3.65.66 \(\int \frac {e^{-\frac {2 (-5 e^{e^{x^2}}-5 x+x^2)}{e^{e^{x^2}}+x}} (2 e^{2 e^{x^2}} \log (x+\log (2))+2 x^2 \log (x+\log (2))+(-2 x^3-2 x^2 \log (2)) \log ^2(x+\log (2))+e^{e^{x^2}} (4 x \log (x+\log (2))+(-4 x^2-4 x \log (2)+e^{x^2} (4 x^4+4 x^3 \log (2))) \log ^2(x+\log (2))))}{x^3+x^2 \log (2)+e^{2 e^{x^2}} (x+\log (2))+e^{e^{x^2}} (2 x^2+2 x \log (2))} \, dx\)

Optimal. Leaf size=30 \[ -5+e^{10-\frac {2 x^2}{e^{e^{x^2}}+x}} \log ^2(x+\log (2)) \]

________________________________________________________________________________________

Rubi [B]  time = 4.23, antiderivative size = 182, normalized size of antiderivative = 6.07, number of steps used = 3, number of rules used = 3, integrand size = 180, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.017, Rules used = {6688, 12, 2288} \begin {gather*} -\frac {e^{\frac {2 \left (5 e^{e^{x^2}}+(5-x) x\right )}{e^{e^{x^2}}+x}} x \left (-e^{x^2+e^{x^2}} x^2 (2 x+\log (4))+e^{e^{x^2}} (2 x+\log (4))+x (x+\log (2))\right ) \log ^2(x+\log (2))}{\left (e^{e^{x^2}}+x\right )^2 \left (\frac {10 e^{x^2+e^{x^2}} x-2 x+5}{e^{e^{x^2}}+x}-\frac {\left (2 e^{x^2+e^{x^2}} x+1\right ) \left (5 e^{e^{x^2}}+(5-x) x\right )}{\left (e^{e^{x^2}}+x\right )^2}\right ) (x+\log (2))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((E^((2*(5*E^E^x^2 + (5 - x)*x))/(E^E^x^2 + x))*x*(x*(x + Log[2]) + E^E^x^2*(2*x + Log[4]) - E^(E^x^2 + x^2)*
x^2*(2*x + Log[4]))*Log[x + Log[2]]^2)/((E^E^x^2 + x)^2*((5 - 2*x + 10*E^(E^x^2 + x^2)*x)/(E^E^x^2 + x) - ((1
+ 2*E^(E^x^2 + x^2)*x)*(5*E^E^x^2 + (5 - x)*x))/(E^E^x^2 + x)^2)*(x + Log[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 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 e^{\frac {10 e^{e^{x^2}}-2 (-5+x) x}{e^{e^{x^2}}+x}} \log (x+\log (2)) \left (\left (e^{e^{x^2}}+x\right )^2+x \left (-x (x+\log (2))-e^{e^{x^2}} (2 x+\log (4))+e^{e^{x^2}+x^2} x^2 (2 x+\log (4))\right ) \log (x+\log (2))\right )}{\left (e^{e^{x^2}}+x\right )^2 (x+\log (2))} \, dx\\ &=2 \int \frac {e^{\frac {10 e^{e^{x^2}}-2 (-5+x) x}{e^{e^{x^2}}+x}} \log (x+\log (2)) \left (\left (e^{e^{x^2}}+x\right )^2+x \left (-x (x+\log (2))-e^{e^{x^2}} (2 x+\log (4))+e^{e^{x^2}+x^2} x^2 (2 x+\log (4))\right ) \log (x+\log (2))\right )}{\left (e^{e^{x^2}}+x\right )^2 (x+\log (2))} \, dx\\ &=-\frac {e^{\frac {2 \left (5 e^{e^{x^2}}+(5-x) x\right )}{e^{e^{x^2}}+x}} x \left (x (x+\log (2))+e^{e^{x^2}} (2 x+\log (4))-e^{e^{x^2}+x^2} x^2 (2 x+\log (4))\right ) \log ^2(x+\log (2))}{\left (e^{e^{x^2}}+x\right )^2 \left (\frac {5-2 x+10 e^{e^{x^2}+x^2} x}{e^{e^{x^2}}+x}-\frac {\left (1+2 e^{e^{x^2}+x^2} x\right ) \left (5 e^{e^{x^2}}+(5-x) x\right )}{\left (e^{e^{x^2}}+x\right )^2}\right ) (x+\log (2))}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [F]  time = 13.36, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{-\frac {2 \left (-5 e^{e^{x^2}}-5 x+x^2\right )}{e^{e^{x^2}}+x}} \left (2 e^{2 e^{x^2}} \log (x+\log (2))+2 x^2 \log (x+\log (2))+\left (-2 x^3-2 x^2 \log (2)\right ) \log ^2(x+\log (2))+e^{e^{x^2}} \left (4 x \log (x+\log (2))+\left (-4 x^2-4 x \log (2)+e^{x^2} \left (4 x^4+4 x^3 \log (2)\right )\right ) \log ^2(x+\log (2))\right )\right )}{x^3+x^2 \log (2)+e^{2 e^{x^2}} (x+\log (2))+e^{e^{x^2}} \left (2 x^2+2 x \log (2)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.65, size = 34, normalized size = 1.13 \begin {gather*} e^{\left (-\frac {2 \, {\left (x^{2} - 5 \, x - 5 \, e^{\left (e^{\left (x^{2}\right )}\right )}\right )}}{x + e^{\left (e^{\left (x^{2}\right )}\right )}}\right )} \log \left (x + \log \relax (2)\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [B]  time = 2.26, size = 74, normalized size = 2.47 \begin {gather*} e^{\left (-x^{2} + \frac {x^{3} + x^{2} e^{\left (e^{\left (x^{2}\right )}\right )} - 2 \, x^{2} + x e^{\left (x^{2}\right )} + 10 \, x + e^{\left (x^{2} + e^{\left (x^{2}\right )}\right )} + 10 \, e^{\left (e^{\left (x^{2}\right )}\right )}}{x + e^{\left (e^{\left (x^{2}\right )}\right )}} - e^{\left (x^{2}\right )}\right )} \log \left (x + \log \relax (2)\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(-x^2 + (x^3 + x^2*e^(e^(x^2)) - 2*x^2 + x*e^(x^2) + 10*x + e^(x^2 + e^(x^2)) + 10*e^(e^(x^2)))/(x + e^(e^(x
^2))) - e^(x^2))*log(x + log(2))^2

