3.89.52 $\int \frac{e^{-e^x{x}^2+\frac{e^{-e^x{x}^2}(4e^{x+e^x{x}^2}+x^2)}{4x}}(5x^2+e^x(-10x^4-5x^5)+e^{e^x{x}^2}(-20x+e^x(-20+20x)))}{4x^3}dx$

Optimal. Leaf size=32 $$\frac{5e^{\frac{e^x+\frac{1}{4}{e}^{-e^x{x}^2}{x}^2}{x}}}{x}$$

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Rubi [F]  time = 6.33, 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 = {} $$\begin{array}{c}\int \frac{\exp (-e^x{x}^2+\frac{e^{-e^x{x}^2}(4e^{x+e^x{x}^2}+x^2)}{4x})(5x^2+e^x(-10x^4-5x^5)+e^{e^x{x}^2}(-20x+e^x(-20+20x)))}{4x^3}dx\end{array}$$

Verification is not applicable to the result.

[In]

Int[(E^(-(E^x*x^2) + (4*E^(x + E^x*x^2) + x^2)/(4*E^(E^x*x^2)*x))*(5*x^2 + E^x*(-10*x^4 - 5*x^5) + E^(E^x*x^2)
*(-20*x + E^x*(-20 + 20*x))))/(4*x^3),x]

[Out]

-5*Defer[Int][E^(E^x/x + x + x/(4*E^(E^x*x^2)))/x^3, x] - 5*Defer[Int][E^(E^x/x + x/(4*E^(E^x*x^2)))/x^2, x] +
5*Defer[Int][E^(E^x/x + x + x/(4*E^(E^x*x^2)))/x^2, x] + (5*Defer[Int][E^(x/(4*E^(E^x*x^2)) + E^x*(x^(-1) - x
^2))/x, x])/4 - (5*Defer[Int][E^(x + x/(4*E^(E^x*x^2)) + E^x*(x^(-1) - x^2))*x, x])/2 - (5*Defer[Int][E^(x + x
/(4*E^(E^x*x^2)) + E^x*(x^(-1) - x^2))*x^2, x])/4

