3.15.86 \(\int \frac {e^{\frac {10-2 e^{e^{\frac {-2+5 x+4 x^2}{x}}}-\log (x^2)}{-5+e^{e^{\frac {-2+5 x+4 x^2}{x}}}}} (-15 x-e^{2 e^{\frac {-2+5 x+4 x^2}{x}}} x+e^{e^{\frac {-2+5 x+4 x^2}{x}}} (8 x+e^{\frac {-2+5 x+4 x^2}{x}} (2+4 x^2) \log (x^2)))}{25 x^3-10 e^{e^{\frac {-2+5 x+4 x^2}{x}}} x^3+e^{2 e^{\frac {-2+5 x+4 x^2}{x}}} x^3} \, dx\)

Optimal. Leaf size=37 \[ \frac {e^{-2+\frac {\log \left (x^2\right )}{5-e^{e^{\frac {-3+x+(1+2 x)^2}{x}}}}}}{x} \]

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Rubi [F]  time = 10.97, 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 {10-2 e^{e^{\frac {-2+5 x+4 x^2}{x}}}-\log \left (x^2\right )}{-5+e^{e^{\frac {-2+5 x+4 x^2}{x}}}}\right ) \left (-15 x-e^{2 e^{\frac {-2+5 x+4 x^2}{x}}} x+e^{e^{\frac {-2+5 x+4 x^2}{x}}} \left (8 x+e^{\frac {-2+5 x+4 x^2}{x}} \left (2+4 x^2\right ) \log \left (x^2\right )\right )\right )}{25 x^3-10 e^{e^{\frac {-2+5 x+4 x^2}{x}}} x^3+e^{2 e^{\frac {-2+5 x+4 x^2}{x}}} x^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((10 - 2*E^E^((-2 + 5*x + 4*x^2)/x) - Log[x^2])/(-5 + E^E^((-2 + 5*x + 4*x^2)/x)))*(-15*x - E^(2*E^((-2
+ 5*x + 4*x^2)/x))*x + E^E^((-2 + 5*x + 4*x^2)/x)*(8*x + E^((-2 + 5*x + 4*x^2)/x)*(2 + 4*x^2)*Log[x^2])))/(25
*x^3 - 10*E^E^((-2 + 5*x + 4*x^2)/x)*x^3 + E^(2*E^((-2 + 5*x + 4*x^2)/x))*x^3),x]

[Out]

