3.94.30 \(\int \frac {e^{\frac {1}{16} (1+8 \log (2 \log (\frac {e^5+e^x}{x}))+16 \log ^2(2 \log (\frac {e^5+e^x}{x})))} (-e^5+e^x (-1+x)+(-4 e^5+e^x (-4+4 x)) \log (2 \log (\frac {e^5+e^x}{x})))}{(2 e^5 x+2 e^x x) \log (\frac {e^5+e^x}{x})} \, dx\)

Optimal. Leaf size=25 \[ e^{\left (-\frac {1}{4}-\log \left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )\right )^2} \]

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Rubi [B]  time = 0.93, antiderivative size = 107, normalized size of antiderivative = 4.28, number of steps used = 3, number of rules used = 3, integrand size = 118, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.025, Rules used = {2274, 12, 2288} \begin {gather*} \frac {\sqrt {2} \left (e^x+e^5\right ) \left (e^x (1-x)+e^5\right ) e^{\frac {1}{16} \left (16 \log ^2\left (2 \log \left (\frac {e^x+e^5}{x}\right )\right )+1\right )} \sqrt {\log \left (\frac {e^x+e^5}{x}\right )}}{\left (\frac {e^x+e^5}{x^2}-\frac {e^x}{x}\right ) x \left (e^x x+e^5 x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((1 + 8*Log[2*Log[(E^5 + E^x)/x]] + 16*Log[2*Log[(E^5 + E^x)/x]]^2)/16)*(-E^5 + E^x*(-1 + x) + (-4*E^5
+ E^x*(-4 + 4*x))*Log[2*Log[(E^5 + E^x)/x]]))/((2*E^5*x + 2*E^x*x)*Log[(E^5 + E^x)/x]),x]

[Out]

(Sqrt[2]*E^((1 + 16*Log[2*Log[(E^5 + E^x)/x]]^2)/16)*(E^5 + E^x)*(E^5 + E^x*(1 - x))*Sqrt[Log[(E^5 + E^x)/x]])
/(((E^5 + E^x)/x^2 - E^x/x)*x*(E^5*x + 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 2274

Int[(u_.)*(F_)^((a_.)*(Log[z_]*(b_.) + (v_.))), x_Symbol] :> Int[u*F^(a*v)*z^(a*b*Log[F]), x] /; FreeQ[{F, a,
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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\sqrt {2} e^{\frac {1}{16} \left (1+16 \log ^2\left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )\right )} \left (-e^5+e^x (-1+x)+\left (-4 e^5+e^x (-4+4 x)\right ) \log \left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )\right )}{\left (2 e^5 x+2 e^x x\right ) \sqrt {\log \left (\frac {e^5+e^x}{x}\right )}} \, dx\\ &=\sqrt {2} \int \frac {e^{\frac {1}{16} \left (1+16 \log ^2\left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )\right )} \left (-e^5+e^x (-1+x)+\left (-4 e^5+e^x (-4+4 x)\right ) \log \left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )\right )}{\left (2 e^5 x+2 e^x x\right ) \sqrt {\log \left (\frac {e^5+e^x}{x}\right )}} \, dx\\ &=\frac {\sqrt {2} e^{\frac {1}{16} \left (1+16 \log ^2\left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )\right )} \left (e^5+e^x\right ) \left (e^5+e^x (1-x)\right ) \sqrt {\log \left (\frac {e^5+e^x}{x}\right )}}{\left (\frac {e^5+e^x}{x^2}-\frac {e^x}{x}\right ) x \left (e^5 x+e^x x\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 45, normalized size = 1.80 \begin {gather*} \sqrt {2} e^{\frac {1}{16}+\log ^2\left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )} \sqrt {\log \left (\frac {e^5+e^x}{x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((1 + 8*Log[2*Log[(E^5 + E^x)/x]] + 16*Log[2*Log[(E^5 + E^x)/x]]^2)/16)*(-E^5 + E^x*(-1 + x) + (-
4*E^5 + E^x*(-4 + 4*x))*Log[2*Log[(E^5 + E^x)/x]]))/((2*E^5*x + 2*E^x*x)*Log[(E^5 + E^x)/x]),x]

[Out]

Sqrt[2]*E^(1/16 + Log[2*Log[(E^5 + E^x)/x]]^2)*Sqrt[Log[(E^5 + E^x)/x]]

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fricas [A]  time = 0.90, size = 33, normalized size = 1.32 \begin {gather*} e^{\left (\log \left (2 \, \log \left (\frac {e^{5} + e^{x}}{x}\right )\right )^{2} + \frac {1}{2} \, \log \left (2 \, \log \left (\frac {e^{5} + e^{x}}{x}\right )\right ) + \frac {1}{16}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x-4)*exp(x)-4*exp(5))*log(2*log((exp(5)+exp(x))/x))+(x-1)*exp(x)-exp(5))*exp(log(2*log((exp(5)+
exp(x))/x))^2+1/2*log(2*log((exp(5)+exp(x))/x))+1/16)/(2*exp(x)*x+2*x*exp(5))/log((exp(5)+exp(x))/x),x, algori
thm="fricas")

[Out]

