\(\int \frac {15 x-50 e^x x+10 e^{2 x} x-5 x^2+(15 x-10 x^2) \log (x)+(-125+50 e^x-5 e^{2 x}+(-15 x+5 x^2) \log (x)) \log (25-10 e^x+e^{2 x}+(3 x-x^2) \log (x)) \log (\log (25-10 e^x+e^{2 x}+(3 x-x^2) \log (x)))}{(-25 x+10 e^x x-e^{2 x} x+(-3 x^2+x^3) \log (x)) \log (25-10 e^x+e^{2 x}+(3 x-x^2) \log (x)) \log (\log (25-10 e^x+e^{2 x}+(3 x-x^2) \log (x))) \log ^2(\frac {\log (\log (25-10 e^x+e^{2 x}+(3 x-x^2) \log (x)))}{x})} \, dx\) [5811]

Optimal result

Integrand size = 234, antiderivative size = 30 \[ \int \frac {15 x-50 e^x x+10 e^{2 x} x-5 x^2+\left (15 x-10 x^2\right ) \log (x)+\left (-125+50 e^x-5 e^{2 x}+\left (-15 x+5 x^2\right ) \log (x)\right ) \log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right )}{\left (-25 x+10 e^x x-e^{2 x} x+\left (-3 x^2+x^3\right ) \log (x)\right ) \log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right )}{x}\right )} \, dx=\frac {5}{\log \left (\frac {\log \left (\log \left (\left (5-e^x\right )^2+(3-x) x \log (x)\right )\right )}{x}\right )} \]

[Out]

5/ln(ln(ln((5-exp(x))^2+(-x+3)*x*ln(x)))/x)

Rubi [F]

\[ \int \frac {15 x-50 e^x x+10 e^{2 x} x-5 x^2+\left (15 x-10 x^2\right ) \log (x)+\left (-125+50 e^x-5 e^{2 x}+\left (-15 x+5 x^2\right ) \log (x)\right ) \log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right )}{\left (-25 x+10 e^x x-e^{2 x} x+\left (-3 x^2+x^3\right ) \log (x)\right ) \log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right )}{x}\right )} \, dx=\int \frac {15 x-50 e^x x+10 e^{2 x} x-5 x^2+\left (15 x-10 x^2\right ) \log (x)+\left (-125+50 e^x-5 e^{2 x}+\left (-15 x+5 x^2\right ) \log (x)\right ) \log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right )}{\left (-25 x+10 e^x x-e^{2 x} x+\left (-3 x^2+x^3\right ) \log (x)\right ) \log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right )}{x}\right )} \, dx \]

[In]

Int[(15*x - 50*E^x*x + 10*E^(2*x)*x - 5*x^2 + (15*x - 10*x^2)*Log[x] + (-125 + 50*E^x - 5*E^(2*x) + (-15*x + 5
*x^2)*Log[x])*Log[25 - 10*E^x + E^(2*x) + (3*x - x^2)*Log[x]]*Log[Log[25 - 10*E^x + E^(2*x) + (3*x - x^2)*Log[
x]]])/((-25*x + 10*E^x*x - E^(2*x)*x + (-3*x^2 + x^3)*Log[x])*Log[25 - 10*E^x + E^(2*x) + (3*x - x^2)*Log[x]]*
Log[Log[25 - 10*E^x + E^(2*x) + (3*x - x^2)*Log[x]]]*Log[Log[Log[25 - 10*E^x + E^(2*x) + (3*x - x^2)*Log[x]]]/
x]^2),x]

[Out]

