3.13.6 \(\int \frac {e^{\log ^2(x^2 \log (x))} ((6-3 \log (3)) \log (x)+(-12+6 \log (3)+(-24+12 \log (3)) \log (x)) \log (x^2 \log (x)))}{x^2 \log (x)} \, dx\)

Optimal. Leaf size=20 \[ \frac {3 e^{\log ^2\left (x^2 \log (x)\right )} (-2+\log (3))}{x} \]

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Rubi [A]  time = 0.13, antiderivative size = 38, normalized size of antiderivative = 1.90, number of steps used = 1, number of rules used = 1, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.019, Rules used = {2288} \begin {gather*} -\frac {3 e^{\log ^2\left (x^2 \log (x)\right )} (2 (2-\log (3)) \log (x)+2-\log (3))}{x+2 x \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^Log[x^2*Log[x]]^2*((6 - 3*Log[3])*Log[x] + (-12 + 6*Log[3] + (-24 + 12*Log[3])*Log[x])*Log[x^2*Log[x]])
)/(x^2*Log[x]),x]

[Out]

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

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 20, normalized size = 1.00 \begin {gather*} \frac {3 e^{\log ^2\left (x^2 \log (x)\right )} (-2+\log (3))}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^Log[x^2*Log[x]]^2*((6 - 3*Log[3])*Log[x] + (-12 + 6*Log[3] + (-24 + 12*Log[3])*Log[x])*Log[x^2*Lo
g[x]]))/(x^2*Log[x]),x]

[Out]

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

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fricas [A]  time = 0.53, size = 19, normalized size = 0.95 \begin {gather*} \frac {3 \, {\left (\log \relax (3) - 2\right )} e^{\left (\log \left (x^{2} \log \relax (x)\right )^{2}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((12*log(3)-24)*log(x)+6*log(3)-12)*log(x^2*log(x))+(-3*log(3)+6)*log(x))*exp(log(x^2*log(x))^2)/x^
2/log(x),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((12*log(3)-24)*log(x)+6*log(3)-12)*log(x^2*log(x))+(-3*log(3)+6)*log(x))*exp(log(x^2*log(x))^2)/x^
2/log(x),x, algorithm="giac")

[Out]

undef

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maple [C]  time = 0.11, size = 158, normalized size = 7.90

method	result	size
risch	\(\frac {3 \left (\ln \relax (3)-2\right ) {\mathrm e}^{\frac {\left (-i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+2 i \pi \mathrm {csgn}\left (i x^{2}\right )^{2} \mathrm {csgn}\left (i x \right )-i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )^{2}+i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{2} \ln \relax (x )\right )^{2}-i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{2} \ln \relax (x )\right ) \mathrm {csgn}\left (i \ln \relax (x )\right )-i \pi \mathrm {csgn}\left (i x^{2} \ln \relax (x )\right )^{3}+i \pi \mathrm {csgn}\left (i x^{2} \ln \relax (x )\right )^{2} \mathrm {csgn}\left (i \ln \relax (x )\right )+4 \ln \relax (x )+2 \ln \left (\ln \relax (x )\right )\right )^{2}}{4}}}{x}\)	\(158\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((12*ln(3)-24)*ln(x)+6*ln(3)-12)*ln(x^2*ln(x))+(-3*ln(3)+6)*ln(x))*exp(ln(x^2*ln(x))^2)/x^2/ln(x),x,metho
d=_RETURNVERBOSE)

[Out]

3*(ln(3)-2)/x*exp(1/4*(-I*Pi*csgn(I*x^2)^3+2*I*Pi*csgn(I*x^2)^2*csgn(I*x)-I*Pi*csgn(I*x^2)*csgn(I*x)^2+I*Pi*cs
gn(I*x^2)*csgn(I*x^2*ln(x))^2-I*Pi*csgn(I*x^2)*csgn(I*x^2*ln(x))*csgn(I*ln(x))-I*Pi*csgn(I*x^2*ln(x))^3+I*Pi*c
sgn(I*x^2*ln(x))^2*csgn(I*ln(x))+4*ln(x)+2*ln(ln(x)))^2)

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maxima [A]  time = 0.64, size = 29, normalized size = 1.45 \begin {gather*} \frac {3 \, {\left (\log \relax (3) - 2\right )} e^{\left (4 \, \log \relax (x)^{2} + 4 \, \log \relax (x) \log \left (\log \relax (x)\right ) + \log \left (\log \relax (x)\right )^{2}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((12*log(3)-24)*log(x)+6*log(3)-12)*log(x^2*log(x))+(-3*log(3)+6)*log(x))*exp(log(x^2*log(x))^2)/x^
2/log(x),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int -\frac {{\mathrm {e}}^{{\ln \left (x^2\,\ln \relax (x)\right )}^2}\,\left (\ln \relax (x)\,\left (3\,\ln \relax (3)-6\right )-\ln \left (x^2\,\ln \relax (x)\right )\,\left (6\,\ln \relax (3)+\ln \relax (x)\,\left (12\,\ln \relax (3)-24\right )-12\right )\right )}{x^2\,\ln \relax (x)} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(log(x^2*log(x))^2)*(log(x)*(3*log(3) - 6) - log(x^2*log(x))*(6*log(3) + log(x)*(12*log(3) - 24) - 12
)))/(x^2*log(x)),x)

[Out]

int(-(exp(log(x^2*log(x))^2)*(log(x)*(3*log(3) - 6) - log(x^2*log(x))*(6*log(3) + log(x)*(12*log(3) - 24) - 12
)))/(x^2*log(x)), x)

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sympy [A]  time = 0.37, size = 19, normalized size = 0.95 \begin {gather*} \frac {\left (-6 + 3 \log {\relax (3 )}\right ) e^{\log {\left (x^{2} \log {\relax (x )} \right )}^{2}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((12*ln(3)-24)*ln(x)+6*ln(3)-12)*ln(x**2*ln(x))+(-3*ln(3)+6)*ln(x))*exp(ln(x**2*ln(x))**2)/x**2/ln(
x),x)

[Out]

(-6 + 3*log(3))*exp(log(x**2*log(x))**2)/x

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