3.17.35 \(\int \frac {e^{e^x} (5-6 x+x^2) \log (16)+(-5+6 x-x^2) \log (25-10 x+x^2)+(2 x+e^{e^x+x} (5 x-x^2) \log (16)) \log (e^{-x} x) \log (\log (e^{-x} x))}{(-5 x+x^2) \log (e^{-x} x)} \, dx\)

Optimal. Leaf size=26 \[ \left (-e^{e^x} \log (16)+\log \left ((-5+x)^2\right )\right ) \log \left (\log \left (e^{-x} x\right )\right ) \]

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Rubi [F]  time = 3.76, 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^x} \left (5-6 x+x^2\right ) \log (16)+\left (-5+6 x-x^2\right ) \log \left (25-10 x+x^2\right )+\left (2 x+e^{e^x+x} \left (5 x-x^2\right ) \log (16)\right ) \log \left (e^{-x} x\right ) \log \left (\log \left (e^{-x} x\right )\right )}{\left (-5 x+x^2\right ) \log \left (e^{-x} x\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^E^x*(5 - 6*x + x^2)*Log[16] + (-5 + 6*x - x^2)*Log[25 - 10*x + x^2] + (2*x + E^(E^x + x)*(5*x - x^2)*Lo
g[16])*Log[x/E^x]*Log[Log[x/E^x]])/((-5*x + x^2)*Log[x/E^x]),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{e^x} \left (5-6 x+x^2\right ) \log (16)+\left (-5+6 x-x^2\right ) \log \left (25-10 x+x^2\right )+\left (2 x+e^{e^x+x} \left (5 x-x^2\right ) \log (16)\right ) \log \left (e^{-x} x\right ) \log \left (\log \left (e^{-x} x\right )\right )}{(-5+x) x \log \left (e^{-x} x\right )} \, dx\\ &=\int \left (-e^{e^x+x} \log (16) \log \left (\log \left (e^{-x} x\right )\right )+\frac {5 e^{e^x} \log (16)-6 e^{e^x} x \log (16)+e^{e^x} x^2 \log (16)-5 \log \left ((-5+x)^2\right )+6 x \log \left ((-5+x)^2\right )-x^2 \log \left ((-5+x)^2\right )+2 x \log \left (e^{-x} x\right ) \log \left (\log \left (e^{-x} x\right )\right )}{(-5+x) x \log \left (e^{-x} x\right )}\right ) \, dx\\ &=-\left (\log (16) \int e^{e^x+x} \log \left (\log \left (e^{-x} x\right )\right ) \, dx\right )+\int \frac {5 e^{e^x} \log (16)-6 e^{e^x} x \log (16)+e^{e^x} x^2 \log (16)-5 \log \left ((-5+x)^2\right )+6 x \log \left ((-5+x)^2\right )-x^2 \log \left ((-5+x)^2\right )+2 x \log \left (e^{-x} x\right ) \log \left (\log \left (e^{-x} x\right )\right )}{(-5+x) x \log \left (e^{-x} x\right )} \, dx\\ &=-\left (\log (16) \int e^{e^x+x} \log \left (\log \left (e^{-x} x\right )\right ) \, dx\right )+\int \frac {-e^{e^x} \left (5-6 x+x^2\right ) \log (16)+\left (5-6 x+x^2\right ) \log \left ((-5+x)^2\right )-2 x \log \left (e^{-x} x\right ) \log \left (\log \left (e^{-x} x\right )\right )}{(5-x) x \log \left (e^{-x} x\right )} \, dx\\ &=-\left (\log (16) \int e^{e^x+x} \log \left (\log \left (e^{-x} x\right )\right ) \, dx\right )+\int \left (\frac {e^{e^x} (-1+x) \log (16)}{x \log \left (e^{-x} x\right )}+\frac {-5 \log \left ((-5+x)^2\right )+6 x \log \left ((-5+x)^2\right )-x^2 \log \left ((-5+x)^2\right )+2 x \log \left (e^{-x} x\right ) \log \left (\log \left (e^{-x} x\right )\right )}{(-5+x) x \log \left (e^{-x} x\right )}\right ) \, dx\\ &=\log (16) \int \frac {e^{e^x} (-1+x)}{x \log \left (e^{-x} x\right )} \, dx-\log (16) \int e^{e^x+x} \log \left (\log \left (e^{-x} x\right )\right ) \, dx+\int \frac {-5 \log \left ((-5+x)^2\right )+6 x \log \left ((-5+x)^2\right )-x^2 \log \left ((-5+x)^2\right )+2 x \log \left (e^{-x} x\right ) \log \left (\log \left (e^{-x} x\right )\right )}{(-5+x) x \log \left (e^{-x} x\right )} \, dx\\ &=\log (16) \int \left (\frac {e^{e^x}}{\log \left (e^{-x} x\right )}-\frac {e^{e^x}}{x \log \left (e^{-x} x\right )}\right ) \, dx-\log (16) \int e^{e^x+x} \log \left (\log \left (e^{-x} x\right )\right ) \, dx+\int \frac {\frac {\left (5-6 x+x^2\right ) \log \left ((-5+x)^2\right )}{x \log \left (e^{-x} x\right )}-2 \log \left (\log \left (e^{-x} x\right )\right )}{5-x} \, dx\\ &=\log (16) \int \frac {e^{e^x}}{\log \left (e^{-x} x\right )} \, dx-\log (16) \int \frac {e^{e^x}}{x \log \left (e^{-x} x\right )} \, dx-\log (16) \int e^{e^x+x} \log \left (\log \left (e^{-x} x\right )\right ) \, dx+\int \left (-\frac {(-1+x) \log \left ((-5+x)^2\right )}{x \log \left (e^{-x} x\right )}+\frac {2 \log \left (\log \left (e^{-x} x\right )\right )}{-5+x}\right ) \, dx\\ &=2 \int \frac {\log \left (\log \left (e^{-x} x\right )\right )}{-5+x} \, dx+\log (16) \int \frac {e^{e^x}}{\log \left (e^{-x} x\right )} \, dx-\log (16) \int \frac {e^{e^x}}{x \log \left (e^{-x} x\right )} \, dx-\log (16) \int e^{e^x+x} \log \left (\log \left (e^{-x} x\right )\right ) \, dx-\int \frac {(-1+x) \log \left ((-5+x)^2\right )}{x \log \left (e^{-x} x\right )} \, dx\\ &=2 \int \frac {\log \left (\log \left (e^{-x} x\right )\right )}{-5+x} \, dx+\log (16) \int \frac {e^{e^x}}{\log \left (e^{-x} x\right )} \, dx-\log (16) \int \frac {e^{e^x}}{x \log \left (e^{-x} x\right )} \, dx-\log (16) \int e^{e^x+x} \log \left (\log \left (e^{-x} x\right )\right ) \, dx-\int \left (\frac {\log \left ((-5+x)^2\right )}{\log \left (e^{-x} x\right )}-\frac {\log \left ((-5+x)^2\right )}{x \log \left (e^{-x} x\right )}\right ) \, dx\\ &=2 \int \frac {\log \left (\log \left (e^{-x} x\right )\right )}{-5+x} \, dx+\log (16) \int \frac {e^{e^x}}{\log \left (e^{-x} x\right )} \, dx-\log (16) \int \frac {e^{e^x}}{x \log \left (e^{-x} x\right )} \, dx-\log (16) \int e^{e^x+x} \log \left (\log \left (e^{-x} x\right )\right ) \, dx-\int \frac {\log \left ((-5+x)^2\right )}{\log \left (e^{-x} x\right )} \, dx+\int \frac {\log \left ((-5+x)^2\right )}{x \log \left (e^{-x} x\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 26, normalized size = 1.00 \begin {gather*} \left (-e^{e^x} \log (16)+\log \left ((-5+x)^2\right )\right ) \log \left (\log \left (e^{-x} x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^E^x*(5 - 6*x + x^2)*Log[16] + (-5 + 6*x - x^2)*Log[25 - 10*x + x^2] + (2*x + E^(E^x + x)*(5*x - x
^2)*Log[16])*Log[x/E^x]*Log[Log[x/E^x]])/((-5*x + x^2)*Log[x/E^x]),x]

