3.46.24 \(\int \frac {e^{e^x} (-1+e^x (-8+x)+e^{x^2} (-e^x+2 x))}{-256+64 x-4 x^2+e^{2 x^2} (-4+\log (3))+(64-16 x+x^2) \log (3)+e^{x^2} (-64+8 x+(16-2 x) \log (3))} \, dx\)

Optimal. Leaf size=26 \[ \frac {e^{e^x}}{\left (8+e^{x^2}-x\right ) (4-\log (3))} \]

________________________________________________________________________________________

Rubi [A]  time = 0.70, antiderivative size = 47, normalized size of antiderivative = 1.81, number of steps used = 3, number of rules used = 3, integrand size = 84, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {6688, 12, 2288} \begin {gather*} \frac {e^{e^x-x} \left (e^{x^2+x}+e^x (8-x)\right )}{\left (e^{x^2}-x+8\right )^2 (4-\log (3))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^E^x*(-1 + E^x*(-8 + x) + E^x^2*(-E^x + 2*x)))/(-256 + 64*x - 4*x^2 + E^(2*x^2)*(-4 + Log[3]) + (64 - 16
*x + x^2)*Log[3] + E^x^2*(-64 + 8*x + (16 - 2*x)*Log[3])),x]

[Out]

(E^(E^x - x)*(E^(x + x^2) + E^x*(8 - x)))/((8 + E^x^2 - x)^2*(4 - Log[3]))

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 {e^{e^x} \left (1+e^{x+x^2}-e^x (-8+x)-2 e^{x^2} x\right )}{\left (8+e^{x^2}-x\right )^2 (4-\log (3))} \, dx\\ &=\frac {\int \frac {e^{e^x} \left (1+e^{x+x^2}-e^x (-8+x)-2 e^{x^2} x\right )}{\left (8+e^{x^2}-x\right )^2} \, dx}{4-\log (3)}\\ &=\frac {e^{e^x-x} \left (e^{x+x^2}+e^x (8-x)\right )}{\left (8+e^{x^2}-x\right )^2 (4-\log (3))}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.13, size = 25, normalized size = 0.96 \begin {gather*} -\frac {e^{e^x}}{\left (8+e^{x^2}-x\right ) (-4+\log (3))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^E^x*(-1 + E^x*(-8 + x) + E^x^2*(-E^x + 2*x)))/(-256 + 64*x - 4*x^2 + E^(2*x^2)*(-4 + Log[3]) + (6
4 - 16*x + x^2)*Log[3] + E^x^2*(-64 + 8*x + (16 - 2*x)*Log[3])),x]

[Out]

-(E^E^x/((8 + E^x^2 - x)*(-4 + Log[3])))

________________________________________________________________________________________

fricas [A]  time = 0.59, size = 28, normalized size = 1.08 \begin {gather*} -\frac {e^{\left (e^{x}\right )}}{{\left (\log \relax (3) - 4\right )} e^{\left (x^{2}\right )} - {\left (x - 8\right )} \log \relax (3) + 4 \, x - 32} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x-exp(x))*exp(x^2)+(-8+x)*exp(x)-1)*exp(exp(x))/((-4+log(3))*exp(x^2)^2+((16-2*x)*log(3)+8*x-64)
*exp(x^2)+(x^2-16*x+64)*log(3)-4*x^2+64*x-256),x, algorithm="fricas")

[Out]

-e^(e^x)/((log(3) - 4)*e^(x^2) - (x - 8)*log(3) + 4*x - 32)

________________________________________________________________________________________

giac [B]  time = 0.21, size = 145, normalized size = 5.58 \begin {gather*} \frac {2 \, x^{2} e^{\left (x + e^{x}\right )} - 16 \, x e^{\left (x + e^{x}\right )} - e^{\left (x + e^{x}\right )}}{2 \, x^{3} e^{x} \log \relax (3) - 8 \, x^{3} e^{x} - 2 \, x^{2} e^{\left (x^{2} + x\right )} \log \relax (3) - 32 \, x^{2} e^{x} \log \relax (3) + 8 \, x^{2} e^{\left (x^{2} + x\right )} + 128 \, x^{2} e^{x} + 16 \, x e^{\left (x^{2} + x\right )} \log \relax (3) + 127 \, x e^{x} \log \relax (3) - 64 \, x e^{\left (x^{2} + x\right )} - 508 \, x e^{x} + e^{\left (x^{2} + x\right )} \log \relax (3) + 8 \, e^{x} \log \relax (3) - 4 \, e^{\left (x^{2} + x\right )} - 32 \, e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x-exp(x))*exp(x^2)+(-8+x)*exp(x)-1)*exp(exp(x))/((-4+log(3))*exp(x^2)^2+((16-2*x)*log(3)+8*x-64)
*exp(x^2)+(x^2-16*x+64)*log(3)-4*x^2+64*x-256),x, algorithm="giac")

