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

Optimal. Leaf size=24 \[ 3+\log \left (\frac {1}{3} \left (\log ^{e^2 x}\left (\frac {9}{x^2}\right )-\log (x)\right )\right ) \]

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Rubi [F]  time = 1.84, 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 {-\log \left (\frac {9}{x^2}\right )+\log ^{e^2 x}\left (\frac {9}{x^2}\right ) \left (-2 e^2 x+e^2 x \log \left (\frac {9}{x^2}\right ) \log \left (\log \left (\frac {9}{x^2}\right )\right )\right )}{x \log ^{1+e^2 x}\left (\frac {9}{x^2}\right )-x \log \left (\frac {9}{x^2}\right ) \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.31, size = 39, normalized size = 1.62 \begin {gather*} \log \left (-\log \left (\frac {9}{x^2}\right )-2 \log ^{e^2 x}\left (\frac {9}{x^2}\right )+2 \left (\frac {1}{2} \log \left (\frac {9}{x^2}\right )+\log (x)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.64, size = 25, normalized size = 1.04 \begin {gather*} \log \left (2 \, \log \left (\frac {9}{x^{2}}\right )^{x e^{2}} - 2 \, \log \relax (3) + \log \left (\frac {9}{x^{2}}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x e^{2} \log \left (\frac {9}{x^{2}}\right ) \log \left (\log \left (\frac {9}{x^{2}}\right )\right ) - 2 \, x e^{2}\right )} \log \left (\frac {9}{x^{2}}\right )^{x e^{2}} - \log \left (\frac {9}{x^{2}}\right )}{x \log \left (\frac {9}{x^{2}}\right )^{x e^{2}} \log \left (\frac {9}{x^{2}}\right ) - x \log \relax (x) \log \left (\frac {9}{x^{2}}\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.30, size = 49, normalized size = 2.04

method	result	size
risch	\(\ln \left (-\ln \relax (x )+\left (2 \ln \relax (3)-2 \ln \relax (x )+\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \left (-\mathrm {csgn}\left (i x^{2}\right )+\mathrm {csgn}\left (i x \right )\right )^{2}}{2}\right )^{{\mathrm e}^{2} x}\right )\)	\(49\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x*exp(2)*ln(9/x^2)*ln(ln(9/x^2))-2*exp(2)*x)*exp(x*exp(2)*ln(ln(9/x^2)))-ln(9/x^2))/(x*ln(9/x^2)*exp(x*e
xp(2)*ln(ln(9/x^2)))-x*ln(9/x^2)*ln(x)),x,method=_RETURNVERBOSE)

[Out]

ln(-ln(x)+(2*ln(3)-2*ln(x)+1/2*I*Pi*csgn(I*x^2)*(-csgn(I*x^2)+csgn(I*x))^2)^(exp(2)*x))

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maxima [C]  time = 0.71, size = 59, normalized size = 2.46 \begin {gather*} {\left (i \, \pi + \log \relax (2)\right )} x e^{2} + \log \left ({\left (e^{\left (i \, \pi x e^{2} + x e^{2} \log \relax (2) + x e^{2} \log \left (-\log \relax (3) + \log \relax (x)\right )\right )} - \log \relax (x)\right )} e^{\left (-i \, \pi x e^{2} - x e^{2} \log \relax (2)\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(I*pi + log(2))*x*e^2 + log((e^(I*pi*x*e^2 + x*e^2*log(2) + x*e^2*log(-log(3) + log(x))) - log(x))*e^(-I*pi*x*
e^2 - x*e^2*log(2)))

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mupad [B]  time = 1.31, size = 17, normalized size = 0.71 \begin {gather*} \ln \left ({\ln \left (\frac {9}{x^2}\right )}^{x\,{\mathrm {e}}^2}-\ln \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(log(9/x^2)^(x*exp(2)) - log(x))

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sympy [A]  time = 0.77, size = 20, normalized size = 0.83 \begin {gather*} \log {\left (e^{x e^{2} \log {\left (- 2 \log {\relax (x )} + \log {\relax (9 )} \right )}} - \log {\relax (x )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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