[Out]

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

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fricas [A]  time = 0.84, size = 37, normalized size = 1.42 \begin {gather*} -{\left (4 \, e^{\left (x + e^{x}\right )} \log \relax (2) - e^{x} \log \left (x^{2} - 10 \, x + 25\right )\right )} e^{\left (-x\right )} \log \left (\log \left (x e^{\left (-x\right )}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*(-x^2+5*x)*log(2)*exp(x)*exp(exp(x))+2*x)*log(x/exp(x))*log(log(x/exp(x)))+4*(x^2-6*x+5)*log(2)*
exp(exp(x))+(-x^2+6*x-5)*log(x^2-10*x+25))/(x^2-5*x)/log(x/exp(x)),x, algorithm="fricas")

[Out]

-(4*e^(x + e^x)*log(2) - e^x*log(x^2 - 10*x + 25))*e^(-x)*log(log(x*e^(-x)))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {4 \, {\left (x^{2} - 6 \, x + 5\right )} e^{\left (e^{x}\right )} \log \relax (2) - 2 \, {\left (2 \, {\left (x^{2} - 5 \, x\right )} e^{\left (x + e^{x}\right )} \log \relax (2) - x\right )} \log \left (x e^{\left (-x\right )}\right ) \log \left (\log \left (x e^{\left (-x\right )}\right )\right ) - {\left (x^{2} - 6 \, x + 5\right )} \log \left (x^{2} - 10 \, x + 25\right )}{{\left (x^{2} - 5 \, x\right )} \log \left (x e^{\left (-x\right )}\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*(-x^2+5*x)*log(2)*exp(x)*exp(exp(x))+2*x)*log(x/exp(x))*log(log(x/exp(x)))+4*(x^2-6*x+5)*log(2)*
exp(exp(x))+(-x^2+6*x-5)*log(x^2-10*x+25))/(x^2-5*x)/log(x/exp(x)),x, algorithm="giac")

