\(\int e^{-2 x} (2 e^x+e^{9 e^{-2 x} x^2 \log ^2(x)} (18 x \log (x)+(18 x-18 x^2) \log ^2(x))) \, dx\) [4301]

Optimal result

Integrand size = 49, antiderivative size = 24 \[ \int e^{-2 x} \left (2 e^x+e^{9 e^{-2 x} x^2 \log ^2(x)} \left (18 x \log (x)+\left (18 x-18 x^2\right ) \log ^2(x)\right )\right ) \, dx=-2 e^{-x}+e^{9 e^{-2 x} x^2 \log ^2(x)} \]

[Out]

-2/exp(x)+exp(9*x^2*ln(x)^2/exp(x)^2)

Rubi [F]

\[ \int e^{-2 x} \left (2 e^x+e^{9 e^{-2 x} x^2 \log ^2(x)} \left (18 x \log (x)+\left (18 x-18 x^2\right ) \log ^2(x)\right )\right ) \, dx=\int e^{-2 x} \left (2 e^x+e^{9 e^{-2 x} x^2 \log ^2(x)} \left (18 x \log (x)+\left (18 x-18 x^2\right ) \log ^2(x)\right )\right ) \, dx \]

[In]

Int[(2*E^x + E^((9*x^2*Log[x]^2)/E^(2*x))*(18*x*Log[x] + (18*x - 18*x^2)*Log[x]^2))/E^(2*x),x]

[Out]

-2/E^x + 18*Defer[Int][E^(-2*x + (9*x^2*Log[x]^2)/E^(2*x))*x*Log[x], x] + 18*Defer[Int][E^(-2*x + (9*x^2*Log[x
]^2)/E^(2*x))*x*Log[x]^2, x] - 18*Defer[Int][E^(-2*x + (9*x^2*Log[x]^2)/E^(2*x))*x^2*Log[x]^2, x]

Rubi steps \begin{align*} \text {integral}& = \int \left (2 e^{-x}-18 e^{-2 x+9 e^{-2 x} x^2 \log ^2(x)} x \log (x) (-1-\log (x)+x \log (x))\right ) \, dx \\ & = 2 \int e^{-x} \, dx-18 \int e^{-2 x+9 e^{-2 x} x^2 \log ^2(x)} x \log (x) (-1-\log (x)+x \log (x)) \, dx \\ & = -2 e^{-x}-18 \int \left (-e^{-2 x+9 e^{-2 x} x^2 \log ^2(x)} x \log (x)+e^{-2 x+9 e^{-2 x} x^2 \log ^2(x)} (-1+x) x \log ^2(x)\right ) \, dx \\ & = -2 e^{-x}+18 \int e^{-2 x+9 e^{-2 x} x^2 \log ^2(x)} x \log (x) \, dx-18 \int e^{-2 x+9 e^{-2 x} x^2 \log ^2(x)} (-1+x) x \log ^2(x) \, dx \\ & = -2 e^{-x}+18 \int e^{-2 x+9 e^{-2 x} x^2 \log ^2(x)} x \log (x) \, dx-18 \int \left (-e^{-2 x+9 e^{-2 x} x^2 \log ^2(x)} x \log ^2(x)+e^{-2 x+9 e^{-2 x} x^2 \log ^2(x)} x^2 \log ^2(x)\right ) \, dx \\ & = -2 e^{-x}+18 \int e^{-2 x+9 e^{-2 x} x^2 \log ^2(x)} x \log (x) \, dx+18 \int e^{-2 x+9 e^{-2 x} x^2 \log ^2(x)} x \log ^2(x) \, dx-18 \int e^{-2 x+9 e^{-2 x} x^2 \log ^2(x)} x^2 \log ^2(x) \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 1.12 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int e^{-2 x} \left (2 e^x+e^{9 e^{-2 x} x^2 \log ^2(x)} \left (18 x \log (x)+\left (18 x-18 x^2\right ) \log ^2(x)\right )\right ) \, dx=2 \left (-e^{-x}+\frac {1}{2} e^{9 e^{-2 x} x^2 \log ^2(x)}\right ) \]

[In]

Integrate[(2*E^x + E^((9*x^2*Log[x]^2)/E^(2*x))*(18*x*Log[x] + (18*x - 18*x^2)*Log[x]^2))/E^(2*x),x]

[Out]

2*(-E^(-x) + E^((9*x^2*Log[x]^2)/E^(2*x))/2)

Maple [A] (verified)

