3.51.51 \(\int \frac {e^x (-8+2 e^3) (i \pi +\log (10))+e^x (-4 x+e^3 x) (i \pi +\log (10)) \log (\frac {5}{x^2})+(-4 x+e^3 x) (i \pi +\log (10)) \log ^2(\frac {5}{x^2})}{e^x x \log (\frac {5}{x^2})+x^2 \log ^2(\frac {5}{x^2})} \, dx\)

Optimal. Leaf size=29 \[ \left (-4+e^3\right ) (i \pi +\log (10)) \log \left (x+\frac {e^x}{\log \left (\frac {5}{x^2}\right )}\right ) \]

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Rubi [F]  time = 1.34, 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^x \left (-8+2 e^3\right ) (i \pi +\log (10))+e^x \left (-4 x+e^3 x\right ) (i \pi +\log (10)) \log \left (\frac {5}{x^2}\right )+\left (-4 x+e^3 x\right ) (i \pi +\log (10)) \log ^2\left (\frac {5}{x^2}\right )}{e^x x \log \left (\frac {5}{x^2}\right )+x^2 \log ^2\left (\frac {5}{x^2}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^x*(-8 + 2*E^3)*(I*Pi + Log[10]) + E^x*(-4*x + E^3*x)*(I*Pi + Log[10])*Log[5/x^2] + (-4*x + E^3*x)*(I*Pi
+ Log[10])*Log[5/x^2]^2)/(E^x*x*Log[5/x^2] + x^2*Log[5/x^2]^2),x]

[Out]

