3.86.43 \(\int \frac {e^{\frac {4+5 x-40 x^2+8 e^x x^2+(13 x^2-e^x x^2) \log (x)-x^2 \log ^2(x)}{-8 x^2+x^2 \log (x)}} (60+35 x-64 x^2-64 e^x x^3+(-8-5 x+16 x^2+16 e^x x^3) \log (x)+(-x^2-e^x x^3) \log ^2(x))}{64 x^3-16 x^3 \log (x)+x^3 \log ^2(x)} \, dx\)

Optimal. Leaf size=34 \[ \frac {e^{5-e^x+\frac {5+\frac {4}{x}}{x-x (9-\log (x))}}}{x} \]

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Rubi [F]  time = 54.51, 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 {\exp \left (\frac {4+5 x-40 x^2+8 e^x x^2+\left (13 x^2-e^x x^2\right ) \log (x)-x^2 \log ^2(x)}{-8 x^2+x^2 \log (x)}\right ) \left (60+35 x-64 x^2-64 e^x x^3+\left (-8-5 x+16 x^2+16 e^x x^3\right ) \log (x)+\left (-x^2-e^x x^3\right ) \log ^2(x)\right )}{64 x^3-16 x^3 \log (x)+x^3 \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((4 + 5*x - 40*x^2 + 8*E^x*x^2 + (13*x^2 - E^x*x^2)*Log[x] - x^2*Log[x]^2)/(-8*x^2 + x^2*Log[x]))*(60 +
35*x - 64*x^2 - 64*E^x*x^3 + (-8 - 5*x + 16*x^2 + 16*E^x*x^3)*Log[x] + (-x^2 - E^x*x^3)*Log[x]^2))/(64*x^3 -
16*x^3*Log[x] + x^3*Log[x]^2),x]

[Out]

