\(\int \frac {-10+e^{2 x^2} (64 x^3-6 x^4+64 x^5-8 x^6)}{40-5 x+e^{2 x^2} (-8 x^4+x^5)} \, dx\) [2572]

Optimal result

Integrand size = 56, antiderivative size = 22 \[ \int \frac {-10+e^{2 x^2} \left (64 x^3-6 x^4+64 x^5-8 x^6\right )}{40-5 x+e^{2 x^2} \left (-8 x^4+x^5\right )} \, dx=\log \left (\frac {(-8+x)^2}{\left (-5+e^{2 x^2} x^4\right )^2}\right ) \]

[Out]

ln((-8+x)^2/(x^4*exp(x^2)^2-5)^2)

Rubi [F]

\[ \int \frac {-10+e^{2 x^2} \left (64 x^3-6 x^4+64 x^5-8 x^6\right )}{40-5 x+e^{2 x^2} \left (-8 x^4+x^5\right )} \, dx=\int \frac {-10+e^{2 x^2} \left (64 x^3-6 x^4+64 x^5-8 x^6\right )}{40-5 x+e^{2 x^2} \left (-8 x^4+x^5\right )} \, dx \]

[In]

Int[(-10 + E^(2*x^2)*(64*x^3 - 6*x^4 + 64*x^5 - 8*x^6))/(40 - 5*x + E^(2*x^2)*(-8*x^4 + x^5)),x]

[Out]

-4*x^2 + 2*Log[8 - x] - 8*Log[x] - 20*Defer[Subst][Defer[Int][(-5 + E^(2*x)*x^2)^(-1), x], x, x^2] - 20*Defer[
Subst][Defer[Int][1/(x*(-5 + E^(2*x)*x^2)), x], x, x^2]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-10+e^{2 x^2} \left (64 x^3-6 x^4+64 x^5-8 x^6\right )}{(8-x) \left (5-e^{2 x^2} x^4\right )} \, dx \\ & = \int \left (-\frac {2 \left (-32+3 x-32 x^2+4 x^3\right )}{(-8+x) x}-\frac {40 \left (1+x^2\right )}{x \left (-5+e^{2 x^2} x^4\right )}\right ) \, dx \\ & = -\left (2 \int \frac {-32+3 x-32 x^2+4 x^3}{(-8+x) x} \, dx\right )-40 \int \frac {1+x^2}{x \left (-5+e^{2 x^2} x^4\right )} \, dx \\ & = -\left (2 \int \left (\frac {1}{8-x}+\frac {4}{x}+4 x\right ) \, dx\right )-20 \text {Subst}\left (\int \frac {1+x}{x \left (-5+e^{2 x} x^2\right )} \, dx,x,x^2\right ) \\ & = -4 x^2+2 \log (8-x)-8 \log (x)-20 \text {Subst}\left (\int \left (\frac {1}{-5+e^{2 x} x^2}+\frac {1}{x \left (-5+e^{2 x} x^2\right )}\right ) \, dx,x,x^2\right ) \\ & = -4 x^2+2 \log (8-x)-8 \log (x)-20 \text {Subst}\left (\int \frac {1}{-5+e^{2 x} x^2} \, dx,x,x^2\right )-20 \text {Subst}\left (\int \frac {1}{x \left (-5+e^{2 x} x^2\right )} \, dx,x,x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 2.37 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.68 \[ \int \frac {-10+e^{2 x^2} \left (64 x^3-6 x^4+64 x^5-8 x^6\right )}{40-5 x+e^{2 x^2} \left (-8 x^4+x^5\right )} \, dx=-4 x^2-4 \text {arctanh}\left (1-\frac {2}{5} e^{2 x^2} x^4\right )+2 \log (8-x)-8 \log (x) \]

[In]

Integrate[(-10 + E^(2*x^2)*(64*x^3 - 6*x^4 + 64*x^5 - 8*x^6))/(40 - 5*x + E^(2*x^2)*(-8*x^4 + x^5)),x]

[Out]

-4*x^2 - 4*ArcTanh[1 - (2*E^(2*x^2)*x^4)/5] + 2*Log[8 - x] - 8*Log[x]

