\(\int \frac {2-x-20 x^2+25 x^3+25 e^{2 x} x^3+e^x (5 x+20 x^2-50 x^3)+(5 x+e^x (-5 x-5 x^2)) \log (x)}{4 x-20 x^2+25 x^3+25 e^{2 x} x^3+e^x (20 x^2-50 x^3)} \, dx\) [2082]

Optimal result

Integrand size = 105, antiderivative size = 18 \[ \int \frac {2-x-20 x^2+25 x^3+25 e^{2 x} x^3+e^x \left (5 x+20 x^2-50 x^3\right )+\left (5 x+e^x \left (-5 x-5 x^2\right )\right ) \log (x)}{4 x-20 x^2+25 x^3+25 e^{2 x} x^3+e^x \left (20 x^2-50 x^3\right )} \, dx=4+x+\frac {\log (x)}{2+5 \left (-1+e^x\right ) x} \]

[Out]

ln(x)/(5*(-1+exp(x))*x+2)+4+x

Rubi [F]

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

[In]

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

[Out]

x + 2*Log[x]*Defer[Int][(2 - 5*x + 5*E^x*x)^(-2), x] + 2*Log[x]*Defer[Int][1/(x*(2 - 5*x + 5*E^x*x)^2), x] - 5
*Log[x]*Defer[Int][x/(2 - 5*x + 5*E^x*x)^2, x] - Log[x]*Defer[Int][(2 - 5*x + 5*E^x*x)^(-1), x] + Defer[Int][1
/(x*(2 - 5*x + 5*E^x*x)), x] - Log[x]*Defer[Int][1/(x*(2 - 5*x + 5*E^x*x)), x] - 2*Defer[Int][Defer[Int][(2 +
5*(-1 + E^x)*x)^(-2), x]/x, x] - 2*Defer[Int][Defer[Int][1/(x*(2 + 5*(-1 + E^x)*x)^2), x]/x, x] + 5*Defer[Int]
[Defer[Int][x/(2 + 5*(-1 + E^x)*x)^2, x]/x, x] + Defer[Int][Defer[Int][(2 + 5*(-1 + E^x)*x)^(-1), x]/x, x] + D
efer[Int][Defer[Int][1/(x*(2 + 5*(-1 + E^x)*x)), x]/x, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {2+\left (-1+5 e^x\right ) x+20 \left (-1+e^x\right ) x^2+25 \left (-1+e^x\right )^2 x^3-5 x \left (-1+e^x (1+x)\right ) \log (x)}{x \left (2+5 \left (-1+e^x\right ) x\right )^2} \, dx \\ & = \int \left (1-\frac {\left (-2-2 x+5 x^2\right ) \log (x)}{x \left (2-5 x+5 e^x x\right )^2}-\frac {-1+\log (x)+x \log (x)}{x \left (2-5 x+5 e^x x\right )}\right ) \, dx \\ & = x-\int \frac {\left (-2-2 x+5 x^2\right ) \log (x)}{x \left (2-5 x+5 e^x x\right )^2} \, dx-\int \frac {-1+\log (x)+x \log (x)}{x \left (2-5 x+5 e^x x\right )} \, dx \\ & = x+(2 \log (x)) \int \frac {1}{\left (2-5 x+5 e^x x\right )^2} \, dx+(2 \log (x)) \int \frac {1}{x \left (2-5 x+5 e^x x\right )^2} \, dx-(5 \log (x)) \int \frac {x}{\left (2-5 x+5 e^x x\right )^2} \, dx-\int \left (-\frac {1}{x \left (2-5 x+5 e^x x\right )}+\frac {\log (x)}{2-5 x+5 e^x x}+\frac {\log (x)}{x \left (2-5 x+5 e^x x\right )}\right ) \, dx+\int \frac {-2 \int \frac {1}{\left (2+5 \left (-1+e^x\right ) x\right )^2} \, dx-2 \int \frac {1}{x \left (2+5 \left (-1+e^x\right ) x\right )^2} \, dx+5 \int \frac {x}{\left (2+5 \left (-1+e^x\right ) x\right )^2} \, dx}{x} \, dx \\ & = x+(2 \log (x)) \int \frac {1}{\left (2-5 x+5 e^x x\right )^2} \, dx+(2 \log (x)) \int \frac {1}{x \left (2-5 x+5 e^x x\right )^2} \, dx-(5 \log (x)) \int \frac {x}{\left (2-5 x+5 e^x x\right )^2} \, dx+\int \frac {1}{x \left (2-5 x+5 e^x x\right )} \, dx-\int \frac {\log (x)}{2-5 x+5 e^x x} \, dx-\int \frac {\log (x)}{x \left (2-5 x+5 e^x x\right )} \, dx+\int \left (-\frac {2 \left (\int \frac {1}{\left (2+5 \left (-1+e^x\right ) x\right )^2} \, dx+\int \frac {1}{x \left (2+5 \left (-1+e^x\right ) x\right )^2} \, dx\right )}{x}+\frac {5 \int \frac {x}{\left (2+5 \left (-1+e^x\right ) x\right )^2} \, dx}{x}\right ) \, dx \\ & = x-2 \int \frac {\int \frac {1}{\left (2+5 \left (-1+e^x\right ) x\right )^2} \, dx+\int \frac {1}{x \left (2+5 \left (-1+e^x\right ) x\right )^2} \, dx}{x} \, dx+5 \int \frac {\int \frac {x}{\left (2+5 \left (-1+e^x\right ) x\right )^2} \, dx}{x} \, dx-\log (x) \int \frac {1}{2-5 x+5 e^x x} \, dx-\log (x) \int \frac {1}{x \left (2-5 x+5 e^x x\right )} \, dx+(2 \log (x)) \int \frac {1}{\left (2-5 x+5 e^x x\right )^2} \, dx+(2 \log (x)) \int \frac {1}{x \left (2-5 x+5 e^x x\right )^2} \, dx-(5 \log (x)) \int \frac {x}{\left (2-5 x+5 e^x x\right )^2} \, dx+\int \frac {1}{x \left (2-5 x+5 e^x x\right )} \, dx+\int \frac {\int \frac {1}{2+5 \left (-1+e^x\right ) x} \, dx}{x} \, dx+\int \frac {\int \frac {1}{x \left (2+5 \left (-1+e^x\right ) x\right )} \, dx}{x} \, dx \\ & = x-2 \int \left (\frac {\int \frac {1}{\left (2+5 \left (-1+e^x\right ) x\right )^2} \, dx}{x}+\frac {\int \frac {1}{x \left (2+5 \left (-1+e^x\right ) x\right )^2} \, dx}{x}\right ) \, dx+5 \int \frac {\int \frac {x}{\left (2+5 \left (-1+e^x\right ) x\right )^2} \, dx}{x} \, dx-\log (x) \int \frac {1}{2-5 x+5 e^x x} \, dx-\log (x) \int \frac {1}{x \left (2-5 x+5 e^x x\right )} \, dx+(2 \log (x)) \int \frac {1}{\left (2-5 x+5 e^x x\right )^2} \, dx+(2 \log (x)) \int \frac {1}{x \left (2-5 x+5 e^x x\right )^2} \, dx-(5 \log (x)) \int \frac {x}{\left (2-5 x+5 e^x x\right )^2} \, dx+\int \frac {1}{x \left (2-5 x+5 e^x x\right )} \, dx+\int \frac {\int \frac {1}{2+5 \left (-1+e^x\right ) x} \, dx}{x} \, dx+\int \frac {\int \frac {1}{x \left (2+5 \left (-1+e^x\right ) x\right )} \, dx}{x} \, dx \\ & = x-2 \int \frac {\int \frac {1}{\left (2+5 \left (-1+e^x\right ) x\right )^2} \, dx}{x} \, dx-2 \int \frac {\int \frac {1}{x \left (2+5 \left (-1+e^x\right ) x\right )^2} \, dx}{x} \, dx+5 \int \frac {\int \frac {x}{\left (2+5 \left (-1+e^x\right ) x\right )^2} \, dx}{x} \, dx-\log (x) \int \frac {1}{2-5 x+5 e^x x} \, dx-\log (x) \int \frac {1}{x \left (2-5 x+5 e^x x\right )} \, dx+(2 \log (x)) \int \frac {1}{\left (2-5 x+5 e^x x\right )^2} \, dx+(2 \log (x)) \int \frac {1}{x \left (2-5 x+5 e^x x\right )^2} \, dx-(5 \log (x)) \int \frac {x}{\left (2-5 x+5 e^x x\right )^2} \, dx+\int \frac {1}{x \left (2-5 x+5 e^x x\right )} \, dx+\int \frac {\int \frac {1}{2+5 \left (-1+e^x\right ) x} \, dx}{x} \, dx+\int \frac {\int \frac {1}{x \left (2+5 \left (-1+e^x\right ) x\right )} \, dx}{x} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {2-x-20 x^2+25 x^3+25 e^{2 x} x^3+e^x \left (5 x+20 x^2-50 x^3\right )+\left (5 x+e^x \left (-5 x-5 x^2\right )\right ) \log (x)}{4 x-20 x^2+25 x^3+25 e^{2 x} x^3+e^x \left (20 x^2-50 x^3\right )} \, dx=x+\frac {\log (x)}{2-5 x+5 e^x x} \]

