3.16.20 \(\int \frac {e^{\frac {-x^2+8 \log (4)}{x}} (5 x^2 \log (2)+40 \log (2) \log (4))}{x^2+2 e^{\frac {-x^2+8 \log (4)}{x}} x^2+e^{\frac {2 (-x^2+8 \log (4))}{x}} x^2} \, dx\)

Optimal. Leaf size=21 \[ \frac {5 \log (2)}{1+e^{-x+\frac {8 \log (4)}{x}}} \]

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Rubi [A]  time = 1.25, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 79, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {6688, 12, 6711, 32} \begin {gather*} \frac {5 \log (2)}{4^{8/x} e^{-x}+1} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*Log[2])/(1 + 4^(8/x)/E^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 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 6688

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

Rule 6711

Int[(u_)*((a_.)*(v_)^(p_.) + (b_.)*(w_)^(q_.))^(m_.), x_Symbol] :> With[{c = Simplify[u/(p*w*D[v, x] - q*v*D[w
, x])]}, Dist[c*p, Subst[Int[(b + a*x^p)^m, x], x, v*w^(m*q + 1)], x] /; FreeQ[c, x]] /; FreeQ[{a, b, m, p, q}
, x] && EqQ[p + q*(m*p + 1), 0] && IntegerQ[p] && IntegerQ[m]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5\ 4^{8/x} e^x \log (2) \left (x^2+8 \log (4)\right )}{\left (4^{8/x}+e^x\right )^2 x^2} \, dx\\ &=(5 \log (2)) \int \frac {4^{8/x} e^x \left (x^2+8 \log (4)\right )}{\left (4^{8/x}+e^x\right )^2 x^2} \, dx\\ &=-\left ((5 \log (2)) \operatorname {Subst}\left (\int \frac {1}{(1+x)^2} \, dx,x,4^{8/x} e^{-x}\right )\right )\\ &=\frac {5 \log (2)}{1+4^{8/x} e^{-x}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.29, size = 24, normalized size = 1.14 \begin {gather*} -\frac {5\ 4^{8/x} \log (2)}{4^{8/x}+e^x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*4^(8/x)*Log[2])/(4^(8/x) + E^x)

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fricas [A]  time = 0.89, size = 22, normalized size = 1.05 \begin {gather*} \frac {5 \, \log \relax (2)}{e^{\left (-\frac {x^{2} - 16 \, \log \relax (2)}{x}\right )} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((80*log(2)^2+5*x^2*log(2))*exp((16*log(2)-x^2)/x)/(x^2*exp((16*log(2)-x^2)/x)^2+2*x^2*exp((16*log(2)
-x^2)/x)+x^2),x, algorithm="fricas")

[Out]

5*log(2)/(e^(-(x^2 - 16*log(2))/x) + 1)

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giac [A]  time = 0.18, size = 22, normalized size = 1.05 \begin {gather*} \frac {5 \, \log \relax (2)}{e^{\left (-\frac {x^{2} - 16 \, \log \relax (2)}{x}\right )} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((80*log(2)^2+5*x^2*log(2))*exp((16*log(2)-x^2)/x)/(x^2*exp((16*log(2)-x^2)/x)^2+2*x^2*exp((16*log(2)
-x^2)/x)+x^2),x, algorithm="giac")

[Out]

5*log(2)/(e^(-(x^2 - 16*log(2))/x) + 1)

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maple [A]  time = 0.07, size = 19, normalized size = 0.90

method	result	size
risch	\(\frac {5 \ln \relax (2)}{65536^{\frac {1}{x}} {\mathrm e}^{-x}+1}\)	\(19\)
norman	\(\frac {5 \ln \relax (2)}{{\mathrm e}^{\frac {16 \ln \relax (2)-x^{2}}{x}}+1}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((80*ln(2)^2+5*x^2*ln(2))*exp((16*ln(2)-x^2)/x)/(x^2*exp((16*ln(2)-x^2)/x)^2+2*x^2*exp((16*ln(2)-x^2)/x)+x^
2),x,method=_RETURNVERBOSE)

[Out]

5*ln(2)/(65536^(1/x)*exp(-x)+1)

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maxima [A]  time = 0.99, size = 18, normalized size = 0.86 \begin {gather*} \frac {5 \, e^{x} \log \relax (2)}{2^{\frac {16}{x}} + e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((80*log(2)^2+5*x^2*log(2))*exp((16*log(2)-x^2)/x)/(x^2*exp((16*log(2)-x^2)/x)^2+2*x^2*exp((16*log(2)
-x^2)/x)+x^2),x, algorithm="maxima")

[Out]

5*e^x*log(2)/(2^(16/x) + e^x)

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mupad [B]  time = 1.12, size = 18, normalized size = 0.86 \begin {gather*} \frac {5\,{\mathrm {e}}^x\,\ln \relax (2)}{{\mathrm {e}}^x+2^{16/x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((16*log(2) - x^2)/x)*(5*x^2*log(2) + 80*log(2)^2))/(2*x^2*exp((16*log(2) - x^2)/x) + x^2*exp((2*(16*l
og(2) - x^2))/x) + x^2),x)

[Out]

(5*exp(x)*log(2))/(exp(x) + 2^(16/x))

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sympy [A]  time = 0.14, size = 17, normalized size = 0.81 \begin {gather*} \frac {5 \log {\relax (2 )}}{e^{\frac {- x^{2} + 16 \log {\relax (2 )}}{x}} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((80*ln(2)**2+5*x**2*ln(2))*exp((16*ln(2)-x**2)/x)/(x**2*exp((16*ln(2)-x**2)/x)**2+2*x**2*exp((16*ln(
2)-x**2)/x)+x**2),x)

[Out]

5*log(2)/(exp((-x**2 + 16*log(2))/x) + 1)

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