-Defer[Int][E^(x + (4 + 5*x - 40*x^2 + 8*E^x*x^2 + (13*x^2 - E^x*x^2)*Log[x] - x^2*Log[x]^2)/(-8*x^2 + x^2*Log
[x])), x] - Defer[Int][E^((4 + 5*x - 40*x^2 + 8*E^x*x^2 + (13*x^2 - E^x*x^2)*Log[x] - x^2*Log[x]^2)/(-8*x^2 +
x^2*Log[x]))/x, x] - 4*Defer[Int][E^((4 + 5*x - 40*x^2 + 8*E^x*x^2 + (13*x^2 - E^x*x^2)*Log[x] - x^2*Log[x]^2)
/(-8*x^2 + x^2*Log[x]))/(x^3*(-8 + Log[x])^2), x] - 5*Defer[Int][E^((4 + 5*x - 40*x^2 + 8*E^x*x^2 + (13*x^2 -
E^x*x^2)*Log[x] - x^2*Log[x]^2)/(-8*x^2 + x^2*Log[x]))/(x^2*(-8 + Log[x])^2), x] - 8*Defer[Int][E^((4 + 5*x -
40*x^2 + 8*E^x*x^2 + (13*x^2 - E^x*x^2)*Log[x] - x^2*Log[x]^2)/(-8*x^2 + x^2*Log[x]))/(x^3*(-8 + Log[x])), x]
- 5*Defer[Int][E^((4 + 5*x - 40*x^2 + 8*E^x*x^2 + (13*x^2 - E^x*x^2)*Log[x] - x^2*Log[x]^2)/(-8*x^2 + x^2*Log[
x]))/(x^2*(-8 + Log[x])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (\frac {4+5 x-40 x^2+8 e^x x^2+\left (13 x^2-e^x x^2\right ) \log (x)-x^2 \log ^2(x)}{-8 x^2+x^2 \log (x)}\right ) \left (60+35 x-64 x^2-64 e^x x^3+\left (-8-5 x+16 x^2+16 e^x x^3\right ) \log (x)+\left (-x^2-e^x x^3\right ) \log ^2(x)\right )}{x^3 (8-\log (x))^2} \, dx\\ &=\int \left (-\exp \left (x+\frac {4+5 x-40 x^2+8 e^x x^2+\left (13 x^2-e^x x^2\right ) \log (x)-x^2 \log ^2(x)}{-8 x^2+x^2 \log (x)}\right )+\frac {60 \exp \left (\frac {4+5 x-40 x^2+8 e^x x^2+\left (13 x^2-e^x x^2\right ) \log (x)-x^2 \log ^2(x)}{-8 x^2+x^2 \log (x)}\right )}{x^3 (-8+\log (x))^2}+\frac {35 \exp \left (\frac {4+5 x-40 x^2+8 e^x x^2+\left (13 x^2-e^x x^2\right ) \log (x)-x^2 \log ^2(x)}{-8 x^2+x^2 \log (x)}\right )}{x^2 (-8+\log (x))^2}-\frac {64 \exp \left (\frac {4+5 x-40 x^2+8 e^x x^2+\left (13 x^2-e^x x^2\right ) \log (x)-x^2 \log ^2(x)}{-8 x^2+x^2 \log (x)}\right )}{x (-8+\log (x))^2}-\frac {8 \exp \left (\frac {4+5 x-40 x^2+8 e^x x^2+\left (13 x^2-e^x x^2\right ) \log (x)-x^2 \log ^2(x)}{-8 x^2+x^2 \log (x)}\right ) \log (x)}{x^3 (-8+\log (x))^2}-\frac {5 \exp \left (\frac {4+5 x-40 x^2+8 e^x x^2+\left (13 x^2-e^x x^2\right ) \log (x)-x^2 \log ^2(x)}{-8 x^2+x^2 \log (x)}\right ) \log (x)}{x^2 (-8+\log (x))^2}+\frac {16 \exp \left (\frac {4+5 x-40 x^2+8 e^x x^2+\left (13 x^2-e^x x^2\right ) \log (x)-x^2 \log ^2(x)}{-8 x^2+x^2 \log (x)}\right ) \log (x)}{x (-8+\log (x))^2}-\frac {\exp \left (\frac {4+5 x-40 x^2+8 e^x x^2+\left (13 x^2-e^x x^2\right ) \log (x)-x^2 \log ^2(x)}{-8 x^2+x^2 \log (x)}\right ) \log ^2(x)}{x (-8+\log (x))^2}\right ) \, dx\\ &=-\left (5 \int \frac {\exp \left (\frac {4+5 x-40 x^2+8 e^x x^2+\left (13 x^2-e^x x^2\right ) \log (x)-x^2 \log ^2(x)}{-8 x^2+x^2 \log (x)}\right ) \log (x)}{x^2 (-8+\log (x))^2} \, dx\right )-8 \int \frac {\exp \left (\frac {4+5 x-40 x^2+8 e^x x^2+\left (13 x^2-e^x x^2\right ) \log (x)-x^2 \log ^2(x)}{-8 x^2+x^2 \log (x)}\right ) \log (x)}{x^3 (-8+\log (x))^2} \, dx+16 \int \frac {\exp \left (\frac {4+5 x-40 x^2+8 e^x x^2+\left (13 x^2-e^x x^2\right ) \log (x)-x^2 \log ^2(x)}{-8 x^2+x^2 \log (x)}\right ) \log (x)}{x (-8+\log (x))^2} \, dx+35 \int \frac {\exp \left (\frac {4+5 x-40 x^2+8 e^x x^2+\left (13 x^2-e^x x^2\right ) \log (x)-x^2 \log ^2(x)}{-8 x^2+x^2 \log (x)}\right )}{x^2 (-8+\log (x))^2} \, dx+60 \int \frac {\exp \left (\frac {4+5 x-40 x^2+8 e^x x^2+\left (13 x^2-e^x x^2\right ) \log (x)-x^2 \log ^2(x)}{-8 x^2+x^2 \log (x)}\right )}{x^3 (-8+\log (x))^2} \, dx-64 \int \frac {\exp \left (\frac {4+5 x-40 x^2+8 e^x x^2+\left (13 x^2-e^x x^2\right ) \log (x)-x^2 \log ^2(x)}{-8 x^2+x^2 \log (x)}\right )}{x (-8+\log (x))^2} \, dx-\int \exp \left (x+\frac {4+5 x-40 x^2+8 e^x x^2+\left (13 x^2-e^x x^2\right ) \log (x)-x^2 \log ^2(x)}{-8 x^2+x^2 \log (x)}\right ) \, dx-\int \frac {\exp \left (\frac {4+5 x-40 x^2+8 e^x x^2+\left (13 x^2-e^x x^2\right ) \log (x)-x^2 \log ^2(x)}{-8 x^2+x^2 \log (x)}\right ) \log ^2(x)}{x (-8+\log (x))^2} \, dx\\ &=-\left (5 \int \left (\frac {8 e^{\frac {4+5 x-40 x^2+8 e^x x^2+\left (13 x^2-e^x x^2\right ) \log (x)-x^2 \log ^2(x)}{-8 x^2+x^2 \log (x)}}}{x^2 (-8+\log (x))^2}+\frac {e^{\frac {4+5 x-40 x^2+8 e^x x^2+\left (13 x^2-e^x x^2\right ) \log (x)-x^2 \log ^2(x)}{-8 x^2+x^2 \log (x)}}}{x^2 (-8+\log (x))}\right ) \, dx\right )-8 \int \left (\frac {8 e^{\frac {4+5 x-40 x^2+8 e^x x^2+\left (13 x^2-e^x x^2\right ) \log (x)-x^2 \log ^2(x)}{-8 x^2+x^2 \log (x)}}}{x^3 (-8+\log (x))^2}+\frac {e^{\frac {4+5 x-40 x^2+8 e^x x^2+\left (13 x^2-e^x x^2\right ) \log (x)-x^2 \log ^2(x)}{-8 x^2+x^2 \log (x)}}}{x^3 (-8+\log (x))}\right ) \, dx+16 \int \left (\frac {8 e^{\frac {4+5 x-40 x^2+8 e^x x^2+\left (13 x^2-e^x x^2\right ) \log (x)-x^2 \log ^2(x)}{-8 x^2+x^2 \log (x)}}}{x (-8+\log (x))^2}+\frac {e^{\frac {4+5 x-40 x^2+8 e^x x^2+\left (13 x^2-e^x x^2\right ) \log (x)-x^2 \log ^2(x)}{-8 x^2+x^2 \log (x)}}}{x (-8+\log (x))}\right ) \, dx+35 \int \frac {e^{\frac {4+5 x-40 x^2+8 e^x x^2+\left (13 x^2-e^x x^2\right ) \log (x)-x^2 \log ^2(x)}{-8 x^2+x^2 \log (x)}}}{x^2 (-8+\log (x))^2} \, dx+60 \int \frac {e^{\frac {4+5 x-40 x^2+8 e^x x^2+\left (13 x^2-e^x x^2\right ) \log (x)-x^2 \log ^2(x)}{-8 x^2+x^2 \log (x)}}}{x^3 (-8+\log (x))^2} \, dx-64 \int \frac {e^{\frac {4+5 x-40 x^2+8 e^x x^2+\left (13 x^2-e^x x^2\right ) \log (x)-x^2 \log ^2(x)}{-8 x^2+x^2 \log (x)}}}{x (-8+\log (x))^2} \, dx-\int e^{x+\frac {4+5 x-40 x^2+8 e^x x^2+\left (13 x^2-e^x x^2\right ) \log (x)-x^2 \log ^2(x)}{-8 x^2+x^2 \log (x)}} \, dx-\int \left (\frac {e^{\frac {4+5 x-40 x^2+8 e^x x^2+\left (13 x^2-e^x x^2\right ) \log (x)-x^2 \log ^2(x)}{-8 x^2+x^2 \log (x)}}}{x}+\frac {64 e^{\frac {4+5 x-40 x^2+8 e^x x^2+\left (13 x^2-e^x x^2\right ) \log (x)-x^2 \log ^2(x)}{-8 x^2+x^2 \log (x)}}}{x (-8+\log (x))^2}+\frac {16 e^{\frac {4+5 x-40 x^2+8 e^x x^2+\left (13 x^2-e^x x^2\right ) \log (x)-x^2 \log ^2(x)}{-8 x^2+x^2 \log (x)}}}{x (-8+\log (x))}\right ) \, dx\\ &=-\left (5 \int \frac {\exp \left (\frac {4+5 x-40 x^2+8 e^x x^2+\left (13 x^2-e^x x^2\right ) \log (x)-x^2 \log ^2(x)}{-8 x^2+x^2 \log (x)}\right )}{x^2 (-8+\log (x))} \, dx\right )-8 \int \frac {\exp \left (\frac {4+5 x-40 x^2+8 e^x x^2+\left (13 x^2-e^x x^2\right ) \log (x)-x^2 \log ^2(x)}{-8 x^2+x^2 \log (x)}\right )}{x^3 (-8+\log (x))} \, dx+35 \int \frac {\exp \left (\frac {4+5 x-40 x^2+8 e^x x^2+\left (13 x^2-e^x x^2\right ) \log (x)-x^2 \log ^2(x)}{-8 x^2+x^2 \log (x)}\right )}{x^2 (-8+\log (x))^2} \, dx-40 \int \frac {\exp \left (\frac {4+5 x-40 x^2+8 e^x x^2+\left (13 x^2-e^x x^2\right ) \log (x)-x^2 \log ^2(x)}{-8 x^2+x^2 \log (x)}\right )}{x^2 (-8+\log (x))^2} \, dx+60 \int \frac {\exp \left (\frac {4+5 x-40 x^2+8 e^x x^2+\left (13 x^2-e^x x^2\right ) \log (x)-x^2 \log ^2(x)}{-8 x^2+x^2 \log (x)}\right )}{x^3 (-8+\log (x))^2} \, dx-64 \int \frac {\exp \left (\frac {4+5 x-40 x^2+8 e^x x^2+\left (13 x^2-e^x x^2\right ) \log (x)-x^2 \log ^2(x)}{-8 x^2+x^2 \log (x)}\right )}{x^3 (-8+\log (x))^2} \, dx-2 \left (64 \int \frac {\exp \left (\frac {4+5 x-40 x^2+8 e^x x^2+\left (13 x^2-e^x x^2\right ) \log (x)-x^2 \log ^2(x)}{-8 x^2+x^2 \log (x)}\right )}{x (-8+\log (x))^2} \, dx\right )+128 \int \frac {\exp \left (\frac {4+5 x-40 x^2+8 e^x x^2+\left (13 x^2-e^x x^2\right ) \log (x)-x^2 \log ^2(x)}{-8 x^2+x^2 \log (x)}\right )}{x (-8+\log (x))^2} \, dx-\int \exp \left (x+\frac {4+5 x-40 x^2+8 e^x x^2+\left (13 x^2-e^x x^2\right ) \log (x)-x^2 \log ^2(x)}{-8 x^2+x^2 \log (x)}\right ) \, dx-\int \frac {\exp \left (\frac {4+5 x-40 x^2+8 e^x x^2+\left (13 x^2-e^x x^2\right ) \log (x)-x^2 \log ^2(x)}{-8 x^2+x^2 \log (x)}\right )}{x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.44, size = 43, normalized size = 1.26 \begin {gather*} \frac {e^{\frac {4+5 x+8 \left (-5+e^x\right ) x^2-\left (-5+e^x\right ) x^2 \log (x)}{x^2 (-8+\log (x))}}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((4 + 5*x - 40*x^2 + 8*E^x*x^2 + (13*x^2 - E^x*x^2)*Log[x] - x^2*Log[x]^2)/(-8*x^2 + x^2*Log[x]))
*(60 + 35*x - 64*x^2 - 64*E^x*x^3 + (-8 - 5*x + 16*x^2 + 16*E^x*x^3)*Log[x] + (-x^2 - E^x*x^3)*Log[x]^2))/(64*
x^3 - 16*x^3*Log[x] + x^3*Log[x]^2),x]

