\(\int \frac {(-10+4 x+e^x (50+10 x-32 x^2+8 x^3)-10 \log (2 x)) \log (\frac {e^x (5 x-2 x^2)-x \log (2 x)}{-5+2 x})}{e^x (25 x-20 x^2+4 x^3)+(-5 x+2 x^2) \log (2 x)} \, dx\) [7651]

Optimal result

Integrand size = 96, antiderivative size = 23 \[ \int \frac {\left (-10+4 x+e^x \left (50+10 x-32 x^2+8 x^3\right )-10 \log (2 x)\right ) \log \left (\frac {e^x \left (5 x-2 x^2\right )-x \log (2 x)}{-5+2 x}\right )}{e^x \left (25 x-20 x^2+4 x^3\right )+\left (-5 x+2 x^2\right ) \log (2 x)} \, dx=\log ^2\left (x \left (-e^x+\frac {\log (2 x)}{5-2 x}\right )\right ) \]

[Out]

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

Rubi [F]

\[ \int \frac {\left (-10+4 x+e^x \left (50+10 x-32 x^2+8 x^3\right )-10 \log (2 x)\right ) \log \left (\frac {e^x \left (5 x-2 x^2\right )-x \log (2 x)}{-5+2 x}\right )}{e^x \left (25 x-20 x^2+4 x^3\right )+\left (-5 x+2 x^2\right ) \log (2 x)} \, dx=\int \frac {\left (-10+4 x+e^x \left (50+10 x-32 x^2+8 x^3\right )-10 \log (2 x)\right ) \log \left (\frac {e^x \left (5 x-2 x^2\right )-x \log (2 x)}{-5+2 x}\right )}{e^x \left (25 x-20 x^2+4 x^3\right )+\left (-5 x+2 x^2\right ) \log (2 x)} \, dx \]

[In]

Int[((-10 + 4*x + E^x*(50 + 10*x - 32*x^2 + 8*x^3) - 10*Log[2*x])*Log[(E^x*(5*x - 2*x^2) - x*Log[2*x])/(-5 + 2
*x)])/(E^x*(25*x - 20*x^2 + 4*x^3) + (-5*x + 2*x^2)*Log[2*x]),x]

[Out]

