3.31.100 \(\int \frac {e^{3+e^{2 x^2}-2 e^{x^2} \log (\frac {5}{3+e^x})+\log ^2(\frac {5}{3+e^x})} (3 e^x+e^{2 x}+2 e^{2 x+x^2}+e^{2 x^2} (12 e^x x+4 e^{2 x} x)+(-2 e^{2 x}+e^{x^2} (-12 e^x x-4 e^{2 x} x)) \log (\frac {5}{3+e^x}))}{12+4 e^x} \, dx\)

Optimal. Leaf size=29 \[ \frac {1}{4} e^{3+x+\left (-e^{x^2}+\log \left (\frac {5}{3+e^x}\right )\right )^2} \]

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Rubi [B]  time = 3.15, antiderivative size = 193, normalized size of antiderivative = 6.66, number of steps used = 1, number of rules used = 1, integrand size = 135, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.007, Rules used = {2288} \begin {gather*} \frac {5^{-2 e^{x^2}} \left (\frac {1}{e^x+3}\right )^{1-2 e^{x^2}} e^{e^{2 x^2}+\log ^2\left (\frac {5}{e^x+3}\right )+3} \left (2 e^{2 x^2} \left (3 e^x x+e^{2 x} x\right )+e^{x^2+2 x}-\left (2 e^{x^2} \left (3 e^x x+e^{2 x} x\right )+e^{2 x}\right ) \log \left (\frac {5}{e^x+3}\right )\right )}{4 \left (2 e^{2 x^2} x+\frac {e^{x^2+x}}{e^x+3}-2 e^{x^2} x \log \left (\frac {5}{e^x+3}\right )-\frac {e^x \log \left (\frac {5}{e^x+3}\right )}{e^x+3}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(3 + E^(2*x^2) - 2*E^x^2*Log[5/(3 + E^x)] + Log[5/(3 + E^x)]^2)*(3*E^x + E^(2*x) + 2*E^(2*x + x^2) + E^
(2*x^2)*(12*E^x*x + 4*E^(2*x)*x) + (-2*E^(2*x) + E^x^2*(-12*E^x*x - 4*E^(2*x)*x))*Log[5/(3 + E^x)]))/(12 + 4*E
^x),x]

[Out]

(E^(3 + E^(2*x^2) + Log[5/(3 + E^x)]^2)*((3 + E^x)^(-1))^(1 - 2*E^x^2)*(E^(2*x + x^2) + 2*E^(2*x^2)*(3*E^x*x +
E^(2*x)*x) - (E^(2*x) + 2*E^x^2*(3*E^x*x + E^(2*x)*x))*Log[5/(3 + E^x)]))/(4*5^(2*E^x^2)*(E^(x + x^2)/(3 + E^
x) + 2*E^(2*x^2)*x - (E^x*Log[5/(3 + E^x)])/(3 + E^x) - 2*E^x^2*x*Log[5/(3 + E^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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {5^{-2 e^{x^2}} e^{3+e^{2 x^2}+\log ^2\left (\frac {5}{3+e^x}\right )} \left (\frac {1}{3+e^x}\right )^{1-2 e^{x^2}} \left (e^{2 x+x^2}+2 e^{2 x^2} \left (3 e^x x+e^{2 x} x\right )-\left (e^{2 x}+2 e^{x^2} \left (3 e^x x+e^{2 x} x\right )\right ) \log \left (\frac {5}{3+e^x}\right )\right )}{4 \left (\frac {e^{x+x^2}}{3+e^x}+2 e^{2 x^2} x-\frac {e^x \log \left (\frac {5}{3+e^x}\right )}{3+e^x}-2 e^{x^2} x \log \left (\frac {5}{3+e^x}\right )\right )}\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 1.55, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{3+e^{2 x^2}-2 e^{x^2} \log \left (\frac {5}{3+e^x}\right )+\log ^2\left (\frac {5}{3+e^x}\right )} \left (3 e^x+e^{2 x}+2 e^{2 x+x^2}+e^{2 x^2} \left (12 e^x x+4 e^{2 x} x\right )+\left (-2 e^{2 x}+e^{x^2} \left (-12 e^x x-4 e^{2 x} x\right )\right ) \log \left (\frac {5}{3+e^x}\right )\right )}{12+4 e^x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(E^(3 + E^(2*x^2) - 2*E^x^2*Log[5/(3 + E^x)] + Log[5/(3 + E^x)]^2)*(3*E^x + E^(2*x) + 2*E^(2*x + x^2
) + E^(2*x^2)*(12*E^x*x + 4*E^(2*x)*x) + (-2*E^(2*x) + E^x^2*(-12*E^x*x - 4*E^(2*x)*x))*Log[5/(3 + E^x)]))/(12
+ 4*E^x),x]