-((4 - E^3)*x*(I*Pi + Log[10])) + (4 - E^3)*(I*Pi + Log[10])*Log[Log[5/x^2]] + 2*(4 - E^3)*(I*Pi + Log[10])*De
fer[Int][(E^x + x*Log[5/x^2])^(-1), x] - (4 - E^3)*(I*Pi + Log[10])*Defer[Int][Log[5/x^2]/(E^x + x*Log[5/x^2])
, x] + (4 - E^3)*(I*Pi + Log[10])*Defer[Int][(x*Log[5/x^2])/(E^x + x*Log[5/x^2]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (4-e^3\right ) (i \pi +\log (10)) \left (-2 e^x-e^x x \log \left (\frac {5}{x^2}\right )-x \log ^2\left (\frac {5}{x^2}\right )\right )}{x \log \left (\frac {5}{x^2}\right ) \left (e^x+x \log \left (\frac {5}{x^2}\right )\right )} \, dx\\ &=\left (\left (4-e^3\right ) (i \pi +\log (10))\right ) \int \frac {-2 e^x-e^x x \log \left (\frac {5}{x^2}\right )-x \log ^2\left (\frac {5}{x^2}\right )}{x \log \left (\frac {5}{x^2}\right ) \left (e^x+x \log \left (\frac {5}{x^2}\right )\right )} \, dx\\ &=\left (\left (4-e^3\right ) (i \pi +\log (10))\right ) \int \left (\frac {-2-x \log \left (\frac {5}{x^2}\right )}{x \log \left (\frac {5}{x^2}\right )}+\frac {2-\log \left (\frac {5}{x^2}\right )+x \log \left (\frac {5}{x^2}\right )}{e^x+x \log \left (\frac {5}{x^2}\right )}\right ) \, dx\\ &=\left (\left (4-e^3\right ) (i \pi +\log (10))\right ) \int \frac {-2-x \log \left (\frac {5}{x^2}\right )}{x \log \left (\frac {5}{x^2}\right )} \, dx+\left (\left (4-e^3\right ) (i \pi +\log (10))\right ) \int \frac {2-\log \left (\frac {5}{x^2}\right )+x \log \left (\frac {5}{x^2}\right )}{e^x+x \log \left (\frac {5}{x^2}\right )} \, dx\\ &=\left (\left (4-e^3\right ) (i \pi +\log (10))\right ) \int \left (-1-\frac {2}{x \log \left (\frac {5}{x^2}\right )}\right ) \, dx+\left (\left (4-e^3\right ) (i \pi +\log (10))\right ) \int \frac {2+(-1+x) \log \left (\frac {5}{x^2}\right )}{e^x+x \log \left (\frac {5}{x^2}\right )} \, dx\\ &=-\left (\left (4-e^3\right ) x (i \pi +\log (10))\right )+\left (\left (4-e^3\right ) (i \pi +\log (10))\right ) \int \left (\frac {2}{e^x+x \log \left (\frac {5}{x^2}\right )}-\frac {\log \left (\frac {5}{x^2}\right )}{e^x+x \log \left (\frac {5}{x^2}\right )}+\frac {x \log \left (\frac {5}{x^2}\right )}{e^x+x \log \left (\frac {5}{x^2}\right )}\right ) \, dx-\left (2 \left (4-e^3\right ) (i \pi +\log (10))\right ) \int \frac {1}{x \log \left (\frac {5}{x^2}\right )} \, dx\\ &=-\left (\left (4-e^3\right ) x (i \pi +\log (10))\right )-\left (\left (4-e^3\right ) (i \pi +\log (10))\right ) \int \frac {\log \left (\frac {5}{x^2}\right )}{e^x+x \log \left (\frac {5}{x^2}\right )} \, dx+\left (\left (4-e^3\right ) (i \pi +\log (10))\right ) \int \frac {x \log \left (\frac {5}{x^2}\right )}{e^x+x \log \left (\frac {5}{x^2}\right )} \, dx+\left (\left (4-e^3\right ) (i \pi +\log (10))\right ) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log \left (\frac {5}{x^2}\right )\right )+\left (2 \left (4-e^3\right ) (i \pi +\log (10))\right ) \int \frac {1}{e^x+x \log \left (\frac {5}{x^2}\right )} \, dx\\ &=-\left (\left (4-e^3\right ) x (i \pi +\log (10))\right )+\left (4-e^3\right ) (i \pi +\log (10)) \log \left (\log \left (\frac {5}{x^2}\right )\right )-\left (\left (4-e^3\right ) (i \pi +\log (10))\right ) \int \frac {\log \left (\frac {5}{x^2}\right )}{e^x+x \log \left (\frac {5}{x^2}\right )} \, dx+\left (\left (4-e^3\right ) (i \pi +\log (10))\right ) \int \frac {x \log \left (\frac {5}{x^2}\right )}{e^x+x \log \left (\frac {5}{x^2}\right )} \, dx+\left (2 \left (4-e^3\right ) (i \pi +\log (10))\right ) \int \frac {1}{e^x+x \log \left (\frac {5}{x^2}\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.21, size = 37, normalized size = 1.28 \begin {gather*} \left (-4+e^3\right ) (i \pi +\log (10)) \left (-\log \left (\log \left (\frac {5}{x^2}\right )\right )+\log \left (e^x+x \log \left (\frac {5}{x^2}\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(-8 + 2*E^3)*(I*Pi + Log[10]) + E^x*(-4*x + E^3*x)*(I*Pi + Log[10])*Log[5/x^2] + (-4*x + E^3*x)
*(I*Pi + Log[10])*Log[5/x^2]^2)/(E^x*x*Log[5/x^2] + x^2*Log[5/x^2]^2),x]

[Out]

(-4 + E^3)*(I*Pi + Log[10])*(-Log[Log[5/x^2]] + Log[E^x + x*Log[5/x^2]])

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fricas [B]  time = 0.73, size = 86, normalized size = 2.97 \begin {gather*} -{\left (4 i \, \pi - i \, \pi e^{3} - {\left (e^{3} - 4\right )} \log \left (10\right )\right )} \log \left (\frac {x \log \left (\frac {5}{x^{2}}\right ) + e^{x}}{x}\right ) + \frac {1}{2} \, {\left (4 i \, \pi - i \, \pi e^{3} - {\left (e^{3} - 4\right )} \log \left (10\right )\right )} \log \left (\frac {5}{x^{2}}\right ) - {\left (-4 i \, \pi + i \, \pi e^{3} + {\left (e^{3} - 4\right )} \log \left (10\right )\right )} \log \left (\log \left (\frac {5}{x^{2}}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(3)-4*x)*(log(10)+I*pi)*log(5/x^2)^2+(x*exp(3)-4*x)*(log(10)+I*pi)*exp(x)*log(5/x^2)+(2*exp(3
)-8)*(log(10)+I*pi)*exp(x))/(x^2*log(5/x^2)^2+x*exp(x)*log(5/x^2)),x, algorithm="fricas")