2*Log[x^2]*Defer[Int][(E^(3 + E^(5 - 2/x + 4*x) - 2/x + 4*x)*(x^2)^(5 - E^E^(5 - 2/x + 4*x))^(-1))/((-5 + E^E^
(5 - 2/x + 4*x))^2*x^3), x] + 4*Log[x^2]*Defer[Int][(E^(3 + E^(5 - 2/x + 4*x) - 2/x + 4*x)*(x^2)^(5 - E^E^(5 -
2/x + 4*x))^(-1))/((-5 + E^E^(5 - 2/x + 4*x))^2*x), x] - Defer[Int][(x^2)^(-1 + (5 - E^E^(5 - 2/x + 4*x))^(-1
)), x]/E^2 - (2*Defer[Int][(x^2)^(-1 + (5 - E^E^(5 - 2/x + 4*x))^(-1))/(-5 + E^E^(5 - 2/x + 4*x)), x])/E^2 - 4
*Defer[Int][Defer[Int][(E^(3 + E^(5 - 2/x + 4*x) - 2/x + 4*x)*(x^2)^(5 - E^E^(5 - 2/x + 4*x))^(-1))/((-5 + E^E
^(5 - 2/x + 4*x))^2*x^3), x]/x, x] - 8*Defer[Int][Defer[Int][(E^(3 + E^(5 - 2/x + 4*x) - 2/x + 4*x)*(x^2)^(5 -
E^E^(5 - 2/x + 4*x))^(-1))/((-5 + E^E^(5 - 2/x + 4*x))^2*x), x]/x, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (x^2\right )^{\frac {1}{5-e^{e^{5-\frac {2}{x}+4 x}}}} \left (-\frac {\left (15-8 e^{e^{5-\frac {2}{x}+4 x}}+e^{2 e^{5-\frac {2}{x}+4 x}}\right ) x}{e^2}+2 e^{3+e^{5-\frac {2}{x}+4 x}-\frac {2}{x}+4 x} \left (1+2 x^2\right ) \log \left (x^2\right )\right )}{\left (5-e^{e^{5-\frac {2}{x}+4 x}}\right )^2 x^3} \, dx\\ &=\int \left (-\frac {\left (-3+e^{e^{5-\frac {2}{x}+4 x}}\right ) \left (x^2\right )^{-1+\frac {1}{5-e^{e^{5-\frac {2}{x}+4 x}}}}}{e^2 \left (-5+e^{e^{5-\frac {2}{x}+4 x}}\right )}+\frac {2 e^{3+e^{5-\frac {2}{x}+4 x}-\frac {2}{x}+4 x} \left (x^2\right )^{\frac {1}{5-e^{e^{5-\frac {2}{x}+4 x}}}} \left (1+2 x^2\right ) \log \left (x^2\right )}{\left (-5+e^{e^{5-\frac {2}{x}+4 x}}\right )^2 x^3}\right ) \, dx\\ &=2 \int \frac {e^{3+e^{5-\frac {2}{x}+4 x}-\frac {2}{x}+4 x} \left (x^2\right )^{\frac {1}{5-e^{e^{5-\frac {2}{x}+4 x}}}} \left (1+2 x^2\right ) \log \left (x^2\right )}{\left (-5+e^{e^{5-\frac {2}{x}+4 x}}\right )^2 x^3} \, dx-\frac {\int \frac {\left (-3+e^{e^{5-\frac {2}{x}+4 x}}\right ) \left (x^2\right )^{-1+\frac {1}{5-e^{e^{5-\frac {2}{x}+4 x}}}}}{-5+e^{e^{5-\frac {2}{x}+4 x}}} \, dx}{e^2}\\ &=-\left (2 \int \frac {2 \left (\int \frac {e^{3+e^{5-\frac {2}{x}+4 x}-\frac {2}{x}+4 x} \left (x^2\right )^{\frac {1}{5-e^{e^{5-\frac {2}{x}+4 x}}}}}{\left (-5+e^{e^{5-\frac {2}{x}+4 x}}\right )^2 x^3} \, dx+2 \int \frac {e^{3+e^{5-\frac {2}{x}+4 x}-\frac {2}{x}+4 x} \left (x^2\right )^{\frac {1}{5-e^{e^{5-\frac {2}{x}+4 x}}}}}{\left (-5+e^{e^{5-\frac {2}{x}+4 x}}\right )^2 x} \, dx\right )}{x} \, dx\right )-\frac {\int \left (\left (x^2\right )^{-1+\frac {1}{5-e^{e^{5-\frac {2}{x}+4 x}}}}+\frac {2 \left (x^2\right )^{-1+\frac {1}{5-e^{e^{5-\frac {2}{x}+4 x}}}}}{-5+e^{e^{5-\frac {2}{x}+4 x}}}\right ) \, dx}{e^2}+\left (2 \log \left (x^2\right )\right ) \int \frac {e^{3+e^{5-\frac {2}{x}+4 x}-\frac {2}{x}+4 x} \left (x^2\right )^{\frac {1}{5-e^{e^{5-\frac {2}{x}+4 x}}}}}{\left (-5+e^{e^{5-\frac {2}{x}+4 x}}\right )^2 x^3} \, dx+\left (4 \log \left (x^2\right )\right ) \int \frac {e^{3+e^{5-\frac {2}{x}+4 x}-\frac {2}{x}+4 x} \left (x^2\right )^{\frac {1}{5-e^{e^{5-\frac {2}{x}+4 x}}}}}{\left (-5+e^{e^{5-\frac {2}{x}+4 x}}\right )^2 x} \, dx\\ &=-\left (4 \int \frac {\int \frac {e^{3+e^{5-\frac {2}{x}+4 x}-\frac {2}{x}+4 x} \left (x^2\right )^{\frac {1}{5-e^{e^{5-\frac {2}{x}+4 x}}}}}{\left (-5+e^{e^{5-\frac {2}{x}+4 x}}\right )^2 x^3} \, dx+2 \int \frac {e^{3+e^{5-\frac {2}{x}+4 x}-\frac {2}{x}+4 x} \left (x^2\right )^{\frac {1}{5-e^{e^{5-\frac {2}{x}+4 x}}}}}{\left (-5+e^{e^{5-\frac {2}{x}+4 x}}\right )^2 