3.17.42 $\int \frac{1}{24}{e}^{-1-\frac{x^2+x^3}{2e}}(12e+48e^{1+\frac{x^2+x^3}{2e}}-12x^2-18x^3+(2x+3x^2)\log (3))dx$

Optimal. Leaf size=33 $$2x-\frac{1}{4}{e}^{-\frac{x(x+x^2)}{2e}}(-2x+\frac{\log (3)}{3})$$

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Rubi [F]  time = 1.55, 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 = {} $$\begin{array}{c}\int \frac{1}{24}{e}^{-1-\frac{x^2+x^3}{2e}}(12e+48e^{1+\frac{x^2+x^3}{2e}}-12x^2-18x^3+(2x+3x^2)\log (3))dx\end{array}$$

Verification is not applicable to the result.

[In]

Int[(E^(-1 - (x^2 + x^3)/(2*E))*(12*E + 48*E^(1 + (x^2 + x^3)/(2*E)) - 12*x^2 - 18*x^3 + (2*x + 3*x^2)*Log[3])
)/24,x]

[Out]

2*x - (E^(-1/2*x^2/E - x^3/(2*E))*Log[3])/12 + Defer[Int][E^(-1/2*(x^2*(1 + x))/E), x]/2 - Defer[Int][E^(-1 -
x^2/(2*E) - x^3/(2*E))*x^2, x]/2 - (3*Defer[Int][E^(-1 - x^2/(2*E) - x^3/(2*E))*x^3, x])/4

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\frac{1}{24}\int e^{-1-\frac{x^2+x^3}{2e}}(12e+48e^{1+\frac{x^2+x^3}{2e}}-12x^2-18x^3+(2x+3x^2)\log (3))dx\\ & =\frac{1}{24}\int e^{-1-\frac{x^2}{2e}-\frac{x^3}{2e}}(12e+48e^{1+\frac{x^2+x^3}{2e}}-12x^2-18x^3+(2x+3x^2)\log (3))dx\\ & =\frac{1}{24}\int (48+12e^{-\frac{x^2}{2e}-\frac{x^3}{2e}}-12e^{-1-\frac{x^2}{2e}-\frac{x^3}{2e}}{x}^2-18e^{-1-\frac{x^2}{2e}-\frac{x^3}{2e}}{x}^3+e^{-1-\frac{x^2}{2e}-\frac{x^3}{2e}}x(2+3x)\log (3))dx\\ & =2x+\frac{1}{2}\int e^{-\frac{x^2}{2e}-\frac{x^3}{2e}}dx-\frac{1}{2}\int e^{-1-\frac{x^2}{2e}-\frac{x^3}{2e}}{x}^{2}dx-\frac{3}{4}\int e^{-1-\frac{x^2}{2e}-\frac{x^3}{2e}}{x}^{3}dx+\frac{1}{24}\log (3)\int e^{-1-\frac{x^2}{2e}-\frac{x^3}{2e}}x(2+3x)dx\\ & =2x-\frac{1}{12}{e}^{-\frac{x^2}{2e}-\frac{x^3}{2e}}\log (3)+\frac{1}{2}\int e^{-\frac{x^2(1+x)}{2e}}dx-\frac{1}{2}\int e^{-1-\frac{x^2}{2e}-\frac{x^3}{2e}}{x}^{2}dx-\frac{3}{4}\int e^{-1-\frac{x^2}{2e}-\frac{x^3}{2e}}{x}^{3}dx\end{array}\end{array}$$

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Mathematica [A]  time = 0.42, size = 48, normalized size = 1.45 $$\begin{array}{c}\frac{1}{24}(48x+\frac{e^{-\frac{x^2(1+x)}{2e}}(36x^2-2\log (9)-2x(-12+\log (27)))}{2+3x})\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(E^(-1 - (x^2 + x^3)/(2*E))*(12*E + 48*E^(1 + (x^2 + x^3)/(2*E)) - 12*x^2 - 18*x^3 + (2*x + 3*x^2)*L
og[3]))/24,x]

[Out]

(48*x + (36*x^2 - 2*Log[9] - 2*x*(-12 + Log[27]))/(E^((x^2*(1 + x))/(2*E))*(2 + 3*x)))/24

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fricas [A]  time = 0.76, size = 49, normalized size = 1.48 $$\begin{array}{c}\frac{1}{12}(6xe+24x{e}^{\left(\frac{1}{2}(x^3+x^2+2e)e^{(-1)}\right)}-e\log (3))e^{(-\frac{1}{2}(x^3+x^2+2e)e^{(-1)})}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/24*(48*exp(1)*exp(1/2*(x^3+x^2)/exp(1))+(3*x^2+2*x)*log(3)+12*exp(1)-18*x^3-12*x^2)/exp(1)/exp(1/2
*(x^3+x^2)/exp(1)),x, algorithm="fricas")