5*Defer[Int][1/(x*Log[Log[Log[(-5 + E^x)^2 - (-3 + x)*x*Log[x]]]/x]^2), x] - 10*Defer[Int][1/(Log[(-5 + E^x)^2
- (-3 + x)*x*Log[x]]*Log[Log[(-5 + E^x)^2 - (-3 + x)*x*Log[x]]]*Log[Log[Log[(-5 + E^x)^2 - (-3 + x)*x*Log[x]]
]/x]^2), x] + 235*Defer[Int][1/(((-5 + E^x)^2 - (-3 + x)*x*Log[x])*Log[(-5 + E^x)^2 - (-3 + x)*x*Log[x]]*Log[L
og[(-5 + E^x)^2 - (-3 + x)*x*Log[x]]]*Log[Log[Log[(-5 + E^x)^2 - (-3 + x)*x*Log[x]]]/x]^2), x] - 50*Defer[Int]
[E^x/(((-5 + E^x)^2 - (-3 + x)*x*Log[x])*Log[(-5 + E^x)^2 - (-3 + x)*x*Log[x]]*Log[Log[(-5 + E^x)^2 - (-3 + x)
*x*Log[x]]]*Log[Log[Log[(-5 + E^x)^2 - (-3 + x)*x*Log[x]]]/x]^2), x] - 5*Defer[Int][x/((-(-5 + E^x)^2 + (-3 +
x)*x*Log[x])*Log[(-5 + E^x)^2 - (-3 + x)*x*Log[x]]*Log[Log[(-5 + E^x)^2 - (-3 + x)*x*Log[x]]]*Log[Log[Log[(-5
+ E^x)^2 - (-3 + x)*x*Log[x]]]/x]^2), x] + 15*Defer[Int][Log[x]/((-(-5 + E^x)^2 + (-3 + x)*x*Log[x])*Log[(-5 +
E^x)^2 - (-3 + x)*x*Log[x]]*Log[Log[(-5 + E^x)^2 - (-3 + x)*x*Log[x]]]*Log[Log[Log[(-5 + E^x)^2 - (-3 + x)*x*
Log[x]]]/x]^2), x] - 40*Defer[Int][(x*Log[x])/((-(-5 + E^x)^2 + (-3 + x)*x*Log[x])*Log[(-5 + E^x)^2 - (-3 + x)
*x*Log[x]]*Log[Log[(-5 + E^x)^2 - (-3 + x)*x*Log[x]]]*Log[Log[Log[(-5 + E^x)^2 - (-3 + x)*x*Log[x]]]/x]^2), x]
+ 10*Defer[Int][(x^2*Log[x])/((-(-5 + E^x)^2 + (-3 + x)*x*Log[x])*Log[(-5 + E^x)^2 - (-3 + x)*x*Log[x]]*Log[L
og[(-5 + E^x)^2 - (-3 + x)*x*Log[x]]]*Log[Log[Log[(-5 + E^x)^2 - (-3 + x)*x*Log[x]]]/x]^2), x]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {5 \left (-47+10 e^x-x+3 \log (x)-8 x \log (x)+2 x^2 \log (x)\right )}{\left (25-10 e^x+e^{2 x}+3 x \log (x)-x^2 \log (x)\right ) \log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right ) \log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right )}{x}\right )}-\frac {5 \left (2 x-\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right ) \log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right )\right )}{x \log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right ) \log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right )}{x}\right )}\right ) \, dx \\ & = -\left (5 \int \frac {-47+10 e^x-x+3 \log (x)-8 x \log (x)+2 x^2 \log (x)}{\left (25-10 e^x+e^{2 x}+3 x \log (x)-x^2 \log (x)\right ) \log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right ) \log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right )}{x}\right )} \, dx\right )-5 \int \frac {2 x-\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right ) \log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right )}{x \log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right ) \log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right )}{x}\right )} \, dx \\ & = -\left (5 \int \left (-\frac {47}{\left (25-10 e^x+e^{2 x}+3 x \log (x)-x^2 \log (x)\right ) \log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right ) \log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right )}{x}\right )}+\frac {10 e^x}{\left (25-10 e^x+e^{2 x}+3 x \log (x)-x^2 \log (x)\right ) \log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right ) \log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right )}{x}\right )}+\frac {x}{\left (-25+10 e^x-e^{2 x}-3 x \log (x)+x^2 \log (x)\right ) \log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right ) \log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right )}{x}\right )}-\frac {3 \log (x)}{\left (-25+10 e^x-e^{2 x}-3 x \log (x)+x^2 \log (x)\right ) \log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right ) \log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right )}{x}\right )}+\frac {8 x \log (x)}{\left (-25+10 e^x-e^{2 x}-3 x \log (x)+x^2 \log (x)\right ) \log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right ) \log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right )}{x}\right )}-\frac {2 x^2 \log (x)}{\left (-25+10 e^x-e^{2 x}-3 x \log (x)+x^2 \log (x)\right ) \log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right ) \log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right )}{x}\right )}\right ) \, dx\right )-5 \int \frac {-\frac {1}{x}+\frac {2}{\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right ) \log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right )}}{\log ^2\left (\frac {\log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right )}{x}\right )} \, dx \\ & = -\left (5 \int \left (-\frac {1}{x \log ^2\left (\frac {\log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right )}{x}\right )}+\frac {2}{\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right ) \log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right )}{x}\right )}\right ) \, dx\right )-5 \int \frac {x}{\left (-25+10 e^x-e^{2 x}-3 x \log (x)+x^2 \log (x)\right ) \log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right ) \log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right )}{x}\right )} \, dx+10 \int \frac {x^2 \log (x)}{\left (-25+10 e^x-e^{2 x}-3 x \log (x)+x^2 \log (x)\right ) \log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right ) \log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right )}{x}\right )} \, dx+15 \int \frac {\log (x)}{\left (-25+10 e^x-e^{2 x}-3 x \log (x)+x^2 \log (x)\right ) \log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right ) \log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right )}{x}\right )} \, dx-40 \int \frac {x \log (x)}{\left (-25+10 e^x-e^{2 x}-3 x \log (x)+x^2 \log (x)\right ) \log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right ) \log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right )}{x}\right )} \, dx-50 \int \frac {e^x}{\left (25-10 e^x+e^{2 x}+3 x \log (x)-x^2 \log (x)\right ) \log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right ) \log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right )}{x}\right )} \, dx+235 \int \frac {1}{\left (25-10 e^x+e^{2 x}+3 x \log (x)-x^2 \log (x)\right ) \log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right ) \log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right )}{x}\right )} \, dx \\ & = 5 \int \frac {1}{x \log ^2\left (\frac {\log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right )}{x}\right )} \, dx-5 \int \frac {x}{\left (-\left (-5+e^x\right )^2+(-3+x) x \log (x)\right ) \log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right ) \log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right )}{x}\right )} \, dx-10 \int \frac {1}{\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right ) \log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right )}{x}\right )} \, dx+10 \int \frac {x^2 \log (x)}{\left (-\left (-5+e^x\right )^2+(-3+x) x \log (x)\right ) \log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right ) \log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right )}{x}\right )} \, dx+15 \int \frac {\log (x)}{\left (-\left (-5+e^x\right )^2+(-3+x) x \log (x)\right ) \log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right ) \log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right )}{x}\right )} \, dx-40 \int \frac {x \log (x)}{\left (-\left (-5+e^x\right )^2+(-3+x) x \log (x)\right ) \log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right ) \log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right )}{x}\right )} \, dx-50 \int \frac {e^x}{\left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right ) \log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right ) \log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right )}{x}\right )} \, dx+235 \int \frac {1}{\left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right ) \log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right ) \log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right )}{x}\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {15 x-50 e^x x+10 e^{2 x} x-5 x^2+\left (15 x-10 x^2\right ) \log (x)+\left (-125+50 e^x-5 e^{2 x}+\left (-15 x+5 x^2\right ) \log (x)\right ) \log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right )}{\left (-25 x+10 e^x x-e^{2 x} x+\left (-3 x^2+x^3\right ) \log (x)\right ) \log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right )}{x}\right )} \, dx=\frac {5}{\log \left (\frac {\log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right )}{x}\right )} \]