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

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giac [A]  time = 0.26, size = 41, normalized size = 1.64 \begin {gather*} e^{\left (\log \left (2 \, \log \left (\frac {e^{5}}{x} + \frac {e^{x}}{x}\right )\right )^{2} + \frac {1}{2} \, \log \left (2 \, \log \left (\frac {e^{5}}{x} + \frac {e^{x}}{x}\right )\right ) + \frac {1}{16}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x-4)*exp(x)-4*exp(5))*log(2*log((exp(5)+exp(x))/x))+(x-1)*exp(x)-exp(5))*exp(log(2*log((exp(5)+
exp(x))/x))^2+1/2*log(2*log((exp(5)+exp(x))/x))+1/16)/(2*exp(x)*x+2*x*exp(5))/log((exp(5)+exp(x))/x),x, algori
thm="giac")

[Out]

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

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maple [C]  time = 0.26, size = 160, normalized size = 6.40

method	result	size
risch	\(\sqrt {-2 \ln \relax (x )+2 \ln \left ({\mathrm e}^{5}+{\mathrm e}^{x}\right )-i \pi \,\mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{5}+{\mathrm e}^{x}\right )}{x}\right ) \left (-\mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{5}+{\mathrm e}^{x}\right )}{x}\right )+\mathrm {csgn}\left (\frac {i}{x}\right )\right ) \left (-\mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{5}+{\mathrm e}^{x}\right )}{x}\right )+\mathrm {csgn}\left (i \left ({\mathrm e}^{5}+{\mathrm e}^{x}\right )\right )\right )}\, {\mathrm e}^{\ln \left (-2 \ln \relax (x )+2 \ln \left ({\mathrm e}^{5}+{\mathrm e}^{x}\right )-i \pi \,\mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{5}+{\mathrm e}^{x}\right )}{x}\right ) \left (-\mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{5}+{\mathrm e}^{x}\right )}{x}\right )+\mathrm {csgn}\left (\frac {i}{x}\right )\right ) \left (-\mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{5}+{\mathrm e}^{x}\right )}{x}\right )+\mathrm {csgn}\left (i \left ({\mathrm e}^{5}+{\mathrm e}^{x}\right )\right )\right )\right )^{2}+\frac {1}{16}}\)	\(160\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((4*x-4)*exp(x)-4*exp(5))*ln(2*ln((exp(5)+exp(x))/x))+(x-1)*exp(x)-exp(5))*exp(ln(2*ln((exp(5)+exp(x))/x)
)^2+1/2*ln(2*ln((exp(5)+exp(x))/x))+1/16)/(2*exp(x)*x+2*x*exp(5))/ln((exp(5)+exp(x))/x),x,method=_RETURNVERBOS
E)

[Out]

(-2*ln(x)+2*ln(exp(5)+exp(x))-I*Pi*csgn(I*(exp(5)+exp(x))/x)*(-csgn(I*(exp(5)+exp(x))/x)+csgn(I/x))*(-csgn(I*(
exp(5)+exp(x))/x)+csgn(I*(exp(5)+exp(x)))))^(1/2)*exp(ln(-2*ln(x)+2*ln(exp(5)+exp(x))-I*Pi*csgn(I*(exp(5)+exp(
x))/x)*(-csgn(I*(exp(5)+exp(x))/x)+csgn(I/x))*(-csgn(I*(exp(5)+exp(x))/x)+csgn(I*(exp(5)+exp(x)))))^2+1/16)

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maxima [B]  time = 0.76, size = 54, normalized size = 2.16 \begin {gather*} \sqrt {2} \sqrt {-\log \relax (x) + \log \left (e^{5} + e^{x}\right )} e^{\left (\log \relax (2)^{2} + 2 \, \log \relax (2) \log \left (-\log \relax (x) + \log \left (e^{5} + e^{x}\right )\right ) + \log \left (-\log \relax (x) + \log \left (e^{5} + e^{x}\right )\right )^{2} + \frac {1}{16}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x-4)*exp(x)-4*exp(5))*log(2*log((exp(5)+exp(x))/x))+(x-1)*exp(x)-exp(5))*exp(log(2*log((exp(5)+
exp(x))/x))^2+1/2*log(2*log((exp(5)+exp(x))/x))+1/16)/(2*exp(x)*x+2*x*exp(5))/log((exp(5)+exp(x))/x),x, algori
thm="maxima")

[Out]

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

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mupad [B]  time = 8.08, size = 39, normalized size = 1.56 \begin {gather*} {\mathrm {e}}^{1/16}\,{\mathrm {e}}^{{\ln \left (2\,\ln \left (\frac {1}{x}\right )+\ln \left ({\left ({\mathrm {e}}^5+{\mathrm {e}}^x\right )}^2\right )\right )}^2}\,\sqrt {2\,\ln \left (\frac {1}{x}\right )+\ln \left ({\left ({\mathrm {e}}^5+{\mathrm {e}}^x\right )}^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(log(2*log((exp(5) + exp(x))/x))/2 + log(2*log((exp(5) + exp(x))/x))^2 + 1/16)*(exp(5) + log(2*log((e
xp(5) + exp(x))/x))*(4*exp(5) - exp(x)*(4*x - 4)) - exp(x)*(x - 1)))/(log((exp(5) + exp(x))/x)*(2*x*exp(5) + 2
*x*exp(x))),x)

[Out]

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

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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-4)*exp(x)-4*exp(5))*ln(2*ln((exp(5)+exp(x))/x))+(x-1)*exp(x)-exp(5))*exp(ln(2*ln((exp(5)+exp(
x))/x))**2+1/2*ln(2*ln((exp(5)+exp(x))/x))+1/16)/(2*exp(x)*x+2*x*exp(5))/ln((exp(5)+exp(x))/x),x)

[Out]

Timed out

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