[Out]

-(4*I*pi - I*pi*e^3 - (e^3 - 4)*log(10))*log((x*log(5/x^2) + e^x)/x) + 1/2*(4*I*pi - I*pi*e^3 - (e^3 - 4)*log(
10))*log(5/x^2) - (-4*I*pi + I*pi*e^3 + (e^3 - 4)*log(10))*log(log(5/x^2))

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giac [B]  time = 0.36, size = 205, normalized size = 7.07 \begin {gather*} i \, \pi e^{3} \log \left (x \log \relax (5) - x \log \left (x^{2}\right ) + e^{x}\right ) + e^{3} \log \relax (5) \log \left (x \log \relax (5) - x \log \left (x^{2}\right ) + e^{x}\right ) + e^{3} \log \relax (2) \log \left (x \log \relax (5) - x \log \left (x^{2}\right ) + e^{x}\right ) - i \, \pi e^{3} \log \left (\log \relax (5) - \log \left (x^{2}\right )\right ) - e^{3} \log \relax (5) \log \left (\log \relax (5) - \log \left (x^{2}\right )\right ) - e^{3} \log \relax (2) \log \left (\log \relax (5) - \log \left (x^{2}\right )\right ) - 4 i \, \pi \log \left (x \log \relax (5) - x \log \left (x^{2}\right ) + e^{x}\right ) - 4 \, \log \relax (5) \log \left (x \log \relax (5) - x \log \left (x^{2}\right ) + e^{x}\right ) - 4 \, \log \relax (2) \log \left (x \log \relax (5) - x \log \left (x^{2}\right ) + e^{x}\right ) + 4 i \, \pi \log \left (\log \relax (5) - \log \left (x^{2}\right )\right ) + 4 \, \log \relax (5) \log \left (\log \relax (5) - \log \left (x^{2}\right )\right ) + 4 \, \log \relax (2) \log \left (\log \relax (5) - \log \left (x^{2}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(3)-4*x)*(log(10)+I*pi)*log(5/x^2)^2+(x*exp(3)-4*x)*(log(10)+I*pi)*exp(x)*log(5/x^2)+(2*exp(3
)-8)*(log(10)+I*pi)*exp(x))/(x^2*log(5/x^2)^2+x*exp(x)*log(5/x^2)),x, algorithm="giac")

[Out]

I*pi*e^3*log(x*log(5) - x*log(x^2) + e^x) + e^3*log(5)*log(x*log(5) - x*log(x^2) + e^x) + e^3*log(2)*log(x*log
(5) - x*log(x^2) + e^x) - I*pi*e^3*log(log(5) - log(x^2)) - e^3*log(5)*log(log(5) - log(x^2)) - e^3*log(2)*log
(log(5) - log(x^2)) - 4*I*pi*log(x*log(5) - x*log(x^2) + e^x) - 4*log(5)*log(x*log(5) - x*log(x^2) + e^x) - 4*
log(2)*log(x*log(5) - x*log(x^2) + e^x) + 4*I*pi*log(log(5) - log(x^2)) + 4*log(5)*log(log(5) - log(x^2)) + 4*
log(2)*log(log(5) - log(x^2))

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maple [B]  time = 0.71, size = 64, normalized size = 2.21