[Out]

(2*x^2*e^(x + e^x) - 16*x*e^(x + e^x) - e^(x + e^x))/(2*x^3*e^x*log(3) - 8*x^3*e^x - 2*x^2*e^(x^2 + x)*log(3)
- 32*x^2*e^x*log(3) + 8*x^2*e^(x^2 + x) + 128*x^2*e^x + 16*x*e^(x^2 + x)*log(3) + 127*x*e^x*log(3) - 64*x*e^(x
^2 + x) - 508*x*e^x + e^(x^2 + x)*log(3) + 8*e^x*log(3) - 4*e^(x^2 + x) - 32*e^x)

________________________________________________________________________________________

maple [F]  time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (2 x -{\mathrm e}^{x}\right ) {\mathrm e}^{x^{2}}+\left (-8+x \right ) {\mathrm e}^{x}-1\right ) {\mathrm e}^{{\mathrm e}^{x}}}{\left (-4+\ln \relax (3)\right ) {\mathrm e}^{2 x^{2}}+\left (\left (16-2 x \right ) \ln \relax (3)+8 x -64\right ) {\mathrm e}^{x^{2}}+\left (x^{2}-16 x +64\right ) \ln \relax (3)-4 x^{2}+64 x -256}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x-exp(x))*exp(x^2)+(-8+x)*exp(x)-1)*exp(exp(x))/((-4+ln(3))*exp(x^2)^2+((16-2*x)*ln(3)+8*x-64)*exp(x^2
)+(x^2-16*x+64)*ln(3)-4*x^2+64*x-256),x)

[Out]

int(((2*x-exp(x))*exp(x^2)+(-8+x)*exp(x)-1)*exp(exp(x))/((-4+ln(3))*exp(x^2)^2+((16-2*x)*ln(3)+8*x-64)*exp(x^2
)+(x^2-16*x+64)*ln(3)-4*x^2+64*x-256),x)

________________________________________________________________________________________

maxima [A]  time = 0.52, size = 28, normalized size = 1.08 \begin {gather*} \frac {e^{\left (e^{x}\right )}}{x {\left (\log \relax (3) - 4\right )} - {\left (\log \relax (3) - 4\right )} e^{\left (x^{2}\right )} - 8 \, \log \relax (3) + 32} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x-exp(x))*exp(x^2)+(-8+x)*exp(x)-1)*exp(exp(x))/((-4+log(3))*exp(x^2)^2+((16-2*x)*log(3)+8*x-64)
*exp(x^2)+(x^2-16*x+64)*log(3)-4*x^2+64*x-256),x, algorithm="maxima")

[Out]

e^(e^x)/(x*(log(3) - 4) - (log(3) - 4)*e^(x^2) - 8*log(3) + 32)

________________________________________________________________________________________

mupad [B]  time = 3.60, size = 22, normalized size = 0.85 \begin {gather*} -\frac {{\mathrm {e}}^{{\mathrm {e}}^x}}{\left (\ln \relax (3)-4\right )\,\left ({\mathrm {e}}^{x^2}-x+8\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp(x))*(exp(x^2)*(2*x - exp(x)) + exp(x)*(x - 8) - 1))/(64*x + exp(2*x^2)*(log(3) - 4) - exp(x^2)*(l
og(3)*(2*x - 16) - 8*x + 64) - 4*x^2 + log(3)*(x^2 - 16*x + 64) - 256),x)

[Out]

-exp(exp(x))/((log(3) - 4)*(exp(x^2) - x + 8))

________________________________________________________________________________________

sympy [A]  time = 0.26, size = 34, normalized size = 1.31 \begin {gather*} \frac {e^{e^{x}}}{- 4 x + x \log {\relax (3 )} - e^{x^{2}} \log {\relax (3 )} + 4 e^{x^{2}} - 8 \log {\relax (3 )} + 32} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x-exp(x))*exp(x**2)+(-8+x)*exp(x)-1)*exp(exp(x))/((-4+ln(3))*exp(x**2)**2+((16-2*x)*ln(3)+8*x-64
)*exp(x**2)+(x**2-16*x+64)*ln(3)-4*x**2+64*x-256),x)

[Out]

exp(exp(x))/(-4*x + x*log(3) - exp(x**2)*log(3) + 4*exp(x**2) - 8*log(3) + 32)

________________________________________________________________________________________