3.61.20 \(\int \frac {e^{9+x^2+\log (x) \log (x^5 \log ^2(4))} (2 x^2+5 \log (x)+\log (x^5 \log ^2(4)))}{x} \, dx\)

Optimal. Leaf size=19 \[ e^{9+x^2+\log (x) \log \left (x^5 \log ^2(4)\right )} \]

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Rubi [F]  time = 0.80, 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^{9+x^2+\log (x) \log \left (x^5 \log ^2(4)\right )} \left (2 x^2+5 \log (x)+\log \left (x^5 \log ^2(4)\right )\right )}{x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(9 + x^2 + Log[x]*Log[x^5*Log[4]^2])*(2*x^2 + 5*Log[x] + Log[x^5*Log[4]^2]))/x,x]

[Out]

2*Defer[Int][E^(9 + x^2 + Log[x]*Log[x^5*Log[4]^2])*x, x] + 5*Defer[Int][(E^(9 + x^2 + Log[x]*Log[x^5*Log[4]^2
])*Log[x])/x, x] + Defer[Int][(E^(9 + x^2 + Log[x]*Log[x^5*Log[4]^2])*Log[x^5*Log[4]^2])/x, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {e^{9+x^2+\log (x) \log \left (x^5 \log ^2(4)\right )} \left (2 x^2+5 \log (x)\right )}{x}+\frac {e^{9+x^2+\log (x) \log \left (x^5 \log ^2(4)\right )} \log \left (x^5 \log ^2(4)\right )}{x}\right ) \, dx\\ &=\int \frac {e^{9+x^2+\log (x) \log \left (x^5 \log ^2(4)\right )} \left (2 x^2+5 \log (x)\right )}{x} \, dx+\int \frac {e^{9+x^2+\log (x) \log \left (x^5 \log ^2(4)\right )} \log \left (x^5 \log ^2(4)\right )}{x} \, dx\\ &=\int \left (2 e^{9+x^2+\log (x) \log \left (x^5 \log ^2(4)\right )} x+\frac {5 e^{9+x^2+\log (x) \log \left (x^5 \log ^2(4)\right )} \log (x)}{x}\right ) \, dx+\int \frac {e^{9+x^2+\log (x) \log \left (x^5 \log ^2(4)\right )} \log \left (x^5 \log ^2(4)\right )}{x} \, dx\\ &=2 \int e^{9+x^2+\log (x) \log \left (x^5 \log ^2(4)\right )} x \, dx+5 \int \frac {e^{9+x^2+\log (x) \log \left (x^5 \log ^2(4)\right )} \log (x)}{x} \, dx+\int \frac {e^{9+x^2+\log (x) \log \left (x^5 \log ^2(4)\right )} \log \left (x^5 \log ^2(4)\right )}{x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.30, size = 20, normalized size = 1.05 \begin {gather*} e^{9+x^2} x^{\log \left (x^5\right )+2 \log (\log (4))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(9 + x^2 + Log[x]*Log[x^5*Log[4]^2])*(2*x^2 + 5*Log[x] + Log[x^5*Log[4]^2]))/x,x]

[Out]

E^(9 + x^2)*x^(Log[x^5] + 2*Log[Log[4]])

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fricas [A]  time = 0.61, size = 22, normalized size = 1.16 \begin {gather*} e^{\left (x^{2} + \log \left (4 \, \log \relax (2)^{2}\right ) \log \relax (x) + 5 \, \log \relax (x)^{2} + 9\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(4*x^5*log(2)^2)+5*log(x)+2*x^2)*exp(log(x)*log(4*x^5*log(2)^2)+x^2+9)/x,x, algorithm="fricas")

[Out]

e^(x^2 + log(4*log(2)^2)*log(x) + 5*log(x)^2 + 9)

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giac [A]  time = 1.03, size = 19, normalized size = 1.00 \begin {gather*} e^{\left (x^{2} + \log \left (4 \, x^{5} \log \relax (2)^{2}\right ) \log \relax (x) + 9\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(4*x^5*log(2)^2)+5*log(x)+2*x^2)*exp(log(x)*log(4*x^5*log(2)^2)+x^2+9)/x,x, algorithm="giac")

[Out]

e^(x^2 + log(4*x^5*log(2)^2)*log(x) + 9)

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maple [A]  time = 0.13, size = 20, normalized size = 1.05

method	result	size
default	\({\mathrm e}^{\ln \relax (x ) \ln \left (4 x^{5} \ln \relax (2)^{2}\right )+x^{2}+9}\)	\(20\)
risch	\(x^{-\frac {i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )}{2}-\frac {i \pi \,\mathrm {csgn}\left (i x^{5}\right )}{2}-\frac {i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{4}\right ) \mathrm {csgn}\left (i x^{5}\right )}{2}+\frac {5 i \pi \,\mathrm {csgn}\left (i x \right )}{2}-\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right )}{2}-\frac {i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{3}\right ) \mathrm {csgn}\left (i x^{4}\right )}{2}+5 \ln \relax (x )+2 \ln \relax (2)+2 \ln \left (\ln \relax (2)\right )} {\mathrm e}^{x^{2}+9}\)	\(124\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((ln(4*x^5*ln(2)^2)+5*ln(x)+2*x^2)*exp(ln(x)*ln(4*x^5*ln(2)^2)+x^2+9)/x,x,method=_RETURNVERBOSE)

[Out]

exp(ln(x)*ln(4*x^5*ln(2)^2)+x^2+9)

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maxima [A]  time = 0.53, size = 25, normalized size = 1.32 \begin {gather*} e^{\left (x^{2} + 2 \, \log \relax (2) \log \relax (x) + 5 \, \log \relax (x)^{2} + 2 \, \log \relax (x) \log \left (\log \relax (2)\right ) + 9\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(4*x^5*log(2)^2)+5*log(x)+2*x^2)*exp(log(x)*log(4*x^5*log(2)^2)+x^2+9)/x,x, algorithm="maxima")

[Out]

e^(x^2 + 2*log(2)*log(x) + 5*log(x)^2 + 2*log(x)*log(log(2)) + 9)

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mupad [B]  time = 4.34, size = 26, normalized size = 1.37 \begin {gather*} x^{2\,\ln \left (\ln \relax (2)\right )}\,x^{\ln \left (x^5\right )}\,x^{2\,\ln \relax (2)}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^9 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(log(4*x^5*log(2)^2)*log(x) + x^2 + 9)*(log(4*x^5*log(2)^2) + 5*log(x) + 2*x^2))/x,x)

[Out]

x^(2*log(log(2)))*x^log(x^5)*x^(2*log(2))*exp(x^2)*exp(9)

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sympy [A]  time = 0.36, size = 22, normalized size = 1.16 \begin {gather*} e^{x^{2} + \left (5 \log {\relax (x )} + \log {\left (4 \log {\relax (2 )}^{2} \right )}\right ) \log {\relax (x )} + 9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((ln(4*x**5*ln(2)**2)+5*ln(x)+2*x**2)*exp(ln(x)*ln(4*x**5*ln(2)**2)+x**2+9)/x,x)

[Out]

exp(x**2 + (5*log(x) + log(4*log(2)**2))*log(x) + 9)

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