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

Optimal. Leaf size=21 \[ \log \left (\frac {\log \left (-2 x^3 \log (3)\right )}{e^x+\log \left (\log \left (x^2\right )\right )}\right ) \]

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Rubi [F]  time = 2.08, 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 {-2 \log \left (-2 x^3 \log (3)\right )+\log \left (x^2\right ) \left (3 e^x-e^x x \log \left (-2 x^3 \log (3)\right )\right )+3 \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{e^x x \log \left (x^2\right ) \log \left (-2 x^3 \log (3)\right )+x \log \left (x^2\right ) \log \left (-2 x^3 \log (3)\right ) \log \left (\log \left (x^2\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

-x + Log[Log[-(x^3*Log[9])]] - 2*Defer[Int][1/(x*Log[x^2]*(E^x + Log[Log[x^2]])), x] + Defer[Int][Log[Log[x^2]
]/(E^x + Log[Log[x^2]]), x]

Rubi steps

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

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Mathematica [A]  time = 0.34, size = 40, normalized size = 1.90 \begin {gather*} \log \left (3 \log \left (x^2\right )+2 \left (-\frac {3}{2} \log \left (x^2\right )+\log \left (-x^3 \log (9)\right )\right )\right )-\log \left (e^x+\log \left (\log \left (x^2\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.61, size = 37, normalized size = 1.76 \begin {gather*} -\log \left (e^{x} + \log \left (\frac {2}{3} \, \log \left (-2 \, x^{3} \log \relax (3)\right ) - \frac {1}{3} \, \log \left (4 \, \log \relax (3)^{2}\right )\right )\right ) + \log \left (\log \left (-2 \, x^{3} \log \relax (3)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(e^x + log(2/3*log(-2*x^3*log(3)) - 1/3*log(4*log(3)^2))) + log(log(-2*x^3*log(3)))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left (x e^{x} \log \left (-2 \, x^{3} \log \relax (3)\right ) - 3 \, e^{x}\right )} \log \left (x^{2}\right ) - 3 \, \log \left (x^{2}\right ) \log \left (\log \left (x^{2}\right )\right ) + 2 \, \log \left (-2 \, x^{3} \log \relax (3)\right )}{x e^{x} \log \left (-2 \, x^{3} \log \relax (3)\right ) \log \left (x^{2}\right ) + x \log \left (-2 \, x^{3} \log \relax (3)\right ) \log \left (x^{2}\right ) \log \left (\log \left (x^{2}\right )\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.71, size = 189, normalized size = 9.00

method	result	size
risch	\(\ln \left (\ln \relax (x )-\frac {i \left (\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )-\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{3}\right )^{2}+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}-\pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )^{2}+2 \pi \mathrm {csgn}\left (i x^{3}\right )^{2}-\pi \mathrm {csgn}\left (i x^{3}\right )^{3}+2 i \ln \relax (2)+2 i \ln \left (\ln \relax (3)\right )-2 \pi \right )}{6}\right )-\ln \left ({\mathrm e}^{x}+\ln \left (2 \ln \relax (x )-\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \left (-\mathrm {csgn}\left (i x^{2}\right )+\mathrm {csgn}\left (i x \right )\right )^{2}}{2}\right )\right )\)	\(189\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(ln(x)-1/6*I*(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x)*csgn(I*x^2)*csgn(I*x^3)-P
i*csgn(I*x)*csgn(I*x^3)^2+Pi*csgn(I*x^2)^3-Pi*csgn(I*x^2)*csgn(I*x^3)^2+2*Pi*csgn(I*x^3)^2-Pi*csgn(I*x^3)^3+2*
I*ln(2)+2*I*ln(ln(3))-2*Pi))-ln(exp(x)+ln(2*ln(x)-1/2*I*Pi*csgn(I*x^2)*(-csgn(I*x^2)+csgn(I*x))^2))

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maxima [A]  time = 0.53, size = 27, normalized size = 1.29 \begin {gather*} -\log \left (e^{x} + \log \relax (2) + \log \left (\log \relax (x)\right )\right ) + \log \left (\frac {1}{3} \, \log \relax (2) + \log \left (-x\right ) + \frac {1}{3} \, \log \left (\log \relax (3)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(e^x + log(2) + log(log(x))) + log(1/3*log(2) + log(-x) + 1/3*log(log(3)))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int \frac {\ln \left (x^2\right )\,\left (3\,{\mathrm {e}}^x-x\,\ln \left (-2\,x^3\,\ln \relax (3)\right )\,{\mathrm {e}}^x\right )-2\,\ln \left (-2\,x^3\,\ln \relax (3)\right )+3\,\ln \left (x^2\right )\,\ln \left (\ln \left (x^2\right )\right )}{x\,\ln \left (-2\,x^3\,\ln \relax (3)\right )\,\ln \left (x^2\right )\,{\mathrm {e}}^x+x\,\ln \left (-2\,x^3\,\ln \relax (3)\right )\,\ln \left (x^2\right )\,\ln \left (\ln \left (x^2\right )\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 0.84, size = 39, normalized size = 1.86 \begin {gather*} - \log {\left (e^{x} + \log {\left (\log {\left (x^{2} \right )} \right )} \right )} + \log {\left (\log {\left (x^{2} \right )} + \frac {2 \log {\left (\log {\relax (3 )} \right )}}{3} + \frac {2 \log {\relax (2 )}}{3} + \frac {2 i \pi }{3} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(exp(x) + log(log(x**2))) + log(log(x**2) + 2*log(log(3))/3 + 2*log(2)/3 + 2*I*pi/3)

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