[Out]

Integrate[(E^(3 + E^(2*x^2) - 2*E^x^2*Log[5/(3 + E^x)] + Log[5/(3 + E^x)]^2)*(3*E^x + E^(2*x) + 2*E^(2*x + x^2
) + E^(2*x^2)*(12*E^x*x + 4*E^(2*x)*x) + (-2*E^(2*x) + E^x^2*(-12*E^x*x - 4*E^(2*x)*x))*Log[5/(3 + E^x)]))/(12
+ 4*E^x), x]

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fricas [B]  time = 0.89, size = 62, normalized size = 2.14 \begin {gather*} \frac {1}{4} \, e^{\left ({\left (e^{\left (4 \, x\right )} \log \left (\frac {5}{e^{x} + 3}\right )^{2} - 2 \, e^{\left (x^{2} + 4 \, x\right )} \log \left (\frac {5}{e^{x} + 3}\right ) + e^{\left (2 \, x^{2} + 4 \, x\right )} + 3 \, e^{\left (4 \, x\right )}\right )} e^{\left (-4 \, x\right )} + x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x*exp(x)^2-12*exp(x)*x)*exp(x^2)-2*exp(x)^2)*log(5/(3+exp(x)))+(4*x*exp(x)^2+12*exp(x)*x)*exp(
x^2)^2+2*exp(x)^2*exp(x^2)+exp(x)^2+3*exp(x))*exp(log(5/(3+exp(x)))^2-2*exp(x^2)*log(5/(3+exp(x)))+exp(x^2)^2+
3)/(4*exp(x)+12),x, algorithm="fricas")

[Out]

1/4*e^((e^(4*x)*log(5/(e^x + 3))^2 - 2*e^(x^2 + 4*x)*log(5/(e^x + 3)) + e^(2*x^2 + 4*x) + 3*e^(4*x))*e^(-4*x)
+ x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x*exp(x)^2-12*exp(x)*x)*exp(x^2)-2*exp(x)^2)*log(5/(3+exp(x)))+(4*x*exp(x)^2+12*exp(x)*x)*exp(
x^2)^2+2*exp(x)^2*exp(x^2)+exp(x)^2+3*exp(x))*exp(log(5/(3+exp(x)))^2-2*exp(x^2)*log(5/(3+exp(x)))+exp(x^2)^2+
3)/(4*exp(x)+12),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.36, size = 50, normalized size = 1.72

method	result	size
risch	\(\frac {\left (3+{\mathrm e}^{x}\right )^{-2 \ln \relax (5)} \left (3+{\mathrm e}^{x}\right )^{2 \,{\mathrm e}^{x^{2}}} \left (\frac {1}{25}\right )^{{\mathrm e}^{x^{2}}} {\mathrm e}^{x +3+\ln \relax (5)^{2}+\ln \left (3+{\mathrm e}^{x}\right )^{2}+{\mathrm e}^{2 x^{2}}}}{4}\)	\(50\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-4*x*exp(x)^2-12*exp(x)*x)*exp(x^2)-2*exp(x)^2)*ln(5/(3+exp(x)))+(4*x*exp(x)^2+12*exp(x)*x)*exp(x^2)^2+
2*exp(x)^2*exp(x^2)+exp(x)^2+3*exp(x))*exp(ln(5/(3+exp(x)))^2-2*exp(x^2)*ln(5/(3+exp(x)))+exp(x^2)^2+3)/(4*exp
(x)+12),x,method=_RETURNVERBOSE)