[In]

Integrate[(15*x - 50*E^x*x + 10*E^(2*x)*x - 5*x^2 + (15*x - 10*x^2)*Log[x] + (-125 + 50*E^x - 5*E^(2*x) + (-15
*x + 5*x^2)*Log[x])*Log[25 - 10*E^x + E^(2*x) + (3*x - x^2)*Log[x]]*Log[Log[25 - 10*E^x + E^(2*x) + (3*x - x^2
)*Log[x]]])/((-25*x + 10*E^x*x - E^(2*x)*x + (-3*x^2 + x^3)*Log[x])*Log[25 - 10*E^x + E^(2*x) + (3*x - x^2)*Lo
g[x]]*Log[Log[25 - 10*E^x + E^(2*x) + (3*x - x^2)*Log[x]]]*Log[Log[Log[25 - 10*E^x + E^(2*x) + (3*x - x^2)*Log
[x]]]/x]^2),x]

[Out]

5/Log[Log[Log[(-5 + E^x)^2 - (-3 + x)*x*Log[x]]]/x]

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.33 (sec) , antiderivative size = 250, normalized size of antiderivative = 8.33

\[\frac {10 i}{\pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (i \ln \left (\ln \left (\left (-x^{2}+3 x \right ) \ln \left (x \right )+{\mathrm e}^{2 x}-10 \,{\mathrm e}^{x}+25\right )\right )\right ) \operatorname {csgn}\left (\frac {i \ln \left (\ln \left (\left (-x^{2}+3 x \right ) \ln \left (x \right )+{\mathrm e}^{2 x}-10 \,{\mathrm e}^{x}+25\right )\right )}{x}\right )-\pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) {\operatorname {csgn}\left (\frac {i \ln \left (\ln \left (\left (-x^{2}+3 x \right ) \ln \left (x \right )+{\mathrm e}^{2 x}-10 \,{\mathrm e}^{x}+25\right )\right )}{x}\right )}^{2}-\pi \,\operatorname {csgn}\left (i \ln \left (\ln \left (\left (-x^{2}+3 x \right ) \ln \left (x \right )+{\mathrm e}^{2 x}-10 \,{\mathrm e}^{x}+25\right )\right )\right ) {\operatorname {csgn}\left (\frac {i \ln \left (\ln \left (\left (-x^{2}+3 x \right ) \ln \left (x \right )+{\mathrm e}^{2 x}-10 \,{\mathrm e}^{x}+25\right )\right )}{x}\right )}^{2}+\pi {\operatorname {csgn}\left (\frac {i \ln \left (\ln \left (\left (-x^{2}+3 x \right ) \ln \left (x \right )+{\mathrm e}^{2 x}-10 \,{\mathrm e}^{x}+25\right )\right )}{x}\right )}^{3}-2 i \ln \left (x \right )+2 i \ln \left (\ln \left (\ln \left (\left (-x^{2}+3 x \right ) \ln \left (x \right )+{\mathrm e}^{2 x}-10 \,{\mathrm e}^{x}+25\right )\right )\right )}\]

[In]

int((((5*x^2-15*x)*ln(x)-5*exp(x)^2+50*exp(x)-125)*ln((-x^2+3*x)*ln(x)+exp(x)^2-10*exp(x)+25)*ln(ln((-x^2+3*x)
*ln(x)+exp(x)^2-10*exp(x)+25))+(-10*x^2+15*x)*ln(x)+10*x*exp(x)^2-50*exp(x)*x-5*x^2+15*x)/((x^3-3*x^2)*ln(x)-x
*exp(x)^2+10*exp(x)*x-25*x)/ln((-x^2+3*x)*ln(x)+exp(x)^2-10*exp(x)+25)/ln(ln((-x^2+3*x)*ln(x)+exp(x)^2-10*exp(
x)+25))/ln(ln(ln((-x^2+3*x)*ln(x)+exp(x)^2-10*exp(x)+25))/x)^2,x)

[Out]

10*I/(Pi*csgn(I/x)*csgn(I*ln(ln((-x^2+3*x)*ln(x)+exp(2*x)-10*exp(x)+25)))*csgn(I/x*ln(ln((-x^2+3*x)*ln(x)+exp(
2*x)-10*exp(x)+25)))-Pi*csgn(I/x)*csgn(I/x*ln(ln((-x^2+3*x)*ln(x)+exp(2*x)-10*exp(x)+25)))^2-Pi*csgn(I*ln(ln((
-x^2+3*x)*ln(x)+exp(2*x)-10*exp(x)+25)))*csgn(I/x*ln(ln((-x^2+3*x)*ln(x)+exp(2*x)-10*exp(x)+25)))^2+Pi*csgn(I/
x*ln(ln((-x^2+3*x)*ln(x)+exp(2*x)-10*exp(x)+25)))^3-2*I*ln(x)+2*I*ln(ln(ln((-x^2+3*x)*ln(x)+exp(2*x)-10*exp(x)
+25))))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {15 x-50 e^x x+10 e^{2 x} x-5 x^2+\left (15 x-10 x^2\right ) \log (x)+\left (-125+50 e^x-5 e^{2 x}+\left (-15 x+5 x^2\right ) \log (x)\right ) \log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right )}{\left (-25 x+10 e^x x-e^{2 x} x+\left (-3 x^2+x^3\right ) \log (x)\right ) \log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right )}{x}\right )} \, dx=\frac {5}{\log \left (\frac {\log \left (\log \left (-{\left (x^{2} - 3 \, x\right )} \log \left (x\right ) + e^{\left (2 \, x\right )} - 10 \, e^{x} + 25\right )\right )}{x}\right )} \]

