3.61.61 \(\int \frac {-40 \log (2)+(x+e^x x) \log ^2(x)}{(40 x \log (2) \log (x)+(x+e^x x+x^2-8 x \log (2)) \log ^2(x)) \log (\frac {40 \log (2)+(1+e^x+x-8 \log (2)) \log (x)}{\log (x)})} \, dx\)

Optimal. Leaf size=20 \[ \log \left (\log \left (1+e^x+x-\frac {8 \log (2) (-5+\log (x))}{\log (x)}\right )\right ) \]

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Rubi [F]  time = 7.67, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-40 \log (2)+\left (x+e^x x\right ) \log ^2(x)}{\left (40 x \log (2) \log (x)+\left (x+e^x x+x^2-8 x \log (2)\right ) \log ^2(x)\right ) \log \left (\frac {40 \log (2)+\left (1+e^x+x-8 \log (2)\right ) \log (x)}{\log (x)}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-40*Log[2] + (x + E^x*x)*Log[x]^2)/((40*x*Log[2]*Log[x] + (x + E^x*x + x^2 - 8*x*Log[2])*Log[x]^2)*Log[(4
0*Log[2] + (1 + E^x + x - 8*Log[2])*Log[x])/Log[x]]),x]

[Out]

Defer[Int][Log[1 + E^x + x - 8*Log[2] + (40*Log[2])/Log[x]]^(-1), x] + 8*Log[2]*Defer[Int][Log[x]/((40*Log[2]
+ E^x*Log[x] + x*Log[x] + (1 - 8*Log[2])*Log[x])*Log[1 + E^x + x - 8*Log[2] + (40*Log[2])/Log[x]]), x] + 40*Lo
g[2]*Defer[Int][1/((-40*Log[2] - (1 + E^x + x - 8*Log[2])*Log[x])*Log[1 + E^x + x - 8*Log[2] + (40*Log[2])/Log
[x]]), x] + 40*Log[2]*Defer[Int][1/(x*Log[x]*(-40*Log[2] - (1 + E^x + x - 8*Log[2])*Log[x])*Log[1 + E^x + x -
8*Log[2] + (40*Log[2])/Log[x]]), x] + Defer[Int][(x*Log[x])/((-40*Log[2] - (1 + E^x + x - 8*Log[2])*Log[x])*Lo
g[1 + E^x + x - 8*Log[2] + (40*Log[2])/Log[x]]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-40 \log (2)+\left (x+e^x x\right ) \log ^2(x)}{x \log (x) \left (40 \log (2)+e^x \log (x)+x \log (x)+(1-8 \log (2)) \log (x)\right ) \log \left (\frac {40 \log (2)+\left (1+e^x+x-8 \log (2)\right ) \log (x)}{\log (x)}\right )} \, dx\\ &=\int \left (\frac {1}{\log \left (1+e^x+x-8 \log (2)+\frac {40 \log (2)}{\log (x)}\right )}+\frac {-40 \log (2)-40 x \log (2) \log (x)-x^2 \log ^2(x)+8 x \log (2) \log ^2(x)}{x \log (x) \left (40 \log (2)+e^x \log (x)+x \log (x)+(1-8 \log (2)) \log (x)\right ) \log \left (1+e^x+x-8 \log (2)+\frac {40 \log (2)}{\log (x)}\right )}\right ) \, dx\\ &=\int \frac {1}{\log \left (1+e^x+x-8 \log (2)+\frac {40 \log (2)}{\log (x)}\right )} \, dx+\int \frac {-40 \log (2)-40 x \log (2) \log (x)-x^2 \log ^2(x)+8 x \log (2) \log ^2(x)}{x \log (x) \left (40 \log (2)+e^x \log (x)+x \log (x)+(1-8 \log (2)) \log (x)\right ) \log \left (1+e^x+x-8 \log (2)+\frac {40 \log (2)}{\log (x)}\right )} \, dx\\ &=\int \left (\frac {40 \log (2)}{\left (-40 \log (2)-e^x \log (x)-x \log (x)-(1-8 \log (2)) \log (x)\right ) \log \left (1+e^x+x-8 \log (2)+\frac {40 \log (2)}{\log (x)}\right )}+\frac {40 \log (2)}{x \log (x) \left (-40 \log (2)-e^x \log (x)-x \log (x)-(1-8 \log (2)) \log (x)\right ) \log \left (1+e^x+x-8 \log (2)+\frac {40 \log (2)}{\log (x)}\right )}+\frac {x \log (x)}{\left (-40 \log (2)-e^x \log (x)-x \log (x)-(1-8 \log (2)) \log (x)\right ) \log \left (1+e^x+x-8 \log (2)+\frac {40 \log (2)}{\log (x)}\right )}+\frac {8 \log (2) \log (x)}{\left (40 \log (2)+e^x \log (x)+x \log (x)+(1-8 \log (2)) \log (x)\right ) \log \left (1+e^x+x-8 \log (2)+\frac {40 \log (2)}{\log (x)}\right )}\right ) \, dx+\int \frac {1}{\log \left (1+e^x+x-8 \log (2)+\frac {40 \log (2)}{\log (x)}\right )} \, dx\\ &=(8 \log (2)) \int \frac {\log (x)}{\left (40 \log (2)+e^x \log (x)+x \log (x)+(1-8 \log (2)) \log (x)\right ) \log \left (1+e^x+x-8 \log (2)+\frac {40 \log (2)}{\log (x)}\right )} \, dx+(40 \log (2)) \int \frac {1}{\left (-40 \log (2)-e^x \log (x)-x \log (x)-(1-8 \log (2)) \log (x)\right ) \log \left (1+e^x+x-8 \log (2)+\frac {40 \log (2)}{\log (x)}\right )} \, dx+(40 \log (2)) \int \frac {1}{x \log (x) \left (-40 \log (2)-e^x \log (x)-x \log (x)-(1-8 \log (2)) \log (x)\right ) \log \left (1+e^x+x-8 \log (2)+\frac {40 \log (2)}{\log (x)}\right )} \, dx+\int \frac {1}{\log \left (1+e^x+x-8 \log (2)+\frac {40 \log (2)}{\log (x)}\right )} \, dx+\int \frac {x \log (x)}{\left (-40 \log (2)-e^x \log (x)-x \log (x)-(1-8 \log (2)) \log (x)\right ) \log \left (1+e^x+x-8 \log (2)+\frac {40 \log (2)}{\log (x)}\right )} \, dx\\ &=(8 \log (2)) \int \frac {\log (x)}{\left (40 \log (2)+e^x \log (x)+x \log (x)+(1-8 \log (2)) \log (x)\right ) \log \left (1+e^x+x-8 \log (2)+\frac {40 \log (2)}{\log (x)}\right )} \, dx+(40 \log (2)) \int \frac {1}{\left (-40 \log (2)-\left (1+e^x+x-8 \log (2)\right ) \log (x)\right ) \log \left (1+e^x+x-8 \log (2)+\frac {40 \log (2)}{\log (x)}\right )} \, dx+(40 \log (2)) \int \frac {1}{x \log (x) \left (-40 \log (2)-\left (1+e^x+x-8 \log (2)\right ) \log (x)\right ) \log \left (1+e^x+x-8 \log (2)+\frac {40 \log (2)}{\log (x)}\right )} \, dx+\int \frac {1}{\log \left (1+e^x+x-8 \log (2)+\frac {40 \log (2)}{\log (x)}\right )} \, dx+\int \frac {x \log (x)}{\left (-40 \log (2)-\left (1+e^x+x-8 \log (2)\right ) \log (x)\right ) \log \left (1+e^x+x-8 \log (2)+\frac {40 \log (2)}{\log (x)}\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.59, size = 20, normalized size = 1.00 \begin {gather*} \log \left (\log \left (1+e^x+x-8 \log (2)+\frac {40 \log (2)}{\log (x)}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-40*Log[2] + (x + E^x*x)*Log[x]^2)/((40*x*Log[2]*Log[x] + (x + E^x*x + x^2 - 8*x*Log[2])*Log[x]^2)*
Log[(40*Log[2] + (1 + E^x + x - 8*Log[2])*Log[x])/Log[x]]),x]

[Out]

Log[Log[1 + E^x + x - 8*Log[2] + (40*Log[2])/Log[x]]]

