3.51.73 \(\int \frac {e^{-\frac {3+e^x}{3 e^{x^2}-6 x}} (-6 e^{2 x^2} \log (5)+(-6 x-24 x^2) \log (5)+e^x (-2 x+2 x^2) \log (5)+e^{x^2} (e^x (-x+2 x^2) \log (5)+(24 x+6 x^2) \log (5)))}{48 e^{2 x^2} x^3-192 e^{x^2} x^4+192 x^5} \, dx\)

Optimal. Leaf size=31 \[ \frac {e^{-\frac {3+e^x}{3 \left (e^{x^2}-2 x\right )}} \log (5)}{16 x^2} \]

________________________________________________________________________________________

Rubi [B]  time = 24.69, antiderivative size = 137, normalized size of antiderivative = 4.42, number of steps used = 3, number of rules used = 3, integrand size = 126, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.024, Rules used = {6741, 12, 2288} \begin {gather*} \frac {e^{-\frac {e^x+3}{3 \left (e^{x^2}-2 x\right )}} \left (-6 e^{x^2} x^2-2 e^x x^2-2 e^{x^2+x} x^2+e^{x^2+x} x+2 e^x x+6 x\right ) \log (5)}{16 \left (e^{x^2}-2 x\right )^2 x^3 \left (\frac {2 \left (e^x+3\right ) \left (1-e^{x^2} x\right )}{\left (e^{x^2}-2 x\right )^2}+\frac {e^x}{e^{x^2}-2 x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-6*E^(2*x^2)*Log[5] + (-6*x - 24*x^2)*Log[5] + E^x*(-2*x + 2*x^2)*Log[5] + E^x^2*(E^x*(-x + 2*x^2)*Log[5]
+ (24*x + 6*x^2)*Log[5]))/(E^((3 + E^x)/(3*E^x^2 - 6*x))*(48*E^(2*x^2)*x^3 - 192*E^x^2*x^4 + 192*x^5)),x]

[Out]

((6*x + 2*E^x*x + E^(x + x^2)*x - 2*E^x*x^2 - 6*E^x^2*x^2 - 2*E^(x + x^2)*x^2)*Log[5])/(16*E^((3 + E^x)/(3*(E^
x^2 - 2*x)))*(E^x^2 - 2*x)^2*x^3*(E^x/(E^x^2 - 2*x) + (2*(3 + E^x)*(1 - E^x^2*x))/(E^x^2 - 2*x)^2))

Rule 12

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

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-\frac {3+e^x}{3 \left (e^{x^2}-2 x\right )}} \left (-6 e^{2 x^2}-6 x-2 e^x x+24 e^{x^2} x-e^{x+x^2} x-24 x^2+2 e^x x^2+6 e^{x^2} x^2+2 e^{x+x^2} x^2\right ) \log (5)}{48 \left (e^{x^2}-2 x\right )^2 x^3} \, dx\\ &=\frac {1}{48} \log (5) \int \frac {e^{-\frac {3+e^x}{3 \left (e^{x^2}-2 x\right )}} \left (-6 e^{2 x^2}-6 x-2 e^x x+24 e^{x^2} x-e^{x+x^2} x-24 x^2+2 e^x x^2+6 e^{x^2} x^2+2 e^{x+x^2} x^2\right )}{\left (e^{x^2}-2 x\right )^2 x^3} \, dx\\ &=\frac {e^{-\frac {3+e^x}{3 \left (e^{x^2}-2 x\right )}} \left (6 x+2 e^x x+e^{x+x^2} x-2 e^x x^2-6 e^{x^2} x^2-2 e^{x+x^2} x^2\right ) \log (5)}{16 \left (e^{x^2}-2 x\right )^2 x^3 \left (\frac {e^x}{e^{x^2}-2 x}+\frac {2 \left (3+e^x\right ) \left (1-e^{x^2} x\right )}{\left (e^{x^2}-2 x\right )^2}\right )}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.16, size = 33, normalized size = 1.06 \begin {gather*} \frac {e^{\frac {-3-e^x}{3 \left (e^{x^2}-2 x\right )}} \log (5)}{16 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-6*E^(2*x^2)*Log[5] + (-6*x - 24*x^2)*Log[5] + E^x*(-2*x + 2*x^2)*Log[5] + E^x^2*(E^x*(-x + 2*x^2)*
Log[5] + (24*x + 6*x^2)*Log[5]))/(E^((3 + E^x)/(3*E^x^2 - 6*x))*(48*E^(2*x^2)*x^3 - 192*E^x^2*x^4 + 192*x^5)),
x]

[Out]

(E^((-3 - E^x)/(3*(E^x^2 - 2*x)))*Log[5])/(16*x^2)

________________________________________________________________________________________

fricas [A]  time = 0.80, size = 26, normalized size = 0.84 \begin {gather*} \frac {e^{\left (\frac {e^{x} + 3}{3 \, {\left (2 \, x - e^{\left (x^{2}\right )}\right )}}\right )} \log \relax (5)}{16 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-6*log(5)*exp(x^2)^2+((2*x^2-x)*log(5)*exp(x)+(6*x^2+24*x)*log(5))*exp(x^2)+(2*x^2-2*x)*log(5)*exp(
x)+(-24*x^2-6*x)*log(5))/(48*x^3*exp(x^2)^2-192*x^4*exp(x^2)+192*x^5)/exp((3+exp(x))/(3*exp(x^2)-6*x)),x, algo
rithm="fricas")

[Out]

