3.89.50 $\int \frac{e^{x-16e^{2+2e^3}{x}^2-32e^{2+2e^3}x\log (3x)-16e^{2+2e^3}{\log }^2(3x)}(x+e^{2+2e^3}(-32x-32x^2)+e^{2+2e^3}(-32-32x)\log (3x))}{x}dx$

Optimal. Leaf size=25 $$16+e^{x-16e^{2+2e^3}(x+\log (3x){)}^2}$$

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Rubi [B]  time = 1.76, antiderivative size = 64, normalized size of antiderivative = 2.56, number of steps used = 1, number of rules used = 1, integrand size = 95, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.011, Rules used = {6706} $$\begin{array}{c}{3}^{-32e^{2+2e^3}x}{x}^{-32e^{2+2e^3}x}\exp (-16e^{2+2e^3}{x}^2+x-16e^{2+2e^3}{\log }^2(3x))\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(E^(x - 16*E^(2 + 2*E^3)*x^2 - 32*E^(2 + 2*E^3)*x*Log[3*x] - 16*E^(2 + 2*E^3)*Log[3*x]^2)*(x + E^(2 + 2*E^
3)*(-32*x - 32*x^2) + E^(2 + 2*E^3)*(-32 - 32*x)*Log[3*x]))/x,x]

[Out]

E^(x - 16*E^(2 + 2*E^3)*x^2 - 16*E^(2 + 2*E^3)*Log[3*x]^2)/(3^(32*E^(2 + 2*E^3)*x)*x^(32*E^(2 + 2*E^3)*x))

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =3^{-32e^{2+2e^3}x}\exp (x-16e^{2+2e^3}{x}^2-16e^{2+2e^3}{\log }^2(3x))x^{-32e^{2+2e^3}x}\end{array}\end{array}$$

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Mathematica [B]  time = 0.16, size = 51, normalized size = 2.04 $$\begin{array}{c}{e}^{x-16e^{2+2e^3}{x}^2-32e^{2+2e^3}x\log (3x)-16e^{2+2e^3}{\log }^2(3x)}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(E^(x - 16*E^(2 + 2*E^3)*x^2 - 32*E^(2 + 2*E^3)*x*Log[3*x] - 16*E^(2 + 2*E^3)*Log[3*x]^2)*(x + E^(2
+ 2*E^3)*(-32*x - 32*x^2) + E^(2 + 2*E^3)*(-32 - 32*x)*Log[3*x]))/x,x]

[Out]

E^(x - 16*E^(2 + 2*E^3)*x^2 - 32*E^(2 + 2*E^3)*x*Log[3*x] - 16*E^(2 + 2*E^3)*Log[3*x]^2)

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fricas [A]  time = 0.52, size = 44, normalized size = 1.76 $$\begin{array}{c}{e}^{(-16x^2{e}^{(2e^3+2)}-32x{e}^{(2e^3+2)}\log \left(3x\right)-16e^{(2e^3+2)}\log {\left(3x\right)}^2+x)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-32*x-32)*exp(exp(3)+1)^2*log(3*x)+(-32*x^2-32*x)*exp(exp(3)+1)^2+x)*exp(-16*exp(exp(3)+1)^2*log(3
*x)^2-32*x*exp(exp(3)+1)^2*log(3*x)-16*x^2*exp(exp(3)+1)^2+x)/x,x, algorithm="fricas")

[Out]

e^(-16*x^2*e^(2*e^3 + 2) - 32*x*e^(2*e^3 + 2)*log(3*x) - 16*e^(2*e^3 + 2)*log(3*x)^2 + x)

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giac [A]  time = 0.37, size = 44, normalized size = 1.76 $$\begin{array}{c}{e}^{(-16x^2{e}^{(2e^3+2)}-32x{e}^{(2e^3+2)}\log \left(3x\right)-16e^{(2e^3+2)}\log {\left(3x\right)}^2+x)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-32*x-32)*exp(exp(3)+1)^2*log(3*x)+(-32*x^2-32*x)*exp(exp(3)+1)^2+x)*exp(-16*exp(exp(3)+1)^2*log(3
*x)^2-32*x*exp(exp(3)+1)^2*log(3*x)-16*x^2*exp(exp(3)+1)^2+x)/x,x, algorithm="giac")

[Out]

e^(-16*x^2*e^(2*e^3 + 2) - 32*x*e^(2*e^3 + 2)*log(3*x) - 16*e^(2*e^3 + 2)*log(3*x)^2 + x)