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fricas [A]  time = 0.83, size = 24, normalized size = 1.20 \begin {gather*} \log \left (\log \left (\frac {{\left (x + e^{x} - 8 \, \log \relax (2) + 1\right )} \log \relax (x) + 40 \, \log \relax (2)}{\log \relax (x)}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*x+x)*log(x)^2-40*log(2))/((exp(x)*x-8*x*log(2)+x^2+x)*log(x)^2+40*x*log(2)*log(x))/log(((ex
p(x)-8*log(2)+x+1)*log(x)+40*log(2))/log(x)),x, algorithm="fricas")

[Out]

log(log(((x + e^x - 8*log(2) + 1)*log(x) + 40*log(2))/log(x)))

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giac [A]  time = 0.20, size = 30, normalized size = 1.50 \begin {gather*} \log \left (-\log \left (x \log \relax (x) + e^{x} \log \relax (x) - 8 \, \log \relax (2) \log \relax (x) + 40 \, \log \relax (2) + \log \relax (x)\right ) + \log \left (\log \relax (x)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*x+x)*log(x)^2-40*log(2))/((exp(x)*x-8*x*log(2)+x^2+x)*log(x)^2+40*x*log(2)*log(x))/log(((ex
p(x)-8*log(2)+x+1)*log(x)+40*log(2))/log(x)),x, algorithm="giac")

[Out]

log(-log(x*log(x) + e^x*log(x) - 8*log(2)*log(x) + 40*log(2) + log(x)) + log(log(x)))

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maple [C]  time = 0.19, size = 299, normalized size = 14.95

method	result	size
risch	\(\ln \left (\ln \left (\left (\ln \relax (x )-5\right ) \ln \relax (2)-\frac {x \ln \relax (x )}{8}-\frac {{\mathrm e}^{x} \ln \relax (x )}{8}-\frac {\ln \relax (x )}{8}\right )+\frac {i \left (-2 \pi \mathrm {csgn}\left (\frac {i \left (-\left (\ln \relax (x )-5\right ) \ln \relax (2)+\frac {x \ln \relax (x )}{8}+\frac {{\mathrm e}^{x} \ln \relax (x )}{8}+\frac {\ln \relax (x )}{8}\right )}{\ln \relax (x )}\right )^{2}-\pi \,\mathrm {csgn}\left (i \left (-\left (\ln \relax (x )-5\right ) \ln \relax (2)+\frac {x \ln \relax (x )}{8}+\frac {{\mathrm e}^{x} \ln \relax (x )}{8}+\frac {\ln \relax (x )}{8}\right )\right ) \mathrm {csgn}\left (\frac {i \left (-\left (\ln \relax (x )-5\right ) \ln \relax (2)+\frac {x \ln \relax (x )}{8}+\frac {{\mathrm e}^{x} \ln \relax (x )}{8}+\frac {\ln \relax (x )}{8}\right )}{\ln \relax (x )}\right )^{2}-\pi \,\mathrm {csgn}\left (i \left (-\left (\ln \relax (x )-5\right ) \ln \relax (2)+\frac {x \ln \relax (x )}{8}+\frac {{\mathrm e}^{x} \ln \relax (x )}{8}+\frac {\ln \relax (x )}{8}\right )\right ) \mathrm {csgn}\left (\frac {i \left (-\left (\ln \relax (x )-5\right ) \ln \relax (2)+\frac {x \ln \relax (x )}{8}+\frac {{\mathrm e}^{x} \ln \relax (x )}{8}+\frac {\ln \relax (x )}{8}\right )}{\ln \relax (x )}\right ) \mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right )-\pi \mathrm {csgn}\left (\frac {i \left (-\left (\ln \relax (x )-5\right ) \ln \relax (2)+\frac {x \ln \relax (x )}{8}+\frac {{\mathrm e}^{x} \ln \relax (x )}{8}+\frac {\ln \relax (x )}{8}\right )}{\ln \relax (x )}\right )^{3}+\pi \mathrm {csgn}\left (\frac {i \left (-\left (\ln \relax (x )-5\right ) \ln \relax (2)+\frac {x \ln \relax (x )}{8}+\frac {{\mathrm e}^{x} \ln \relax (x )}{8}+\frac {\ln \relax (x )}{8}\right )}{\ln \relax (x )}\right )^{2} \mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right )-6 i \ln \relax (2)+2 i \ln \left (\ln \relax (x )\right )+2 \pi \right )}{2}\right )\)	\(299\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(x)*x+x)*ln(x)^2-40*ln(2))/((exp(x)*x-8*x*ln(2)+x^2+x)*ln(x)^2+40*x*ln(2)*ln(x))/ln(((exp(x)-8*ln(2)+
x+1)*ln(x)+40*ln(2))/ln(x)),x,method=_RETURNVERBOSE)

