3.55.74 \(\int \frac {e^{-3+(-e^x-x+2 \log (5)) \log ^2(x)} ((-2 e^x-2 x+4 \log (5)) \log (x)+(-x-e^x x) \log ^2(x))}{x} \, dx\)

Optimal. Leaf size=19 \[ e^{-3-\left (e^x+x-2 \log (5)\right ) \log ^2(x)} \]

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Rubi [F]  time = 10.07, 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 {e^{-3+\left (-e^x-x+2 \log (5)\right ) \log ^2(x)} \left (\left (-2 e^x-2 x+4 \log (5)\right ) \log (x)+\left (-x-e^x x\right ) \log ^2(x)\right )}{x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

-2*Defer[Int][E^(-3 + (-E^x - x + 2*Log[5])*Log[x]^2)*Log[x], x] + 4*Log[5]*Defer[Int][(E^(-3 + (-E^x - x + 2*
Log[5])*Log[x]^2)*Log[x])/x, x] - 2*Defer[Int][(E^(-3 + x + (-E^x - x + 2*Log[5])*Log[x]^2)*Log[x])/x, x] - De
fer[Int][E^(-3 + (-E^x - x + 2*Log[5])*Log[x]^2)*Log[x]^2, x] - Defer[Int][E^(-3 + x + (-E^x - x + 2*Log[5])*L
og[x]^2)*Log[x]^2, x]

Rubi steps

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

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Mathematica [A]  time = 0.67, size = 19, normalized size = 1.00 \begin {gather*} e^{-3-\left (e^x+x-2 \log (5)\right ) \log ^2(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(-3 - (E^x + x - 2*Log[5])*Log[x]^2)

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fricas [A]  time = 1.17, size = 17, normalized size = 0.89 \begin {gather*} e^{\left (-{\left (x + e^{x} - 2 \, \log \relax (5)\right )} \log \relax (x)^{2} - 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)*x-x)*log(x)^2+(-2*exp(x)+4*log(5)-2*x)*log(x))*exp((-exp(x)+2*log(5)-x)*log(x)^2-3)/x,x, a
lgorithm="fricas")

[Out]

e^(-(x + e^x - 2*log(5))*log(x)^2 - 3)

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giac [A]  time = 0.17, size = 26, normalized size = 1.37 \begin {gather*} e^{\left (-x \log \relax (x)^{2} - e^{x} \log \relax (x)^{2} + 2 \, \log \relax (5) \log \relax (x)^{2} - 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)*x-x)*log(x)^2+(-2*exp(x)+4*log(5)-2*x)*log(x))*exp((-exp(x)+2*log(5)-x)*log(x)^2-3)/x,x, a
lgorithm="giac")

[Out]

e^(-x*log(x)^2 - e^x*log(x)^2 + 2*log(5)*log(x)^2 - 3)

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maple [A]  time = 0.12, size = 26, normalized size = 1.37

method	result	size
risch	\(25^{\ln \relax (x )^{2}} {\mathrm e}^{-3-{\mathrm e}^{x} \ln \relax (x )^{2}-x \ln \relax (x )^{2}}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-exp(x)*x-x)*ln(x)^2+(-2*exp(x)+4*ln(5)-2*x)*ln(x))*exp((-exp(x)+2*ln(5)-x)*ln(x)^2-3)/x,x,method=_RETUR
NVERBOSE)

[Out]

25^(ln(x)^2)*exp(-3-exp(x)*ln(x)^2-x*ln(x)^2)

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maxima [A]  time = 0.73, size = 26, normalized size = 1.37 \begin {gather*} e^{\left (-x \log \relax (x)^{2} - e^{x} \log \relax (x)^{2} + 2 \, \log \relax (5) \log \relax (x)^{2} - 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)*x-x)*log(x)^2+(-2*exp(x)+4*log(5)-2*x)*log(x))*exp((-exp(x)+2*log(5)-x)*log(x)^2-3)/x,x, a
lgorithm="maxima")

[Out]

e^(-x*log(x)^2 - e^x*log(x)^2 + 2*log(5)*log(x)^2 - 3)

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mupad [B]  time = 3.76, size = 28, normalized size = 1.47 \begin {gather*} 5^{2\,{\ln \relax (x)}^2}\,{\mathrm {e}}^{-3}\,{\mathrm {e}}^{-x\,{\ln \relax (x)}^2}\,{\mathrm {e}}^{-{\mathrm {e}}^x\,{\ln \relax (x)}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5^(2*log(x)^2)*exp(-3)*exp(-x*log(x)^2)*exp(-exp(x)*log(x)^2)

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sympy [A]  time = 0.55, size = 17, normalized size = 0.89 \begin {gather*} e^{\left (- x - e^{x} + 2 \log {\relax (5 )}\right ) \log {\relax (x )}^{2} - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)*x-x)*ln(x)**2+(-2*exp(x)+4*ln(5)-2*x)*ln(x))*exp((-exp(x)+2*ln(5)-x)*ln(x)**2-3)/x,x)

[Out]

exp((-x - exp(x) + 2*log(5))*log(x)**2 - 3)

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