[In]

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

[Out]

x + Log[x]/(2 - 5*x + 5*E^x*x)

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00

method	result	size
risch	\(\frac {\ln \left (x \right )}{5 \,{\mathrm e}^{x} x -5 x +2}+x\)	\(18\)
parallelrisch	\(\frac {25 \,{\mathrm e}^{x} x^{2}-25 x^{2}+10 x +5 \ln \left (x \right )}{25 \,{\mathrm e}^{x} x -25 x +10}\)	\(35\)

[In]

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

[Out]

1/(5*exp(x)*x-5*x+2)*ln(x)+x

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.72 \[ \int \frac {2-x-20 x^2+25 x^3+25 e^{2 x} x^3+e^x \left (5 x+20 x^2-50 x^3\right )+\left (5 x+e^x \left (-5 x-5 x^2\right )\right ) \log (x)}{4 x-20 x^2+25 x^3+25 e^{2 x} x^3+e^x \left (20 x^2-50 x^3\right )} \, dx=\frac {5 \, x^{2} e^{x} - 5 \, x^{2} + 2 \, x + \log \left (x\right )}{5 \, x e^{x} - 5 \, x + 2} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {2-x-20 x^2+25 x^3+25 e^{2 x} x^3+e^x \left (5 x+20 x^2-50 x^3\right )+\left (5 x+e^x \left (-5 x-5 x^2\right )\right ) \log (x)}{4 x-20 x^2+25 x^3+25 e^{2 x} x^3+e^x \left (20 x^2-50 x^3\right )} \, dx=x + \frac {\log {\left (x \right )}}{5 x e^{x} - 5 x + 2} \]

[In]

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

[Out]

x + log(x)/(5*x*exp(x) - 5*x + 2)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.72 \[ \int \frac {2-x-20 x^2+25 x^3+25 e^{2 x} x^3+e^x \left (5 x+20 x^2-50 x^3\right )+\left (5 x+e^x \left (-5 x-5 x^2\right )\right ) \log (x)}{4 x-20 x^2+25 x^3+25 e^{2 x} x^3+e^x \left (20 x^2-50 x^3\right )} \, dx=\frac {5 \, x^{2} e^{x} - 5 \, x^{2} + 2 \, x + \log \left (x\right )}{5 \, x e^{x} - 5 \, x + 2} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.72 \[ \int \frac {2-x-20 x^2+25 x^3+25 e^{2 x} x^3+e^x \left (5 x+20 x^2-50 x^3\right )+\left (5 x+e^x \left (-5 x-5 x^2\right )\right ) \log (x)}{4 x-20 x^2+25 x^3+25 e^{2 x} x^3+e^x \left (20 x^2-50 x^3\right )} \, dx=\frac {5 \, x^{2} e^{x} - 5 \, x^{2} + 2 \, x + \log \left (x\right )}{5 \, x e^{x} - 5 \, x + 2} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 8.85 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {2-x-20 x^2+25 x^3+25 e^{2 x} x^3+e^x \left (5 x+20 x^2-50 x^3\right )+\left (5 x+e^x \left (-5 x-5 x^2\right )\right ) \log (x)}{4 x-20 x^2+25 x^3+25 e^{2 x} x^3+e^x \left (20 x^2-50 x^3\right )} \, dx=x+\frac {\ln \left (x\right )}{5\,x\,{\mathrm {e}}^x-5\,x+2} \]

[In]

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

[Out]

x + log(x)/(5*x*exp(x) - 5*x + 2)