3.96.43 \(\int \frac {1}{10} e^{2-2 e^x} (4+20 e^{-2+2 e^x} x-8 e^x x-2 \log (6 x)+(-1+2 e^x x) \log ^2(6 x)) \, dx\)

Optimal. Leaf size=26 \[ x^2-\frac {1}{10} e^{2-2 e^x} x \left (-4+\log ^2(6 x)\right ) \]

________________________________________________________________________________________

Rubi [F]  time = 1.45, 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}{10} e^{2-2 e^x} \left (4+20 e^{-2+2 e^x} x-8 e^x x-2 \log (6 x)+\left (-1+2 e^x x\right ) \log ^2(6 x)\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(2 - 2*E^x)*(4 + 20*E^(-2 + 2*E^x)*x - 8*E^x*x - 2*Log[6*x] + (-1 + 2*E^x*x)*Log[6*x]^2))/10,x]

[Out]

x^2 + (2*E^2*ExpIntegralEi[-2*E^x])/5 - (E^2*ExpIntegralEi[-2*E^x]*Log[6*x])/5 - (4*Defer[Int][E^(-2*(-1 + E^x
) + x)*x, x])/5 + (E^2*Defer[Int][ExpIntegralEi[-2*E^x]/x, x])/5 - Defer[Int][Log[6*x]^2/E^(2*(-1 + E^x)), x]/
10 + Defer[Int][E^(-2*(-1 + E^x) + x)*x*Log[6*x]^2, x]/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{10} \int e^{2-2 e^x} \left (4+20 e^{-2+2 e^x} x-8 e^x x-2 \log (6 x)+\left (-1+2 e^x x\right ) \log ^2(6 x)\right ) \, dx\\ &=\frac {1}{10} \int e^{-2 \left (-1+e^x\right )} \left (4+20 e^{-2+2 e^x} x-8 e^x x-2 \log (6 x)+\left (-1+2 e^x x\right ) \log ^2(6 x)\right ) \, dx\\ &=\frac {1}{10} \int \left (4 e^{-2 \left (-1+e^x\right )}+20 x-8 e^{-2 \left (-1+e^x\right )+x} x-2 e^{-2 \left (-1+e^x\right )} \log (6 x)+e^{-2 \left (-1+e^x\right )} \left (-1+2 e^x x\right ) \log ^2(6 x)\right ) \, dx\\ &=x^2+\frac {1}{10} \int e^{-2 \left (-1+e^x\right )} \left (-1+2 e^x x\right ) \log ^2(6 x) \, dx-\frac {1}{5} \int e^{-2 \left (-1+e^x\right )} \log (6 x) \, dx+\frac {2}{5} \int e^{-2 \left (-1+e^x\right )} \, dx-\frac {4}{5} \int e^{-2 \left (-1+e^x\right )+x} x \, dx\\ &=x^2-\frac {1}{5} e^2 \text {Ei}\left (-2 e^x\right ) \log (6 x)+\frac {1}{10} \int \left (-e^{-2 \left (-1+e^x\right )} \log ^2(6 x)+2 e^{-2 \left (-1+e^x\right )+x} x \log ^2(6 x)\right ) \, dx+\frac {1}{5} \int \frac {e^2 \text {Ei}\left (-2 e^x\right )}{x} \, dx+\frac {2}{5} \operatorname {Subst}\left (\int \frac {e^{2-2 x}}{x} \, dx,x,e^x\right )-\frac {4}{5} \int e^{-2 \left (-1+e^x\right )+x} x \, dx\\ &=x^2+\frac {2}{5} e^2 \text {Ei}\left (-2 e^x\right )-\frac {1}{5} e^2 \text {Ei}\left (-2 e^x\right ) \log (6 x)-\frac {1}{10} \int e^{-2 \left (-1+e^x\right )} \log ^2(6 x) \, dx+\frac {1}{5} \int e^{-2 \left (-1+e^x\right )+x} x \log ^2(6 x) \, dx-\frac {4}{5} \int e^{-2 \left (-1+e^x\right )+x} x \, dx+\frac {1}{5} e^2 \int \frac {\text {Ei}\left (-2 e^x\right )}{x} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 1.50, size = 39, normalized size = 1.50 \begin {gather*} \frac {1}{10} e^{-2 e^x} x \left (4 e^2+10 e^{2 e^x} x-e^2 \log ^2(6 x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(2 - 2*E^x)*(4 + 20*E^(-2 + 2*E^x)*x - 8*E^x*x - 2*Log[6*x] + (-1 + 2*E^x*x)*Log[6*x]^2))/10,x]