1/16*e^(1/3*(e^x + 3)/(2*x - e^(x^2)))*log(5)/x^2

________________________________________________________________________________________

giac [A]  time = 0.79, size = 26, normalized size = 0.84 \begin {gather*} \frac {e^{\left (\frac {e^{x} + 3}{3 \, {\left (2 \, x - e^{\left (x^{2}\right )}\right )}}\right )} \log \relax (5)}{16 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-6*log(5)*exp(x^2)^2+((2*x^2-x)*log(5)*exp(x)+(6*x^2+24*x)*log(5))*exp(x^2)+(2*x^2-2*x)*log(5)*exp(
x)+(-24*x^2-6*x)*log(5))/(48*x^3*exp(x^2)^2-192*x^4*exp(x^2)+192*x^5)/exp((3+exp(x))/(3*exp(x^2)-6*x)),x, algo
rithm="giac")

[Out]

1/16*e^(1/3*(e^x + 3)/(2*x - e^(x^2)))*log(5)/x^2

________________________________________________________________________________________

maple [A]  time = 0.11, size = 27, normalized size = 0.87

method	result	size
risch	\(\frac {\ln \relax (5) {\mathrm e}^{\frac {3+{\mathrm e}^{x}}{-3 \,{\mathrm e}^{x^{2}}+6 x}}}{16 x^{2}}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-6*ln(5)*exp(x^2)^2+((2*x^2-x)*ln(5)*exp(x)+(6*x^2+24*x)*ln(5))*exp(x^2)+(2*x^2-2*x)*ln(5)*exp(x)+(-24*x^
2-6*x)*ln(5))/(48*x^3*exp(x^2)^2-192*x^4*exp(x^2)+192*x^5)/exp((3+exp(x))/(3*exp(x^2)-6*x)),x,method=_RETURNVE
RBOSE)

[Out]

1/16*ln(5)/x^2*exp(1/3*(3+exp(x))/(-exp(x^2)+2*x))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {1}{48} \, \int \frac {{\left (2 \, {\left (x^{2} - x\right )} e^{x} \log \relax (5) + {\left ({\left (2 \, x^{2} - x\right )} e^{x} \log \relax (5) + 6 \, {\left (x^{2} + 4 \, x\right )} \log \relax (5)\right )} e^{\left (x^{2}\right )} - 6 \, {\left (4 \, x^{2} + x\right )} \log \relax (5) - 6 \, e^{\left (2 \, x^{2}\right )} \log \relax (5)\right )} e^{\left (\frac {e^{x} + 3}{3 \, {\left (2 \, x - e^{\left (x^{2}\right )}\right )}}\right )}}{4 \, x^{5} - 4 \, x^{4} e^{\left (x^{2}\right )} + x^{3} e^{\left (2 \, x^{2}\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-6*log(5)*exp(x^2)^2+((2*x^2-x)*log(5)*exp(x)+(6*x^2+24*x)*log(5))*exp(x^2)+(2*x^2-2*x)*log(5)*exp(
x)+(-24*x^2-6*x)*log(5))/(48*x^3*exp(x^2)^2-192*x^4*exp(x^2)+192*x^5)/exp((3+exp(x))/(3*exp(x^2)-6*x)),x, algo
rithm="maxima")

[Out]

1/48*integrate((2*(x^2 - x)*e^x*log(5) + ((2*x^2 - x)*e^x*log(5) + 6*(x^2 + 4*x)*log(5))*e^(x^2) - 6*(4*x^2 +
x)*log(5) - 6*e^(2*x^2)*log(5))*e^(1/3*(e^x + 3)/(2*x - e^(x^2)))/(4*x^5 - 4*x^4*e^(x^2) + x^3*e^(2*x^2)), x)

________________________________________________________________________________________

mupad [B]  time = 3.91, size = 36, normalized size = 1.16 \begin {gather*} \frac {{\mathrm {e}}^{\frac {1}{2\,x-{\mathrm {e}}^{x^2}}}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^x}{3\,\left (2\,x-{\mathrm {e}}^{x^2}\right )}}\,\ln \relax (5)}{16\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((exp(x) + 3)/(6*x - 3*exp(x^2)))*(log(5)*(6*x + 24*x^2) + 6*exp(2*x^2)*log(5) - exp(x^2)*(log(5)*(24
*x + 6*x^2) - exp(x)*log(5)*(x - 2*x^2)) + exp(x)*log(5)*(2*x - 2*x^2)))/(48*x^3*exp(2*x^2) - 192*x^4*exp(x^2)
+ 192*x^5),x)

[Out]

(exp(1/(2*x - exp(x^2)))*exp(exp(x)/(3*(2*x - exp(x^2))))*log(5))/(16*x^2)

________________________________________________________________________________________

sympy [A]  time = 0.70, size = 24, normalized size = 0.77 \begin {gather*} \frac {e^{- \frac {e^{x} + 3}{- 6 x + 3 e^{x^{2}}}} \log {\relax (5 )}}{16 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-6*ln(5)*exp(x**2)**2+((2*x**2-x)*ln(5)*exp(x)+(6*x**2+24*x)*ln(5))*exp(x**2)+(2*x**2-2*x)*ln(5)*ex
p(x)+(-24*x**2-6*x)*ln(5))/(48*x**3*exp(x**2)**2-192*x**4*exp(x**2)+192*x**5)/exp((3+exp(x))/(3*exp(x**2)-6*x)
),x)

[Out]

exp(-(exp(x) + 3)/(-6*x + 3*exp(x**2)))*log(5)/(16*x**2)

________________________________________________________________________________________