x} \, dx}{x} \, dx\right )-\frac {\int \left (x^2\right )^{-1+\frac {1}{5-e^{e^{5-\frac {2}{x}+4 x}}}} \, dx}{e^2}-\frac {2 \int \frac {\left (x^2\right )^{-1+\frac {1}{5-e^{e^{5-\frac {2}{x}+4 x}}}}}{-5+e^{e^{5-\frac {2}{x}+4 x}}} \, dx}{e^2}+\left (2 \log \left (x^2\right )\right ) \int \frac {e^{3+e^{5-\frac {2}{x}+4 x}-\frac {2}{x}+4 x} \left (x^2\right )^{\frac {1}{5-e^{e^{5-\frac {2}{x}+4 x}}}}}{\left (-5+e^{e^{5-\frac {2}{x}+4 x}}\right )^2 x^3} \, dx+\left (4 \log \left (x^2\right )\right ) \int \frac {e^{3+e^{5-\frac {2}{x}+4 x}-\frac {2}{x}+4 x} \left (x^2\right )^{\frac {1}{5-e^{e^{5-\frac {2}{x}+4 x}}}}}{\left (-5+e^{e^{5-\frac {2}{x}+4 x}}\right )^2 x} \, dx\\ &=-\left (4 \int \left (\frac {\int \frac {e^{3+e^{5-\frac {2}{x}+4 x}-\frac {2}{x}+4 x} \left (x^2\right )^{\frac {1}{5-e^{e^{5-\frac {2}{x}+4 x}}}}}{\left (-5+e^{e^{5-\frac {2}{x}+4 x}}\right )^2 x^3} \, dx}{x}+\frac {2 \int \frac {e^{3+e^{5-\frac {2}{x}+4 x}-\frac {2}{x}+4 x} \left (x^2\right )^{\frac {1}{5-e^{e^{5-\frac {2}{x}+4 x}}}}}{\left (-5+e^{e^{5-\frac {2}{x}+4 x}}\right )^2 x} \, dx}{x}\right ) \, dx\right )-\frac {\int \left (x^2\right )^{-1+\frac {1}{5-e^{e^{5-\frac {2}{x}+4 x}}}} \, dx}{e^2}-\frac {2 \int \frac {\left (x^2\right )^{-1+\frac {1}{5-e^{e^{5-\frac {2}{x}+4 x}}}}}{-5+e^{e^{5-\frac {2}{x}+4 x}}} \, dx}{e^2}+\left (2 \log \left (x^2\right )\right ) \int \frac {e^{3+e^{5-\frac {2}{x}+4 x}-\frac {2}{x}+4 x} \left (x^2\right )^{\frac {1}{5-e^{e^{5-\frac {2}{x}+4 x}}}}}{\left (-5+e^{e^{5-\frac {2}{x}+4 x}}\right )^2 x^3} \, dx+\left (4 \log \left (x^2\right )\right ) \int \frac {e^{3+e^{5-\frac {2}{x}+4 x}-\frac {2}{x}+4 x} \left (x^2\right )^{\frac {1}{5-e^{e^{5-\frac {2}{x}+4 x}}}}}{\left (-5+e^{e^{5-\frac {2}{x}+4 x}}\right )^2 x} \, dx\\ &=-\left (4 \int \frac {\int \frac {e^{3+e^{5-\frac {2}{x}+4 x}-\frac {2}{x}+4 x} \left (x^2\right )^{\frac {1}{5-e^{e^{5-\frac {2}{x}+4 x}}}}}{\left (-5+e^{e^{5-\frac {2}{x}+4 x}}\right )^2 x^3} \, dx}{x} \, dx\right )-8 \int \frac {\int \frac {e^{3+e^{5-\frac {2}{x}+4 x}-\frac {2}{x}+4 x} \left (x^2\right )^{\frac {1}{5-e^{e^{5-\frac {2}{x}+4 x}}}}}{\left (-5+e^{e^{5-\frac {2}{x}+4 x}}\right )^2 x} \, dx}{x} \, dx-\frac {\int \left (x^2\right )^{-1+\frac {1}{5-e^{e^{5-\frac {2}{x}+4 x}}}} \, dx}{e^2}-\frac {2 \int \frac {\left (x^2\right )^{-1+\frac {1}{5-e^{e^{5-\frac {2}{x}+4 x}}}}}{-5+e^{e^{5-\frac {2}{x}+4 x}}} \, dx}{e^2}+\left (2 \log \left (x^2\right )\right ) \int \frac {e^{3+e^{5-\frac {2}{x}+4 x}-\frac {2}{x}+4 x} \left (x^2\right )^{\frac {1}{5-e^{e^{5-\frac {2}{x}+4 x}}}}}{\left (-5+e^{e^{5-\frac {2}{x}+4 x}}\right )^2 x^3} \, dx+\left (4 \log \left (x^2\right )\right ) \int \frac {e^{3+e^{5-\frac {2}{x}+4 x}-\frac {2}{x}+4 x} \left (x^2\right )^{\frac {1}{5-e^{e^{5-\frac {2}{x}+4 x}}}}}{\left (-5+e^{e^{5-\frac {2}{x}+4 x}}\right )^2 x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.40, size = 31, normalized size = 0.84 \begin {gather*} \frac {\left (x^2\right )^{\frac {1}{5-e^{e^{5-\frac {2}{x}+4 x}}}}}{e^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((10 - 2*E^E^((-2 + 5*x + 4*x^2)/x) - Log[x^2])/(-5 + E^E^((-2 + 5*x + 4*x^2)/x)))*(-15*x - E^(2*
E^((-2 + 5*x + 4*x^2)/x))*x + E^E^((-2 + 5*x + 4*x^2)/x)*(8*x + E^((-2 + 5*x + 4*x^2)/x)*(2 + 4*x^2)*Log[x^2])
))/(25*x^3 - 10*E^E^((-2 + 5*x + 4*x^2)/x)*x^3 + E^(2*E^((-2 + 5*x + 4*x^2)/x))*x^3),x]

