3.20.71 \(\int \frac {x+e^x (-2 x+x^2) \log (5 x) \log (\log (5 x))+2 x \log (5 x) \log (\log (5 x)) \log (\frac {x^2}{\log (\log (5 x))})}{(1-2 e^x+e^{2 x}) \log (5 x) \log (\log (5 x))+(2-2 e^x) \log (5 x) \log (\log (5 x)) \log (\frac {x^2}{\log (\log (5 x))})+\log (5 x) \log (\log (5 x)) \log ^2(\frac {x^2}{\log (\log (5 x))})} \, dx\)

Optimal. Leaf size=25 \[ \frac {x^2}{1-e^x+\log \left (\frac {x^2}{\log (\log (5 x))}\right )} \]

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Rubi [F]  time = 3.83, 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+e^x \left (-2 x+x^2\right ) \log (5 x) \log (\log (5 x))+2 x \log (5 x) \log (\log (5 x)) \log \left (\frac {x^2}{\log (\log (5 x))}\right )}{\left (1-2 e^x+e^{2 x}\right ) \log (5 x) \log (\log (5 x))+\left (2-2 e^x\right ) \log (5 x) \log (\log (5 x)) \log \left (\frac {x^2}{\log (\log (5 x))}\right )+\log (5 x) \log (\log (5 x)) \log ^2\left (\frac {x^2}{\log (\log (5 x))}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(x + E^x*(-2*x + x^2)*Log[5*x]*Log[Log[5*x]] + 2*x*Log[5*x]*Log[Log[5*x]]*Log[x^2/Log[Log[5*x]]])/((1 - 2*
E^x + E^(2*x))*Log[5*x]*Log[Log[5*x]] + (2 - 2*E^x)*Log[5*x]*Log[Log[5*x]]*Log[x^2/Log[Log[5*x]]] + Log[5*x]*L
og[Log[5*x]]*Log[x^2/Log[Log[5*x]]]^2),x]

[Out]

-2*Defer[Int][x/(-1 + E^x - Log[x^2/Log[Log[5*x]]])^2, x] + Defer[Int][x^2/(-1 + E^x - Log[x^2/Log[Log[5*x]]])
^2, x] + Defer[Int][x/(Log[5*x]*Log[Log[5*x]]*(-1 + E^x - Log[x^2/Log[Log[5*x]]])^2), x] - 2*Defer[Int][x/(-1
+ E^x - Log[x^2/Log[Log[5*x]]]), x] + Defer[Int][x^2/(-1 + E^x - Log[x^2/Log[Log[5*x]]]), x] + Defer[Int][(x^2
*Log[x^2/Log[Log[5*x]]])/(-1 + E^x - Log[x^2/Log[Log[5*x]]])^2, x]

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 25, normalized size = 1.00 \begin {gather*} \frac {x^2}{1-e^x+\log \left (\frac {x^2}{\log (\log (5 x))}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x + E^x*(-2*x + x^2)*Log[5*x]*Log[Log[5*x]] + 2*x*Log[5*x]*Log[Log[5*x]]*Log[x^2/Log[Log[5*x]]])/((
1 - 2*E^x + E^(2*x))*Log[5*x]*Log[Log[5*x]] + (2 - 2*E^x)*Log[5*x]*Log[Log[5*x]]*Log[x^2/Log[Log[5*x]]] + Log[
5*x]*Log[Log[5*x]]*Log[x^2/Log[Log[5*x]]]^2),x]

[Out]

x^2/(1 - E^x + Log[x^2/Log[Log[5*x]]])

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fricas [A]  time = 0.66, size = 25, normalized size = 1.00 \begin {gather*} -\frac {x^{2}}{e^{x} - \log \left (\frac {x^{2}}{\log \left (\log \left (5 \, x\right )\right )}\right ) - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*log(5*x)*log(log(5*x))*log(x^2/log(log(5*x)))+(x^2-2*x)*exp(x)*log(5*x)*log(log(5*x))+x)/(log(5
*x)*log(log(5*x))*log(x^2/log(log(5*x)))^2+(-2*exp(x)+2)*log(5*x)*log(log(5*x))*log(x^2/log(log(5*x)))+(exp(x)
^2-2*exp(x)+1)*log(5*x)*log(log(5*x))),x, algorithm="fricas")

