3.69.12 $\int \frac{e^{\frac{3e^{x^2}+2x-3\log (x)}{6x}}(-1+e^{x^2}(-1+2x^2)+\log (x))}{8x^2}dx$

Optimal. Leaf size=28 $$\frac{1}{4}{e}^{\frac{e^{x^2}+\frac{2x}{3}-\log (x)}{2x}}$$

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Rubi [A]  time = 0.54, antiderivative size = 33, normalized size of antiderivative = 1.18, number of steps used = 2, number of rules used = 2, integrand size = 48, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.042, Rules used = {12, 6706} $$\begin{array}{c}\frac{1}{4}{e}^{\frac{3e^{x^2}+2x}{6x}}{x}^{-\frac{1}{2}/x}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(E^((3*E^x^2 + 2*x - 3*Log[x])/(6*x))*(-1 + E^x^2*(-1 + 2*x^2) + Log[x]))/(8*x^2),x]

[Out]

E^((3*E^x^2 + 2*x)/(6*x))/(4*x^(1/(2*x)))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\frac{1}{8}\int \frac{e^{\frac{3e^{x^2}+2x-3\log (x)}{6x}}(-1+e^{x^2}(-1+2x^2)+\log (x))}{x^2}dx\\ & =\frac{1}{4}{e}^{\frac{3e^{x^2}+2x}{6x}}{x}^{-\frac{1}{2}/x}\end{array}\end{array}$$

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Mathematica [A]  time = 0.62, size = 31, normalized size = 1.11 $$\begin{array}{c}\frac{1}{4}{e}^{\frac{1}{3}+\frac{e^{x^2}}{2x}}{x}^{-\frac{1}{2}/x}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(E^((3*E^x^2 + 2*x - 3*Log[x])/(6*x))*(-1 + E^x^2*(-1 + 2*x^2) + Log[x]))/(8*x^2),x]

[Out]

E^(1/3 + E^x^2/(2*x))/(4*x^(1/(2*x)))

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fricas [A]  time = 0.45, size = 22, normalized size = 0.79 $$\begin{array}{c}\frac{1}{4}{e}^{\left(\frac{2x+3e^{\left(x^2\right)}-3\log (x)}{6x}\right)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*(log(x)+(2*x^2-1)*exp(x^2)-1)*exp(1/6*(-3*log(x)+3*exp(x^2)+2*x)/x)/x^2,x, algorithm="fricas")

[Out]

1/4*e^(1/6*(2*x + 3*e^(x^2) - 3*log(x))/x)

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giac [A]  time = 0.15, size = 21, normalized size = 0.75 $$\begin{array}{c}\frac{1}{4}{e}^{(\frac{e^{\left(x^2\right)}}{2x}-\frac{\log (x)}{2x}+\frac{1}{3})}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*(log(x)+(2*x^2-1)*exp(x^2)-1)*exp(1/6*(-3*log(x)+3*exp(x^2)+2*x)/x)/x^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.04, size = 23, normalized size = 0.82

method	result	size
risch	$\frac{{\mathrm{e}}^{-\frac{3\ln (x)-3{\mathrm{e}}^{x^2}-2x}{6x}}}{4}$	$23$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/8*(ln(x)+(2*x^2-1)*exp(x^2)-1)*exp(1/6*(-3*ln(x)+3*exp(x^2)+2*x)/x)/x^2,x,method=_RETURNVERBOSE)

[Out]

1/4*exp(-1/6*(3*ln(x)-3*exp(x^2)-2*x)/x)

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maxima [A]  time = 0.50, size = 21, normalized size = 0.75 $$\begin{array}{c}\frac{1}{4}{e}^{(\frac{e^{\left(x^2\right)}}{2x}-\frac{\log (x)}{2x}+\frac{1}{3})}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*(log(x)+(2*x^2-1)*exp(x^2)-1)*exp(1/6*(-3*log(x)+3*exp(x^2)+2*x)/x)/x^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.13, size = 23, normalized size = 0.82 $$\begin{array}{c}\frac{{\mathrm{e}}^{\frac{{\mathrm{e}}^{x^2}}{2x}+\frac{1}{3}}}{4x^{\frac{1}{2x}}}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((x/3 + exp(x^2)/2 - log(x)/2)/x)*(log(x) + exp(x^2)*(2*x^2 - 1) - 1))/(8*x^2),x)

[Out]

exp(exp(x^2)/(2*x) + 1/3)/(4*x^(1/(2*x)))

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sympy [A]  time = 0.51, size = 19, normalized size = 0.68 $$\begin{array}{c}\frac{e^{\frac{\frac{x}{3}+\frac{e^{x^2}}{2}-\frac{\log (x)}{2}}{x}}}{4}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*(ln(x)+(2*x**2-1)*exp(x**2)-1)*exp(1/6*(-3*ln(x)+3*exp(x**2)+2*x)/x)/x**2,x)

[Out]

exp((x/3 + exp(x**2)/2 - log(x)/2)/x)/4

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