[In]

integrate((((5*x^2-15*x)*log(x)-5*exp(x)^2+50*exp(x)-125)*log((-x^2+3*x)*log(x)+exp(x)^2-10*exp(x)+25)*log(log
((-x^2+3*x)*log(x)+exp(x)^2-10*exp(x)+25))+(-10*x^2+15*x)*log(x)+10*x*exp(x)^2-50*exp(x)*x-5*x^2+15*x)/((x^3-3
*x^2)*log(x)-x*exp(x)^2+10*exp(x)*x-25*x)/log((-x^2+3*x)*log(x)+exp(x)^2-10*exp(x)+25)/log(log((-x^2+3*x)*log(
x)+exp(x)^2-10*exp(x)+25))/log(log(log((-x^2+3*x)*log(x)+exp(x)^2-10*exp(x)+25))/x)^2,x, algorithm="fricas")

[Out]

5/log(log(log(-(x^2 - 3*x)*log(x) + e^(2*x) - 10*e^x + 25))/x)

Sympy [F(-1)]

Timed out. \[ \int \frac {15 x-50 e^x x+10 e^{2 x} x-5 x^2+\left (15 x-10 x^2\right ) \log (x)+\left (-125+50 e^x-5 e^{2 x}+\left (-15 x+5 x^2\right ) \log (x)\right ) \log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right )}{\left (-25 x+10 e^x x-e^{2 x} x+\left (-3 x^2+x^3\right ) \log (x)\right ) \log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right )}{x}\right )} \, dx=\text {Timed out} \]

[In]

integrate((((5*x**2-15*x)*ln(x)-5*exp(x)**2+50*exp(x)-125)*ln((-x**2+3*x)*ln(x)+exp(x)**2-10*exp(x)+25)*ln(ln(
(-x**2+3*x)*ln(x)+exp(x)**2-10*exp(x)+25))+(-10*x**2+15*x)*ln(x)+10*x*exp(x)**2-50*exp(x)*x-5*x**2+15*x)/((x**
3-3*x**2)*ln(x)-x*exp(x)**2+10*exp(x)*x-25*x)/ln((-x**2+3*x)*ln(x)+exp(x)**2-10*exp(x)+25)/ln(ln((-x**2+3*x)*l
n(x)+exp(x)**2-10*exp(x)+25))/ln(ln(ln((-x**2+3*x)*ln(x)+exp(x)**2-10*exp(x)+25))/x)**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.65 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10 \[ \int \frac {15 x-50 e^x x+10 e^{2 x} x-5 x^2+\left (15 x-10 x^2\right ) \log (x)+\left (-125+50 e^x-5 e^{2 x}+\left (-15 x+5 x^2\right ) \log (x)\right ) \log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right )}{\left (-25 x+10 e^x x-e^{2 x} x+\left (-3 x^2+x^3\right ) \log (x)\right ) \log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right )}{x}\right )} \, dx=-\frac {5}{\log \left (x\right ) - \log \left (\log \left (\log \left (-{\left (x^{2} - 3 \, x\right )} \log \left (x\right ) + e^{\left (2 \, x\right )} - 10 \, e^{x} + 25\right )\right )\right )} \]

