3.3.86 \(\int \frac {e^{\frac {5+(-2-36 x^2-e^{2 x} x^2-24 x^3-4 x^4+e^x (12 x^2+4 x^3)) \log (\frac {1}{2} (4+e^x))}{\log (\frac {1}{2} (4+e^x))}} (-5 e^x+(-288 x-288 x^2-64 x^3+e^{3 x} (-2 x-2 x^2)+e^x (24 x+24 x^2)+e^{2 x} (16 x+16 x^2+4 x^3)) \log ^2(\frac {1}{2} (4+e^x)))}{(4+e^x) \log ^2(\frac {1}{2} (4+e^x))} \, dx\)

Optimal. Leaf size=35 \[ e^{-2-x^2 \left (6-e^x+2 x\right )^2+\frac {5}{\log \left (\frac {1}{2} \left (4+e^x\right )\right )}} \]

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Rubi [A]  time = 6.59, antiderivative size = 49, normalized size of antiderivative = 1.40, number of steps used = 2, number of rules used = 2, integrand size = 171, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.012, Rules used = {6688, 6706} \begin {gather*} \exp \left (-4 x^4-4 \left (6-e^x\right ) x^3-\left (6-e^x\right )^2 x^2+\frac {5}{\log \left (\frac {1}{2} \left (e^x+4\right )\right )}-2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((5 + (-2 - 36*x^2 - E^(2*x)*x^2 - 24*x^3 - 4*x^4 + E^x*(12*x^2 + 4*x^3))*Log[(4 + E^x)/2])/Log[(4 + E^
x)/2])*(-5*E^x + (-288*x - 288*x^2 - 64*x^3 + E^(3*x)*(-2*x - 2*x^2) + E^x*(24*x + 24*x^2) + E^(2*x)*(16*x + 1
6*x^2 + 4*x^3))*Log[(4 + E^x)/2]^2))/((4 + E^x)*Log[(4 + E^x)/2]^2),x]

[Out]

E^(-2 - (6 - E^x)^2*x^2 - 4*(6 - E^x)*x^3 - 4*x^4 + 5/Log[(4 + E^x)/2])

Rule 6688

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

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (-2-\left (-6+e^x\right )^2 x^2+4 \left (-6+e^x\right ) x^3-4 x^4+\frac {5}{\log \left (\frac {1}{2} \left (4+e^x\right )\right )}\right ) \left (-5 e^x-2 \left (4+e^x\right ) x \left (e^{2 x} (1+x)-2 e^x \left (6+6 x+x^2\right )+4 \left (9+9 x+2 x^2\right )\right ) \log ^2\left (\frac {1}{2} \left (4+e^x\right )\right )\right )}{\left (4+e^x\right ) \log ^2\left (\frac {1}{2} \left (4+e^x\right )\right )} \, dx\\ &=\exp \left (-2-\left (6-e^x\right )^2 x^2-4 \left (6-e^x\right ) x^3-4 x^4+\frac {5}{\log \left (\frac {1}{2} \left (4+e^x\right )\right )}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.39, size = 45, normalized size = 1.29 \begin {gather*} e^{-2-\left (-6+e^x\right )^2 x^2+4 \left (-6+e^x\right ) x^3-4 x^4+\frac {5}{\log \left (\frac {1}{2} \left (4+e^x\right )\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((5 + (-2 - 36*x^2 - E^(2*x)*x^2 - 24*x^3 - 4*x^4 + E^x*(12*x^2 + 4*x^3))*Log[(4 + E^x)/2])/Log[(
4 + E^x)/2])*(-5*E^x + (-288*x - 288*x^2 - 64*x^3 + E^(3*x)*(-2*x - 2*x^2) + E^x*(24*x + 24*x^2) + E^(2*x)*(16
*x + 16*x^2 + 4*x^3))*Log[(4 + E^x)/2]^2))/((4 + E^x)*Log[(4 + E^x)/2]^2),x]

[Out]