[Out]

1/4*(3+exp(x))^(-2*ln(5))*(3+exp(x))^(2*exp(x^2))*(1/25)^exp(x^2)*exp(x+3+ln(5)^2+ln(3+exp(x))^2+exp(2*x^2))

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maxima [B]  time = 0.67, size = 51, normalized size = 1.76 \begin {gather*} \frac {1}{4} \, e^{\left (-2 \, e^{\left (x^{2}\right )} \log \relax (5) + \log \relax (5)^{2} + 2 \, e^{\left (x^{2}\right )} \log \left (e^{x} + 3\right ) - 2 \, \log \relax (5) \log \left (e^{x} + 3\right ) + \log \left (e^{x} + 3\right )^{2} + x + e^{\left (2 \, x^{2}\right )} + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x*exp(x)^2-12*exp(x)*x)*exp(x^2)-2*exp(x)^2)*log(5/(3+exp(x)))+(4*x*exp(x)^2+12*exp(x)*x)*exp(
x^2)^2+2*exp(x)^2*exp(x^2)+exp(x)^2+3*exp(x))*exp(log(5/(3+exp(x)))^2-2*exp(x^2)*log(5/(3+exp(x)))+exp(x^2)^2+
3)/(4*exp(x)+12),x, algorithm="maxima")

[Out]

1/4*e^(-2*e^(x^2)*log(5) + log(5)^2 + 2*e^(x^2)*log(e^x + 3) - 2*log(5)*log(e^x + 3) + log(e^x + 3)^2 + x + e^
(2*x^2) + 3)

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mupad [B]  time = 1.98, size = 48, normalized size = 1.66 \begin {gather*} \frac {{\left (\frac {1}{25}\right )}^{{\mathrm {e}}^{x^2}}\,{\mathrm {e}}^{{\mathrm {e}}^{2\,x^2}}\,{\mathrm {e}}^{{\ln \relax (5)}^2}\,{\mathrm {e}}^3\,{\mathrm {e}}^{{\ln \left ({\mathrm {e}}^x+3\right )}^2}\,{\mathrm {e}}^x\,{\left ({\mathrm {e}}^x+3\right )}^{2\,{\mathrm {e}}^{x^2}-2\,\ln \relax (5)}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp(2*x^2) - 2*exp(x^2)*log(5/(exp(x) + 3)) + log(5/(exp(x) + 3))^2 + 3)*(exp(2*x) + 3*exp(x) + exp(2
*x^2)*(4*x*exp(2*x) + 12*x*exp(x)) - log(5/(exp(x) + 3))*(2*exp(2*x) + exp(x^2)*(4*x*exp(2*x) + 12*x*exp(x)))
+ 2*exp(2*x)*exp(x^2)))/(4*exp(x) + 12),x)

[Out]

((1/25)^exp(x^2)*exp(exp(2*x^2))*exp(log(5)^2)*exp(3)*exp(log(exp(x) + 3)^2)*exp(x)*(exp(x) + 3)^(2*exp(x^2) -
2*log(5)))/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((((-4*x*exp(x)**2-12*exp(x)*x)*exp(x**2)-2*exp(x)**2)*ln(5/(3+exp(x)))+(4*x*exp(x)**2+12*exp(x)*x)*e
xp(x**2)**2+2*exp(x)**2*exp(x**2)+exp(x)**2+3*exp(x))*exp(ln(5/(3+exp(x)))**2-2*exp(x**2)*ln(5/(3+exp(x)))+exp
(x**2)**2+3)/(4*exp(x)+12),x)

[Out]

Timed out

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