[Out]

(x^2)^(5 - E^E^(5 - 2/x + 4*x))^(-1)/(E^2*x)

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fricas [A]  time = 0.64, size = 51, normalized size = 1.38 \begin {gather*} \frac {e^{\left (-\frac {2 \, e^{\left (e^{\left (\frac {4 \, x^{2} + 5 \, x - 2}{x}\right )}\right )} + \log \left (x^{2}\right ) - 10}{e^{\left (e^{\left (\frac {4 \, x^{2} + 5 \, x - 2}{x}\right )}\right )} - 5}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*exp(exp((4*x^2+5*x-2)/x))^2+((4*x^2+2)*exp((4*x^2+5*x-2)/x)*log(x^2)+8*x)*exp(exp((4*x^2+5*x-2)/
x))-15*x)*exp((-2*exp(exp((4*x^2+5*x-2)/x))-log(x^2)+10)/(exp(exp((4*x^2+5*x-2)/x))-5))/(x^3*exp(exp((4*x^2+5*
x-2)/x))^2-10*x^3*exp(exp((4*x^2+5*x-2)/x))+25*x^3),x, algorithm="fricas")

[Out]

e^(-(2*e^(e^((4*x^2 + 5*x - 2)/x)) + log(x^2) - 10)/(e^(e^((4*x^2 + 5*x - 2)/x)) - 5))/x