[Out]

integrate((4*(x^2 - 6*x + 5)*e^(e^x)*log(2) - 2*(2*(x^2 - 5*x)*e^(x + e^x)*log(2) - x)*log(x*e^(-x))*log(log(x
*e^(-x))) - (x^2 - 6*x + 5)*log(x^2 - 10*x + 25))/((x^2 - 5*x)*log(x*e^(-x))), x)

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maple [C]  time = 0.42, size = 220, normalized size = 8.46

method	result	size
risch	\(\left (-4 \ln \relax (2) {\mathrm e}^{{\mathrm e}^{x}}+2 \ln \left (x -5\right )\right ) \ln \left (\ln \relax (x )-\ln \left ({\mathrm e}^{x}\right )-\frac {i \pi \,\mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right ) \left (-\mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )+\mathrm {csgn}\left (i x \right )\right ) \left (-\mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )+\mathrm {csgn}\left (i {\mathrm e}^{-x}\right )\right )}{2}\right )-\frac {i \pi \,\mathrm {csgn}\left (i \left (x -5\right )^{2}\right ) \left (\mathrm {csgn}\left (i \left (x -5\right )\right )^{2}-2 \,\mathrm {csgn}\left (i \left (x -5\right )^{2}\right ) \mathrm {csgn}\left (i \left (x -5\right )\right )+\mathrm {csgn}\left (i \left (x -5\right )^{2}\right )^{2}\right ) \ln \left (\ln \left ({\mathrm e}^{x}\right )+\frac {i \left (\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )-\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2}-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2}+\pi \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{3}+2 i \ln \relax (x )\right )}{2}\right )}{2}\)	\(220\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*(-x^2+5*x)*ln(2)*exp(x)*exp(exp(x))+2*x)*ln(x/exp(x))*ln(ln(x/exp(x)))+4*(x^2-6*x+5)*ln(2)*exp(exp(x))
+(-x^2+6*x-5)*ln(x^2-10*x+25))/(x^2-5*x)/ln(x/exp(x)),x,method=_RETURNVERBOSE)

[Out]

(-4*ln(2)*exp(exp(x))+2*ln(x-5))*ln(ln(x)-ln(exp(x))-1/2*I*Pi*csgn(I*x*exp(-x))*(-csgn(I*x*exp(-x))+csgn(I*x))
*(-csgn(I*x*exp(-x))+csgn(I*exp(-x))))-1/2*I*Pi*csgn(I*(x-5)^2)*(csgn(I*(x-5))^2-2*csgn(I*(x-5)^2)*csgn(I*(x-5
))+csgn(I*(x-5)^2)^2)*ln(ln(exp(x))+1/2*I*(Pi*csgn(I*x)*csgn(I*exp(-x))*csgn(I*x*exp(-x))-Pi*csgn(I*x)*csgn(I*
x*exp(-x))^2-Pi*csgn(I*exp(-x))*csgn(I*x*exp(-x))^2+Pi*csgn(I*x*exp(-x))^3+2*I*ln(x)))

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maxima [A]  time = 0.70, size = 23, normalized size = 0.88 \begin {gather*} -2 \, {\left (2 \, e^{\left (e^{x}\right )} \log \relax (2) - \log \left (x - 5\right )\right )} \log \left (-x + \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*(-x^2+5*x)*log(2)*exp(x)*exp(exp(x))+2*x)*log(x/exp(x))*log(log(x/exp(x)))+4*(x^2-6*x+5)*log(2)*
exp(exp(x))+(-x^2+6*x-5)*log(x^2-10*x+25))/(x^2-5*x)/log(x/exp(x)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.71, size = 32, normalized size = 1.23 \begin {gather*} \ln \left (x^2-10\,x+25\right )\,\ln \left (\ln \relax (x)-x\right )-4\,{\mathrm {e}}^{{\mathrm {e}}^x}\,\ln \relax (2)\,\ln \left (\ln \relax (x)-x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*exp(exp(x))*log(2)*(x^2 - 6*x + 5) - log(x^2 - 10*x + 25)*(x^2 - 6*x + 5) + log(x*exp(-x))*log(log(x*e
xp(-x)))*(2*x + 4*exp(exp(x))*exp(x)*log(2)*(5*x - x^2)))/(log(x*exp(-x))*(5*x - x^2)),x)

[Out]

log(x^2 - 10*x + 25)*log(log(x) - x) - 4*exp(exp(x))*log(2)*log(log(x) - x)

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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**2+5*x)*ln(2)*exp(x)*exp(exp(x))+2*x)*ln(x/exp(x))*ln(ln(x/exp(x)))+4*(x**2-6*x+5)*ln(2)*exp
(exp(x))+(-x**2+6*x-5)*ln(x**2-10*x+25))/(x**2-5*x)/ln(x/exp(x)),x)

[Out]

Timed out

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