[In]

integrate((((5*x^2-15*x)*log(x)-5*exp(x)^2+50*exp(x)-125)*log((-x^2+3*x)*log(x)+exp(x)^2-10*exp(x)+25)*log(log
((-x^2+3*x)*log(x)+exp(x)^2-10*exp(x)+25))+(-10*x^2+15*x)*log(x)+10*x*exp(x)^2-50*exp(x)*x-5*x^2+15*x)/((x^3-3
*x^2)*log(x)-x*exp(x)^2+10*exp(x)*x-25*x)/log((-x^2+3*x)*log(x)+exp(x)^2-10*exp(x)+25)/log(log((-x^2+3*x)*log(
x)+exp(x)^2-10*exp(x)+25))/log(log(log((-x^2+3*x)*log(x)+exp(x)^2-10*exp(x)+25))/x)^2,x, algorithm="maxima")

[Out]

-5/(log(x) - log(log(log(-(x^2 - 3*x)*log(x) + e^(2*x) - 10*e^x + 25))))

Giac [A] (verification not implemented)

none

Time = 3.32 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {15 x-50 e^x x+10 e^{2 x} x-5 x^2+\left (15 x-10 x^2\right ) \log (x)+\left (-125+50 e^x-5 e^{2 x}+\left (-15 x+5 x^2\right ) \log (x)\right ) \log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right )}{\left (-25 x+10 e^x x-e^{2 x} x+\left (-3 x^2+x^3\right ) \log (x)\right ) \log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right )}{x}\right )} \, dx=-\frac {5}{\log \left (x\right ) - \log \left (\log \left (\log \left (-x^{2} \log \left (x\right ) + 3 \, x \log \left (x\right ) + e^{\left (2 \, x\right )} - 10 \, e^{x} + 25\right )\right )\right )} \]

[In]

integrate((((5*x^2-15*x)*log(x)-5*exp(x)^2+50*exp(x)-125)*log((-x^2+3*x)*log(x)+exp(x)^2-10*exp(x)+25)*log(log
((-x^2+3*x)*log(x)+exp(x)^2-10*exp(x)+25))+(-10*x^2+15*x)*log(x)+10*x*exp(x)^2-50*exp(x)*x-5*x^2+15*x)/((x^3-3
*x^2)*log(x)-x*exp(x)^2+10*exp(x)*x-25*x)/log((-x^2+3*x)*log(x)+exp(x)^2-10*exp(x)+25)/log(log((-x^2+3*x)*log(
x)+exp(x)^2-10*exp(x)+25))/log(log(log((-x^2+3*x)*log(x)+exp(x)^2-10*exp(x)+25))/x)^2,x, algorithm="giac")

[Out]

-5/(log(x) - log(log(log(-x^2*log(x) + 3*x*log(x) + e^(2*x) - 10*e^x + 25))))

Mupad [B] (verification not implemented)

Time = 14.49 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10 \[ \int \frac {15 x-50 e^x x+10 e^{2 x} x-5 x^2+\left (15 x-10 x^2\right ) \log (x)+\left (-125+50 e^x-5 e^{2 x}+\left (-15 x+5 x^2\right ) \log (x)\right ) \log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right )}{\left (-25 x+10 e^x x-e^{2 x} x+\left (-3 x^2+x^3\right ) \log (x)\right ) \log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right )}{x}\right )} \, dx=\frac {5}{\ln \left (\frac {\ln \left (\ln \left ({\mathrm {e}}^{2\,x}-10\,{\mathrm {e}}^x+\ln \left (x\right )\,\left (3\,x-x^2\right )+25\right )\right )}{x}\right )} \]

[In]

int(-(15*x + 10*x*exp(2*x) + log(x)*(15*x - 10*x^2) - 50*x*exp(x) - 5*x^2 - log(log(exp(2*x) - 10*exp(x) + log
(x)*(3*x - x^2) + 25))*log(exp(2*x) - 10*exp(x) + log(x)*(3*x - x^2) + 25)*(5*exp(2*x) - 50*exp(x) + log(x)*(1
5*x - 5*x^2) + 125))/(log(log(log(exp(2*x) - 10*exp(x) + log(x)*(3*x - x^2) + 25))/x)^2*log(log(exp(2*x) - 10*
exp(x) + log(x)*(3*x - x^2) + 25))*log(exp(2*x) - 10*exp(x) + log(x)*(3*x - x^2) + 25)*(25*x + log(x)*(3*x^2 -
x^3) + x*exp(2*x) - 10*x*exp(x))),x)

[Out]

5/log(log(log(exp(2*x) - 10*exp(x) + log(x)*(3*x - x^2) + 25))/x)