3.76.38 \(\int \frac {2-5 e^x x+e^{e^{16 e^{-x}} x} (-5 e^x+e^{16 e^{-x}} (2-32 e^{-x} x))}{e^{e^{16 e^{-x}} x}+x} \, dx\)

Optimal. Leaf size=28 \[ 5 \left (25-e^x\right )+\log \left (\left (e^{e^{16 e^{-x}} x}+x\right )^2\right ) \]

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Rubi [F]  time = 1.83, 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 {2-5 e^x x+e^{e^{16 e^{-x}} x} \left (-5 e^x+e^{16 e^{-x}} \left (2-32 e^{-x} x\right )\right )}{e^{e^{16 e^{-x}} x}+x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(2 - 5*E^x*x + E^(E^(16/E^x)*x)*(-5*E^x + E^(16/E^x)*(2 - (32*x)/E^x)))/(E^(E^(16/E^x)*x) + x),x]

[Out]

-5*E^x - 2*ExpIntegralEi[16/E^x] - 32*Defer[Int][E^(16/E^x - x)*x, x] + 2*Defer[Int][(E^(E^(16/E^x)*x) + x)^(-
1), x] - 2*Defer[Int][(E^(16/E^x)*x)/(E^(E^(16/E^x)*x) + x), x] + 32*Defer[Int][(E^(16/E^x - x)*x^2)/(E^(E^(16
/E^x)*x) + x), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-e^{-x} \left (5 e^{2 x}-2 e^{16 e^{-x}+x}+32 e^{16 e^{-x}} x\right )-\frac {2 e^{-x} \left (-e^x+e^{16 e^{-x}+x} x-16 e^{16 e^{-x}} x^2\right )}{e^{e^{16 e^{-x}} x}+x}\right ) \, dx\\ &=-\left (2 \int \frac {e^{-x} \left (-e^x+e^{16 e^{-x}+x} x-16 e^{16 e^{-x}} x^2\right )}{e^{e^{16 e^{-x}} x}+x} \, dx\right )-\int e^{-x} \left (5 e^{2 x}-2 e^{16 e^{-x}+x}+32 e^{16 e^{-x}} x\right ) \, dx\\ &=-\left (2 \int \left (-\frac {1}{e^{e^{16 e^{-x}} x}+x}+\frac {e^{16 e^{-x}} x}{e^{e^{16 e^{-x}} x}+x}-\frac {16 e^{16 e^{-x}-x} x^2}{e^{e^{16 e^{-x}} x}+x}\right ) \, dx\right )-\int \left (-2 e^{16 e^{-x}}+5 e^x+32 e^{16 e^{-x}-x} x\right ) \, dx\\ &=2 \int e^{16 e^{-x}} \, dx+2 \int \frac {1}{e^{e^{16 e^{-x}} x}+x} \, dx-2 \int \frac {e^{16 e^{-x}} x}{e^{e^{16 e^{-x}} x}+x} \, dx-5 \int e^x \, dx-32 \int e^{16 e^{-x}-x} x \, dx+32 \int \frac {e^{16 e^{-x}-x} x^2}{e^{e^{16 e^{-x}} x}+x} \, dx\\ &=-5 e^x+2 \int \frac {1}{e^{e^{16 e^{-x}} x}+x} \, dx-2 \int \frac {e^{16 e^{-x}} x}{e^{e^{16 e^{-x}} x}+x} \, dx-2 \operatorname {Subst}\left (\int \frac {e^{16 x}}{x} \, dx,x,e^{-x}\right )-32 \int e^{16 e^{-x}-x} x \, dx+32 \int \frac {e^{16 e^{-x}-x} x^2}{e^{e^{16 e^{-x}} x}+x} \, dx\\ &=-5 e^x-2 \text {Ei}\left (16 e^{-x}\right )+2 \int \frac {1}{e^{e^{16 e^{-x}} x}+x} \, dx-2 \int \frac {e^{16 e^{-x}} x}{e^{e^{16 e^{-x}} x}+x} \, dx-32 \int e^{16 e^{-x}-x} x \, dx+32 \int \frac {e^{16 e^{-x}-x} x^2}{e^{e^{16 e^{-x}} x}+x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.86, size = 24, normalized size = 0.86 \begin {gather*} -5 e^x+2 \log \left (e^{e^{16 e^{-x}} x}+x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 - 5*E^x*x + E^(E^(16/E^x)*x)*(-5*E^x + E^(16/E^x)*(2 - (32*x)/E^x)))/(E^(E^(16/E^x)*x) + x),x]

[Out]

-5*E^x + 2*Log[E^(E^(16/E^x)*x) + x]

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fricas [A]  time = 0.45, size = 37, normalized size = 1.32 \begin {gather*} 2 \, {\left (e^{\left (-x + 4 \, \log \relax (2)\right )} \log \left (x + e^{\left (x e^{\left (e^{\left (-x + 4 \, \log \relax (2)\right )}\right )}\right )}\right ) - 40\right )} e^{\left (x - 4 \, \log \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(e^(-x + 4*log(2))*log(x + e^(x*e^(e^(-x + 4*log(2))))) - 40)*e^(x - 4*log(2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.08, size = 21, normalized size = 0.75

method	result	size
norman	\(-5 \,{\mathrm e}^{x}+2 \ln \left ({\mathrm e}^{x \,{\mathrm e}^{16 \,{\mathrm e}^{-x}}}+x \right )\)	\(21\)
risch	\(-5 \,{\mathrm e}^{x}+2 \ln \left ({\mathrm e}^{x \,{\mathrm e}^{16 \,{\mathrm e}^{-x}}}+x \right )\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-2*x*exp(4*ln(2)-x)+2)*exp(exp(4*ln(2)-x))-5*exp(x))*exp(x*exp(exp(4*ln(2)-x)))-5*exp(x)*x+2)/(exp(x*ex
p(exp(4*ln(2)-x)))+x),x,method=_RETURNVERBOSE)

[Out]

-5*exp(x)+2*ln(exp(x*exp(16/exp(x)))+x)

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maxima [A]  time = 0.41, size = 20, normalized size = 0.71 \begin {gather*} -5 \, e^{x} + 2 \, \log \left (x + e^{\left (x e^{\left (16 \, e^{\left (-x\right )}\right )}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5*e^x + 2*log(x + e^(x*e^(16*e^(-x))))

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mupad [B]  time = 5.11, size = 20, normalized size = 0.71 \begin {gather*} 2\,\ln \left (x+{\mathrm {e}}^{x\,{\mathrm {e}}^{16\,{\mathrm {e}}^{-x}}}\right )-5\,{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*log(x + exp(x*exp(16*exp(-x)))) - 5*exp(x)

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sympy [A]  time = 0.26, size = 19, normalized size = 0.68 \begin {gather*} - 5 e^{x} + 2 \log {\left (x + e^{x e^{16 e^{- x}}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x*exp(4*ln(2)-x)+2)*exp(exp(4*ln(2)-x))-5*exp(x))*exp(x*exp(exp(4*ln(2)-x)))-5*exp(x)*x+2)/(ex
p(x*exp(exp(4*ln(2)-x)))+x),x)

[Out]

-5*exp(x) + 2*log(x + exp(x*exp(16*exp(-x))))

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