x)+\log \left (9 e^{-x}\right ) \left (2 x^3-2 x^3 \log (2 x)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^x*(-1 + 2*x)*Log[3] + E^x*(-2 + x)*Log[3]*Log[9/E^x] + E^x*(2 - x)*Log[3]*Log[2*x])/(x^3 + x^3*Log[9/E^
x]^2 - 2*x^3*Log[2*x] + x^3*Log[2*x]^2 + Log[9/E^x]*(2*x^3 - 2*x^3*Log[2*x])),x]

[Out]

Log[3]*Defer[Int][E^x/(x^3*(1 + Log[9/E^x] - Log[2*x])^2), x] + Log[3]*Defer[Int][E^x/(x^2*(1 + Log[9/E^x] - L
og[2*x])^2), x] - 2*Log[3]*Defer[Int][E^x/(x^3*(1 + Log[9/E^x] - Log[2*x])), x] + Log[3]*Defer[Int][E^x/(x^2*(
1 + Log[9/E^x] - Log[2*x])), x]

Rubi steps

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

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Mathematica [A]  time = 0.57, size = 28, normalized size = 0.93 \begin {gather*} -\frac {e^x \log (3)}{x^2 \left (-1-\log \left (9 e^{-x}\right )+\log (2 x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(-1 + 2*x)*Log[3] + E^x*(-2 + x)*Log[3]*Log[9/E^x] + E^x*(2 - x)*Log[3]*Log[2*x])/(x^3 + x^3*Lo
g[9/E^x]^2 - 2*x^3*Log[2*x] + x^3*Log[2*x]^2 + Log[9/E^x]*(2*x^3 - 2*x^3*Log[2*x])),x]

[Out]

-((E^x*Log[3])/(x^2*(-1 - Log[9/E^x] + Log[2*x])))

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fricas [A]  time = 0.93, size = 32, normalized size = 1.07 \begin {gather*} -\frac {e^{x} \log \relax (3)}{x^{3} - 2 \, x^{2} \log \relax (3) + x^{2} \log \left (2 \, x\right ) - x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x-2)*log(3)*exp(x)*log(9/exp(x))+(2-x)*log(3)*exp(x)*log(2*x)+(2*x-1)*log(3)*exp(x))/(x^3*log(9/ex
p(x))^2+(-2*x^3*log(2*x)+2*x^3)*log(9/exp(x))+x^3*log(2*x)^2-2*x^3*log(2*x)+x^3),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.34, size = 36, normalized size = 1.20 \begin {gather*} -\frac {e^{x} \log \relax (3)}{x^{3} - 2 \, x^{2} \log \relax (3) + x^{2} \log \relax (2) + x^{2} \log \relax (x) - x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x-2)*log(3)*exp(x)*log(9/exp(x))+(2-x)*log(3)*exp(x)*log(2*x)+(2*x-1)*log(3)*exp(x))/(x^3*log(9/ex
p(x))^2+(-2*x^3*log(2*x)+2*x^3)*log(9/exp(x))+x^3*log(2*x)^2-2*x^3*log(2*x)+x^3),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.17, size = 29, normalized size = 0.97

method	result	size
risch	\(\frac {2 \ln \relax (3) {\mathrm e}^{x}}{x^{2} \left (2+4 \ln \relax (3)-2 \ln \left (2 x \right )-2 \ln \left ({\mathrm e}^{x}\right )\right )}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x-2)*ln(3)*exp(x)*ln(9/exp(x))+(2-x)*ln(3)*exp(x)*ln(2*x)+(2*x-1)*ln(3)*exp(x))/(x^3*ln(9/exp(x))^2+(-2*
x^3*ln(2*x)+2*x^3)*ln(9/exp(x))+x^3*ln(2*x)^2-2*x^3*ln(2*x)+x^3),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.76, size = 33, normalized size = 1.10 \begin {gather*} -\frac {e^{x} \log \relax (3)}{x^{3} - x^{2} {\left (2 \, \log \relax (3) - \log \relax (2) + 1\right )} + x^{2} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x-2)*log(3)*exp(x)*log(9/exp(x))+(2-x)*log(3)*exp(x)*log(2*x)+(2*x-1)*log(3)*exp(x))/(x^3*log(9/ex
p(x))^2+(-2*x^3*log(2*x)+2*x^3)*log(9/exp(x))+x^3*log(2*x)^2-2*x^3*log(2*x)+x^3),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.92, size = 50, normalized size = 1.67 \begin {gather*} \frac {{\mathrm {e}}^x\,\left (\ln \relax (3)-2\,\ln \left (2\,x\right )\,\ln \relax (3)+\ln \left (2\,x\right )\,\ln \relax (9)+x\,\ln \relax (3)\right )}{x^2\,\left (x+1\right )\,\left (\ln \left (9\,{\mathrm {e}}^{-x}\right )-\ln \left (2\,x\right )+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*log(3)*(2*x - 1) + log(9*exp(-x))*exp(x)*log(3)*(x - 2) - log(2*x)*exp(x)*log(3)*(x - 2))/(x^3*log
(9*exp(-x))^2 - 2*x^3*log(2*x) - log(9*exp(-x))*(2*x^3*log(2*x) - 2*x^3) + x^3 + x^3*log(2*x)^2),x)

[Out]

(exp(x)*(log(3) - 2*log(2*x)*log(3) + log(2*x)*log(9) + x*log(3)))/(x^2*(x + 1)*(log(9*exp(-x)) - log(2*x) + 1
))

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sympy [A]  time = 0.42, size = 31, normalized size = 1.03 \begin {gather*} - \frac {e^{x} \log {\relax (3 )}}{x^{3} + x^{2} \log {\left (2 x \right )} - 2 x^{2} \log {\relax (3 )} - x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x-2)*ln(3)*exp(x)*ln(9/exp(x))+(2-x)*ln(3)*exp(x)*ln(2*x)+(2*x-1)*ln(3)*exp(x))/(x**3*ln(9/exp(x))
**2+(-2*x**3*ln(2*x)+2*x**3)*ln(9/exp(x))+x**3*ln(2*x)**2-2*x**3*ln(2*x)+x**3),x)

[Out]

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

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