________________________________________________________________________________________

maple [A]  time = 0.15, size = 35, normalized size = 1.17




method result size



risch \(\ln \left (\ln \relax (2)+x \right )^{2} {\mathrm e}^{-\frac {2 \left (-5 \,{\mathrm e}^{{\mathrm e}^{x^{2}}}+x^{2}-5 x \right )}{{\mathrm e}^{{\mathrm e}^{x^{2}}}+x}}\) \(35\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*ln(ln(2)+x)*exp(exp(x^2))^2+(((4*x^3*ln(2)+4*x^4)*exp(x^2)-4*x*ln(2)-4*x^2)*ln(ln(2)+x)^2+4*x*ln(ln(2)+
x))*exp(exp(x^2))+(-2*x^2*ln(2)-2*x^3)*ln(ln(2)+x)^2+2*x^2*ln(ln(2)+x))/((ln(2)+x)*exp(exp(x^2))^2+(2*x*ln(2)+
2*x^2)*exp(exp(x^2))+x^2*ln(2)+x^3)/exp((-5*exp(exp(x^2))+x^2-5*x)/(exp(exp(x^2))+x))^2,x,method=_RETURNVERBOS
E)

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {{\mathrm {e}}^{\frac {2\,\left (5\,x+5\,{\mathrm {e}}^{{\mathrm {e}}^{x^2}}-x^2\right )}{x+{\mathrm {e}}^{{\mathrm {e}}^{x^2}}}}\,\left ({\mathrm {e}}^{{\mathrm {e}}^{x^2}}\,\left ({\ln \left (x+\ln \relax (2)\right )}^2\,\left (4\,x\,\ln \relax (2)-{\mathrm {e}}^{x^2}\,\left (4\,x^4+4\,\ln \relax (2)\,x^3\right )+4\,x^2\right )-4\,x\,\ln \left (x+\ln \relax (2)\right )\right )-2\,{\mathrm {e}}^{2\,{\mathrm {e}}^{x^2}}\,\ln \left (x+\ln \relax (2)\right )+{\ln \left (x+\ln \relax (2)\right )}^2\,\left (2\,x^3+2\,\ln \relax (2)\,x^2\right )-2\,x^2\,\ln \left (x+\ln \relax (2)\right )\right )}{x^2\,\ln \relax (2)+{\mathrm {e}}^{2\,{\mathrm {e}}^{x^2}}\,\left (x+\ln \relax (2)\right )+x^3+{\mathrm {e}}^{{\mathrm {e}}^{x^2}}\,\left (2\,x^2+2\,\ln \relax (2)\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*ln(ln(2)+x)*exp(exp(x**2))**2+(((4*x**3*ln(2)+4*x**4)*exp(x**2)-4*x*ln(2)-4*x**2)*ln(ln(2)+x)**2+
4*x*ln(ln(2)+x))*exp(exp(x**2))+(-2*x**2*ln(2)-2*x**3)*ln(ln(2)+x)**2+2*x**2*ln(ln(2)+x))/((ln(2)+x)*exp(exp(x
**2))**2+(2*x*ln(2)+2*x**2)*exp(exp(x**2))+x**2*ln(2)+x**3)/exp((-5*exp(exp(x**2))+x**2-5*x)/(exp(exp(x**2))+x
))**2,x)

[Out]

Timed out

________________________________________________________________________________________