[Out]

E^((4 + 5*x + 8*(-5 + E^x)*x^2 - (-5 + E^x)*x^2*Log[x])/(x^2*(-8 + Log[x])))/x

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fricas [A]  time = 0.90, size = 57, normalized size = 1.68 \begin {gather*} e^{\left (-\frac {x^{2} \log \relax (x)^{2} - 8 \, x^{2} e^{x} + 40 \, x^{2} + {\left (x^{2} e^{x} - 13 \, x^{2}\right )} \log \relax (x) - 5 \, x - 4}{x^{2} \log \relax (x) - 8 \, x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)*x^3-x^2)*log(x)^2+(16*exp(x)*x^3+16*x^2-5*x-8)*log(x)-64*exp(x)*x^3-64*x^2+35*x+60)*exp((-
x^2*log(x)^2+(-exp(x)*x^2+13*x^2)*log(x)+8*exp(x)*x^2-40*x^2+5*x+4)/(x^2*log(x)-8*x^2))/(x^3*log(x)^2-16*x^3*l
og(x)+64*x^3),x, algorithm="fricas")

[Out]

e^(-(x^2*log(x)^2 - 8*x^2*e^x + 40*x^2 + (x^2*e^x - 13*x^2)*log(x) - 5*x - 4)/(x^2*log(x) - 8*x^2))

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giac [B]  time = 0.64, size = 142, normalized size = 4.18 \begin {gather*} e^{\left (-\frac {x^{2} e^{x} \log \relax (x)}{x^{2} \log \relax (x) - 8 \, x^{2}} - \frac {x^{2} \log \relax (x)^{2}}{x^{2} \log \relax (x) - 8 \, x^{2}} + \frac {8 \, x^{2} e^{x}}{x^{2} \log \relax (x) - 8 \, x^{2}} + \frac {13 \, x^{2} \log \relax (x)}{x^{2} \log \relax (x) - 8 \, x^{2}} - \frac {40 \, x^{2}}{x^{2} \log \relax (x) - 8 \, x^{2}} + \frac {5 \, x}{x^{2} \log \relax (x) - 8 \, x^{2}} + \frac {4}{x^{2} \log \relax (x) - 8 \, x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)*x^3-x^2)*log(x)^2+(16*exp(x)*x^3+16*x^2-5*x-8)*log(x)-64*exp(x)*x^3-64*x^2+35*x+60)*exp((-
x^2*log(x)^2+(-exp(x)*x^2+13*x^2)*log(x)+8*exp(x)*x^2-40*x^2+5*x+4)/(x^2*log(x)-8*x^2))/(x^3*log(x)^2-16*x^3*l
og(x)+64*x^3),x, algorithm="giac")

[Out]

e^(-x^2*e^x*log(x)/(x^2*log(x) - 8*x^2) - x^2*log(x)^2/(x^2*log(x) - 8*x^2) + 8*x^2*e^x/(x^2*log(x) - 8*x^2) +
13*x^2*log(x)/(x^2*log(x) - 8*x^2) - 40*x^2/(x^2*log(x) - 8*x^2) + 5*x/(x^2*log(x) - 8*x^2) + 4/(x^2*log(x) -
8*x^2))

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maple [A]  time = 0.08, size = 53, normalized size = 1.56

method	result	size
risch	\({\mathrm e}^{-\frac {x^{2} \ln \relax (x )^{2}+x^{2} {\mathrm e}^{x} \ln \relax (x )-13 x^{2} \ln \relax (x )-8 \,{\mathrm e}^{x} x^{2}+40 x^{2}-5 x -4}{x^{2} \left (\ln \relax (x )-8\right )}}\)	\(53\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-exp(x)*x^3-x^2)*ln(x)^2+(16*exp(x)*x^3+16*x^2-5*x-8)*ln(x)-64*exp(x)*x^3-64*x^2+35*x+60)*exp((-x^2*ln(x
)^2+(-exp(x)*x^2+13*x^2)*ln(x)+8*exp(x)*x^2-40*x^2+5*x+4)/(x^2*ln(x)-8*x^2))/(x^3*ln(x)^2-16*x^3*ln(x)+64*x^3)
,x,method=_RETURNVERBOSE)

[Out]

exp(-(x^2*ln(x)^2+x^2*exp(x)*ln(x)-13*x^2*ln(x)-8*exp(x)*x^2+40*x^2-5*x-4)/x^2/(ln(x)-8))