[Out]

(x*(4*E^2 + 10*E^(2*E^x)*x - E^2*Log[6*x]^2))/(10*E^(2*E^x))

________________________________________________________________________________________

fricas [A]  time = 0.67, size = 34, normalized size = 1.31 \begin {gather*} \frac {1}{10} \, {\left (10 \, x^{2} e^{\left (2 \, e^{x} - 2\right )} - x \log \left (6 \, x\right )^{2} + 4 \, x\right )} e^{\left (-2 \, e^{x} + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*(20*x*exp(exp(x)-1)^2+(2*exp(x)*x-1)*log(6*x)^2-2*log(6*x)-8*exp(x)*x+4)/exp(exp(x)-1)^2,x, alg
orithm="fricas")

[Out]

1/10*(10*x^2*e^(2*e^x - 2) - x*log(6*x)^2 + 4*x)*e^(-2*e^x + 2)

________________________________________________________________________________________

giac [B]  time = 0.15, size = 68, normalized size = 2.62 \begin {gather*} -\frac {1}{10} \, {\left (x e^{\left (x - 2 \, e^{x} + 2\right )} \log \relax (6)^{2} + 2 \, x e^{\left (x - 2 \, e^{x} + 2\right )} \log \relax (6) \log \relax (x) + x e^{\left (x - 2 \, e^{x} + 2\right )} \log \relax (x)^{2} - 10 \, x^{2} e^{x} - 4 \, x e^{\left (x - 2 \, e^{x} + 2\right )}\right )} e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*(20*x*exp(exp(x)-1)^2+(2*exp(x)*x-1)*log(6*x)^2-2*log(6*x)-8*exp(x)*x+4)/exp(exp(x)-1)^2,x, alg
orithm="giac")

[Out]

-1/10*(x*e^(x - 2*e^x + 2)*log(6)^2 + 2*x*e^(x - 2*e^x + 2)*log(6)*log(x) + x*e^(x - 2*e^x + 2)*log(x)^2 - 10*
x^2*e^x - 4*x*e^(x - 2*e^x + 2))*e^(-x)

________________________________________________________________________________________

maple [A]  time = 0.10, size = 27, normalized size = 1.04

method	result	size
risch	\(x^{2}+\frac {\left (-x \ln \left (6 x \right )^{2}+4 x \right ) {\mathrm e}^{-2 \,{\mathrm e}^{x}+2}}{10}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/10*(20*x*exp(exp(x)-1)^2+(2*exp(x)*x-1)*ln(6*x)^2-2*ln(6*x)-8*exp(x)*x+4)/exp(exp(x)-1)^2,x,method=_RETU
RNVERBOSE)

[Out]

x^2+1/10*(-x*ln(6*x)^2+4*x)*exp(-2*exp(x)+2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*(20*x*exp(exp(x)-1)^2+(2*exp(x)*x-1)*log(6*x)^2-2*log(6*x)-8*exp(x)*x+4)/exp(exp(x)-1)^2,x, alg
orithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 9.32, size = 25, normalized size = 0.96 \begin {gather*} {\mathrm {e}}^{2-2\,{\mathrm {e}}^x}\,\left (\frac {2\,x}{5}-\frac {x\,{\ln \left (6\,x\right )}^2}{10}\right )+x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(2 - 2*exp(x))*((log(6*x)^2*(2*x*exp(x) - 1))/10 - log(6*x)/5 - (4*x*exp(x))/5 + 2*x*exp(2*exp(x) - 2)
+ 2/5),x)

[Out]

exp(2 - 2*exp(x))*((2*x)/5 - (x*log(6*x)^2)/10) + x^2

________________________________________________________________________________________

sympy [A]  time = 10.26, size = 24, normalized size = 0.92 \begin {gather*} x^{2} + \frac {\left (- x \log {\left (6 x \right )}^{2} + 4 x\right ) e^{2 - 2 e^{x}}}{10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*(20*x*exp(exp(x)-1)**2+(2*exp(x)*x-1)*ln(6*x)**2-2*ln(6*x)-8*exp(x)*x+4)/exp(exp(x)-1)**2,x)

[Out]

x**2 + (-x*log(6*x)**2 + 4*x)*exp(2 - 2*exp(x))/10

________________________________________________________________________________________