[Out]

ln(ln((ln(x)-5)*ln(2)-1/8*x*ln(x)-1/8*exp(x)*ln(x)-1/8*ln(x))+1/2*I*(-2*Pi*csgn(I*(-(ln(x)-5)*ln(2)+1/8*x*ln(x
)+1/8*exp(x)*ln(x)+1/8*ln(x))/ln(x))^2-Pi*csgn(I*(-(ln(x)-5)*ln(2)+1/8*x*ln(x)+1/8*exp(x)*ln(x)+1/8*ln(x)))*cs
gn(I*(-(ln(x)-5)*ln(2)+1/8*x*ln(x)+1/8*exp(x)*ln(x)+1/8*ln(x))/ln(x))^2-Pi*csgn(I*(-(ln(x)-5)*ln(2)+1/8*x*ln(x
)+1/8*exp(x)*ln(x)+1/8*ln(x)))*csgn(I*(-(ln(x)-5)*ln(2)+1/8*x*ln(x)+1/8*exp(x)*ln(x)+1/8*ln(x))/ln(x))*csgn(I/
ln(x))-Pi*csgn(I*(-(ln(x)-5)*ln(2)+1/8*x*ln(x)+1/8*exp(x)*ln(x)+1/8*ln(x))/ln(x))^3+Pi*csgn(I*(-(ln(x)-5)*ln(2
)+1/8*x*ln(x)+1/8*exp(x)*ln(x)+1/8*ln(x))/ln(x))^2*csgn(I/ln(x))-6*I*ln(2)+2*I*ln(ln(x))+2*Pi))

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maxima [A]  time = 0.50, size = 25, normalized size = 1.25 \begin {gather*} \log \left (\log \left ({\left (x + e^{x} - 8 \, \log \relax (2) + 1\right )} \log \relax (x) + 40 \, \log \relax (2)\right ) - \log \left (\log \relax (x)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*x+x)*log(x)^2-40*log(2))/((exp(x)*x-8*x*log(2)+x^2+x)*log(x)^2+40*x*log(2)*log(x))/log(((ex
p(x)-8*log(2)+x+1)*log(x)+40*log(2))/log(x)),x, algorithm="maxima")

[Out]

log(log((x + e^x - 8*log(2) + 1)*log(x) + 40*log(2)) - log(log(x)))

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mupad [B]  time = 4.97, size = 24, normalized size = 1.20 \begin {gather*} \ln \left (\ln \left (\frac {40\,\ln \relax (2)+\ln \relax (x)\,\left (x-8\,\ln \relax (2)+{\mathrm {e}}^x+1\right )}{\ln \relax (x)}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(40*log(2) - log(x)^2*(x + x*exp(x)))/(log((40*log(2) + log(x)*(x - 8*log(2) + exp(x) + 1))/log(x))*(log(
x)^2*(x - 8*x*log(2) + x*exp(x) + x^2) + 40*x*log(2)*log(x))),x)

[Out]

log(log((40*log(2) + log(x)*(x - 8*log(2) + exp(x) + 1))/log(x)))

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sympy [A]  time = 5.32, size = 26, normalized size = 1.30 \begin {gather*} \log {\left (\log {\left (\frac {\left (x + e^{x} - 8 \log {\relax (2 )} + 1\right ) \log {\relax (x )} + 40 \log {\relax (2 )}}{\log {\relax (x )}} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*x+x)*ln(x)**2-40*ln(2))/((exp(x)*x-8*x*ln(2)+x**2+x)*ln(x)**2+40*x*ln(2)*ln(x))/ln(((exp(x)
-8*ln(2)+x+1)*ln(x)+40*ln(2))/ln(x)),x)

[Out]

log(log(((x + exp(x) - 8*log(2) + 1)*log(x) + 40*log(2))/log(x)))

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