3.44.71 \(\int \frac {e^{e^8} (20-4 e^x)+e^{e^8} (-20+4 e^x) \log (x)+e^{e^8} (10+e^x (-2+2 x)) \log (x) \log (\frac {x}{\log (x)}) \log (\log ^2(\frac {x}{\log (x)}))}{(25-10 e^x+e^{2 x}) \log (x) \log (\frac {x}{\log (x)}) \log ^2(\log ^2(\frac {x}{\log (x)}))} \, dx\)

Optimal. Leaf size=29 \[ \frac {2 e^{e^8} x}{\left (5-e^x\right ) \log \left (\log ^2\left (\frac {x}{\log (x)}\right )\right )} \]

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Rubi [F]  time = 5.26, 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 {e^{e^8} \left (20-4 e^x\right )+e^{e^8} \left (-20+4 e^x\right ) \log (x)+e^{e^8} \left (10+e^x (-2+2 x)\right ) \log (x) \log \left (\frac {x}{\log (x)}\right ) \log \left (\log ^2\left (\frac {x}{\log (x)}\right )\right )}{\left (25-10 e^x+e^{2 x}\right ) \log (x) \log \left (\frac {x}{\log (x)}\right ) \log ^2\left (\log ^2\left (\frac {x}{\log (x)}\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^E^8*(20 - 4*E^x) + E^E^8*(-20 + 4*E^x)*Log[x] + E^E^8*(10 + E^x*(-2 + 2*x))*Log[x]*Log[x/Log[x]]*Log[Lo
g[x/Log[x]]^2])/((25 - 10*E^x + E^(2*x))*Log[x]*Log[x/Log[x]]*Log[Log[x/Log[x]]^2]^2),x]

[Out]

