\(\int \frac {25 x^2-6 e^x x^3+e^{2 x^2} (25-5 e^x x)+e^{x^2} (-50 x+e^x (x+11 x^2-2 x^3))}{25 e^{2 x^2} x-50 e^{x^2} x^2+25 x^3} \, dx\) [7307]

Optimal result

Integrand size = 85, antiderivative size = 28 \[ \int \frac {25 x^2-6 e^x x^3+e^{2 x^2} \left (25-5 e^x x\right )+e^{x^2} \left (-50 x+e^x \left (x+11 x^2-2 x^3\right )\right )}{25 e^{2 x^2} x-50 e^{x^2} x^2+25 x^3} \, dx=\frac {5}{2}-\frac {1}{25} e^x \left (5+\frac {x}{-e^{x^2}+x}\right )+\log (x) \]

[Out]

ln(x)-exp(x)*(1/25*x/(-exp(x^2)+x)+1/5)+5/2

Rubi [F]

\[ \int \frac {25 x^2-6 e^x x^3+e^{2 x^2} \left (25-5 e^x x\right )+e^{x^2} \left (-50 x+e^x \left (x+11 x^2-2 x^3\right )\right )}{25 e^{2 x^2} x-50 e^{x^2} x^2+25 x^3} \, dx=\int \frac {25 x^2-6 e^x x^3+e^{2 x^2} \left (25-5 e^x x\right )+e^{x^2} \left (-50 x+e^x \left (x+11 x^2-2 x^3\right )\right )}{25 e^{2 x^2} x-50 e^{x^2} x^2+25 x^3} \, dx \]

[In]

Int[(25*x^2 - 6*E^x*x^3 + E^(2*x^2)*(25 - 5*E^x*x) + E^x^2*(-50*x + E^x*(x + 11*x^2 - 2*x^3)))/(25*E^(2*x^2)*x
- 50*E^x^2*x^2 + 25*x^3),x]

[Out]

-1/5*E^x + Log[x] + Defer[Int][E^x/(E^x^2 - x), x]/25 + Defer[Int][(E^x*x)/(E^x^2 - x)^2, x]/25 + Defer[Int][(
E^x*x)/(E^x^2 - x), x]/25 - (2*Defer[Int][(E^x*x^2)/(E^x^2 - x), x])/25 - (2*Defer[Int][(E^x*x^3)/(E^x^2 - x)^
2, x])/25

Rubi steps \begin{align*} \text {integral}& = \int \frac {25 x^2-6 e^x x^3+e^{2 x^2} \left (25-5 e^x x\right )+e^{x^2} \left (-50 x+e^x \left (x+11 x^2-2 x^3\right )\right )}{25 \left (e^{x^2}-x\right )^2 x} \, dx \\ & = \frac {1}{25} \int \frac {25 x^2-6 e^x x^3+e^{2 x^2} \left (25-5 e^x x\right )+e^{x^2} \left (-50 x+e^x \left (x+11 x^2-2 x^3\right )\right )}{\left (e^{x^2}-x\right )^2 x} \, dx \\ & = \frac {1}{25} \int \left (-\frac {5 \left (-5+e^x x\right )}{x}-\frac {e^x x \left (-1+2 x^2\right )}{\left (e^{x^2}-x\right )^2}-\frac {e^x \left (-1-x+2 x^2\right )}{e^{x^2}-x}\right ) \, dx \\ & = -\left (\frac {1}{25} \int \frac {e^x x \left (-1+2 x^2\right )}{\left (e^{x^2}-x\right )^2} \, dx\right )-\frac {1}{25} \int \frac {e^x \left (-1-x+2 x^2\right )}{e^{x^2}-x} \, dx-\frac {1}{5} \int \frac {-5+e^x x}{x} \, dx \\ & = -\left (\frac {1}{25} \int \left (-\frac {e^x}{e^{x^2}-x}-\frac {e^x x}{e^{x^2}-x}+\frac {2 e^x x^2}{e^{x^2}-x}\right ) \, dx\right )-\frac {1}{25} \int \left (-\frac {e^x x}{\left (e^{x^2}-x\right )^2}+\frac {2 e^x x^3}{\left (e^{x^2}-x\right )^2}\right ) \, dx-\frac {1}{5} \int \left (e^x-\frac {5}{x}\right ) \, dx \\ & = \log (x)+\frac {1}{25} \int \frac {e^x}{e^{x^2}-x} \, dx+\frac {1}{25} \int \frac {e^x x}{\left (e^{x^2}-x\right )^2} \, dx+\frac {1}{25} \int \frac {e^x x}{e^{x^2}-x} \, dx-\frac {2}{25} \int \frac {e^x x^2}{e^{x^2}-x} \, dx-\frac {2}{25} \int \frac {e^x x^3}{\left (e^{x^2}-x\right )^2} \, dx-\frac {\int e^x \, dx}{5} \\ & = -\frac {e^x}{5}+\log (x)+\frac {1}{25} \int \frac {e^x}{e^{x^2}-x} \, dx+\frac {1}{25} \int \frac {e^x x}{\left (e^{x^2}-x\right )^2} \, dx+\frac {1}{25} \int \frac {e^x x}{e^{x^2}-x} \, dx-\frac {2}{25} \int \frac {e^x x^2}{e^{x^2}-x} \, dx-\frac {2}{25} \int \frac {e^x x^3}{\left (e^{x^2}-x\right )^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 9.71 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.25 \[ \int \frac {25 x^2-6 e^x x^3+e^{2 x^2} \left (25-5 e^x x\right )+e^{x^2} \left (-50 x+e^x \left (x+11 x^2-2 x^3\right )\right )}{25 e^{2 x^2} x-50 e^{x^2} x^2+25 x^3} \, dx=\frac {1}{25} \left (\frac {e^x \left (-5 e^{x^2}+6 x\right )}{e^{x^2}-x}+25 \log (x)\right ) \]