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giac [B]  time = 0.78, size = 100, normalized size = 2.70 \begin {gather*} \frac {e^{\left (\frac {e^{\left (\frac {4 \, x^{2} + 5 \, x - 2}{x} + e^{\left (\frac {4 \, x^{2} + 5 \, x - 2}{x}\right )}\right )} - 5 \, e^{\left (\frac {4 \, x^{2} + 5 \, x - 2}{x}\right )} - \log \left (x^{2}\right )}{e^{\left (e^{\left (\frac {4 \, x^{2} + 5 \, x - 2}{x}\right )}\right )} - 5} - e^{\left (\frac {4 \, x^{2} + 5 \, x - 2}{x}\right )} - 2\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*exp(exp((4*x^2+5*x-2)/x))^2+((4*x^2+2)*exp((4*x^2+5*x-2)/x)*log(x^2)+8*x)*exp(exp((4*x^2+5*x-2)/
x))-15*x)*exp((-2*exp(exp((4*x^2+5*x-2)/x))-log(x^2)+10)/(exp(exp((4*x^2+5*x-2)/x))-5))/(x^3*exp(exp((4*x^2+5*
x-2)/x))^2-10*x^3*exp(exp((4*x^2+5*x-2)/x))+25*x^3),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.34, size = 101, normalized size = 2.73

method	result	size
risch	\(\frac {{\mathrm e}^{-\frac {-i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+2 i \pi \mathrm {csgn}\left (i x^{2}\right )^{2} \mathrm {csgn}\left (i x \right )-i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )^{2}+4 \ln \relax (x )+4 \,{\mathrm e}^{{\mathrm e}^{\frac {4 x^{2}+5 x -2}{x}}}-20}{2 \left ({\mathrm e}^{{\mathrm e}^{\frac {4 x^{2}+5 x -2}{x}}}-5\right )}}}{x}\)	\(101\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x*exp(exp((4*x^2+5*x-2)/x))^2+((4*x^2+2)*exp((4*x^2+5*x-2)/x)*ln(x^2)+8*x)*exp(exp((4*x^2+5*x-2)/x))-15*
x)*exp((-2*exp(exp((4*x^2+5*x-2)/x))-ln(x^2)+10)/(exp(exp((4*x^2+5*x-2)/x))-5))/(x^3*exp(exp((4*x^2+5*x-2)/x))
^2-10*x^3*exp(exp((4*x^2+5*x-2)/x))+25*x^3),x,method=_RETURNVERBOSE)

[Out]

1/x*exp(-1/2*(-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+4*ln(x)+4*exp(ex
p((4*x^2+5*x-2)/x))-20)/(exp(exp((4*x^2+5*x-2)/x))-5))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\int \frac {{\left (x e^{\left (2 \, e^{\left (\frac {4 \, x^{2} + 5 \, x - 2}{x}\right )}\right )} - 2 \, {\left ({\left (2 \, x^{2} + 1\right )} e^{\left (\frac {4 \, x^{2} + 5 \, x - 2}{x}\right )} \log \left (x^{2}\right ) + 4 \, x\right )} e^{\left (e^{\left (\frac {4 \, x^{2} + 5 \, x - 2}{x}\right )}\right )} + 15 \, x\right )} e^{\left (-\frac {2 \, e^{\left (e^{\left (\frac {4 \, x^{2} + 5 \, x - 2}{x}\right )}\right )} + \log \left (x^{2}\right ) - 10}{e^{\left (e^{\left (\frac {4 \, x^{2} + 5 \, x - 2}{x}\right )}\right )} - 5}\right )}}{x^{3} e^{\left (2 \, e^{\left (\frac {4 \, x^{2} + 5 \, x - 2}{x}\right )}\right )} - 10 \, x^{3} e^{\left (e^{\left (\frac {4 \, x^{2} + 5 \, x - 2}{x}\right )}\right )} + 25 \, x^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*exp(exp((4*x^2+5*x-2)/x))^2+((4*x^2+2)*exp((4*x^2+5*x-2)/x)*log(x^2)+8*x)*exp(exp((4*x^2+5*x-2)/
x))-15*x)*exp((-2*exp(exp((4*x^2+5*x-2)/x))-log(x^2)+10)/(exp(exp((4*x^2+5*x-2)/x))-5))/(x^3*exp(exp((4*x^2+5*
x-2)/x))^2-10*x^3*exp(exp((4*x^2+5*x-2)/x))+25*x^3),x, algorithm="maxima")