2*Defer[Int][Log[-(E^x*x) - (x*Log[2*x])/(-5 + 2*x)], x] + 2*Defer[Int][Log[-(E^x*x) - (x*Log[2*x])/(-5 + 2*x)
]/x, x] + 2*Defer[Int][Log[-(E^x*x) - (x*Log[2*x])/(-5 + 2*x)]/(x*(-5*E^x + 2*E^x*x + Log[2*x])), x] - 2*Defer
[Int][(Log[2*x]*Log[-(E^x*x) - (x*Log[2*x])/(-5 + 2*x)])/(-5*E^x + 2*E^x*x + Log[2*x]), x] - 4*Defer[Int][(Log
[2*x]*Log[-(E^x*x) - (x*Log[2*x])/(-5 + 2*x)])/((-5 + 2*x)*(-5*E^x + 2*E^x*x + Log[2*x])), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\left (-10+4 x+e^x \left (50+10 x-32 x^2+8 x^3\right )-10 \log (2 x)\right ) \log \left (\frac {e^x \left (5 x-2 x^2\right )-x \log (2 x)}{-5+2 x}\right )}{(5-2 x) x \left (5 e^x-2 e^x x-\log (2 x)\right )} \, dx \\ & = \int \left (\frac {2 (1+x) \log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{x}-\frac {2 \left (5-2 x-3 x \log (2 x)+2 x^2 \log (2 x)\right ) \log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{x (-5+2 x) \left (-5 e^x+2 e^x x+\log (2 x)\right )}\right ) \, dx \\ & = 2 \int \frac {(1+x) \log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{x} \, dx-2 \int \frac {\left (5-2 x-3 x \log (2 x)+2 x^2 \log (2 x)\right ) \log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{x (-5+2 x) \left (-5 e^x+2 e^x x+\log (2 x)\right )} \, dx \\ & = 2 \int \left (\log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )+\frac {\log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{x}\right ) \, dx-2 \int \left (-\frac {\left (5-2 x-3 x \log (2 x)+2 x^2 \log (2 x)\right ) \log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{5 x \left (-5 e^x+2 e^x x+\log (2 x)\right )}+\frac {2 \left (5-2 x-3 x \log (2 x)+2 x^2 \log (2 x)\right ) \log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{5 (-5+2 x) \left (-5 e^x+2 e^x x+\log (2 x)\right )}\right ) \, dx \\ & = \frac {2}{5} \int \frac {\left (5-2 x-3 x \log (2 x)+2 x^2 \log (2 x)\right ) \log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{x \left (-5 e^x+2 e^x x+\log (2 x)\right )} \, dx-\frac {4}{5} \int \frac {\left (5-2 x-3 x \log (2 x)+2 x^2 \log (2 x)\right ) \log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{(-5+2 x) \left (-5 e^x+2 e^x x+\log (2 x)\right )} \, dx+2 \int \log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right ) \, dx+2 \int \frac {\log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{x} \, dx \\ & = \frac {2}{5} \int \left (-\frac {2 \log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{-5 e^x+2 e^x x+\log (2 x)}+\frac {5 \log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{x \left (-5 e^x+2 e^x x+\log (2 x)\right )}-\frac {3 \log (2 x) \log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{-5 e^x+2 e^x x+\log (2 x)}+\frac {2 x \log (2 x) \log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{-5 e^x+2 e^x x+\log (2 x)}\right ) \, dx-\frac {4}{5} \int \left (\frac {5 \log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{(-5+2 x) \left (-5 e^x+2 e^x x+\log (2 x)\right )}-\frac {2 x \log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{(-5+2 x) \left (-5 e^x+2 e^x x+\log (2 x)\right )}-\frac {3 x \log (2 x) \log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{(-5+2 x) \left (-5 e^x+2 e^x x+\log (2 x)\right )}+\frac {2 x^2 \log (2 x) \log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{(-5+2 x) \left (-5 e^x+2 e^x x+\log (2 x)\right )}\right ) \, dx+2 \int \log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right ) \, dx+2 \int \frac {\log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{x} \, dx \\ & = -\left (\frac {4}{5} \int \frac {\log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{-5 e^x+2 e^x x+\log (2 x)} \, dx\right )+\frac {4}{5} \int \frac {x \log (2 x) \log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{-5 e^x+2 e^x x+\log (2 x)} \, dx-\frac {6}{5} \int \frac {\log (2 x) \log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{-5 e^x+2 e^x x+\log (2 x)} \, dx+\frac {8}{5} \int \frac {x \log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{(-5+2 x) \left (-5 e^x+2 e^x x+\log (2 x)\right )} \, dx-\frac {8}{5} \int \frac {x^2 \log (2 x) \log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{(-5+2 x) \left (-5 e^x+2 e^x x+\log (2 x)\right )} \, dx+2 \int \log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right ) \, dx+2 \int \frac {\log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{x} \, dx+2 \int \frac {\log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{x \left (-5 e^x+2 e^x x+\log (2 x)\right )} \, dx+\frac {12}{5} \int \frac {x \log (2 x) \log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{(-5+2 x) \left (-5 e^x+2 e^x x+\log (2 x)\right )} \, dx-4 \int \frac {\log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{(-5+2 x) \left (-5 e^x+2 e^x x+\log (2 x)\right )} \, dx \\ & = -\left (\frac {4}{5} \int \frac {\log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{-5 e^x+2 e^x x+\log (2 x)} \, dx\right )+\frac {4}{5} \int \frac {x \log (2 x) \log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{-5 e^x+2 e^x x+\log (2 x)} \, dx-\frac {6}{5} \int \frac {\log (2 x) \log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{-5 e^x+2 e^x x+\log (2 x)} \, dx+\frac {8}{5} \int \left (\frac {\log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{2 \left (-5 e^x+2 e^x x+\log (2 x)\right )}+\frac {5 \log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{2 (-5+2 x) \left (-5 e^x+2 e^x x+\log (2 x)\right )}\right ) \, dx-\frac {8}{5} \int \left (\frac {5 \log (2 x) \log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{4 \left (-5 e^x+2 e^x x+\log (2 x)\right )}+\frac {x \log (2 x) \log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{2 \left (-5 e^x+2 e^x x+\log (2 x)\right )}+\frac {25 \log (2 x) \log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{4 (-5+2 x) \left (-5 e^x+2 e^x x+\log (2 x)\right )}\right ) \, dx+2 \int \log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right ) \, dx+2 \int \frac {\log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{x} \, dx+2 \int \frac {\log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{x \left (-5 e^x+2 e^x x+\log (2 x)\right )} \, dx+\frac {12}{5} \int \left (\frac {\log (2 x) \log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{2 \left (-5 e^x+2 e^x x+\log (2 x)\right )}+\frac {5 \log (2 x) \log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{2 (-5+2 x) \left (-5 e^x+2 e^x x+\log (2 x)\right )}\right ) \, dx-4 \int \frac {\log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{(-5+2 x) \left (-5 e^x+2 e^x x+\log (2 x)\right )} \, dx \\ & = 2 \int \log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right ) \, dx+2 \int \frac {\log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{x} \, dx+2 \int \frac {\log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{x \left (-5 e^x+2 e^x x+\log (2 x)\right )} \, dx-2 \int \frac {\log (2 x) \log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{-5 e^x+2 e^x x+\log (2 x)} \, dx+6 \int \frac {\log (2 x) \log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{(-5+2 x) \left (-5 e^x+2 e^x x+\log (2 x)\right )} \, dx-10 \int \frac {\log (2 x) \log \left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right )}{(-5+2 x) \left (-5 e^x+2 e^x x+\log (2 x)\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {\left (-10+4 x+e^x \left (50+10 x-32 x^2+8 x^3\right )-10 \log (2 x)\right ) \log \left (\frac {e^x \left (5 x-2 x^2\right )-x \log (2 x)}{-5+2 x}\right )}{e^x \left (25 x-20 x^2+4 x^3\right )+\left (-5 x+2 x^2\right ) \log (2 x)} \, dx=\log ^2\left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right ) \]