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maple [A]  time = 0.11, size = 45, normalized size = 1.80

method	result	size
norman	${\mathrm{e}}^{-16{\mathrm{e}}^{2{\mathrm{e}}^3+2}\ln {\left(3x\right)}^2-32x{\mathrm{e}}^{2{\mathrm{e}}^3+2}\ln \left(3x\right)-16x^2{\mathrm{e}}^{2{\mathrm{e}}^3+2}+x}$	$45$
risch	${\left(3x\right)}^{-32x{\mathrm{e}}^{2{\mathrm{e}}^3+2}}{\mathrm{e}}^{-16{\mathrm{e}}^{2{\mathrm{e}}^3+2}\ln {\left(3x\right)}^2-16x^2{\mathrm{e}}^{2{\mathrm{e}}^3+2}+x}$	$46$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-32*x-32)*exp(exp(3)+1)^2*ln(3*x)+(-32*x^2-32*x)*exp(exp(3)+1)^2+x)*exp(-16*exp(exp(3)+1)^2*ln(3*x)^2-32
*x*exp(exp(3)+1)^2*ln(3*x)-16*x^2*exp(exp(3)+1)^2+x)/x,x,method=_RETURNVERBOSE)

[Out]

exp(-16*exp(exp(3)+1)^2*ln(3*x)^2-32*x*exp(exp(3)+1)^2*ln(3*x)-16*x^2*exp(exp(3)+1)^2+x)

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maxima [B]  time = 0.71, size = 78, normalized size = 3.12 $$\begin{array}{c}{e}^{(-16x^2{e}^{(2e^3+2)}-32x{e}^{(2e^3+2)}\log (3)-16e^{(2e^3+2)}\log (3{)}^2-32x{e}^{(2e^3+2)}\log (x)-32e^{(2e^3+2)}\log (3)\log (x)-16e^{(2e^3+2)}\log (x{)}^2+x)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-32*x-32)*exp(exp(3)+1)^2*log(3*x)+(-32*x^2-32*x)*exp(exp(3)+1)^2+x)*exp(-16*exp(exp(3)+1)^2*log(3
*x)^2-32*x*exp(exp(3)+1)^2*log(3*x)-16*x^2*exp(exp(3)+1)^2+x)/x,x, algorithm="maxima")

[Out]

e^(-16*x^2*e^(2*e^3 + 2) - 32*x*e^(2*e^3 + 2)*log(3) - 16*e^(2*e^3 + 2)*log(3)^2 - 32*x*e^(2*e^3 + 2)*log(x) -
32*e^(2*e^3 + 2)*log(3)*log(x) - 16*e^(2*e^3 + 2)*log(x)^2 + x)

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mupad [B]  time = 5.45, size = 87, normalized size = 3.48 $$\begin{array}{c}\frac{{\mathrm{e}}^{-16{\mathrm{e}}^{2{\mathrm{e}}^3}{\mathrm{e}}^2{\ln (3)}^2}{\mathrm{e}}^{-16x^2{\mathrm{e}}^{2{\mathrm{e}}^3}{\mathrm{e}}^2}{\mathrm{e}}^{-16{\mathrm{e}}^{2{\mathrm{e}}^3}{\mathrm{e}}^2{\ln (x)}^2}{\mathrm{e}}^x}{3^{32x{\mathrm{e}}^{2{\mathrm{e}}^3}{\mathrm{e}}^2}{x}^{32{\mathrm{e}}^{2{\mathrm{e}}^3}{\mathrm{e}}^2\ln (3)}{x}^{32x{\mathrm{e}}^{2{\mathrm{e}}^3}{\mathrm{e}}^2}}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x - 16*log(3*x)^2*exp(2*exp(3) + 2) - 16*x^2*exp(2*exp(3) + 2) - 32*x*log(3*x)*exp(2*exp(3) + 2))*(e
xp(2*exp(3) + 2)*(32*x + 32*x^2) - x + log(3*x)*exp(2*exp(3) + 2)*(32*x + 32)))/x,x)

[Out]

(exp(-16*exp(2*exp(3))*exp(2)*log(3)^2)*exp(-16*x^2*exp(2*exp(3))*exp(2))*exp(-16*exp(2*exp(3))*exp(2)*log(x)^
2)*exp(x))/(3^(32*x*exp(2*exp(3))*exp(2))*x^(32*exp(2*exp(3))*exp(2)*log(3))*x^(32*x*exp(2*exp(3))*exp(2)))

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sympy [B]  time = 0.44, size = 49, normalized size = 1.96 $$\begin{array}{c}{e}^{-16x^2{e}^{2+2e^3}-32x{e}^{2+2e^3}\log \left(3x\right)+x-16e^{2+2e^3}\log {\left(3x\right)}^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-32*x-32)*exp(exp(3)+1)**2*ln(3*x)+(-32*x**2-32*x)*exp(exp(3)+1)**2+x)*exp(-16*exp(exp(3)+1)**2*ln
(3*x)**2-32*x*exp(exp(3)+1)**2*ln(3*x)-16*x**2*exp(exp(3)+1)**2+x)/x,x)

[Out]

exp(-16*x**2*exp(2 + 2*exp(3)) - 32*x*exp(2 + 2*exp(3))*log(3*x) + x - 16*exp(2 + 2*exp(3))*log(3*x)**2)

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