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\frac{1}{4}\int \frac{\exp (-e^x{x}^2+\frac{e^{-e^x{x}^2}(4e^{x+e^x{x}^2}+x^2)}{4x})(5x^2+e^x(-10x^4-5x^5)+e^{e^x{x}^2}(-20x+e^x(-20+20x)))}{x^3}dx\\ & =\frac{1}{4}\int (\frac{20\exp \left(\frac{e^{-e^x{x}^2}(4e^{x+e^x{x}^2}+x^2)}{4x}\right)(-e^x-x+e^{x}x)}{x^3}-\frac{5\exp (-e^x{x}^2+\frac{e^{-e^x{x}^2}(4e^{x+e^x{x}^2}+x^2)}{4x})(-1+2e^x{x}^2+e^x{x}^3)}{x})dx\\ & =-(\frac{5}{4}\int \frac{\exp (-e^x{x}^2+\frac{e^{-e^x{x}^2}(4e^{x+e^x{x}^2}+x^2)}{4x})(-1+2e^x{x}^2+e^x{x}^3)}{x}dx)+5\int \frac{\exp \left(\frac{e^{-e^x{x}^2}(4e^{x+e^x{x}^2}+x^2)}{4x}\right)(-e^x-x+e^{x}x)}{x^3}dx\\ & =-(\frac{5}{4}\int \frac{e^{\frac{1}{4}{e}^{-e^x{x}^2}x+e^x(\frac{1}{x}-x^2)}(-1+e^x{x}^2(2+x))}{x}dx)+5\int \frac{e^{\frac{e^x}{x}+\frac{1}{4}{e}^{-e^x{x}^2}x}(e^x(-1+x)-x)}{x^3}dx\\ & =-(\frac{5}{4}\int (-\frac{e^{\frac{1}{4}{e}^{-e^x{x}^2}x+e^x(\frac{1}{x}-x^2)}}{x}+e^{x+\frac{1}{4}{e}^{-e^x{x}^2}x+e^x(\frac{1}{x}-x^2)}x(2+x))dx)+5\int (\frac{e^{\frac{e^x}{x}+x+\frac{1}{4}{e}^{-e^x{x}^2}x}(-1+x)}{x^3}-\frac{e^{\frac{e^x}{x}+\frac{1}{4}{e}^{-e^x{x}^2}x}}{x^2})dx\\ & =\frac{5}{4}\int \frac{e^{\frac{1}{4}{e}^{-e^x{x}^2}x+e^x(\frac{1}{x}-x^2)}}{x}dx-\frac{5}{4}\int e^{x+\frac{1}{4}{e}^{-e^x{x}^2}x+e^x(\frac{1}{x}-x^2)}x(2+x)dx+5\int \frac{e^{\frac{e^x}{x}+x+\frac{1}{4}{e}^{-e^x{x}^2}x}(-1+x)}{x^3}dx-5\int \frac{e^{\frac{e^x}{x}+\frac{1}{4}{e}^{-e^x{x}^2}x}}{x^2}dx\\ & =\frac{5}{4}\int \frac{e^{\frac{1}{4}{e}^{-e^x{x}^2}x+e^x(\frac{1}{x}-x^2)}}{x}dx-\frac{5}{4}\int (2e^{x+\frac{1}{4}{e}^{-e^x{x}^2}x+e^x(\frac{1}{x}-x^2)}x+e^{x+\frac{1}{4}{e}^{-e^x{x}^2}x+e^x(\frac{1}{x}-x^2)}{x}^2)dx+5\int (-\frac{e^{\frac{e^x}{x}+x+\frac{1}{4}{e}^{-e^x{x}^2}x}}{x^3}+\frac{e^{\frac{e^x}{x}+x+\frac{1}{4}{e}^{-e^x{x}^2}x}}{x^2})dx-5\int \frac{e^{\frac{e^x}{x}+\frac{1}{4}{e}^{-e^x{x}^2}x}}{x^2}dx\\ & =\frac{5}{4}\int \frac{e^{\frac{1}{4}{e}^{-e^x{x}^2}x+e^x(\frac{1}{x}-x^2)}}{x}dx-\frac{5}{4}\int e^{x+\frac{1}{4}{e}^{-e^x{x}^2}x+e^x(\frac{1}{x}-x^2)}{x}^{2}dx-\frac{5}{2}\int e^{x+\frac{1}{4}{e}^{-e^x{x}^2}x+e^x(\frac{1}{x}-x^2)}xdx-5\int \frac{e^{\frac{e^x}{x}+x+\frac{1}{4}{e}^{-e^x{x}^2}x}}{x^3}dx-5\int \frac{e^{\frac{e^x}{x}+\frac{1}{4}{e}^{-e^x{x}^2}x}}{x^2}dx+5\int \frac{e^{\frac{e^x}{x}+x+\frac{1}{4}{e}^{-e^x{x}^2}x}}{x^2}dx\end{array}\end{array}$$

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Mathematica [A]  time = 4.18, size = 30, normalized size = 0.94 $$\begin{array}{c}\frac{5e^{\frac{e^x}{x}+\frac{1}{4}{e}^{-e^x{x}^2}x}}{x}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(E^(-(E^x*x^2) + (4*E^(x + E^x*x^2) + x^2)/(4*E^(E^x*x^2)*x))*(5*x^2 + E^x*(-10*x^4 - 5*x^5) + E^(E^
x*x^2)*(-20*x + E^x*(-20 + 20*x))))/(4*x^3),x]

[Out]

(5*E^(E^x/x + x/(4*E^(E^x*x^2))))/x

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fricas [A]  time = 0.48, size = 55, normalized size = 1.72 $$\begin{array}{c}\frac{5e^{(x^2{e}^x+\frac{(x^2{e}^x-4(x^3-1)e^{(x^2{e}^x+2x)})e^{(-x^2{e}^x-x)}}{4x})}}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(((20*x-20)*exp(x)-20*x)*exp(exp(x)*x^2)+(-5*x^5-10*x^4)*exp(x)+5*x^2)*exp(1/4*(4*exp(x)*exp(exp
(x)*x^2)+x^2)/x/exp(exp(x)*x^2))/x^3/exp(exp(x)*x^2),x, algorithm="fricas")