[Out]

-x^2/(e^x - log(x^2/log(log(5*x))) - 1)

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giac [B]  time = 1.73, size = 397, normalized size = 15.88 \begin {gather*} \frac {x^{3} e^{x} \log \relax (5) \log \left (5 \, x\right ) \log \left (\log \left (5 \, x\right )\right ) + x^{3} e^{x} \log \left (5 \, x\right ) \log \relax (x) \log \left (\log \left (5 \, x\right )\right ) - 2 \, x^{2} \log \relax (5) \log \left (5 \, x\right ) \log \left (\log \left (5 \, x\right )\right ) - 2 \, x^{2} \log \left (5 \, x\right ) \log \relax (x) \log \left (\log \left (5 \, x\right )\right ) + x^{2} \log \left (5 \, x\right )}{2 \, x e^{x} \log \relax (5) \log \left (5 \, x\right ) \log \relax (x) \log \left (\log \left (5 \, x\right )\right ) + 2 \, x e^{x} \log \left (5 \, x\right ) \log \relax (x)^{2} \log \left (\log \left (5 \, x\right )\right ) - x e^{x} \log \relax (5) \log \left (5 \, x\right ) \log \left (\log \left (5 \, x\right )\right ) \log \left (\log \left (\log \left (5 \, x\right )\right )\right ) - x e^{x} \log \left (5 \, x\right ) \log \relax (x) \log \left (\log \left (5 \, x\right )\right ) \log \left (\log \left (\log \left (5 \, x\right )\right )\right ) - x e^{\left (2 \, x\right )} \log \relax (5) \log \left (5 \, x\right ) \log \left (\log \left (5 \, x\right )\right ) + x e^{x} \log \relax (5) \log \left (5 \, x\right ) \log \left (\log \left (5 \, x\right )\right ) - x e^{\left (2 \, x\right )} \log \left (5 \, x\right ) \log \relax (x) \log \left (\log \left (5 \, x\right )\right ) + x e^{x} \log \left (5 \, x\right ) \log \relax (x) \log \left (\log \left (5 \, x\right )\right ) + 2 \, e^{x} \log \relax (5) \log \left (5 \, x\right ) \log \left (\log \left (5 \, x\right )\right ) + 2 \, e^{x} \log \left (5 \, x\right ) \log \relax (x) \log \left (\log \left (5 \, x\right )\right ) - 4 \, \log \relax (5) \log \left (5 \, x\right ) \log \relax (x) \log \left (\log \left (5 \, x\right )\right ) - 4 \, \log \left (5 \, x\right ) \log \relax (x)^{2} \log \left (\log \left (5 \, x\right )\right ) + 2 \, \log \relax (5) \log \left (5 \, x\right ) \log \left (\log \left (5 \, x\right )\right ) \log \left (\log \left (\log \left (5 \, x\right )\right )\right ) + 2 \, \log \left (5 \, x\right ) \log \relax (x) \log \left (\log \left (5 \, x\right )\right ) \log \left (\log \left (\log \left (5 \, x\right )\right )\right ) - 2 \, \log \relax (5) \log \left (5 \, x\right ) \log \left (\log \left (5 \, x\right )\right ) - 2 \, \log \left (5 \, x\right ) \log \relax (x) \log \left (\log \left (5 \, x\right )\right ) - e^{x} \log \relax (5) - e^{x} \log \relax (x) + 2 \, \log \relax (5) \log \relax (x) + 2 \, \log \relax (x)^{2} - \log \relax (5) \log \left (\log \left (\log \left (5 \, x\right )\right )\right ) - \log \relax (x) \log \left (\log \left (\log \left (5 \, x\right )\right )\right ) + \log \relax (5) + \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*log(5*x)*log(log(5*x))*log(x^2/log(log(5*x)))+(x^2-2*x)*exp(x)*log(5*x)*log(log(5*x))+x)/(log(5
*x)*log(log(5*x))*log(x^2/log(log(5*x)))^2+(-2*exp(x)+2)*log(5*x)*log(log(5*x))*log(x^2/log(log(5*x)))+(exp(x)
^2-2*exp(x)+1)*log(5*x)*log(log(5*x))),x, algorithm="giac")