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maxima [B]  time = 0.57, size = 82, normalized size = 2.41 \begin {gather*} e^{\left (-\frac {e^{x} \log \relax (x)}{\log \relax (x) - 8} - \frac {\log \relax (x)^{2}}{\log \relax (x) - 8} + \frac {8 \, e^{x}}{\log \relax (x) - 8} + \frac {13 \, \log \relax (x)}{\log \relax (x) - 8} + \frac {4}{x^{2} \log \relax (x) - 8 \, x^{2}} + \frac {5}{x \log \relax (x) - 8 \, x} - \frac {40}{\log \relax (x) - 8}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)*x^3-x^2)*log(x)^2+(16*exp(x)*x^3+16*x^2-5*x-8)*log(x)-64*exp(x)*x^3-64*x^2+35*x+60)*exp((-
x^2*log(x)^2+(-exp(x)*x^2+13*x^2)*log(x)+8*exp(x)*x^2-40*x^2+5*x+4)/(x^2*log(x)-8*x^2))/(x^3*log(x)^2-16*x^3*l
og(x)+64*x^3),x, algorithm="maxima")

[Out]

e^(-e^x*log(x)/(log(x) - 8) - log(x)^2/(log(x) - 8) + 8*e^x/(log(x) - 8) + 13*log(x)/(log(x) - 8) + 4/(x^2*log
(x) - 8*x^2) + 5/(x*log(x) - 8*x) - 40/(log(x) - 8))

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mupad [B]  time = 5.53, size = 117, normalized size = 3.44 \begin {gather*} \frac {{\mathrm {e}}^{\frac {5\,x}{x^2\,\ln \relax (x)-8\,x^2}}\,{\mathrm {e}}^{\frac {8\,x^2\,{\mathrm {e}}^x}{x^2\,\ln \relax (x)-8\,x^2}}\,{\mathrm {e}}^{-\frac {40\,x^2}{x^2\,\ln \relax (x)-8\,x^2}}\,{\mathrm {e}}^{\frac {4}{x^2\,\ln \relax (x)-8\,x^2}}\,{\mathrm {e}}^{-\frac {x^2\,{\ln \relax (x)}^2}{x^2\,\ln \relax (x)-8\,x^2}}}{x^{\frac {{\mathrm {e}}^x-13}{\ln \relax (x)-8}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((5*x + 8*x^2*exp(x) - x^2*log(x)^2 - 40*x^2 - log(x)*(x^2*exp(x) - 13*x^2) + 4)/(x^2*log(x) - 8*x^2)
)*(64*x^3*exp(x) - 35*x + log(x)*(5*x - 16*x^3*exp(x) - 16*x^2 + 8) + 64*x^2 + log(x)^2*(x^3*exp(x) + x^2) - 6
0))/(x^3*log(x)^2 - 16*x^3*log(x) + 64*x^3),x)

[Out]

(exp((5*x)/(x^2*log(x) - 8*x^2))*exp((8*x^2*exp(x))/(x^2*log(x) - 8*x^2))*exp(-(40*x^2)/(x^2*log(x) - 8*x^2))*
exp(4/(x^2*log(x) - 8*x^2))*exp(-(x^2*log(x)^2)/(x^2*log(x) - 8*x^2)))/x^((exp(x) - 13)/(log(x) - 8))

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sympy [B]  time = 1.19, size = 54, normalized size = 1.59 \begin {gather*} e^{\frac {8 x^{2} e^{x} - x^{2} \log {\relax (x )}^{2} - 40 x^{2} + 5 x + \left (- x^{2} e^{x} + 13 x^{2}\right ) \log {\relax (x )} + 4}{x^{2} \log {\relax (x )} - 8 x^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)*x**3-x**2)*ln(x)**2+(16*exp(x)*x**3+16*x**2-5*x-8)*ln(x)-64*exp(x)*x**3-64*x**2+35*x+60)*e
xp((-x**2*ln(x)**2+(-exp(x)*x**2+13*x**2)*ln(x)+8*exp(x)*x**2-40*x**2+5*x+4)/(x**2*ln(x)-8*x**2))/(x**3*ln(x)*
*2-16*x**3*ln(x)+64*x**3),x)

[Out]

exp((8*x**2*exp(x) - x**2*log(x)**2 - 40*x**2 + 5*x + (-x**2*exp(x) + 13*x**2)*log(x) + 4)/(x**2*log(x) - 8*x*
*2))

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