Time = 1.56 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92

method	result	size
risch	\(-2 \,{\mathrm e}^{-x}+{\mathrm e}^{9 x^{2} \ln \left (x \right )^{2} {\mathrm e}^{-2 x}}\)	\(22\)
parallelrisch	\({\mathrm e}^{-x} \left (-2+{\mathrm e}^{9 x^{2} \ln \left (x \right )^{2} {\mathrm e}^{-2 x}} {\mathrm e}^{x}\right )\)	\(25\)

[In]

int((((-18*x^2+18*x)*ln(x)^2+18*x*ln(x))*exp(9*x^2*ln(x)^2/exp(x)^2)+2*exp(x))/exp(x)^2,x,method=_RETURNVERBOS
E)

[Out]

-2*exp(-x)+exp(9*x^2*ln(x)^2*exp(-2*x))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int e^{-2 x} \left (2 e^x+e^{9 e^{-2 x} x^2 \log ^2(x)} \left (18 x \log (x)+\left (18 x-18 x^2\right ) \log ^2(x)\right )\right ) \, dx={\left (e^{\left (9 \, x^{2} e^{\left (-2 \, x\right )} \log \left (x\right )^{2} + x\right )} - 2\right )} e^{\left (-x\right )} \]

[In]

integrate((((-18*x^2+18*x)*log(x)^2+18*x*log(x))*exp(9*x^2*log(x)^2/exp(x)^2)+2*exp(x))/exp(x)^2,x, algorithm=
"fricas")

[Out]

(e^(9*x^2*e^(-2*x)*log(x)^2 + x) - 2)*e^(-x)

Sympy [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int e^{-2 x} \left (2 e^x+e^{9 e^{-2 x} x^2 \log ^2(x)} \left (18 x \log (x)+\left (18 x-18 x^2\right ) \log ^2(x)\right )\right ) \, dx=e^{9 x^{2} e^{- 2 x} \log {\left (x \right )}^{2}} - 2 e^{- x} \]

[In]

integrate((((-18*x**2+18*x)*ln(x)**2+18*x*ln(x))*exp(9*x**2*ln(x)**2/exp(x)**2)+2*exp(x))/exp(x)**2,x)

[Out]

exp(9*x**2*exp(-2*x)*log(x)**2) - 2*exp(-x)

Maxima [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int e^{-2 x} \left (2 e^x+e^{9 e^{-2 x} x^2 \log ^2(x)} \left (18 x \log (x)+\left (18 x-18 x^2\right ) \log ^2(x)\right )\right ) \, dx=e^{\left (9 \, x^{2} e^{\left (-2 \, x\right )} \log \left (x\right )^{2}\right )} - 2 \, e^{\left (-x\right )} \]

[In]

integrate((((-18*x^2+18*x)*log(x)^2+18*x*log(x))*exp(9*x^2*log(x)^2/exp(x)^2)+2*exp(x))/exp(x)^2,x, algorithm=
"maxima")

[Out]

e^(9*x^2*e^(-2*x)*log(x)^2) - 2*e^(-x)

Giac [F]

\[ \int e^{-2 x} \left (2 e^x+e^{9 e^{-2 x} x^2 \log ^2(x)} \left (18 x \log (x)+\left (18 x-18 x^2\right ) \log ^2(x)\right )\right ) \, dx=\int { -2 \, {\left (9 \, {\left ({\left (x^{2} - x\right )} \log \left (x\right )^{2} - x \log \left (x\right )\right )} e^{\left (9 \, x^{2} e^{\left (-2 \, x\right )} \log \left (x\right )^{2}\right )} - e^{x}\right )} e^{\left (-2 \, x\right )} \,d x } \]

[In]

integrate((((-18*x^2+18*x)*log(x)^2+18*x*log(x))*exp(9*x^2*log(x)^2/exp(x)^2)+2*exp(x))/exp(x)^2,x, algorithm=
"giac")

[Out]

integrate(-2*(9*((x^2 - x)*log(x)^2 - x*log(x))*e^(9*x^2*e^(-2*x)*log(x)^2) - e^x)*e^(-2*x), x)

Mupad [B] (verification not implemented)

Time = 11.10 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int e^{-2 x} \left (2 e^x+e^{9 e^{-2 x} x^2 \log ^2(x)} \left (18 x \log (x)+\left (18 x-18 x^2\right ) \log ^2(x)\right )\right ) \, dx={\mathrm {e}}^{9\,x^2\,{\mathrm {e}}^{-2\,x}\,{\ln \left (x\right )}^2}-2\,{\mathrm {e}}^{-x} \]

[In]

int(exp(-2*x)*(2*exp(x) + exp(9*x^2*exp(-2*x)*log(x)^2)*(log(x)^2*(18*x - 18*x^2) + 18*x*log(x))),x)

[Out]

exp(9*x^2*exp(-2*x)*log(x)^2) - 2*exp(-x)