[Out]

(x^3*e^x*log(5)*log(5*x)*log(log(5*x)) + x^3*e^x*log(5*x)*log(x)*log(log(5*x)) - 2*x^2*log(5)*log(5*x)*log(log
(5*x)) - 2*x^2*log(5*x)*log(x)*log(log(5*x)) + x^2*log(5*x))/(2*x*e^x*log(5)*log(5*x)*log(x)*log(log(5*x)) + 2
*x*e^x*log(5*x)*log(x)^2*log(log(5*x)) - x*e^x*log(5)*log(5*x)*log(log(5*x))*log(log(log(5*x))) - x*e^x*log(5*
x)*log(x)*log(log(5*x))*log(log(log(5*x))) - x*e^(2*x)*log(5)*log(5*x)*log(log(5*x)) + x*e^x*log(5)*log(5*x)*l
og(log(5*x)) - x*e^(2*x)*log(5*x)*log(x)*log(log(5*x)) + x*e^x*log(5*x)*log(x)*log(log(5*x)) + 2*e^x*log(5)*lo
g(5*x)*log(log(5*x)) + 2*e^x*log(5*x)*log(x)*log(log(5*x)) - 4*log(5)*log(5*x)*log(x)*log(log(5*x)) - 4*log(5*
x)*log(x)^2*log(log(5*x)) + 2*log(5)*log(5*x)*log(log(5*x))*log(log(log(5*x))) + 2*log(5*x)*log(x)*log(log(5*x
))*log(log(log(5*x))) - 2*log(5)*log(5*x)*log(log(5*x)) - 2*log(5*x)*log(x)*log(log(5*x)) - e^x*log(5) - e^x*l
og(x) + 2*log(5)*log(x) + 2*log(x)^2 - log(5)*log(log(log(5*x))) - log(x)*log(log(log(5*x))) + log(5) + log(x)
)

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maple [F]  time = 0.14, size = 0, normalized size = 0.00 \[\int \frac {2 x \ln \left (5 x \right ) \ln \left (\ln \left (5 x \right )\right ) \ln \left (\frac {x^{2}}{\ln \left (\ln \left (5 x \right )\right )}\right )+\left (x^{2}-2 x \right ) {\mathrm e}^{x} \ln \left (5 x \right ) \ln \left (\ln \left (5 x \right )\right )+x}{\ln \left (5 x \right ) \ln \left (\ln \left (5 x \right )\right ) \ln \left (\frac {x^{2}}{\ln \left (\ln \left (5 x \right )\right )}\right )^{2}+\left (-2 \,{\mathrm e}^{x}+2\right ) \ln \left (5 x \right ) \ln \left (\ln \left (5 x \right )\right ) \ln \left (\frac {x^{2}}{\ln \left (\ln \left (5 x \right )\right )}\right )+\left ({\mathrm e}^{2 x}-2 \,{\mathrm e}^{x}+1\right ) \ln \left (5 x \right ) \ln \left (\ln \left (5 x \right )\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x*ln(5*x)*ln(ln(5*x))*ln(x^2/ln(ln(5*x)))+(x^2-2*x)*exp(x)*ln(5*x)*ln(ln(5*x))+x)/(ln(5*x)*ln(ln(5*x))*
ln(x^2/ln(ln(5*x)))^2+(-2*exp(x)+2)*ln(5*x)*ln(ln(5*x))*ln(x^2/ln(ln(5*x)))+(exp(x)^2-2*exp(x)+1)*ln(5*x)*ln(l
n(5*x))),x)

[Out]

int((2*x*ln(5*x)*ln(ln(5*x))*ln(x^2/ln(ln(5*x)))+(x^2-2*x)*exp(x)*ln(5*x)*ln(ln(5*x))+x)/(ln(5*x)*ln(ln(5*x))*
ln(x^2/ln(ln(5*x)))^2+(-2*exp(x)+2)*ln(5*x)*ln(ln(5*x))*ln(x^2/ln(ln(5*x)))+(exp(x)^2-2*exp(x)+1)*ln(5*x)*ln(l
n(5*x))),x)

