\(\int \frac {e^{\frac {3 e^{x^2}+2 x-3 \log (x)}{6 x}} (-1+e^{x^2} (-1+2 x^2)+\log (x))}{8 x^2} \, dx\) [6812]

Optimal result

Integrand size = 48, antiderivative size = 28 \[ \int \frac {e^{\frac {3 e^{x^2}+2 x-3 \log (x)}{6 x}} \left (-1+e^{x^2} \left (-1+2 x^2\right )+\log (x)\right )}{8 x^2} \, dx=\frac {1}{4} e^{\frac {e^{x^2}+\frac {2 x}{3}-\log (x)}{2 x}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.18, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {12, 6838} \[ \int \frac {e^{\frac {3 e^{x^2}+2 x-3 \log (x)}{6 x}} \left (-1+e^{x^2} \left (-1+2 x^2\right )+\log (x)\right )}{8 x^2} \, dx=\frac {1}{4} e^{\frac {3 e^{x^2}+2 x}{6 x}} x^{\left .-\frac {1}{2}\right /x} \]

[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 6838

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{align*} \text {integral}& = \frac {1}{8} \int \frac {e^{\frac {3 e^{x^2}+2 x-3 \log (x)}{6 x}} \left (-1+e^{x^2} \left (-1+2 x^2\right )+\log (x)\right )}{x^2} \, dx \\ & = \frac {1}{4} e^{\frac {3 e^{x^2}+2 x}{6 x}} x^{\left .-\frac {1}{2}\right /x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.66 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {e^{\frac {3 e^{x^2}+2 x-3 \log (x)}{6 x}} \left (-1+e^{x^2} \left (-1+2 x^2\right )+\log (x)\right )}{8 x^2} \, dx=\frac {1}{4} e^{\frac {1}{3}+\frac {e^{x^2}}{2 x}} x^{\left .-\frac {1}{2}\right /x} \]

[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)))

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82

method	result	size
risch	\(\frac {{\mathrm e}^{-\frac {3 \ln \left (x \right )-3 \,{\mathrm e}^{x^{2}}-2 x}{6 x}}}{4}\)	\(23\)
parallelrisch	\(\frac {{\mathrm e}^{\frac {-3 \ln \left (x \right )+3 \,{\mathrm e}^{x^{2}}+2 x}{6 x}}}{4}\)	\(23\)

[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)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {e^{\frac {3 e^{x^2}+2 x-3 \log (x)}{6 x}} \left (-1+e^{x^2} \left (-1+2 x^2\right )+\log (x)\right )}{8 x^2} \, dx=\frac {1}{4} \, e^{\left (\frac {2 \, x + 3 \, e^{\left (x^{2}\right )} - 3 \, \log \left (x\right )}{6 \, x}\right )} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {e^{\frac {3 e^{x^2}+2 x-3 \log (x)}{6 x}} \left (-1+e^{x^2} \left (-1+2 x^2\right )+\log (x)\right )}{8 x^2} \, dx=\frac {e^{\frac {\frac {x}{3} + \frac {e^{x^{2}}}{2} - \frac {\log {\left (x \right )}}{2}}{x}}}{4} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75 \[ \int \frac {e^{\frac {3 e^{x^2}+2 x-3 \log (x)}{6 x}} \left (-1+e^{x^2} \left (-1+2 x^2\right )+\log (x)\right )}{8 x^2} \, dx=\frac {1}{4} \, e^{\left (\frac {e^{\left (x^{2}\right )}}{2 \, x} - \frac {\log \left (x\right )}{2 \, x} + \frac {1}{3}\right )} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75 \[ \int \frac {e^{\frac {3 e^{x^2}+2 x-3 \log (x)}{6 x}} \left (-1+e^{x^2} \left (-1+2 x^2\right )+\log (x)\right )}{8 x^2} \, dx=\frac {1}{4} \, e^{\left (\frac {e^{\left (x^{2}\right )}}{2 \, x} - \frac {\log \left (x\right )}{2 \, x} + \frac {1}{3}\right )} \]

[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)

Mupad [B] (verification not implemented)

Time = 12.40 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82 \[ \int \frac {e^{\frac {3 e^{x^2}+2 x-3 \log (x)}{6 x}} \left (-1+e^{x^2} \left (-1+2 x^2\right )+\log (x)\right )}{8 x^2} \, dx=\frac {{\mathrm {e}}^{\frac {{\mathrm {e}}^{x^2}}{2\,x}+\frac {1}{3}}}{4\,x^{\frac {1}{2\,x}}} \]

[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)))