3.17.87 $\int e^{e^{2(-5-x)\log (x)+2\log (x)\log (3e^x{\log }^2(x))}{x}^2+2(-5-x)\log (x)+2\log (x)\log (3e^x{\log }^2(x))}(-4x-2x^2+2x\log (3e^x{\log }^2(x)))dx$

Optimal. Leaf size=21 $$e^{x^{-8-2x+2\log (3e^x{\log }^2(x))}}$$

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Rubi [F]  time = 3.37, 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 = {} $$\begin{array}{c}\int \exp (\exp (2(-5-x)\log (x)+2\log (x)\log (3e^x{\log }^2(x)))x^2+2(-5-x)\log (x)+2\log (x)\log (3e^x{\log }^2(x)))(-4x-2x^2+2x\log (3e^x{\log }^2(x)))dx\end{array}$$

Verification is not applicable to the result.

[In]

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

[Out]

-4*Defer[Int][E^(E^(2*(-5 - x)*Log[x] + 2*Log[x]*Log[3*E^x*Log[x]^2])*x^2 + 2*(-5 - x)*Log[x] + 2*Log[x]*Log[3
*E^x*Log[x]^2])*x, x] - 2*Defer[Int][E^(x^(-8 - 2*x + 2*Log[3*E^x*Log[x]^2]) + 2*Log[x]*Log[3*E^x*Log[x]^2])*x
^(-8 - 2*x), x] + 2*Defer[Int][E^(E^(2*(-5 - x)*Log[x] + 2*Log[x]*Log[3*E^x*Log[x]^2])*x^2 + 2*(-5 - x)*Log[x]
+ 2*Log[x]*Log[3*E^x*Log[x]^2])*x*Log[3*E^x*Log[x]^2], x]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int 2\exp (\exp (2(-5-x)\log (x)+2\log (x)\log (3e^x{\log }^2(x)))x^2+2(-5-x)\log (x)+2\log (x)\log (3e^x{\log }^2(x)))x(-2-x+\log (3e^x{\log }^2(x)))dx\\ & =2\int \exp (\exp (2(-5-x)\log (x)+2\log (x)\log (3e^x{\log }^2(x)))x^2+2(-5-x)\log (x)+2\log (x)\log (3e^x{\log }^2(x)))x(-2-x+\log (3e^x{\log }^2(x)))dx\\ & =2\int (-\exp (\exp (2(-5-x)\log (x)+2\log (x)\log (3e^x{\log }^2(x)))x^2+2(-5-x)\log (x)+2\log (x)\log (3e^x{\log }^2(x)))x(2+x)+\exp (\exp (2(-5-x)\log (x)+2\log (x)\log (3e^x{\log }^2(x)))x^2+2(-5-x)\log (x)+2\log (x)\log (3e^x{\log }^2(x)))x\log (3e^x{\log }^2(x)))dx\\ & =-(2\int \exp (\exp (2(-5-x)\log (x)+2\log (x)\log (3e^x{\log }^2(x)))x^2+2(-5-x)\log (x)+2\log (x)\log (3e^x{\log }^2(x)))x(2+x)dx)+2\int \exp (\exp (2(-5-x)\log (x)+2\log (x)\log (3e^x{\log }^2(x)))x^2+2(-5-x)\log (x)+2\log (x)\log (3e^x{\log }^2(x)))x\log (3e^x{\log }^2(x))dx\\ & =-(2\int (2\exp (\exp (2(-5-x)\log (x)+2\log (x)\log (3e^x{\log }^2(x)))x^2+2(-5-x)\log (x)+2\log (x)\log (3e^x{\log }^2(x)))x+\exp (\exp (2(-5-x)\log (x)+2\log (x)\log (3e^x{\log }^2(x)))x^2+2(-5-x)\log (x)+2\log (x)\log (3e^x{\log }^2(x)))x^2)dx)+2\int \exp (\exp (2(-5-x)\log (x)+2\log (x)\log (3e^x{\log }^2(x)))x^2+2(-5-x)\log (x)+2\log (x)\log (3e^x{\log }^2(x)))x\log (3e^x{\log }^2(x))dx\\ & =-(2\int \exp (\exp (2(-5-x)\log (x)+2\log (x)\log (3e^x{\log }^2(x)))x^2+2(-5-x)\log (x)+2\log (x)\log (3e^x{\log }^2(x)))x^{2}dx)+2\int \exp (\exp (2(-5-x)\log (x)+2\log (x)\log (3e^x{\log }^2(x)))x^2+2(-5-x)\log (x)+2\log (x)\log (3e^x{\log }^2(x)))x\log (3e^x{\log }^2(x))dx-4\int \exp (\exp (2(-5-x)\log (x)+2\log (x)\log (3e^x{\log }^2(x)))x^2+2(-5-x)\log (x)+2\log (x)\log (3e^x{\log }^2(x)))xdx\\ & =-(2\int \exp (x^{-8-2x+2\log (3e^x{\log }^2(x))}+2\log (x)\log (3e^x{\log }^2(x)))x^{-8-2x}dx)+2\int \exp (\exp (2(-5-x)\log (x)+2\log (x)\log (3e^x{\log }^2(x)))x^2+2(-5-x)\log (x)+2\log (x)\log (3e^x{\log }^2(x)))x\log (3e^x{\log }^2(x))dx-4\int \exp (\exp (2(-5-x)\log (x)+2\log (x)\log (3e^x{\log }^2(x)))x^2+2(-5-x)\log (x)+2\log (x)\log (3e^x{\log }^2(x)))xdx\end{array}\end{array}$$

