3.62.64 \(\int \frac {e^x+e^x \log (\frac {x}{2})-18 \log (\frac {x}{2}) \log (\frac {x \log (\frac {x}{2})}{-4+\log (2)})+e^x (-1+x) \log (\frac {x}{2}) \log (\frac {x \log (\frac {x}{2})}{-4+\log (2)}) \log (\log (\frac {x \log (\frac {x}{2})}{-4+\log (2)}))}{6 x^2 \log (\frac {x}{2}) \log (\frac {x \log (\frac {x}{2})}{-4+\log (2)})} \, dx\)

Optimal. Leaf size=32 \[ \frac {6+\frac {1}{3} e^x \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right )}{2 x} \]

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Rubi [A]  time = 1.51, antiderivative size = 35, normalized size of antiderivative = 1.09, number of steps used = 5, number of rules used = 4, integrand size = 113, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.035, Rules used = {12, 6688, 6742, 2288} \begin {gather*} \frac {3}{x}+\frac {e^x \log \left (\log \left (-\frac {x \log \left (\frac {x}{2}\right )}{4-\log (2)}\right )\right )}{6 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x + E^x*Log[x/2] - 18*Log[x/2]*Log[(x*Log[x/2])/(-4 + Log[2])] + E^x*(-1 + x)*Log[x/2]*Log[(x*Log[x/2])
/(-4 + Log[2])]*Log[Log[(x*Log[x/2])/(-4 + Log[2])]])/(6*x^2*Log[x/2]*Log[(x*Log[x/2])/(-4 + Log[2])]),x]

[Out]

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{6} \int \frac {e^x+e^x \log \left (\frac {x}{2}\right )-18 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )+e^x (-1+x) \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right ) \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right )}{x^2 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )} \, dx\\ &=\frac {1}{6} \int \frac {e^x+\log \left (\frac {x}{2}\right ) \left (e^x+\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right ) \left (-18+e^x (-1+x) \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right )\right )\right )}{x^2 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )} \, dx\\ &=\frac {1}{6} \int \left (-\frac {18}{x^2}+\frac {e^x \left (1+\log \left (\frac {x}{2}\right )-\log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right ) \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right )+x \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right ) \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right )\right )}{x^2 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )}\right ) \, dx\\ &=\frac {3}{x}+\frac {1}{6} \int \frac {e^x \left (1+\log \left (\frac {x}{2}\right )-\log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right ) \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right )+x \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right ) \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right )\right )}{x^2 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )} \, dx\\ &=\frac {3}{x}+\frac {e^x \log \left (\log \left (-\frac {x \log \left (\frac {x}{2}\right )}{4-\log (2)}\right )\right )}{6 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.27, size = 29, normalized size = 0.91 \begin {gather*} \frac {18+e^x \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right )}{6 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x + E^x*Log[x/2] - 18*Log[x/2]*Log[(x*Log[x/2])/(-4 + Log[2])] + E^x*(-1 + x)*Log[x/2]*Log[(x*Log
[x/2])/(-4 + Log[2])]*Log[Log[(x*Log[x/2])/(-4 + Log[2])]])/(6*x^2*Log[x/2]*Log[(x*Log[x/2])/(-4 + Log[2])]),x
]

[Out]

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

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fricas [A]  time = 0.58, size = 24, normalized size = 0.75 \begin {gather*} \frac {e^{x} \log \left (\log \left (\frac {x \log \left (\frac {1}{2} \, x\right )}{\log \relax (2) - 4}\right )\right ) + 18}{6 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*((x-1)*exp(x)*log(1/2*x)*log(x*log(1/2*x)/(log(2)-4))*log(log(x*log(1/2*x)/(log(2)-4)))-18*log(1
/2*x)*log(x*log(1/2*x)/(log(2)-4))+exp(x)*log(1/2*x)+exp(x))/x^2/log(1/2*x)/log(x*log(1/2*x)/(log(2)-4)),x, al
gorithm="fricas")

[Out]

1/6*(e^x*log(log(x*log(1/2*x)/(log(2) - 4))) + 18)/x

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giac [A]  time = 1.01, size = 29, normalized size = 0.91 \begin {gather*} \frac {e^{x} \log \left (\log \relax (x) - \log \left (\log \relax (2) - 4\right ) + \log \left (-\log \relax (2) + \log \relax (x)\right )\right ) + 18}{6 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*((x-1)*exp(x)*log(1/2*x)*log(x*log(1/2*x)/(log(2)-4))*log(log(x*log(1/2*x)/(log(2)-4)))-18*log(1
/2*x)*log(x*log(1/2*x)/(log(2)-4))+exp(x)*log(1/2*x)+exp(x))/x^2/log(1/2*x)/log(x*log(1/2*x)/(log(2)-4)),x, al
gorithm="giac")

[Out]

