3.5.58 \(\int \frac {e^{-1+e^{x^2}} (-10+e^{x^2} (40 x-20 x^2+e^x (8 x-4 x^2)))+e^{-1+e^{x^2}+x^2} (10 x+2 e^x x) \log (25+10 e^x+e^{2 x})}{5+e^x} \, dx\)

Optimal. Leaf size=23 \[ e^{-1+e^{x^2}} \left (4-2 x+\log \left (\left (5+e^x\right )^2\right )\right ) \]

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Rubi [B]  time = 0.62, antiderivative size = 74, normalized size of antiderivative = 3.22, number of steps used = 3, number of rules used = 3, integrand size = 85, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.035, Rules used = {6688, 12, 2288} \begin {gather*} \frac {e^{-x^2+e^{x^2}-1} \left (10 e^{x^2} (2-x) x+2 e^{x^2+x} (2-x) x+e^{x^2} \left (e^x+5\right ) x \log \left (\left (e^x+5\right )^2\right )\right )}{\left (e^x+5\right ) x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-1 + E^x^2)*(-10 + E^x^2*(40*x - 20*x^2 + E^x*(8*x - 4*x^2))) + E^(-1 + E^x^2 + x^2)*(10*x + 2*E^x*x)*
Log[25 + 10*E^x + E^(2*x)])/(5 + E^x),x]

[Out]

(E^(-1 + E^x^2 - x^2)*(10*E^x^2*(2 - x)*x + 2*E^(x + x^2)*(2 - x)*x + E^x^2*(5 + E^x)*x*Log[(5 + E^x)^2]))/((5
+ E^x)*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 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^{-1+e^{x^2}} \left (-5-10 e^{x^2} (-2+x) x-2 e^{x+x^2} (-2+x) x+e^{x^2} \left (5+e^x\right ) x \log \left (\left (5+e^x\right )^2\right )\right )}{5+e^x} \, dx\\ &=2 \int \frac {e^{-1+e^{x^2}} \left (-5-10 e^{x^2} (-2+x) x-2 e^{x+x^2} (-2+x) x+e^{x^2} \left (5+e^x\right ) x \log \left (\left (5+e^x\right )^2\right )\right )}{5+e^x} \, dx\\ &=\frac {e^{-1+e^{x^2}-x^2} \left (10 e^{x^2} (2-x) x+2 e^{x+x^2} (2-x) x+e^{x^2} \left (5+e^x\right ) x \log \left (\left (5+e^x\right )^2\right )\right )}{\left (5+e^x\right ) x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.31, size = 23, normalized size = 1.00 \begin {gather*} e^{-1+e^{x^2}} \left (4-2 x+\log \left (\left (5+e^x\right )^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-1 + E^x^2)*(-10 + E^x^2*(40*x - 20*x^2 + E^x*(8*x - 4*x^2))) + E^(-1 + E^x^2 + x^2)*(10*x + 2*E
^x*x)*Log[25 + 10*E^x + E^(2*x)])/(5 + E^x),x]

[Out]

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

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fricas [B]  time = 0.59, size = 47, normalized size = 2.04 \begin {gather*} -{\left (2 \, {\left (x - 2\right )} e^{\left (x^{2} + e^{\left (x^{2}\right )} - 1\right )} - e^{\left (x^{2} + e^{\left (x^{2}\right )} - 1\right )} \log \left (e^{\left (2 \, x\right )} + 10 \, e^{x} + 25\right )\right )} e^{\left (-x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)*x+10*x)*exp(x^2)*exp(exp(x^2)-1)*log(exp(x)^2+10*exp(x)+25)+(((-4*x^2+8*x)*exp(x)-20*x^2+
40*x)*exp(x^2)-10)*exp(exp(x^2)-1))/(exp(x)+5),x, algorithm="fricas")

[Out]

