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

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

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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 = {} $$\begin{array}{c}\int \frac{e^{e^x}(5-6x+x^2)\log (16)+(-5+6x-x^2)\log (25-10x+x^2)+(2x+e^{e^x+x}(5x-x^2)\log (16))\log \left(e^{-x}x\right)\log (\log \left(e^{-x}x\right))}{(-5x+x^2)\log \left(e^{-x}x\right)}dx\end{array}$$

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{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{e^{e^x}(5-6x+x^2)\log (16)+(-5+6x-x^2)\log (25-10x+x^2)+(2x+e^{e^x+x}(5x-x^2)\log (16))\log \left(e^{-x}x\right)\log (\log \left(e^{-x}x\right))}{(-5+x)x\log \left(e^{-x}x\right)}dx\\ & =\int (-e^{e^x+x}\log (16)\log (\log \left(e^{-x}x\right))+\frac{5e^{e^x}\log (16)-6e^{e^x}x\log (16)+e^{e^x}{x}^2\log (16)-5\log ((-5+x{)}^2)+6x\log ((-5+x{)}^2)-x^2\log ((-5+x{)}^2)+2x\log \left(e^{-x}x\right)\log (\log \left(e^{-x}x\right))}{(-5+x)x\log \left(e^{-x}x\right)})dx\\ & =-(\log (16)\int e^{e^x+x}\log (\log \left(e^{-x}x\right))dx)+\int \frac{5e^{e^x}\log (16)-6e^{e^x}x\log (16)+e^{e^x}{x}^2\log (16)-5\log ((-5+x{)}^2)+6x\log ((-5+x{)}^2)-x^2\log ((-5+x{)}^2)+2x\log \left(e^{-x}x\right)\log (\log \left(e^{-x}x\right))}{(-5+x)x\log \left(e^{-x}x\right)}dx\\ & =-(\log (16)\int e^{e^x+x}\log (\log \left(e^{-x}x\right))dx)+\int \frac{-e^{e^x}(5-6x+x^2)\log (16)+(5-6x+x^2)\log ((-5+x{)}^2)-2x\log \left(e^{-x}x\right)\log (\log \left(e^{-x}x\right))}{(5-x)x\log \left(e^{-x}x\right)}dx\\ & =-(\log (16)\int e^{e^x+x}\log (\log \left(e^{-x}x\right))dx)+\int (\frac{e^{e^x}(-1+x)\log (16)}{x\log \left(e^{-x}x\right)}+\frac{-5\log ((-5+x{)}^2)+6x\log ((-5+x{)}^2)-x^2\log ((-5+x{)}^2)+2x\log \left(e^{-x}x\right)\log (\log \left(e^{-x}x\right))}{(-5+x)x\log \left(e^{-x}x\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 (\log \left(e^{-x}x\right))dx+\int \frac{-5\log ((-5+x{)}^2)+6x\log ((-5+x{)}^2)-x^2\log ((-5+x{)}^2)+2x\log \left(e^{-x}x\right)\log (\log \left(e^{-x}x\right))}{(-5+x)x\log \left(e^{-x}x\right)}dx\\ & =\log (16)\int (\frac{e^{e^x}}{\log \left(e^{-x}x\right)}-\frac{e^{e^x}}{x\log \left(e^{-x}x\right)})dx-\log (16)\int e^{e^x+x}\log (\log \left(e^{-x}x\right))dx+\int \frac{\frac{(5-6x+x^2)\log ((-5+x{)}^2)}{x\log \left(e^{-x}x\right)}-2\log (\log \left(e^{-x}x\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 (\log \left(e^{-x}x\right))dx+\int (-\frac{(-1+x)\log ((-5+x{)}^2)}{x\log \left(e^{-x}x\right)}+\frac{2\log (\log \left(e^{-x}x\right))}{-5+x})dx\\ & =2\int \frac{\log (\log \left(e^{-x}x\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 (\log \left(e^{-x}x\right))dx-\int \frac{(-1+x)\log ((-5+x{)}^2)}{x\log \left(e^{-x}x\right)}dx\\ & =2\int \frac{\log (\log \left(e^{-x}x\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 (\log \left(e^{-x}x\right))dx-\int (\frac{\log ((-5+x{)}^2)}{\log \left(e^{-x}x\right)}-\frac{\log ((-5+x{)}^2)}{x\log \left(e^{-x}x\right)})dx\\ & =2\int \frac{\log (\log \left(e^{-x}x\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 (\log \left(e^{-x}x\right))dx-\int \frac{\log ((-5+x{)}^2)}{\log \left(e^{-x}x\right)}dx+\int \frac{\log ((-5+x{)}^2)}{x\log \left(e^{-x}x\right)}dx\end{array}\end{array}$$

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

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 $$\begin{array}{c}-(4e^{(x+e^x)}\log (2)-e^x\log (x^2-10x+25))e^{(-x)}\log (\log \left(x{e}^{(-x)}\right))\end{array}$$

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 $$\begin{array}{c}\int \frac{4(x^2-6x+5)e^{\left(e^x\right)}\log (2)-2(2(x^2-5x)e^{(x+e^x)}\log (2)-x)\log \left(x{e}^{(-x)}\right)\log (\log \left(x{e}^{(-x)}\right))-(x^2-6x+5)\log (x^2-10x+25)}{(x^2-5x)\log \left(x{e}^{(-x)}\right)}dx\end{array}$$

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	$(-4\ln (2){\mathrm{e}}^{{\mathrm{e}}^x}+2\ln (x-5))\ln (\ln (x)-\ln \left({\mathrm{e}}^x\right)-\frac{i\pi \mathrm{csgn}\left(ix{\mathrm{e}}^{-x}\right)(-\mathrm{csgn}\left(ix{\mathrm{e}}^{-x}\right)+\mathrm{csgn}\left(ix\right))(-\mathrm{csgn}\left(ix{\mathrm{e}}^{-x}\right)+\mathrm{csgn}\left(i{\mathrm{e}}^{-x}\right))}{2})-\frac{i\pi \mathrm{csgn}\left(i{(x-5)}^2\right)(\mathrm{csgn}{\left(i(x-5)\right)}^2-2\mathrm{csgn}\left(i{(x-5)}^2\right)\mathrm{csgn}\left(i(x-5)\right)+\mathrm{csgn}{\left(i{(x-5)}^2\right)}^2)\ln (\ln \left({\mathrm{e}}^x\right)+\frac{i(\pi \mathrm{csgn}\left(ix\right)\mathrm{csgn}\left(i{\mathrm{e}}^{-x}\right)\mathrm{csgn}\left(ix{\mathrm{e}}^{-x}\right)-\pi \mathrm{csgn}\left(ix\right)\mathrm{csgn}{\left(ix{\mathrm{e}}^{-x}\right)}^2-\pi \mathrm{csgn}\left(i{\mathrm{e}}^{-x}\right)\mathrm{csgn}{\left(ix{\mathrm{e}}^{-x}\right)}^2+\pi \mathrm{csgn}{\left(ix{\mathrm{e}}^{-x}\right)}^3+2i\ln (x))}{2})}{2}$	$220$

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 $$\begin{array}{c}-2(2e^{\left(e^x\right)}\log (2)-\log (x-5))\log (-x+\log (x))\end{array}$$

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 $$\begin{array}{c}\ln (x^2-10x+25)\ln (\ln (x)-x)-4{\mathrm{e}}^{{\mathrm{e}}^x}\ln (2)\ln (\ln (x)-x)\end{array}$$

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 $$\begin{array}{c}\text{Timed out}\end{array}$$

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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