4*E^E^8*Defer[Int][1/((-5 + E^x)*Log[x/Log[x]]*Log[Log[x/Log[x]]^2]^2), x] - 4*E^E^8*Defer[Int][1/((-5 + E^x)*
Log[x]*Log[x/Log[x]]*Log[Log[x/Log[x]]^2]^2), x] - 2*E^E^8*Defer[Int][1/((-5 + E^x)*Log[Log[x/Log[x]]^2]), x]
+ 10*E^E^8*Defer[Int][x/((-5 + E^x)^2*Log[Log[x/Log[x]]^2]), x] + 2*E^E^8*Defer[Int][x/((-5 + E^x)*Log[Log[x/L
og[x]]^2]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{e^8} \left (20-4 e^x\right )+e^{e^8} \left (-20+4 e^x\right ) \log (x)+e^{e^8} \left (10+e^x (-2+2 x)\right ) \log (x) \log \left (\frac {x}{\log (x)}\right ) \log \left (\log ^2\left (\frac {x}{\log (x)}\right )\right )}{\left (5-e^x\right )^2 \log (x) \log \left (\frac {x}{\log (x)}\right ) \log ^2\left (\log ^2\left (\frac {x}{\log (x)}\right )\right )} \, dx\\ &=\int \left (\frac {10 e^{e^8} x}{\left (-5+e^x\right )^2 \log \left (\log ^2\left (\frac {x}{\log (x)}\right )\right )}+\frac {2 e^{e^8} \left (-2+2 \log (x)-\log (x) \log \left (\frac {x}{\log (x)}\right ) \log \left (\log ^2\left (\frac {x}{\log (x)}\right )\right )+x \log (x) \log \left (\frac {x}{\log (x)}\right ) \log \left (\log ^2\left (\frac {x}{\log (x)}\right )\right )\right )}{\left (-5+e^x\right ) \log (x) \log \left (\frac {x}{\log (x)}\right ) \log ^2\left (\log ^2\left (\frac {x}{\log (x)}\right )\right )}\right ) \, dx\\ &=\left (2 e^{e^8}\right ) \int \frac {-2+2 \log (x)-\log (x) \log \left (\frac {x}{\log (x)}\right ) \log \left (\log ^2\left (\frac {x}{\log (x)}\right )\right )+x \log (x) \log \left (\frac {x}{\log (x)}\right ) \log \left (\log ^2\left (\frac {x}{\log (x)}\right )\right )}{\left (-5+e^x\right ) \log (x) \log \left (\frac {x}{\log (x)}\right ) \log ^2\left (\log ^2\left (\frac {x}{\log (x)}\right )\right )} \, dx+\left (10 e^{e^8}\right ) \int \frac {x}{\left (-5+e^x\right )^2 \log \left (\log ^2\left (\frac {x}{\log (x)}\right )\right )} \, dx\\ &=\left (2 e^{e^8}\right ) \int \frac {2-\log (x) \left (2+(-1+x) \log \left (\frac {x}{\log (x)}\right ) \log \left (\log ^2\left (\frac {x}{\log (x)}\right )\right )\right )}{\left (5-e^x\right ) \log (x) \log \left (\frac {x}{\log (x)}\right ) \log ^2\left (\log ^2\left (\frac {x}{\log (x)}\right )\right )} \, dx+\left (10 e^{e^8}\right ) \int \frac {x}{\left (-5+e^x\right )^2 \log \left (\log ^2\left (\frac {x}{\log (x)}\right )\right )} \, dx\\ &=\left (2 e^{e^8}\right ) \int \left (\frac {2}{\left (-5+e^x\right ) \log \left (\frac {x}{\log (x)}\right ) \log ^2\left (\log ^2\left (\frac {x}{\log (x)}\right )\right )}-\frac {2}{\left (-5+e^x\right ) \log (x) \log \left (\frac {x}{\log (x)}\right ) \log ^2\left (\log ^2\left (\frac {x}{\log (x)}\right )\right )}-\frac {1}{\left (-5+e^x\right ) \log \left (\log ^2\left (\frac {x}{\log (x)}\right )\right )}+\frac {x}{\left (-5+e^x\right ) \log \left (\log ^2\left (\frac {x}{\log (x)}\right )\right )}\right ) \, dx+\left (10 e^{e^8}\right ) \int \frac {x}{\left (-5+e^x\right )^2 \log \left (\log ^2\left (\frac {x}{\log (x)}\right )\right )} \, dx\\ &=-\left (\left (2 e^{e^8}\right ) \int \frac {1}{\left (-5+e^x\right ) \log \left (\log ^2\left (\frac {x}{\log (x)}\right )\right )} \, dx\right )+\left (2 e^{e^8}\right ) \int \frac {x}{\left (-5+e^x\right ) \log \left (\log ^2\left (\frac {x}{\log (x)}\right )\right )} \, dx+\left (4 e^{e^8}\right ) \int \frac {1}{\left (-5+e^x\right ) \log \left (\frac {x}{\log (x)}\right ) \log ^2\left (\log ^2\left (\frac {x}{\log (x)}\right )\right )} \, dx-\left (4 e^{e^8}\right ) \int \frac {1}{\left (-5+e^x\right ) \log (x) \log \left (\frac {x}{\log (x)}\right ) \log ^2\left (\log ^2\left (\frac {x}{\log (x)}\right )\right )} \, dx+\left (10 e^{e^8}\right ) \int \frac {x}{\left (-5+e^x\right )^2 \log \left (\log ^2\left (\frac {x}{\log (x)}\right )\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.47, size = 27, normalized size = 0.93 \begin {gather*} -\frac {2 e^{e^8} x}{\left (-5+e^x\right ) \log \left (\log ^2\left (\frac {x}{\log (x)}\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^E^8*(20 - 4*E^x) + E^E^8*(-20 + 4*E^x)*Log[x] + E^E^8*(10 + E^x*(-2 + 2*x))*Log[x]*Log[x/Log[x]]*
Log[Log[x/Log[x]]^2])/((25 - 10*E^x + E^(2*x))*Log[x]*Log[x/Log[x]]*Log[Log[x/Log[x]]^2]^2),x]

[Out]

(-2*E^E^8*x)/((-5 + E^x)*Log[Log[x/Log[x]]^2])

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fricas [A]  time = 0.64, size = 24, normalized size = 0.83 \begin {gather*} -\frac {2 \, x e^{\left (e^{8}\right )}}{{\left (e^{x} - 5\right )} \log \left (\log \left (\frac {x}{\log \relax (x)}\right )^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x-2)*exp(x)+10)*exp(exp(8))*log(x)*log(x/log(x))*log(log(x/log(x))^2)+(4*exp(x)-20)*exp(exp(8))
*log(x)+(-4*exp(x)+20)*exp(exp(8)))/(exp(x)^2-10*exp(x)+25)/log(x)/log(x/log(x))/log(log(x/log(x))^2)^2,x, alg
orithm="fricas")

[Out]

-2*x*e^(e^8)/((e^x - 5)*log(log(x/log(x))^2))

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giac [B]  time = 95.32, size = 50, normalized size = 1.72 \begin {gather*} -\frac {2 \, x e^{\left (e^{8}\right )}}{e^{x} \log \left (\log \relax (x)^{2} - 2 \, \log \relax (x) \log \left (\log \relax (x)\right ) + \log \left (\log \relax (x)\right )^{2}\right ) - 5 \, \log \left (\log \relax (x)^{2} - 2 \, \log \relax (x) \log \left (\log \relax (x)\right ) + \log \left (\log \relax (x)\right )^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x-2)*exp(x)+10)*exp(exp(8))*log(x)*log(x/log(x))*log(log(x/log(x))^2)+(4*exp(x)-20)*exp(exp(8))
*log(x)+(-4*exp(x)+20)*exp(exp(8)))/(exp(x)^2-10*exp(x)+25)/log(x)/log(x/log(x))/log(log(x/log(x))^2)^2,x, alg
orithm="giac")