[In]

Integrate[((-10 + 4*x + E^x*(50 + 10*x - 32*x^2 + 8*x^3) - 10*Log[2*x])*Log[(E^x*(5*x - 2*x^2) - x*Log[2*x])/(
-5 + 2*x)])/(E^x*(25*x - 20*x^2 + 4*x^3) + (-5*x + 2*x^2)*Log[2*x]),x]

[Out]

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

Maple [A] (verified)

Time = 41.41 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.39

method	result	size
parallelrisch	\({\ln \left (\frac {-x \ln \left (2 x \right )+\left (-2 x^{2}+5 x \right ) {\mathrm e}^{x}}{-5+2 x}\right )}^{2}\)	\(32\)

[In]

int((-10*ln(2*x)+(8*x^3-32*x^2+10*x+50)*exp(x)+4*x-10)*ln((-x*ln(2*x)+(-2*x^2+5*x)*exp(x))/(-5+2*x))/((2*x^2-5
*x)*ln(2*x)+(4*x^3-20*x^2+25*x)*exp(x)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.35 \[ \int \frac {\left (-10+4 x+e^x \left (50+10 x-32 x^2+8 x^3\right )-10 \log (2 x)\right ) \log \left (\frac {e^x \left (5 x-2 x^2\right )-x \log (2 x)}{-5+2 x}\right )}{e^x \left (25 x-20 x^2+4 x^3\right )+\left (-5 x+2 x^2\right ) \log (2 x)} \, dx=\log \left (-\frac {{\left (2 \, x^{2} - 5 \, x\right )} e^{x} + x \log \left (2 \, x\right )}{2 \, x - 5}\right )^{2} \]

[In]

integrate((-10*log(2*x)+(8*x^3-32*x^2+10*x+50)*exp(x)+4*x-10)*log((-x*log(2*x)+(-2*x^2+5*x)*exp(x))/(-5+2*x))/
((2*x^2-5*x)*log(2*x)+(4*x^3-20*x^2+25*x)*exp(x)),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.85 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {\left (-10+4 x+e^x \left (50+10 x-32 x^2+8 x^3\right )-10 \log (2 x)\right ) \log \left (\frac {e^x \left (5 x-2 x^2\right )-x \log (2 x)}{-5+2 x}\right )}{e^x \left (25 x-20 x^2+4 x^3\right )+\left (-5 x+2 x^2\right ) \log (2 x)} \, dx=\log {\left (\frac {- x \log {\left (2 x \right )} + \left (- 2 x^{2} + 5 x\right ) e^{x}}{2 x - 5} \right )}^{2} \]

[In]

integrate((-10*ln(2*x)+(8*x**3-32*x**2+10*x+50)*exp(x)+4*x-10)*ln((-x*ln(2*x)+(-2*x**2+5*x)*exp(x))/(-5+2*x))/
((2*x**2-5*x)*ln(2*x)+(4*x**3-20*x**2+25*x)*exp(x)),x)

[Out]

