3.21.17 \(\int \frac {18-9 \log (x^2)+(-x+e^{e^2} x-9 \log (x^2)) \log (\frac {x-e^{e^2} x+9 \log (x^2)}{x})}{x^3-e^{e^2} x^3+9 x^2 \log (x^2)} \, dx\)

Optimal. Leaf size=23 \[ \frac {\log \left (1-e^{e^2}+\frac {9 \log \left (x^2\right )}{x}\right )}{x} \]

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Rubi [F]  time = 1.69, 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 {18-9 \log \left (x^2\right )+\left (-x+e^{e^2} x-9 \log \left (x^2\right )\right ) \log \left (\frac {x-e^{e^2} x+9 \log \left (x^2\right )}{x}\right )}{x^3-e^{e^2} x^3+9 x^2 \log \left (x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(18 - 9*Log[x^2] + (-x + E^E^2*x - 9*Log[x^2])*Log[(x - E^E^2*x + 9*Log[x^2])/x])/(x^3 - E^E^2*x^3 + 9*x^2
*Log[x^2]),x]

[Out]

x^(-1) + 18*Defer[Int][1/(x^2*((1 - E^E^2)*x + 9*Log[x^2])), x] + (1 - E^E^2)*Defer[Int][1/(x*((1 - E^E^2)*x +
9*Log[x^2])), x] - Defer[Int][Log[1 - E^E^2 + (9*Log[x^2])/x]/x^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {18-9 \log \left (x^2\right )+\left (-x+e^{e^2} x-9 \log \left (x^2\right )\right ) \log \left (\frac {x-e^{e^2} x+9 \log \left (x^2\right )}{x}\right )}{\left (1-e^{e^2}\right ) x^3+9 x^2 \log \left (x^2\right )} \, dx\\ &=\int \frac {18-9 \log \left (x^2\right )+\left (-x+e^{e^2} x-9 \log \left (x^2\right )\right ) \log \left (\frac {x-e^{e^2} x+9 \log \left (x^2\right )}{x}\right )}{x^2 \left (\left (1-e^{e^2}\right ) x+9 \log \left (x^2\right )\right )} \, dx\\ &=\int \left (\frac {9 \left (2-\log \left (x^2\right )\right )}{x^2 \left (\left (1-e^{e^2}\right ) x+9 \log \left (x^2\right )\right )}-\frac {\log \left (1-e^{e^2}+\frac {9 \log \left (x^2\right )}{x}\right )}{x^2}\right ) \, dx\\ &=9 \int \frac {2-\log \left (x^2\right )}{x^2 \left (\left (1-e^{e^2}\right ) x+9 \log \left (x^2\right )\right )} \, dx-\int \frac {\log \left (1-e^{e^2}+\frac {9 \log \left (x^2\right )}{x}\right )}{x^2} \, dx\\ &=9 \int \left (-\frac {1}{9 x^2}+\frac {18+\left (1-e^{e^2}\right ) x}{9 x^2 \left (\left (1-e^{e^2}\right ) x+9 \log \left (x^2\right )\right )}\right ) \, dx-\int \frac {\log \left (1-e^{e^2}+\frac {9 \log \left (x^2\right )}{x}\right )}{x^2} \, dx\\ &=\frac {1}{x}+\int \frac {18+\left (1-e^{e^2}\right ) x}{x^2 \left (\left (1-e^{e^2}\right ) x+9 \log \left (x^2\right )\right )} \, dx-\int \frac {\log \left (1-e^{e^2}+\frac {9 \log \left (x^2\right )}{x}\right )}{x^2} \, dx\\ &=\frac {1}{x}+\int \left (\frac {18}{x^2 \left (\left (1-e^{e^2}\right ) x+9 \log \left (x^2\right )\right )}+\frac {1-e^{e^2}}{x \left (\left (1-e^{e^2}\right ) x+9 \log \left (x^2\right )\right )}\right ) \, dx-\int \frac {\log \left (1-e^{e^2}+\frac {9 \log \left (x^2\right )}{x}\right )}{x^2} \, dx\\ &=\frac {1}{x}+18 \int \frac {1}{x^2 \left (\left (1-e^{e^2}\right ) x+9 \log \left (x^2\right )\right )} \, dx+\left (1-e^{e^2}\right ) \int \frac {1}{x \left (\left (1-e^{e^2}\right ) x+9 \log \left (x^2\right )\right )} \, dx-\int \frac {\log \left (1-e^{e^2}+\frac {9 \log \left (x^2\right )}{x}\right )}{x^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.52, size = 23, normalized size = 1.00 \begin {gather*} \frac {\log \left (1-e^{e^2}+\frac {9 \log \left (x^2\right )}{x}\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(18 - 9*Log[x^2] + (-x + E^E^2*x - 9*Log[x^2])*Log[(x - E^E^2*x + 9*Log[x^2])/x])/(x^3 - E^E^2*x^3 +
9*x^2*Log[x^2]),x]