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Mathematica [A]  time = 1.01, size = 21, normalized size = 1.00 $$\begin{array}{c}{e}^{x^{-8-2x+2\log (3e^x{\log }^2(x))}}\end{array}$$

Antiderivative was successfully verified.

[In]

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

[Out]

E^x^(-8 - 2*x + 2*Log[3*E^x*Log[x]^2])

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fricas [A]  time = 0.89, size = 27, normalized size = 1.29 $$\begin{array}{c}{e}^{\left(x^2{e}^{(-2(x+5)\log (x)+2\log (3e^x\log (x{)}^2)\log (x))}\right)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*log(3*exp(x)*log(x)^2)-2*x^2-4*x)*exp(log(x)*log(3*exp(x)*log(x)^2)+(-x-5)*log(x))^2*exp(x^2*ex
p(log(x)*log(3*exp(x)*log(x)^2)+(-x-5)*log(x))^2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 $$\begin{array}{c}\mathit{undef}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*log(3*exp(x)*log(x)^2)-2*x^2-4*x)*exp(log(x)*log(3*exp(x)*log(x)^2)+(-x-5)*log(x))^2*exp(x^2*ex
p(log(x)*log(3*exp(x)*log(x)^2)+(-x-5)*log(x))^2),x, algorithm="giac")

[Out]

undef

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maple [C]  time = 0.36, size = 104, normalized size = 4.95

method	result	size
risch	${\mathrm{e}}^{x^2{x}^{-i\mathrm{csgn}(i\ln (x{)}^2)\pi +2i\pi \mathrm{csgn}(i\ln (x))-i\pi \mathrm{csgn}(i\ln (x{)}^2)\mathrm{csgn}(i{\mathrm{e}}^x\ln (x{)}^2)\mathrm{csgn}\left(i{\mathrm{e}}^x\right)-i\mathrm{csgn}(i{\mathrm{e}}^x\ln (x{)}^2)\pi +4\ln (\ln (x))+2\ln \left(3{\mathrm{e}}^x\right)+i\pi \mathrm{csgn}\left(i{\mathrm{e}}^x\right)}{x}^{-2x-10}}$	$104$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(x^2*(x^(-1/2*I*csgn(I*ln(x)^2)*Pi+I*Pi*csgn(I*ln(x))-1/2*I*Pi*csgn(I*ln(x)^2)*csgn(I*exp(x)*ln(x)^2)*csgn(
I*exp(x))-1/2*I*csgn(I*exp(x)*ln(x)^2)*Pi+1/2*I*Pi*csgn(I*exp(x))+2*ln(ln(x))+ln(exp(x))+ln(3)))^2*(x^(-x-5))^
2)

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maxima [A]  time = 1.13, size = 20, normalized size = 0.95 $$\begin{array}{c}{e}^{\left(\frac{e^{(2\log (3)\log (x)+4\log (x)\log (\log (x)))}}{x^8}\right)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*log(3*exp(x)*log(x)^2)-2*x^2-4*x)*exp(log(x)*log(3*exp(x)*log(x)^2)+(-x-5)*log(x))^2*exp(x^2*ex
p(log(x)*log(3*exp(x)*log(x)^2)+(-x-5)*log(x))^2),x, algorithm="maxima")

[Out]

e^(e^(2*log(3)*log(x) + 4*log(x)*log(log(x)))/x^8)

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mupad [B]  time = 1.23, size = 20, normalized size = 0.95 $$\begin{array}{c}{\mathrm{e}}^{\frac{x^{2\ln (3)}{x}^{2\ln \left({\ln (x)}^2\right)}}{x^8}}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp((x^(2*log(3))*x^(2*log(log(x)^2)))/x^8)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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