method	result	size
norman	\(\left (-i \pi \,{\mathrm e}^{3}+4 i \pi -{\mathrm e}^{3} \ln \left (10\right )+4 \ln \left (10\right )\right ) \ln \left (\ln \left (\frac {5}{x^{2}}\right )\right )+\left (i \pi \,{\mathrm e}^{3}-4 i \pi +{\mathrm e}^{3} \ln \left (10\right )-4 \ln \left (10\right )\right ) \ln \left (x \ln \left (\frac {5}{x^{2}}\right )+{\mathrm e}^{x}\right )\)	\(64\)
default	\(-i \pi \ln \left (-\ln \left (\frac {5}{x^{2}}\right )\right ) {\mathrm e}^{3}+4 i \pi \ln \left (-\ln \left (\frac {5}{x^{2}}\right )\right )-\ln \left (-\ln \left (\frac {5}{x^{2}}\right )\right ) {\mathrm e}^{3} \ln \left (10\right )+4 \ln \left (-\ln \left (\frac {5}{x^{2}}\right )\right ) \ln \left (10\right )+i \pi \ln \left (2 x \ln \relax (x )-x \left (\ln \left (\frac {5}{x^{2}}\right )+2 \ln \relax (x )\right )-{\mathrm e}^{x}\right ) {\mathrm e}^{3}-4 i \pi \ln \left (2 x \ln \relax (x )-x \left (\ln \left (\frac {5}{x^{2}}\right )+2 \ln \relax (x )\right )-{\mathrm e}^{x}\right )+\ln \left (2 x \ln \relax (x )-x \left (\ln \left (\frac {5}{x^{2}}\right )+2 \ln \relax (x )\right )-{\mathrm e}^{x}\right ) {\mathrm e}^{3} \ln \left (10\right )-4 \ln \left (2 x \ln \relax (x )-x \left (\ln \left (\frac {5}{x^{2}}\right )+2 \ln \relax (x )\right )-{\mathrm e}^{x}\right ) \ln \left (10\right )\)	\(177\)
risch	\(\ln \relax (x ) \ln \relax (5) {\mathrm e}^{3}+\ln \relax (x ) {\mathrm e}^{3} \ln \relax (2)-4 i \pi \ln \relax (x )-4 \ln \relax (5) \ln \relax (x )-4 \ln \relax (2) \ln \relax (x )+i \pi \ln \relax (x ) {\mathrm e}^{3}+i \pi \,{\mathrm e}^{3} \ln \left (\ln \relax (x )-\frac {i \left (\pi x \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi x \mathrm {csgn}\left (i x^{2}\right )^{3}-2 i \ln \relax (5) x -2 i {\mathrm e}^{x}\right )}{4 x}\right )+4 i \pi \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^{2}\right )^{3}-2 i \ln \relax (5)\right )}{4}\right )+\ln \relax (5) {\mathrm e}^{3} \ln \left (\ln \relax (x )-\frac {i \left (\pi x \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi x \mathrm {csgn}\left (i x^{2}\right )^{3}-2 i \ln \relax (5) x -2 i {\mathrm e}^{x}\right )}{4 x}\right )+{\mathrm e}^{3} \ln \relax (2) \ln \left (\ln \relax (x )-\frac {i \left (\pi x \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi x \mathrm {csgn}\left (i x^{2}\right )^{3}-2 i \ln \relax (5) x -2 i {\mathrm e}^{x}\right )}{4 x}\right )-4 \ln \relax (5) \ln \left (\ln \relax (x )-\frac {i \left (\pi x \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi x \mathrm {csgn}\left (i x^{2}\right )^{3}-2 i \ln \relax (5) x -2 i {\mathrm e}^{x}\right )}{4 x}\right )-4 \ln \relax (2) \ln \left (\ln \relax (x )-\frac {i \left (\pi x \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi x \mathrm {csgn}\left (i x^{2}\right )^{3}-2 i \ln \relax (5) x -2 i {\mathrm e}^{x}\right )}{4 x}\right )-4 i \pi \ln \left (\ln \relax (x )-\frac {i \left (\pi x \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi x \mathrm {csgn}\left (i x^{2}\right )^{3}-2 i \ln \relax (5) x -2 i {\mathrm e}^{x}\right )}{4 x}\right )-i \pi \,{\mathrm e}^{3} \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^{2}\right )^{3}-2 i \ln \relax (5)\right )}{4}\right )-\ln \relax (5) {\mathrm e}^{3} \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^{2}\right )^{3}-2 i \ln \relax (5)\right )}{4}\right )-{\mathrm e}^{3} \ln \relax (2) \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^{2}\right )^{3}-2 i \ln \relax (5)\right )}{4}\right )+4 \ln \relax (5) \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^{2}\right )^{3}-2 i \ln \relax (5)\right )}{4}\right )+4 \ln \relax (2) \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^{2}\right )^{3}-2 i \ln \relax (5)\right )}{4}\right )\)	\(856\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x*exp(3)-4*x)*(ln(10)+I*Pi)*ln(5/x^2)^2+(x*exp(3)-4*x)*(ln(10)+I*Pi)*exp(x)*ln(5/x^2)+(2*exp(3)-8)*(ln(1
0)+I*Pi)*exp(x))/(x^2*ln(5/x^2)^2+x*exp(x)*ln(5/x^2)),x,method=_RETURNVERBOSE)