[Out]

5*e^(x^2*e^x + 1/4*(x^2*e^x - 4*(x^3 - 1)*e^(x^2*e^x + 2*x))*e^(-x^2*e^x - x)/x)/x

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 $$\begin{array}{c}\int \frac{5(x^2+4((x-1)e^x-x)e^{\left(x^2{e}^x\right)}-(x^5+2x^4)e^x)e^{(-x^2{e}^x+\frac{(x^2+4e^{(x^2{e}^x+x)})e^{(-x^2{e}^x)}}{4x})}}{4x^3}dx\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(((20*x-20)*exp(x)-20*x)*exp(exp(x)*x^2)+(-5*x^5-10*x^4)*exp(x)+5*x^2)*exp(1/4*(4*exp(x)*exp(exp
(x)*x^2)+x^2)/x/exp(exp(x)*x^2))/x^3/exp(exp(x)*x^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.08, size = 35, normalized size = 1.09

method	result	size
risch	$\frac{5{\mathrm{e}}^{\frac{(4{\mathrm{e}}^{x({\mathrm{e}}^{x}x+1)}+x^2){\mathrm{e}}^{-{\mathrm{e}}^x{x}^2}}{4x}}}{x}$	$35$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(((20*x-20)*exp(x)-20*x)*exp(exp(x)*x^2)+(-5*x^5-10*x^4)*exp(x)+5*x^2)*exp(1/4*(4*exp(x)*exp(exp(x)*x^
2)+x^2)/x/exp(exp(x)*x^2))/x^3/exp(exp(x)*x^2),x,method=_RETURNVERBOSE)

[Out]

5/x*exp(1/4*(4*exp(x*(exp(x)*x+1))+x^2)/x*exp(-exp(x)*x^2))

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maxima [A]  time = 0.54, size = 24, normalized size = 0.75 $$\begin{array}{c}\frac{5e^{(\frac{1}{4}x{e}^{(-x^2{e}^x)}+\frac{e^x}{x})}}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(((20*x-20)*exp(x)-20*x)*exp(exp(x)*x^2)+(-5*x^5-10*x^4)*exp(x)+5*x^2)*exp(1/4*(4*exp(x)*exp(exp
(x)*x^2)+x^2)/x/exp(exp(x)*x^2))/x^3/exp(exp(x)*x^2),x, algorithm="maxima")

[Out]

5*e^(1/4*x*e^(-x^2*e^x) + e^x/x)/x

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mupad [B]  time = 5.29, size = 24, normalized size = 0.75 $$\begin{array}{c}\frac{5{\mathrm{e}}^{\frac{{\mathrm{e}}^x}{x}+\frac{x{\mathrm{e}}^{-x^2{\mathrm{e}}^x}}{4}}}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((exp(-x^2*exp(x))*(exp(x^2*exp(x))*exp(x) + x^2/4))/x)*exp(-x^2*exp(x))*(exp(x)*(10*x^4 + 5*x^5) + e
xp(x^2*exp(x))*(20*x - exp(x)*(20*x - 20)) - 5*x^2))/(4*x^3),x)

[Out]

(5*exp(exp(x)/x + (x*exp(-x^2*exp(x)))/4))/x

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sympy [A]  time = 0.52, size = 31, normalized size = 0.97 $$\begin{array}{c}\frac{5e^{\frac{(\frac{x^2}{4}+e^x{e}^{x^2{e}^x})e^{-x^2{e}^x}}{x}}}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(((20*x-20)*exp(x)-20*x)*exp(exp(x)*x**2)+(-5*x**5-10*x**4)*exp(x)+5*x**2)*exp(1/4*(4*exp(x)*exp
(exp(x)*x**2)+x**2)/x/exp(exp(x)*x**2))/x**3/exp(exp(x)*x**2),x)

[Out]

5*exp((x**2/4 + exp(x)*exp(x**2*exp(x)))*exp(-x**2*exp(x))/x)/x

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