3.96.95 \(\int \frac {1}{5} e^{-x} (e^{\frac {2 e^{-x} x}{5}} (5 e^x+2 x-2 x^2)+e^{(-4+e^5)^x} (-5 e^x-5 (e (-4+e^5))^x x \log (-4+e^5))) \, dx\)

Optimal. Leaf size=26 \[ \left (-e^{\left (-4+e^5\right )^x}+e^{\frac {2 e^{-x} x}{5}}\right ) x \]

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Rubi [F]  time = 0.78, 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 {1}{5} e^{-x} \left (e^{\frac {2 e^{-x} x}{5}} \left (5 e^x+2 x-2 x^2\right )+e^{\left (-4+e^5\right )^x} \left (-5 e^x-5 \left (e \left (-4+e^5\right )\right )^x x \log \left (-4+e^5\right )\right )\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [C]  time = 0.64, size = 85, normalized size = 3.27 \begin {gather*} \frac {1}{5} \left (5 e^{\frac {2 e^{-x} x}{5}} x-\frac {5 \text {Ei}\left (\left (-4+e^5\right )^x\right )}{\log \left (-4+e^5\right )}\right )-\left (-\frac {\text {Ei}\left (\left (-4+e^5\right )^x\right )}{\log ^2\left (-4+e^5\right )}+\frac {e^{\left (-4+e^5\right )^x} x}{\log \left (-4+e^5\right )}\right ) \log \left (-4+e^5\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.48, size = 21, normalized size = 0.81 \begin {gather*} x e^{\left (\frac {2}{5} \, x e^{\left (-x\right )}\right )} - x e^{\left ({\left (e^{5} - 4\right )}^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-5*x*exp(x)*log(exp(5)-4)*exp(x*log(exp(5)-4))-5*exp(x))*exp(exp(x*log(exp(5)-4)))+(5*exp(x)-2
*x^2+2*x)*exp(1/5*x/exp(x))^2)/exp(x),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-5*x*exp(x)*log(exp(5)-4)*exp(x*log(exp(5)-4))-5*exp(x))*exp(exp(x*log(exp(5)-4)))+(5*exp(x)-2
*x^2+2*x)*exp(1/5*x/exp(x))^2)/exp(x),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.07, size = 22, normalized size = 0.85

method	result	size
risch	\({\mathrm e}^{\frac {2 x \,{\mathrm e}^{-x}}{5}} x -x \,{\mathrm e}^{\left ({\mathrm e}^{5}-4\right )^{x}}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((-5*x*exp(x)*ln(exp(5)-4)*exp(x*ln(exp(5)-4))-5*exp(x))*exp(exp(x*ln(exp(5)-4)))+(5*exp(x)-2*x^2+2*x)
*exp(1/5*x/exp(x))^2)/exp(x),x,method=_RETURNVERBOSE)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-5*x*exp(x)*log(exp(5)-4)*exp(x*log(exp(5)-4))-5*exp(x))*exp(exp(x*log(exp(5)-4)))+(5*exp(x)-2
*x^2+2*x)*exp(1/5*x/exp(x))^2)/exp(x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 9.23, size = 21, normalized size = 0.81 \begin {gather*} -x\,\left ({\mathrm {e}}^{{\left ({\mathrm {e}}^5-4\right )}^x}-{\mathrm {e}}^{\frac {2\,x\,{\mathrm {e}}^{-x}}{5}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-5*x*exp(x)*ln(exp(5)-4)*exp(x*ln(exp(5)-4))-5*exp(x))*exp(exp(x*ln(exp(5)-4)))+(5*exp(x)-2*x*
*2+2*x)*exp(1/5*x/exp(x))**2)/exp(x),x)

[Out]

Timed out

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