[Out]

1/12*(6*x*e + 24*x*e^(1/2*(x^3 + x^2 + 2*e)*e^(-1)) - e*log(3))*e^(-1/2*(x^3 + x^2 + 2*e)*e^(-1))

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giac [A]  time = 0.35, size = 35, normalized size = 1.06 $$\begin{array}{c}\frac{1}{2}x{e}^{(-\frac{1}{2}(x^3+x^2)e^{(-1)})}-\frac{1}{12}{e}^{(-\frac{1}{2}(x^3+x^2)e^{(-1)})}\log (3)+2x\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/24*(48*exp(1)*exp(1/2*(x^3+x^2)/exp(1))+(3*x^2+2*x)*log(3)+12*exp(1)-18*x^3-12*x^2)/exp(1)/exp(1/2
*(x^3+x^2)/exp(1)),x, algorithm="giac")

[Out]

1/2*x*e^(-1/2*(x^3 + x^2)*e^(-1)) - 1/12*e^(-1/2*(x^3 + x^2)*e^(-1))*log(3) + 2*x

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maple [A]  time = 0.10, size = 36, normalized size = 1.09

method	result	size
risch	$2x+\frac{(-2\mathrm{e}\ln (3)+12x\mathrm{e}){\mathrm{e}}^{-1-\frac{x^3{\mathrm{e}}^{-1}}{2}-\frac{x^2{\mathrm{e}}^{-1}}{2}}}{24}$	$36$
norman	$(\frac{x}{2}+2x{\mathrm{e}}^{\frac{(x^3+x^2){\mathrm{e}}^{-1}}{2}}-\frac{\ln (3)}{12}){\mathrm{e}}^{-\frac{(x^3+x^2){\mathrm{e}}^{-1}}{2}}$	$43$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/24*(48*exp(1)*exp(1/2*(x^3+x^2)/exp(1))+(3*x^2+2*x)*ln(3)+12*exp(1)-18*x^3-12*x^2)/exp(1)/exp(1/2*(x^3+x
^2)/exp(1)),x,method=_RETURNVERBOSE)

[Out]

2*x+1/24*(-2*exp(1)*ln(3)+12*x*exp(1))*exp(-1-1/2*x^3*exp(-1)-1/2*x^2*exp(-1))

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maxima [A]  time = 0.83, size = 30, normalized size = 0.91 $$\begin{array}{c}\frac{1}{12}(6x-\log (3))e^{(-\frac{1}{2}{x}^3{e}^{(-1)}-\frac{1}{2}{x}^2{e}^{(-1)})}+2x\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/24*(48*exp(1)*exp(1/2*(x^3+x^2)/exp(1))+(3*x^2+2*x)*log(3)+12*exp(1)-18*x^3-12*x^2)/exp(1)/exp(1/2
*(x^3+x^2)/exp(1)),x, algorithm="maxima")

[Out]

1/12*(6*x - log(3))*e^(-1/2*x^3*e^(-1) - 1/2*x^2*e^(-1)) + 2*x

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mupad [B]  time = 1.11, size = 43, normalized size = 1.30 $$\begin{array}{c}2x-\frac{{\mathrm{e}}^{-\frac{{\mathrm{e}}^{-1}{x}^3}{2}-\frac{{\mathrm{e}}^{-1}{x}^2}{2}}\ln (3)}{12}+\frac{x{\mathrm{e}}^{-\frac{{\mathrm{e}}^{-1}{x}^3}{2}-\frac{{\mathrm{e}}^{-1}{x}^2}{2}}}{2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-1)*exp(-exp(-1)*(x^2/2 + x^3/2))*(exp(1)/2 + (log(3)*(2*x + 3*x^2))/24 + 2*exp(1)*exp(exp(-1)*(x^2/2
+ x^3/2)) - x^2/2 - (3*x^3)/4),x)

[Out]

2*x - (exp(- (x^2*exp(-1))/2 - (x^3*exp(-1))/2)*log(3))/12 + (x*exp(- (x^2*exp(-1))/2 - (x^3*exp(-1))/2))/2

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sympy [A]  time = 0.18, size = 26, normalized size = 0.79 $$\begin{array}{c}2x+\frac{(6x-\log (3))e^{-\frac{\frac{x^3}{2}+\frac{x^2}{2}}{e}}}{12}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/24*(48*exp(1)*exp(1/2*(x**3+x**2)/exp(1))+(3*x**2+2*x)*ln(3)+12*exp(1)-18*x**3-12*x**2)/exp(1)/exp
(1/2*(x**3+x**2)/exp(1)),x)

[Out]

2*x + (6*x - log(3))*exp(-(x**3/2 + x**2/2)*exp(-1))/12

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