[Out]

-integrate((x*e^(2*e^((4*x^2 + 5*x - 2)/x)) - 2*((2*x^2 + 1)*e^((4*x^2 + 5*x - 2)/x)*log(x^2) + 4*x)*e^(e^((4*
x^2 + 5*x - 2)/x)) + 15*x)*e^(-(2*e^(e^((4*x^2 + 5*x - 2)/x)) + log(x^2) - 10)/(e^(e^((4*x^2 + 5*x - 2)/x)) -
5))/(x^3*e^(2*e^((4*x^2 + 5*x - 2)/x)) - 10*x^3*e^(e^((4*x^2 + 5*x - 2)/x)) + 25*x^3), x)

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mupad [B]  time = 1.74, size = 84, normalized size = 2.27 \begin {gather*} \frac {{\mathrm {e}}^{\frac {10}{{\mathrm {e}}^{{\mathrm {e}}^{4\,x}\,{\mathrm {e}}^5\,{\mathrm {e}}^{-\frac {2}{x}}}-5}}\,{\mathrm {e}}^{-\frac {2\,{\mathrm {e}}^{{\mathrm {e}}^{4\,x}\,{\mathrm {e}}^5\,{\mathrm {e}}^{-\frac {2}{x}}}}{{\mathrm {e}}^{{\mathrm {e}}^{4\,x}\,{\mathrm {e}}^5\,{\mathrm {e}}^{-\frac {2}{x}}}-5}}}{x\,{\left (x^2\right )}^{\frac {1}{{\mathrm {e}}^{{\mathrm {e}}^{4\,x}\,{\mathrm {e}}^5\,{\mathrm {e}}^{-\frac {2}{x}}}-5}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(log(x^2) + 2*exp(exp((5*x + 4*x^2 - 2)/x)) - 10)/(exp(exp((5*x + 4*x^2 - 2)/x)) - 5))*(15*x - exp(
exp((5*x + 4*x^2 - 2)/x))*(8*x + log(x^2)*exp((5*x + 4*x^2 - 2)/x)*(4*x^2 + 2)) + x*exp(2*exp((5*x + 4*x^2 - 2
)/x))))/(x^3*exp(2*exp((5*x + 4*x^2 - 2)/x)) - 10*x^3*exp(exp((5*x + 4*x^2 - 2)/x)) + 25*x^3),x)

[Out]

(exp(10/(exp(exp(4*x)*exp(5)*exp(-2/x)) - 5))*exp(-(2*exp(exp(4*x)*exp(5)*exp(-2/x)))/(exp(exp(4*x)*exp(5)*exp
(-2/x)) - 5)))/(x*(x^2)^(1/(exp(exp(4*x)*exp(5)*exp(-2/x)) - 5)))

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sympy [A]  time = 21.59, size = 42, normalized size = 1.14 \begin {gather*} \frac {e^{\frac {- 2 e^{e^{\frac {4 x^{2} + 5 x - 2}{x}}} - \log {\left (x^{2} \right )} + 10}{e^{e^{\frac {4 x^{2} + 5 x - 2}{x}}} - 5}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*exp(exp((4*x**2+5*x-2)/x))**2+((4*x**2+2)*exp((4*x**2+5*x-2)/x)*ln(x**2)+8*x)*exp(exp((4*x**2+5*
x-2)/x))-15*x)*exp((-2*exp(exp((4*x**2+5*x-2)/x))-ln(x**2)+10)/(exp(exp((4*x**2+5*x-2)/x))-5))/(x**3*exp(exp((
4*x**2+5*x-2)/x))**2-10*x**3*exp(exp((4*x**2+5*x-2)/x))+25*x**3),x)

[Out]

exp((-2*exp(exp((4*x**2 + 5*x - 2)/x)) - log(x**2) + 10)/(exp(exp((4*x**2 + 5*x - 2)/x)) - 5))/x

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