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05

method	result	size
norman	\(2 \ln \left (-8+x \right )-2 \ln \left ({\mathrm e}^{2 x^{2}} x^{4}-5\right )\)	\(23\)
parallelrisch	\(2 \ln \left (-8+x \right )-2 \ln \left ({\mathrm e}^{2 x^{2}} x^{4}-5\right )\)	\(23\)
risch	\(-8 \ln \left (x \right )+2 \ln \left (-8+x \right )-2 \ln \left ({\mathrm e}^{2 x^{2}}-\frac {5}{x^{4}}\right )\)	\(27\)

[In]

int(((-8*x^6+64*x^5-6*x^4+64*x^3)*exp(x^2)^2-10)/((x^5-8*x^4)*exp(x^2)^2-5*x+40),x,method=_RETURNVERBOSE)

[Out]

2*ln(-8+x)-2*ln(x^4*exp(x^2)^2-5)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.36 \[ \int \frac {-10+e^{2 x^2} \left (64 x^3-6 x^4+64 x^5-8 x^6\right )}{40-5 x+e^{2 x^2} \left (-8 x^4+x^5\right )} \, dx=2 \, \log \left (x - 8\right ) - 8 \, \log \left (x\right ) - 2 \, \log \left (\frac {x^{4} e^{\left (2 \, x^{2}\right )} - 5}{x^{4}}\right ) \]

[In]

integrate(((-8*x^6+64*x^5-6*x^4+64*x^3)*exp(x^2)^2-10)/((x^5-8*x^4)*exp(x^2)^2-5*x+40),x, algorithm="fricas")

[Out]

2*log(x - 8) - 8*log(x) - 2*log((x^4*e^(2*x^2) - 5)/x^4)

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {-10+e^{2 x^2} \left (64 x^3-6 x^4+64 x^5-8 x^6\right )}{40-5 x+e^{2 x^2} \left (-8 x^4+x^5\right )} \, dx=- 8 \log {\left (x \right )} + 2 \log {\left (x - 8 \right )} - 2 \log {\left (e^{2 x^{2}} - \frac {5}{x^{4}} \right )} \]

[In]

integrate(((-8*x**6+64*x**5-6*x**4+64*x**3)*exp(x**2)**2-10)/((x**5-8*x**4)*exp(x**2)**2-5*x+40),x)

[Out]

-8*log(x) + 2*log(x - 8) - 2*log(exp(2*x**2) - 5/x**4)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.36 \[ \int \frac {-10+e^{2 x^2} \left (64 x^3-6 x^4+64 x^5-8 x^6\right )}{40-5 x+e^{2 x^2} \left (-8 x^4+x^5\right )} \, dx=2 \, \log \left (x - 8\right ) - 8 \, \log \left (x\right ) - 2 \, \log \left (\frac {x^{4} e^{\left (2 \, x^{2}\right )} - 5}{x^{4}}\right ) \]

[In]

integrate(((-8*x^6+64*x^5-6*x^4+64*x^3)*exp(x^2)^2-10)/((x^5-8*x^4)*exp(x^2)^2-5*x+40),x, algorithm="maxima")

[Out]

2*log(x - 8) - 8*log(x) - 2*log((x^4*e^(2*x^2) - 5)/x^4)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {-10+e^{2 x^2} \left (64 x^3-6 x^4+64 x^5-8 x^6\right )}{40-5 x+e^{2 x^2} \left (-8 x^4+x^5\right )} \, dx=-2 \, \log \left (x^{4} e^{\left (2 \, x^{2}\right )} - 5\right ) + 2 \, \log \left (x - 8\right ) \]

[In]

integrate(((-8*x^6+64*x^5-6*x^4+64*x^3)*exp(x^2)^2-10)/((x^5-8*x^4)*exp(x^2)^2-5*x+40),x, algorithm="giac")

[Out]

-2*log(x^4*e^(2*x^2) - 5) + 2*log(x - 8)

Mupad [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {-10+e^{2 x^2} \left (64 x^3-6 x^4+64 x^5-8 x^6\right )}{40-5 x+e^{2 x^2} \left (-8 x^4+x^5\right )} \, dx=2\,\ln \left (x-8\right )-2\,\ln \left (x^4\,{\mathrm {e}}^{2\,x^2}-5\right ) \]

[In]

int(-(exp(2*x^2)*(64*x^3 - 6*x^4 + 64*x^5 - 8*x^6) - 10)/(5*x + exp(2*x^2)*(8*x^4 - x^5) - 40),x)

[Out]

2*log(x - 8) - 2*log(x^4*exp(2*x^2) - 5)