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

Maxima [F]

\[ \int \frac {\left (-10+4 x+e^x \left (50+10 x-32 x^2+8 x^3\right )-10 \log (2 x)\right ) \log \left (\frac {e^x \left (5 x-2 x^2\right )-x \log (2 x)}{-5+2 x}\right )}{e^x \left (25 x-20 x^2+4 x^3\right )+\left (-5 x+2 x^2\right ) \log (2 x)} \, dx=\int { \frac {2 \, {\left ({\left (4 \, x^{3} - 16 \, x^{2} + 5 \, x + 25\right )} e^{x} + 2 \, x - 5 \, \log \left (2 \, x\right ) - 5\right )} \log \left (-\frac {{\left (2 \, x^{2} - 5 \, x\right )} e^{x} + x \log \left (2 \, x\right )}{2 \, x - 5}\right )}{{\left (4 \, x^{3} - 20 \, x^{2} + 25 \, x\right )} e^{x} + {\left (2 \, x^{2} - 5 \, x\right )} \log \left (2 \, x\right )} \,d x } \]

[In]

integrate((-10*log(2*x)+(8*x^3-32*x^2+10*x+50)*exp(x)+4*x-10)*log((-x*log(2*x)+(-2*x^2+5*x)*exp(x))/(-5+2*x))/
((2*x^2-5*x)*log(2*x)+(4*x^3-20*x^2+25*x)*exp(x)),x, algorithm="maxima")

[Out]

2*integrate(((4*x^3 - 16*x^2 + 5*x + 25)*e^x + 2*x - 5*log(2*x) - 5)*log(-((2*x^2 - 5*x)*e^x + x*log(2*x))/(2*
x - 5))/((4*x^3 - 20*x^2 + 25*x)*e^x + (2*x^2 - 5*x)*log(2*x)), x)

Giac [F]

\[ \int \frac {\left (-10+4 x+e^x \left (50+10 x-32 x^2+8 x^3\right )-10 \log (2 x)\right ) \log \left (\frac {e^x \left (5 x-2 x^2\right )-x \log (2 x)}{-5+2 x}\right )}{e^x \left (25 x-20 x^2+4 x^3\right )+\left (-5 x+2 x^2\right ) \log (2 x)} \, dx=\int { \frac {2 \, {\left ({\left (4 \, x^{3} - 16 \, x^{2} + 5 \, x + 25\right )} e^{x} + 2 \, x - 5 \, \log \left (2 \, x\right ) - 5\right )} \log \left (-\frac {{\left (2 \, x^{2} - 5 \, x\right )} e^{x} + x \log \left (2 \, x\right )}{2 \, x - 5}\right )}{{\left (4 \, x^{3} - 20 \, x^{2} + 25 \, x\right )} e^{x} + {\left (2 \, x^{2} - 5 \, x\right )} \log \left (2 \, x\right )} \,d x } \]

[In]

integrate((-10*log(2*x)+(8*x^3-32*x^2+10*x+50)*exp(x)+4*x-10)*log((-x*log(2*x)+(-2*x^2+5*x)*exp(x))/(-5+2*x))/
((2*x^2-5*x)*log(2*x)+(4*x^3-20*x^2+25*x)*exp(x)),x, algorithm="giac")

[Out]

integrate(2*((4*x^3 - 16*x^2 + 5*x + 25)*e^x + 2*x - 5*log(2*x) - 5)*log(-((2*x^2 - 5*x)*e^x + x*log(2*x))/(2*
x - 5))/((4*x^3 - 20*x^2 + 25*x)*e^x + (2*x^2 - 5*x)*log(2*x)), x)

Mupad [B] (verification not implemented)

Time = 15.33 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.39 \[ \int \frac {\left (-10+4 x+e^x \left (50+10 x-32 x^2+8 x^3\right )-10 \log (2 x)\right ) \log \left (\frac {e^x \left (5 x-2 x^2\right )-x \log (2 x)}{-5+2 x}\right )}{e^x \left (25 x-20 x^2+4 x^3\right )+\left (-5 x+2 x^2\right ) \log (2 x)} \, dx={\ln \left (-\frac {x\,\ln \left (2\,x\right )-{\mathrm {e}}^x\,\left (5\,x-2\,x^2\right )}{2\,x-5}\right )}^2 \]

[In]

int(-(log(-(x*log(2*x) - exp(x)*(5*x - 2*x^2))/(2*x - 5))*(4*x - 10*log(2*x) + exp(x)*(10*x - 32*x^2 + 8*x^3 +
50) - 10))/(log(2*x)*(5*x - 2*x^2) - exp(x)*(25*x - 20*x^2 + 4*x^3)),x)

[Out]

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