3.22.47 \(\int \frac {e^x (25+25 x-25 x^2) \log (x)+50 x^2 \log (\log (x))-25 x \log (x) \log ^2(\log (x))}{-e^x x^2 \log (x)+x^3 \log (x) \log ^2(\log (x))} \, dx\)

Optimal. Leaf size=23 \[ -1+25 \left (\frac {1}{x}+\log \left (-\frac {e^x}{x}+\log ^2(\log (x))\right )\right ) \]

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Rubi [F]  time = 1.83, 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 {e^x \left (25+25 x-25 x^2\right ) \log (x)+50 x^2 \log (\log (x))-25 x \log (x) \log ^2(\log (x))}{-e^x x^2 \log (x)+x^3 \log (x) \log ^2(\log (x))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^x*(25 + 25*x - 25*x^2)*Log[x] + 50*x^2*Log[Log[x]] - 25*x*Log[x]*Log[Log[x]]^2)/(-(E^x*x^2*Log[x]) + x^
3*Log[x]*Log[Log[x]]^2),x]

[Out]

25/x + 25*x - 25*Log[x] + 50*Defer[Int][Log[Log[x]]/(Log[x]*(-E^x + x*Log[Log[x]]^2)), x] + 25*Defer[Int][Log[
Log[x]]^2/(-E^x + x*Log[Log[x]]^2), x] - 25*Defer[Int][(x*Log[Log[x]]^2)/(-E^x + x*Log[Log[x]]^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-e^x \left (25+25 x-25 x^2\right ) \log (x)-50 x^2 \log (\log (x))+25 x \log (x) \log ^2(\log (x))}{x^2 \log (x) \left (e^x-x \log ^2(\log (x))\right )} \, dx\\ &=\int \left (\frac {25 \left (-1-x+x^2\right )}{x^2}-\frac {25 \log (\log (x)) (-2-\log (x) \log (\log (x))+x \log (x) \log (\log (x)))}{\log (x) \left (-e^x+x \log ^2(\log (x))\right )}\right ) \, dx\\ &=25 \int \frac {-1-x+x^2}{x^2} \, dx-25 \int \frac {\log (\log (x)) (-2-\log (x) \log (\log (x))+x \log (x) \log (\log (x)))}{\log (x) \left (-e^x+x \log ^2(\log (x))\right )} \, dx\\ &=25 \int \left (1-\frac {1}{x^2}-\frac {1}{x}\right ) \, dx-25 \int \left (-\frac {2 \log (\log (x))}{\log (x) \left (-e^x+x \log ^2(\log (x))\right )}-\frac {\log ^2(\log (x))}{-e^x+x \log ^2(\log (x))}+\frac {x \log ^2(\log (x))}{-e^x+x \log ^2(\log (x))}\right ) \, dx\\ &=\frac {25}{x}+25 x-25 \log (x)+25 \int \frac {\log ^2(\log (x))}{-e^x+x \log ^2(\log (x))} \, dx-25 \int \frac {x \log ^2(\log (x))}{-e^x+x \log ^2(\log (x))} \, dx+50 \int \frac {\log (\log (x))}{\log (x) \left (-e^x+x \log ^2(\log (x))\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.35, size = 25, normalized size = 1.09 \begin {gather*} \frac {25}{x}-25 \log (x)+25 \log \left (e^x-x \log ^2(\log (x))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(25 + 25*x - 25*x^2)*Log[x] + 50*x^2*Log[Log[x]] - 25*x*Log[x]*Log[Log[x]]^2)/(-(E^x*x^2*Log[x]
) + x^3*Log[x]*Log[Log[x]]^2),x]

[Out]

25/x - 25*Log[x] + 25*Log[E^x - x*Log[Log[x]]^2]

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fricas [A]  time = 0.72, size = 26, normalized size = 1.13 \begin {gather*} \frac {25 \, {\left (x \log \left (\frac {x \log \left (\log \relax (x)\right )^{2} - e^{x}}{x}\right ) + 1\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-25*x*log(x)*log(log(x))^2+50*x^2*log(log(x))+(-25*x^2+25*x+25)*exp(x)*log(x))/(x^3*log(x)*log(log(
x))^2-x^2*exp(x)*log(x)),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.43, size = 27, normalized size = 1.17 \begin {gather*} \frac {25 \, {\left (x \log \left (x \log \left (\log \relax (x)\right )^{2} - e^{x}\right ) - x \log \relax (x) + 1\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-25*x*log(x)*log(log(x))^2+50*x^2*log(log(x))+(-25*x^2+25*x+25)*exp(x)*log(x))/(x^3*log(x)*log(log(
x))^2-x^2*exp(x)*log(x)),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.03, size = 23, normalized size = 1.00

method	result	size
risch	\(\frac {25}{x}+25 \ln \left (\ln \left (\ln \relax (x )\right )^{2}-\frac {{\mathrm e}^{x}}{x}\right )\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-25*x*ln(x)*ln(ln(x))^2+50*x^2*ln(ln(x))+(-25*x^2+25*x+25)*exp(x)*ln(x))/(x^3*ln(x)*ln(ln(x))^2-x^2*exp(x
)*ln(x)),x,method=_RETURNVERBOSE)

[Out]

25/x+25*ln(ln(ln(x))^2-exp(x)/x)

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maxima [A]  time = 0.90, size = 25, normalized size = 1.09 \begin {gather*} \frac {25}{x} + 25 \, \log \left (\frac {x \log \left (\log \relax (x)\right )^{2} - e^{x}}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-25*x*log(x)*log(log(x))^2+50*x^2*log(log(x))+(-25*x^2+25*x+25)*exp(x)*log(x))/(x^3*log(x)*log(log(
x))^2-x^2*exp(x)*log(x)),x, algorithm="maxima")

[Out]

25/x + 25*log((x*log(log(x))^2 - e^x)/x)

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mupad [B]  time = 1.39, size = 24, normalized size = 1.04 \begin {gather*} 25\,\ln \left (\frac {{\mathrm {e}}^x-x\,{\ln \left (\ln \relax (x)\right )}^2}{x}\right )+\frac {25}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((50*x^2*log(log(x)) - 25*x*log(log(x))^2*log(x) + exp(x)*log(x)*(25*x - 25*x^2 + 25))/(x^3*log(log(x))^2*l
og(x) - x^2*exp(x)*log(x)),x)

[Out]

25*log((exp(x) - x*log(log(x))^2)/x) + 25/x

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sympy [A]  time = 0.39, size = 22, normalized size = 0.96 \begin {gather*} - 25 \log {\relax (x )} + 25 \log {\left (- x \log {\left (\log {\relax (x )} \right )}^{2} + e^{x} \right )} + \frac {25}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-25*x*ln(x)*ln(ln(x))**2+50*x**2*ln(ln(x))+(-25*x**2+25*x+25)*exp(x)*ln(x))/(x**3*ln(x)*ln(ln(x))**
2-x**2*exp(x)*ln(x)),x)

[Out]

-25*log(x) + 25*log(-x*log(log(x))**2 + exp(x)) + 25/x

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