-(2*(x - 2)*e^(x^2 + e^(x^2) - 1) - e^(x^2 + e^(x^2) - 1)*log(e^(2*x) + 10*e^x + 25))*e^(-x^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, {\left ({\left (x e^{x} + 5 \, x\right )} e^{\left (x^{2} + e^{\left (x^{2}\right )} - 1\right )} \log \left (e^{\left (2 \, x\right )} + 10 \, e^{x} + 25\right ) - {\left (2 \, {\left (5 \, x^{2} + {\left (x^{2} - 2 \, x\right )} e^{x} - 10 \, x\right )} e^{\left (x^{2}\right )} + 5\right )} e^{\left (e^{\left (x^{2}\right )} - 1\right )}\right )}}{e^{x} + 5}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)*x+10*x)*exp(x^2)*exp(exp(x^2)-1)*log(exp(x)^2+10*exp(x)+25)+(((-4*x^2+8*x)*exp(x)-20*x^2+
40*x)*exp(x^2)-10)*exp(exp(x^2)-1))/(exp(x)+5),x, algorithm="giac")

[Out]

integrate(2*((x*e^x + 5*x)*e^(x^2 + e^(x^2) - 1)*log(e^(2*x) + 10*e^x + 25) - (2*(5*x^2 + (x^2 - 2*x)*e^x - 10
*x)*e^(x^2) + 5)*e^(e^(x^2) - 1))/(e^x + 5), x)

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maple [C]  time = 0.13, size = 94, normalized size = 4.09

method	result	size
risch	\(2 \,{\mathrm e}^{{\mathrm e}^{x^{2}}-1} \ln \left ({\mathrm e}^{x}+5\right )-\frac {\left (i \pi \mathrm {csgn}\left (i \left ({\mathrm e}^{x}+5\right )\right )^{2} \mathrm {csgn}\left (i \left ({\mathrm e}^{x}+5\right )^{2}\right )-2 i \pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{x}+5\right )\right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{x}+5\right )^{2}\right )^{2}+i \pi \mathrm {csgn}\left (i \left ({\mathrm e}^{x}+5\right )^{2}\right )^{3}+4 x -8\right ) {\mathrm e}^{{\mathrm e}^{x^{2}}-1}}{2}\)	\(94\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*exp(x)*x+10*x)*exp(x^2)*exp(exp(x^2)-1)*ln(exp(x)^2+10*exp(x)+25)+(((-4*x^2+8*x)*exp(x)-20*x^2+40*x)*e
xp(x^2)-10)*exp(exp(x^2)-1))/(exp(x)+5),x,method=_RETURNVERBOSE)

[Out]

2*exp(exp(x^2)-1)*ln(exp(x)+5)-1/2*(I*Pi*csgn(I*(exp(x)+5))^2*csgn(I*(exp(x)+5)^2)-2*I*Pi*csgn(I*(exp(x)+5))*c
sgn(I*(exp(x)+5)^2)^2+I*Pi*csgn(I*(exp(x)+5)^2)^3+4*x-8)*exp(exp(x^2)-1)

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maxima [A]  time = 0.50, size = 19, normalized size = 0.83 \begin {gather*} -2 \, {\left (x - \log \left (e^{x} + 5\right ) - 2\right )} e^{\left (e^{\left (x^{2}\right )} - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)*x+10*x)*exp(x^2)*exp(exp(x^2)-1)*log(exp(x)^2+10*exp(x)+25)+(((-4*x^2+8*x)*exp(x)-20*x^2+
40*x)*exp(x^2)-10)*exp(exp(x^2)-1))/(exp(x)+5),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.66, size = 24, normalized size = 1.04 \begin {gather*} {\mathrm {e}}^{{\mathrm {e}}^{x^2}-1}\,\left (\ln \left ({\mathrm {e}}^{2\,x}+10\,{\mathrm {e}}^x+25\right )-2\,x+4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp(x^2) - 1)*(exp(x^2)*(40*x + exp(x)*(8*x - 4*x^2) - 20*x^2) - 10) + log(exp(2*x) + 10*exp(x) + 25)
*exp(exp(x^2) - 1)*exp(x^2)*(10*x + 2*x*exp(x)))/(exp(x) + 5),x)

[Out]

exp(exp(x^2) - 1)*(log(exp(2*x) + 10*exp(x) + 25) - 2*x + 4)

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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*exp(x)*x+10*x)*exp(x**2)*exp(exp(x**2)-1)*ln(exp(x)**2+10*exp(x)+25)+(((-4*x**2+8*x)*exp(x)-20*x
**2+40*x)*exp(x**2)-10)*exp(exp(x**2)-1))/(exp(x)+5),x)

[Out]

Timed out

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