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maxima [A]  time = 0.59, size = 22, normalized size = 0.88 \begin {gather*} -\frac {x^{2}}{e^{x} - 2 \, \log \relax (x) + \log \left (\log \left (\log \relax (5) + \log \relax (x)\right )\right ) - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*log(5*x)*log(log(5*x))*log(x^2/log(log(5*x)))+(x^2-2*x)*exp(x)*log(5*x)*log(log(5*x))+x)/(log(5
*x)*log(log(5*x))*log(x^2/log(log(5*x)))^2+(-2*exp(x)+2)*log(5*x)*log(log(5*x))*log(x^2/log(log(5*x)))+(exp(x)
^2-2*exp(x)+1)*log(5*x)*log(log(5*x))),x, algorithm="maxima")

[Out]

-x^2/(e^x - 2*log(x) + log(log(log(5) + log(x))) - 1)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {x+2\,x\,\ln \left (5\,x\right )\,\ln \left (\ln \left (5\,x\right )\right )\,\ln \left (\frac {x^2}{\ln \left (\ln \left (5\,x\right )\right )}\right )-\ln \left (5\,x\right )\,{\mathrm {e}}^x\,\ln \left (\ln \left (5\,x\right )\right )\,\left (2\,x-x^2\right )}{\ln \left (5\,x\right )\,\ln \left (\ln \left (5\,x\right )\right )\,{\ln \left (\frac {x^2}{\ln \left (\ln \left (5\,x\right )\right )}\right )}^2-\ln \left (5\,x\right )\,\ln \left (\ln \left (5\,x\right )\right )\,\left (2\,{\mathrm {e}}^x-2\right )\,\ln \left (\frac {x^2}{\ln \left (\ln \left (5\,x\right )\right )}\right )+\ln \left (5\,x\right )\,\ln \left (\ln \left (5\,x\right )\right )\,\left ({\mathrm {e}}^{2\,x}-2\,{\mathrm {e}}^x+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + 2*x*log(5*x)*log(log(5*x))*log(x^2/log(log(5*x))) - log(5*x)*exp(x)*log(log(5*x))*(2*x - x^2))/(log(5
*x)*log(log(5*x))*(exp(2*x) - 2*exp(x) + 1) + log(5*x)*log(log(5*x))*log(x^2/log(log(5*x)))^2 - log(5*x)*log(l
og(5*x))*log(x^2/log(log(5*x)))*(2*exp(x) - 2)),x)

[Out]

int((x + 2*x*log(5*x)*log(log(5*x))*log(x^2/log(log(5*x))) - log(5*x)*exp(x)*log(log(5*x))*(2*x - x^2))/(log(5
*x)*log(log(5*x))*(exp(2*x) - 2*exp(x) + 1) + log(5*x)*log(log(5*x))*log(x^2/log(log(5*x)))^2 - log(5*x)*log(l
og(5*x))*log(x^2/log(log(5*x)))*(2*exp(x) - 2)), x)

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sympy [A]  time = 0.54, size = 20, normalized size = 0.80 \begin {gather*} - \frac {x^{2}}{e^{x} - \log {\left (\frac {x^{2}}{\log {\left (\log {\left (5 x \right )} \right )}} \right )} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*ln(5*x)*ln(ln(5*x))*ln(x**2/ln(ln(5*x)))+(x**2-2*x)*exp(x)*ln(5*x)*ln(ln(5*x))+x)/(ln(5*x)*ln(l
n(5*x))*ln(x**2/ln(ln(5*x)))**2+(-2*exp(x)+2)*ln(5*x)*ln(ln(5*x))*ln(x**2/ln(ln(5*x)))+(exp(x)**2-2*exp(x)+1)*
ln(5*x)*ln(ln(5*x))),x)

[Out]

-x**2/(exp(x) - log(x**2/log(log(5*x))) - 1)

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