[In]

Integrate[(25*x^2 - 6*E^x*x^3 + E^(2*x^2)*(25 - 5*E^x*x) + E^x^2*(-50*x + E^x*(x + 11*x^2 - 2*x^3)))/(25*E^(2*
x^2)*x - 50*E^x^2*x^2 + 25*x^3),x]

[Out]

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

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82

method	result	size
risch	\(\ln \left (x \right )-\frac {{\mathrm e}^{x}}{5}-\frac {x \,{\mathrm e}^{x}}{25 \left (-{\mathrm e}^{x^{2}}+x \right )}\)	\(23\)
norman	\(\frac {-\frac {6 \,{\mathrm e}^{x} x}{25}+\frac {{\mathrm e}^{x} {\mathrm e}^{x^{2}}}{5}}{-{\mathrm e}^{x^{2}}+x}+\ln \left (x \right )\)	\(29\)
parts	\(\frac {-\frac {6 \,{\mathrm e}^{x} x}{25}+\frac {{\mathrm e}^{x} {\mathrm e}^{x^{2}}}{5}}{-{\mathrm e}^{x^{2}}+x}+\ln \left (x \right )\)	\(29\)
parallelrisch	\(\frac {25 x \ln \left (x \right )-25 \,{\mathrm e}^{x^{2}} \ln \left (x \right )-6 \,{\mathrm e}^{x} x +5 \,{\mathrm e}^{x} {\mathrm e}^{x^{2}}}{-25 \,{\mathrm e}^{x^{2}}+25 x}\)	\(40\)

[In]

int(((-5*exp(x)*x+25)*exp(x^2)^2+((-2*x^3+11*x^2+x)*exp(x)-50*x)*exp(x^2)-6*exp(x)*x^3+25*x^2)/(25*x*exp(x^2)^
2-50*x^2*exp(x^2)+25*x^3),x,method=_RETURNVERBOSE)

[Out]

ln(x)-1/5*exp(x)-1/25*x*exp(x)/(-exp(x^2)+x)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int \frac {25 x^2-6 e^x x^3+e^{2 x^2} \left (25-5 e^x x\right )+e^{x^2} \left (-50 x+e^x \left (x+11 x^2-2 x^3\right )\right )}{25 e^{2 x^2} x-50 e^{x^2} x^2+25 x^3} \, dx=-\frac {6 \, x e^{x} - 25 \, {\left (x - e^{\left (x^{2}\right )}\right )} \log \left (x\right ) - 5 \, e^{\left (x^{2} + x\right )}}{25 \, {\left (x - e^{\left (x^{2}\right )}\right )}} \]