[Out]

-2*x*e^(e^8)/(e^x*log(log(x)^2 - 2*log(x)*log(log(x)) + log(log(x))^2) - 5*log(log(x)^2 - 2*log(x)*log(log(x))
+ log(log(x))^2))

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maple [C]  time = 0.72, size = 710, normalized size = 24.48

method	result	size
risch	\(-\frac {4 i {\mathrm e}^{{\mathrm e}^{8}} x}{\left ({\mathrm e}^{x}-5\right ) \left (2 \pi \mathrm {csgn}\left (i \left (\pi \,\mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right )-\pi \mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right )^{2} \mathrm {csgn}\left (i x \right )-\pi \mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right )^{2} \mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right )+\pi \mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right )^{3}+2 i \ln \relax (x )-2 i \ln \left (\ln \relax (x )\right )\right )^{2}\right )^{2}+\pi \mathrm {csgn}\left (i \left (\pi \,\mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right )-\pi \mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right )^{2} \mathrm {csgn}\left (i x \right )-\pi \mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right )^{2} \mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right )+\pi \mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right )^{3}+2 i \ln \relax (x )-2 i \ln \left (\ln \relax (x )\right )\right )\right )^{2} \mathrm {csgn}\left (i \left (\pi \,\mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right )-\pi \mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right )^{2} \mathrm {csgn}\left (i x \right )-\pi \mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right )^{2} \mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right )+\pi \mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right )^{3}+2 i \ln \relax (x )-2 i \ln \left (\ln \relax (x )\right )\right )^{2}\right )-2 \pi \,\mathrm {csgn}\left (i \left (\pi \,\mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right )-\pi \mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right )^{2} \mathrm {csgn}\left (i x \right )-\pi \mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right )^{2} \mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right )+\pi \mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right )^{3}+2 i \ln \relax (x )-2 i \ln \left (\ln \relax (x )\right )\right )\right ) \mathrm {csgn}\left (i \left (\pi \,\mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right )-\pi \mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right )^{2} \mathrm {csgn}\left (i x \right )-\pi \mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right )^{2} \mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right )+\pi \mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right )^{3}+2 i \ln \relax (x )-2 i \ln \left (\ln \relax (x )\right )\right )^{2}\right )^{2}-\pi \mathrm {csgn}\left (i \left (\pi \,\mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right )-\pi \mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right )^{2} \mathrm {csgn}\left (i x \right )-\pi \mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right )^{2} \mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right )+\pi \mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right )^{3}+2 i \ln \relax (x )-2 i \ln \left (\ln \relax (x )\right )\right )^{2}\right )^{3}-2 \pi -4 i \ln \relax (2)+4 i \ln \left (\pi \,\mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right )-\pi \mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right )^{2} \mathrm {csgn}\left (i x \right )-\pi \mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right )^{2} \mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right )+\pi \mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right )^{3}+2 i \ln \relax (x )-2 i \ln \left (\ln \relax (x )\right )\right )\right )}\)	\(710\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*x-2)*exp(x)+10)*exp(exp(8))*ln(x)*ln(x/ln(x))*ln(ln(x/ln(x))^2)+(4*exp(x)-20)*exp(exp(8))*ln(x)+(-4*e
xp(x)+20)*exp(exp(8)))/(exp(x)^2-10*exp(x)+25)/ln(x)/ln(x/ln(x))/ln(ln(x/ln(x))^2)^2,x,method=_RETURNVERBOSE)

[Out]