E^(-2 - (-6 + E^x)^2*x^2 + 4*(-6 + E^x)*x^3 - 4*x^4 + 5/Log[(4 + E^x)/2])

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fricas [A]  time = 1.38, size = 60, normalized size = 1.71 \begin {gather*} e^{\left (-\frac {{\left (4 \, x^{4} + 24 \, x^{3} + x^{2} e^{\left (2 \, x\right )} + 36 \, x^{2} - 4 \, {\left (x^{3} + 3 \, x^{2}\right )} e^{x} + 2\right )} \log \left (\frac {1}{2} \, e^{x} + 2\right ) - 5}{\log \left (\frac {1}{2} \, e^{x} + 2\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2-2*x)*exp(x)^3+(4*x^3+16*x^2+16*x)*exp(x)^2+(24*x^2+24*x)*exp(x)-64*x^3-288*x^2-288*x)*log(
1/2*exp(x)+2)^2-5*exp(x))*exp(((-exp(x)^2*x^2+(4*x^3+12*x^2)*exp(x)-4*x^4-24*x^3-36*x^2-2)*log(1/2*exp(x)+2)+5
)/log(1/2*exp(x)+2))/(exp(x)+4)/log(1/2*exp(x)+2)^2,x, algorithm="fricas")

[Out]

e^(-((4*x^4 + 24*x^3 + x^2*e^(2*x) + 36*x^2 - 4*(x^3 + 3*x^2)*e^x + 2)*log(1/2*e^x + 2) - 5)/log(1/2*e^x + 2))

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2-2*x)*exp(x)^3+(4*x^3+16*x^2+16*x)*exp(x)^2+(24*x^2+24*x)*exp(x)-64*x^3-288*x^2-288*x)*log(
1/2*exp(x)+2)^2-5*exp(x))*exp(((-exp(x)^2*x^2+(4*x^3+12*x^2)*exp(x)-4*x^4-24*x^3-36*x^2-2)*log(1/2*exp(x)+2)+5
)/log(1/2*exp(x)+2))/(exp(x)+4)/log(1/2*exp(x)+2)^2,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Evaluation time: 0.78Unable to divide, perhaps due to rounding error%%%{-8000,[0,3,3]%%%}+%%%{-36000,[0,3,2
]%%%}+%%%{-

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maple [B]  time = 0.12, size = 103, normalized size = 2.94

method	result	size
risch	\({\mathrm e}^{\frac {4 \,{\mathrm e}^{x} \ln \left (\frac {{\mathrm e}^{x}}{2}+2\right ) x^{3}-4 \ln \left (\frac {{\mathrm e}^{x}}{2}+2\right ) x^{4}-{\mathrm e}^{2 x} \ln \left (\frac {{\mathrm e}^{x}}{2}+2\right ) x^{2}+12 \,{\mathrm e}^{x} \ln \left (\frac {{\mathrm e}^{x}}{2}+2\right ) x^{2}-24 \ln \left (\frac {{\mathrm e}^{x}}{2}+2\right ) x^{3}-36 \ln \left (\frac {{\mathrm e}^{x}}{2}+2\right ) x^{2}-2 \ln \left (\frac {{\mathrm e}^{x}}{2}+2\right )+5}{\ln \left (\frac {{\mathrm e}^{x}}{2}+2\right )}}\)	\(103\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-2*x^2-2*x)*exp(x)^3+(4*x^3+16*x^2+16*x)*exp(x)^2+(24*x^2+24*x)*exp(x)-64*x^3-288*x^2-288*x)*ln(1/2*exp
(x)+2)^2-5*exp(x))*exp(((-exp(x)^2*x^2+(4*x^3+12*x^2)*exp(x)-4*x^4-24*x^3-36*x^2-2)*ln(1/2*exp(x)+2)+5)/ln(1/2
*exp(x)+2))/(exp(x)+4)/ln(1/2*exp(x)+2)^2,x,method=_RETURNVERBOSE)