[Out]

(-I*Pi*exp(3)+4*I*Pi-exp(3)*ln(10)+4*ln(10))*ln(ln(5/x^2))+(I*Pi*exp(3)-4*I*Pi+exp(3)*ln(10)-4*ln(10))*ln(x*ln
(5/x^2)+exp(x))

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maxima [B]  time = 0.49, size = 74, normalized size = 2.55 \begin {gather*} {\left (-4 i \, \pi + {\left (i \, \pi + \log \relax (5) + \log \relax (2)\right )} e^{3} - 4 \, \log \relax (5) - 4 \, \log \relax (2)\right )} \log \left (x \log \relax (5) - 2 \, x \log \relax (x) + e^{x}\right ) + {\left (4 i \, \pi + {\left (-i \, \pi - \log \relax (5) - \log \relax (2)\right )} e^{3} + 4 \, \log \relax (5) + 4 \, \log \relax (2)\right )} \log \left (-\frac {1}{2} \, \log \relax (5) + \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(3)-4*x)*(log(10)+I*pi)*log(5/x^2)^2+(x*exp(3)-4*x)*(log(10)+I*pi)*exp(x)*log(5/x^2)+(2*exp(3
)-8)*(log(10)+I*pi)*exp(x))/(x^2*log(5/x^2)^2+x*exp(x)*log(5/x^2)),x, algorithm="maxima")

[Out]

(-4*I*pi + (I*pi + log(5) + log(2))*e^3 - 4*log(5) - 4*log(2))*log(x*log(5) - 2*x*log(x) + e^x) + (4*I*pi + (-
I*pi - log(5) - log(2))*e^3 + 4*log(5) + 4*log(2))*log(-1/2*log(5) + log(x))

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mupad [B]  time = 4.24, size = 35, normalized size = 1.21 \begin {gather*} -\left (\ln \left (10\right )+\Pi \,1{}\mathrm {i}\right )\,\left ({\mathrm {e}}^3-4\right )\,\left (\ln \left (\ln \left (\frac {5}{x^2}\right )\right )-\ln \left ({\mathrm {e}}^x+x\,\ln \left (\frac {5}{x^2}\right )\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(5/x^2)^2*(4*x - x*exp(3))*(Pi*1i + log(10)) - exp(x)*(2*exp(3) - 8)*(Pi*1i + log(10)) + exp(x)*log(5
/x^2)*(4*x - x*exp(3))*(Pi*1i + log(10)))/(x^2*log(5/x^2)^2 + x*exp(x)*log(5/x^2)),x)

[Out]

-(Pi*1i + log(10))*(exp(3) - 4)*(log(log(5/x^2)) - log(exp(x) + x*log(5/x^2)))

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sympy [B]  time = 2.08, size = 53, normalized size = 1.83 \begin {gather*} - \left (-4 + e^{3}\right ) \left (\log {\left (10 \right )} + i \pi \right ) \log {\left (\log {\left (\frac {1}{x^{2}} \right )} + \log {\relax (5 )} \right )} + \left (-4 + e^{3}\right ) \left (\log {\left (10 \right )} + i \pi \right ) \log {\left (x \log {\left (\frac {1}{x^{2}} \right )} + x \log {\relax (5 )} + e^{x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(3)-4*x)*(ln(10)+I*pi)*ln(5/x**2)**2+(x*exp(3)-4*x)*(ln(10)+I*pi)*exp(x)*ln(5/x**2)+(2*exp(3)
-8)*(ln(10)+I*pi)*exp(x))/(x**2*ln(5/x**2)**2+x*exp(x)*ln(5/x**2)),x)

[Out]

-(-4 + exp(3))*(log(10) + I*pi)*log(log(x**(-2)) + log(5)) + (-4 + exp(3))*(log(10) + I*pi)*log(x*log(x**(-2))
+ x*log(5) + exp(x))

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