-4*I*exp(exp(8))*x/(exp(x)-5)/(2*Pi*csgn(I*(Pi*csgn(I*x/ln(x))*csgn(I*x)*csgn(I/ln(x))-Pi*csgn(I*x/ln(x))^2*cs
gn(I*x)-Pi*csgn(I*x/ln(x))^2*csgn(I/ln(x))+Pi*csgn(I*x/ln(x))^3+2*I*ln(x)-2*I*ln(ln(x)))^2)^2+Pi*csgn(I*(Pi*cs
gn(I*x/ln(x))*csgn(I*x)*csgn(I/ln(x))-Pi*csgn(I*x/ln(x))^2*csgn(I*x)-Pi*csgn(I*x/ln(x))^2*csgn(I/ln(x))+Pi*csg
n(I*x/ln(x))^3+2*I*ln(x)-2*I*ln(ln(x))))^2*csgn(I*(Pi*csgn(I*x/ln(x))*csgn(I*x)*csgn(I/ln(x))-Pi*csgn(I*x/ln(x
))^2*csgn(I*x)-Pi*csgn(I*x/ln(x))^2*csgn(I/ln(x))+Pi*csgn(I*x/ln(x))^3+2*I*ln(x)-2*I*ln(ln(x)))^2)-2*Pi*csgn(I
*(Pi*csgn(I*x/ln(x))*csgn(I*x)*csgn(I/ln(x))-Pi*csgn(I*x/ln(x))^2*csgn(I*x)-Pi*csgn(I*x/ln(x))^2*csgn(I/ln(x))
+Pi*csgn(I*x/ln(x))^3+2*I*ln(x)-2*I*ln(ln(x))))*csgn(I*(Pi*csgn(I*x/ln(x))*csgn(I*x)*csgn(I/ln(x))-Pi*csgn(I*x
/ln(x))^2*csgn(I*x)-Pi*csgn(I*x/ln(x))^2*csgn(I/ln(x))+Pi*csgn(I*x/ln(x))^3+2*I*ln(x)-2*I*ln(ln(x)))^2)^2-Pi*c
sgn(I*(Pi*csgn(I*x/ln(x))*csgn(I*x)*csgn(I/ln(x))-Pi*csgn(I*x/ln(x))^2*csgn(I*x)-Pi*csgn(I*x/ln(x))^2*csgn(I/l
n(x))+Pi*csgn(I*x/ln(x))^3+2*I*ln(x)-2*I*ln(ln(x)))^2)^3-2*Pi-4*I*ln(2)+4*I*ln(Pi*csgn(I*x/ln(x))*csgn(I*x)*cs
gn(I/ln(x))-Pi*csgn(I*x/ln(x))^2*csgn(I*x)-Pi*csgn(I*x/ln(x))^2*csgn(I/ln(x))+Pi*csgn(I*x/ln(x))^3+2*I*ln(x)-2
*I*ln(ln(x))))

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maxima [A]  time = 0.51, size = 23, normalized size = 0.79 \begin {gather*} -\frac {x e^{\left (e^{8}\right )}}{{\left (e^{x} - 5\right )} \log \left (\log \relax (x) - \log \left (\log \relax (x)\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x-2)*exp(x)+10)*exp(exp(8))*log(x)*log(x/log(x))*log(log(x/log(x))^2)+(4*exp(x)-20)*exp(exp(8))
*log(x)+(-4*exp(x)+20)*exp(exp(8)))/(exp(x)^2-10*exp(x)+25)/log(x)/log(x/log(x))/log(log(x/log(x))^2)^2,x, alg
orithm="maxima")

[Out]

-x*e^(e^8)/((e^x - 5)*log(log(x) - log(log(x))))

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mupad [B]  time = 5.32, size = 24, normalized size = 0.83 \begin {gather*} -\frac {2\,x\,{\mathrm {e}}^{{\mathrm {e}}^8}}{\ln \left ({\ln \left (\frac {x}{\ln \relax (x)}\right )}^2\right )\,\left ({\mathrm {e}}^x-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp(8))*log(x)*(4*exp(x) - 20) - exp(exp(8))*(4*exp(x) - 20) + log(log(x/log(x))^2)*exp(exp(8))*log(x
/log(x))*log(x)*(exp(x)*(2*x - 2) + 10))/(log(log(x/log(x))^2)^2*log(x/log(x))*log(x)*(exp(2*x) - 10*exp(x) +
25)),x)

[Out]

-(2*x*exp(exp(8)))/(log(log(x/log(x))^2)*(exp(x) - 5))

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sympy [A]  time = 0.64, size = 34, normalized size = 1.17 \begin {gather*} - \frac {2 x e^{e^{8}}}{e^{x} \log {\left (\log {\left (\frac {x}{\log {\relax (x )}} \right )}^{2} \right )} - 5 \log {\left (\log {\left (\frac {x}{\log {\relax (x )}} \right )}^{2} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x-2)*exp(x)+10)*exp(exp(8))*ln(x)*ln(x/ln(x))*ln(ln(x/ln(x))**2)+(4*exp(x)-20)*exp(exp(8))*ln(x
)+(-4*exp(x)+20)*exp(exp(8)))/(exp(x)**2-10*exp(x)+25)/ln(x)/ln(x/ln(x))/ln(ln(x/ln(x))**2)**2,x)

[Out]

-2*x*exp(exp(8))/(exp(x)*log(log(x/log(x))**2) - 5*log(log(x/log(x))**2))

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