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

Optimal. Leaf size=21 \[ \frac {x+\frac {\log ^2(x)}{-4+e^{-x}}}{\log (x)} \]

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Rubi [F]  time = 0.62, 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 {-x+8 e^x x-16 e^{2 x} x+\left (x-8 e^x x+16 e^{2 x} x\right ) \log (x)+\left (e^x-4 e^{2 x}\right ) \log ^2(x)+e^x x \log ^3(x)}{\left (x-8 e^x x+16 e^{2 x} x\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 22, normalized size = 1.05 \begin {gather*} \frac {x}{\log (x)}+\frac {e^x \log (x)}{1-4 e^x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x/Log[x] + (E^x*Log[x])/(1 - 4*E^x)

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fricas [A]  time = 0.64, size = 28, normalized size = 1.33 \begin {gather*} -\frac {e^{x} \log \relax (x)^{2} - 4 \, x e^{x} + x}{{\left (4 \, e^{x} - 1\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.33, size = 29, normalized size = 1.38 \begin {gather*} -\frac {e^{x} \log \relax (x)^{2} - 4 \, x e^{x} + x}{4 \, e^{x} \log \relax (x) - \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 24, normalized size = 1.14

method	result	size
risch	\(-\frac {\ln \relax (x )}{4 \left (4 \,{\mathrm e}^{x}-1\right )}-\frac {\ln \relax (x )}{4}+\frac {x}{\ln \relax (x )}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/(4*exp(x)-1)*ln(x)-1/4*ln(x)+x/ln(x)

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maxima [A]  time = 0.45, size = 35, normalized size = 1.67 \begin {gather*} \frac {16 \, x e^{x} - \log \relax (x)^{2} - 4 \, x}{4 \, {\left (4 \, e^{x} \log \relax (x) - \log \relax (x)\right )}} - \frac {1}{4} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(16*x*e^x - log(x)^2 - 4*x)/(4*e^x*log(x) - log(x)) - 1/4*log(x)

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mupad [B]  time = 0.40, size = 28, normalized size = 1.33 \begin {gather*} -\frac {{\mathrm {e}}^x\,{\ln \relax (x)}^2+x-4\,x\,{\mathrm {e}}^x}{\ln \relax (x)\,\left (4\,{\mathrm {e}}^x-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.29, size = 19, normalized size = 0.90 \begin {gather*} \frac {x}{\log {\relax (x )}} - \frac {\log {\relax (x )}}{4} - \frac {\log {\relax (x )}}{16 e^{x} - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/log(x) - log(x)/4 - log(x)/(16*exp(x) - 4)

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