[Out]

exp((4*exp(x)*ln(1/2*exp(x)+2)*x^3-4*ln(1/2*exp(x)+2)*x^4-exp(2*x)*ln(1/2*exp(x)+2)*x^2+12*exp(x)*ln(1/2*exp(x
)+2)*x^2-24*ln(1/2*exp(x)+2)*x^3-36*ln(1/2*exp(x)+2)*x^2-2*ln(1/2*exp(x)+2)+5)/ln(1/2*exp(x)+2))

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maxima [A]  time = 0.98, size = 55, normalized size = 1.57 \begin {gather*} e^{\left (-4 \, x^{4} + 4 \, x^{3} e^{x} - 24 \, x^{3} - x^{2} e^{\left (2 \, x\right )} + 12 \, x^{2} e^{x} - 36 \, x^{2} - \frac {5}{\log \relax (2) - \log \left (e^{x} + 4\right )} - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2-2*x)*exp(x)^3+(4*x^3+16*x^2+16*x)*exp(x)^2+(24*x^2+24*x)*exp(x)-64*x^3-288*x^2-288*x)*log(
1/2*exp(x)+2)^2-5*exp(x))*exp(((-exp(x)^2*x^2+(4*x^3+12*x^2)*exp(x)-4*x^4-24*x^3-36*x^2-2)*log(1/2*exp(x)+2)+5
)/log(1/2*exp(x)+2))/(exp(x)+4)/log(1/2*exp(x)+2)^2,x, algorithm="maxima")

[Out]

e^(-4*x^4 + 4*x^3*e^x - 24*x^3 - x^2*e^(2*x) + 12*x^2*e^x - 36*x^2 - 5/(log(2) - log(e^x + 4)) - 2)

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mupad [B]  time = 0.60, size = 59, normalized size = 1.69 \begin {gather*} {\mathrm {e}}^{\frac {5}{\ln \left (\frac {{\mathrm {e}}^x}{2}+2\right )}}\,{\mathrm {e}}^{-2}\,{\mathrm {e}}^{4\,x^3\,{\mathrm {e}}^x}\,{\mathrm {e}}^{12\,x^2\,{\mathrm {e}}^x}\,{\mathrm {e}}^{-4\,x^4}\,{\mathrm {e}}^{-24\,x^3}\,{\mathrm {e}}^{-36\,x^2}\,{\mathrm {e}}^{-x^2\,{\mathrm {e}}^{2\,x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(log(exp(x)/2 + 2)*(x^2*exp(2*x) - exp(x)*(12*x^2 + 4*x^3) + 36*x^2 + 24*x^3 + 4*x^4 + 2) - 5)/log(
exp(x)/2 + 2))*(5*exp(x) + log(exp(x)/2 + 2)^2*(288*x + exp(3*x)*(2*x + 2*x^2) - exp(2*x)*(16*x + 16*x^2 + 4*x
^3) - exp(x)*(24*x + 24*x^2) + 288*x^2 + 64*x^3)))/(log(exp(x)/2 + 2)^2*(exp(x) + 4)),x)

[Out]

exp(5/log(exp(x)/2 + 2))*exp(-2)*exp(4*x^3*exp(x))*exp(12*x^2*exp(x))*exp(-4*x^4)*exp(-24*x^3)*exp(-36*x^2)*ex
p(-x^2*exp(2*x))

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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*x**2-2*x)*exp(x)**3+(4*x**3+16*x**2+16*x)*exp(x)**2+(24*x**2+24*x)*exp(x)-64*x**3-288*x**2-288
*x)*ln(1/2*exp(x)+2)**2-5*exp(x))*exp(((-exp(x)**2*x**2+(4*x**3+12*x**2)*exp(x)-4*x**4-24*x**3-36*x**2-2)*ln(1
/2*exp(x)+2)+5)/ln(1/2*exp(x)+2))/(exp(x)+4)/ln(1/2*exp(x)+2)**2,x)

[Out]

Timed out

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