1/6*(e^x*log(log(x) - log(log(2) - 4) + log(-log(2) + log(x))) + 18)/x

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {\left (x -1\right ) {\mathrm e}^{x} \ln \left (\frac {x}{2}\right ) \ln \left (\frac {x \ln \left (\frac {x}{2}\right )}{\ln \relax (2)-4}\right ) \ln \left (\ln \left (\frac {x \ln \left (\frac {x}{2}\right )}{\ln \relax (2)-4}\right )\right )-18 \ln \left (\frac {x}{2}\right ) \ln \left (\frac {x \ln \left (\frac {x}{2}\right )}{\ln \relax (2)-4}\right )+{\mathrm e}^{x} \ln \left (\frac {x}{2}\right )+{\mathrm e}^{x}}{6 x^{2} \ln \left (\frac {x}{2}\right ) \ln \left (\frac {x \ln \left (\frac {x}{2}\right )}{\ln \relax (2)-4}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/6*((x-1)*exp(x)*ln(1/2*x)*ln(x*ln(1/2*x)/(ln(2)-4))*ln(ln(x*ln(1/2*x)/(ln(2)-4)))-18*ln(1/2*x)*ln(x*ln(1
/2*x)/(ln(2)-4))+exp(x)*ln(1/2*x)+exp(x))/x^2/ln(1/2*x)/ln(x*ln(1/2*x)/(ln(2)-4)),x)

[Out]

int(1/6*((x-1)*exp(x)*ln(1/2*x)*ln(x*ln(1/2*x)/(ln(2)-4))*ln(ln(x*ln(1/2*x)/(ln(2)-4)))-18*ln(1/2*x)*ln(x*ln(1
/2*x)/(ln(2)-4))+exp(x)*ln(1/2*x)+exp(x))/x^2/ln(1/2*x)/ln(x*ln(1/2*x)/(ln(2)-4)),x)

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maxima [A]  time = 0.48, size = 32, normalized size = 1.00 \begin {gather*} \frac {e^{x} \log \left (\log \relax (x) - \log \left (\log \relax (2) - 4\right ) + \log \left (-\log \relax (2) + \log \relax (x)\right )\right )}{6 \, x} + \frac {3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*((x-1)*exp(x)*log(1/2*x)*log(x*log(1/2*x)/(log(2)-4))*log(log(x*log(1/2*x)/(log(2)-4)))-18*log(1
/2*x)*log(x*log(1/2*x)/(log(2)-4))+exp(x)*log(1/2*x)+exp(x))/x^2/log(1/2*x)/log(x*log(1/2*x)/(log(2)-4)),x, al
gorithm="maxima")

[Out]

1/6*e^x*log(log(x) - log(log(2) - 4) + log(-log(2) + log(x)))/x + 3/x

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {\frac {{\mathrm {e}}^x}{6}-3\,\ln \left (\frac {x}{2}\right )\,\ln \left (\frac {x\,\ln \left (\frac {x}{2}\right )}{\ln \relax (2)-4}\right )+\frac {\ln \left (\frac {x}{2}\right )\,{\mathrm {e}}^x}{6}+\frac {\ln \left (\ln \left (\frac {x\,\ln \left (\frac {x}{2}\right )}{\ln \relax (2)-4}\right )\right )\,\ln \left (\frac {x}{2}\right )\,{\mathrm {e}}^x\,\ln \left (\frac {x\,\ln \left (\frac {x}{2}\right )}{\ln \relax (2)-4}\right )\,\left (x-1\right )}{6}}{x^2\,\ln \left (\frac {x}{2}\right )\,\ln \left (\frac {x\,\ln \left (\frac {x}{2}\right )}{\ln \relax (2)-4}\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)/6 - 3*log(x/2)*log((x*log(x/2))/(log(2) - 4)) + (log(x/2)*exp(x))/6 + (log(log((x*log(x/2))/(log(2
) - 4)))*log(x/2)*exp(x)*log((x*log(x/2))/(log(2) - 4))*(x - 1))/6)/(x^2*log(x/2)*log((x*log(x/2))/(log(2) - 4
))),x)

[Out]

int((exp(x)/6 - 3*log(x/2)*log((x*log(x/2))/(log(2) - 4)) + (log(x/2)*exp(x))/6 + (log(log((x*log(x/2))/(log(2
) - 4)))*log(x/2)*exp(x)*log((x*log(x/2))/(log(2) - 4))*(x - 1))/6)/(x^2*log(x/2)*log((x*log(x/2))/(log(2) - 4
))), x)

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sympy [A]  time = 0.75, size = 24, normalized size = 0.75 \begin {gather*} \frac {e^{x} \log {\left (\log {\left (\frac {x \log {\left (\frac {x}{2} \right )}}{-4 + \log {\relax (2 )}} \right )} \right )}}{6 x} + \frac {3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*((x-1)*exp(x)*ln(1/2*x)*ln(x*ln(1/2*x)/(ln(2)-4))*ln(ln(x*ln(1/2*x)/(ln(2)-4)))-18*ln(1/2*x)*ln(
x*ln(1/2*x)/(ln(2)-4))+exp(x)*ln(1/2*x)+exp(x))/x**2/ln(1/2*x)/ln(x*ln(1/2*x)/(ln(2)-4)),x)

[Out]

exp(x)*log(log(x*log(x/2)/(-4 + log(2))))/(6*x) + 3/x

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