[In]

integrate(((-5*exp(x)*x+25)*exp(x^2)^2+((-2*x^3+11*x^2+x)*exp(x)-50*x)*exp(x^2)-6*exp(x)*x^3+25*x^2)/(25*x*exp
(x^2)^2-50*x^2*exp(x^2)+25*x^3),x, algorithm="fricas")

[Out]

-1/25*(6*x*e^x - 25*(x - e^(x^2))*log(x) - 5*e^(x^2 + x))/(x - e^(x^2))

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {25 x^2-6 e^x x^3+e^{2 x^2} \left (25-5 e^x x\right )+e^{x^2} \left (-50 x+e^x \left (x+11 x^2-2 x^3\right )\right )}{25 e^{2 x^2} x-50 e^{x^2} x^2+25 x^3} \, dx=\frac {x e^{x}}{- 25 x + 25 e^{x^{2}}} - \frac {e^{x}}{5} + \log {\left (x \right )} \]

[In]

integrate(((-5*exp(x)*x+25)*exp(x**2)**2+((-2*x**3+11*x**2+x)*exp(x)-50*x)*exp(x**2)-6*exp(x)*x**3+25*x**2)/(2
5*x*exp(x**2)**2-50*x**2*exp(x**2)+25*x**3),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {25 x^2-6 e^x x^3+e^{2 x^2} \left (25-5 e^x x\right )+e^{x^2} \left (-50 x+e^x \left (x+11 x^2-2 x^3\right )\right )}{25 e^{2 x^2} x-50 e^{x^2} x^2+25 x^3} \, dx=-\frac {6 \, x e^{x} - 5 \, e^{\left (x^{2} + x\right )}}{25 \, {\left (x - e^{\left (x^{2}\right )}\right )}} + \log \left (x\right ) \]

[In]

integrate(((-5*exp(x)*x+25)*exp(x^2)^2+((-2*x^3+11*x^2+x)*exp(x)-50*x)*exp(x^2)-6*exp(x)*x^3+25*x^2)/(25*x*exp
(x^2)^2-50*x^2*exp(x^2)+25*x^3),x, algorithm="maxima")

[Out]

-1/25*(6*x*e^x - 5*e^(x^2 + x))/(x - e^(x^2)) + log(x)

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int \frac {25 x^2-6 e^x x^3+e^{2 x^2} \left (25-5 e^x x\right )+e^{x^2} \left (-50 x+e^x \left (x+11 x^2-2 x^3\right )\right )}{25 e^{2 x^2} x-50 e^{x^2} x^2+25 x^3} \, dx=-\frac {6 \, x e^{x} - 25 \, x \log \left (x\right ) + 25 \, e^{\left (x^{2}\right )} \log \left (x\right ) - 5 \, e^{\left (x^{2} + x\right )}}{25 \, {\left (x - e^{\left (x^{2}\right )}\right )}} \]

[In]

integrate(((-5*exp(x)*x+25)*exp(x^2)^2+((-2*x^3+11*x^2+x)*exp(x)-50*x)*exp(x^2)-6*exp(x)*x^3+25*x^2)/(25*x*exp
(x^2)^2-50*x^2*exp(x^2)+25*x^3),x, algorithm="giac")

[Out]

-1/25*(6*x*e^x - 25*x*log(x) + 25*e^(x^2)*log(x) - 5*e^(x^2 + x))/(x - e^(x^2))

Mupad [B] (verification not implemented)

Time = 12.65 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {25 x^2-6 e^x x^3+e^{2 x^2} \left (25-5 e^x x\right )+e^{x^2} \left (-50 x+e^x \left (x+11 x^2-2 x^3\right )\right )}{25 e^{2 x^2} x-50 e^{x^2} x^2+25 x^3} \, dx=\ln \left (x\right )+\frac {5\,{\mathrm {e}}^{x^2+x}-6\,x\,{\mathrm {e}}^x}{25\,x-25\,{\mathrm {e}}^{x^2}} \]

[In]

int(-(exp(2*x^2)*(5*x*exp(x) - 25) + 6*x^3*exp(x) + exp(x^2)*(50*x - exp(x)*(x + 11*x^2 - 2*x^3)) - 25*x^2)/(2
5*x*exp(2*x^2) - 50*x^2*exp(x^2) + 25*x^3),x)

[Out]

log(x) + (5*exp(x + x^2) - 6*x*exp(x))/(25*x - 25*exp(x^2))