[Out]

Log[1 - E^E^2 + (9*Log[x^2])/x]/x

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fricas [A]  time = 0.53, size = 25, normalized size = 1.09 \begin {gather*} \frac {\log \left (-\frac {x e^{\left (e^{2}\right )} - x - 9 \, \log \left (x^{2}\right )}{x}\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-9*log(x^2)+x*exp(exp(2))-x)*log((9*log(x^2)-x*exp(exp(2))+x)/x)-9*log(x^2)+18)/(9*x^2*log(x^2)-x^
3*exp(exp(2))+x^3),x, algorithm="fricas")

[Out]

log(-(x*e^(e^2) - x - 9*log(x^2))/x)/x

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giac [A]  time = 0.38, size = 24, normalized size = 1.04 \begin {gather*} \frac {\log \left (-x e^{\left (e^{2}\right )} + x + 9 \, \log \left (x^{2}\right )\right ) - \log \relax (x)}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-9*log(x^2)+x*exp(exp(2))-x)*log((9*log(x^2)-x*exp(exp(2))+x)/x)-9*log(x^2)+18)/(9*x^2*log(x^2)-x^
3*exp(exp(2))+x^3),x, algorithm="giac")

[Out]

(log(-x*e^(e^2) + x + 9*log(x^2)) - log(x))/x

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[\int \frac {\left (-9 \ln \left (x^{2}\right )+x \,{\mathrm e}^{{\mathrm e}^{2}}-x \right ) \ln \left (\frac {9 \ln \left (x^{2}\right )-x \,{\mathrm e}^{{\mathrm e}^{2}}+x}{x}\right )-9 \ln \left (x^{2}\right )+18}{9 x^{2} \ln \left (x^{2}\right )-x^{3} {\mathrm e}^{{\mathrm e}^{2}}+x^{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-9*ln(x^2)+x*exp(exp(2))-x)*ln((9*ln(x^2)-x*exp(exp(2))+x)/x)-9*ln(x^2)+18)/(9*x^2*ln(x^2)-x^3*exp(exp(2
))+x^3),x)

[Out]

int(((-9*ln(x^2)+x*exp(exp(2))-x)*ln((9*ln(x^2)-x*exp(exp(2))+x)/x)-9*ln(x^2)+18)/(9*x^2*ln(x^2)-x^3*exp(exp(2
))+x^3),x)

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maxima [A]  time = 0.49, size = 23, normalized size = 1.00 \begin {gather*} \frac {\log \left (-x {\left (e^{\left (e^{2}\right )} - 1\right )} + 18 \, \log \relax (x)\right ) - \log \relax (x)}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-9*log(x^2)+x*exp(exp(2))-x)*log((9*log(x^2)-x*exp(exp(2))+x)/x)-9*log(x^2)+18)/(9*x^2*log(x^2)-x^
3*exp(exp(2))+x^3),x, algorithm="maxima")

[Out]

(log(-x*(e^(e^2) - 1) + 18*log(x)) - log(x))/x

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mupad [B]  time = 1.50, size = 21, normalized size = 0.91 \begin {gather*} \frac {\ln \left (\frac {x+\ln \left (x^{18}\right )-x\,{\mathrm {e}}^{{\mathrm {e}}^2}}{x}\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(9*log(x^2) + log((x + 9*log(x^2) - x*exp(exp(2)))/x)*(x + 9*log(x^2) - x*exp(exp(2))) - 18)/(9*x^2*log(x
^2) - x^3*exp(exp(2)) + x^3),x)

[Out]

log((x + log(x^18) - x*exp(exp(2)))/x)/x

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sympy [A]  time = 0.44, size = 19, normalized size = 0.83 \begin {gather*} \frac {\log {\left (\frac {- x e^{e^{2}} + x + 9 \log {\left (x^{2} \right )}}{x} \right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-9*ln(x**2)+x*exp(exp(2))-x)*ln((9*ln(x**2)-x*exp(exp(2))+x)/x)-9*ln(x**2)+18)/(9*x**2*ln(x**2)-x*
*3*exp(exp(2))+x**3),x)

[Out]

log((-x*exp(exp(2)) + x + 9*log(x**2))/x)/x

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