\(\int \frac {-8-3 e^{10} x^3+e^{10+x} x^3}{4 x-e^{15} x^3+e^{10+x} x^3+e^{10} (4 x^3-3 x^4)} \, dx\) [9542]

Optimal result

Integrand size = 58, antiderivative size = 22 \[ \int \frac {-8-3 e^{10} x^3+e^{10+x} x^3}{4 x-e^{15} x^3+e^{10+x} x^3+e^{10} \left (4 x^3-3 x^4\right )} \, dx=\log \left (-4+e^5-e^x-\frac {4}{e^{10} x^2}+3 x\right ) \]

[Out]

ln(exp(5)+3*x-4-exp(x)-4/x^2/exp(5)^2)

Rubi [F]

\[ \int \frac {-8-3 e^{10} x^3+e^{10+x} x^3}{4 x-e^{15} x^3+e^{10+x} x^3+e^{10} \left (4 x^3-3 x^4\right )} \, dx=\int \frac {-8-3 e^{10} x^3+e^{10+x} x^3}{4 x-e^{15} x^3+e^{10+x} x^3+e^{10} \left (4 x^3-3 x^4\right )} \, dx \]

[In]

Int[(-8 - 3*E^10*x^3 + E^(10 + x)*x^3)/(4*x - E^15*x^3 + E^(10 + x)*x^3 + E^10*(4*x^3 - 3*x^4)),x]

[Out]

x - E^10*(7 - E^5)*Defer[Int][x^2/(4 + E^(10 + x)*x^2 + 4*E^10*(1 - E^5/4)*x^2 - 3*E^10*x^3), x] + 3*E^10*Defe
r[Int][x^3/(4 + E^(10 + x)*x^2 + 4*E^10*(1 - E^5/4)*x^2 - 3*E^10*x^3), x] + 4*Defer[Int][(-4 - E^(10 + x)*x^2
- 4*E^10*(1 - E^5/4)*x^2 + 3*E^10*x^3)^(-1), x] + 8*Defer[Int][1/(x*(-4 - E^(10 + x)*x^2 - 4*E^10*(1 - E^5/4)*
x^2 + 3*E^10*x^3)), x]

Rubi steps \begin{align*} \text {integral}& = \int \left (1+\frac {-8-4 x-e^{10} \left (7-e^5\right ) x^3+3 e^{10} x^4}{x \left (4+e^{10+x} x^2+4 e^{10} \left (1-\frac {e^5}{4}\right ) x^2-3 e^{10} x^3\right )}\right ) \, dx \\ & = x+\int \frac {-8-4 x-e^{10} \left (7-e^5\right ) x^3+3 e^{10} x^4}{x \left (4+e^{10+x} x^2+4 e^{10} \left (1-\frac {e^5}{4}\right ) x^2-3 e^{10} x^3\right )} \, dx \\ & = x+\int \left (\frac {e^{10} \left (-7+e^5\right ) x^2}{4+e^{10+x} x^2+4 e^{10} \left (1-\frac {e^5}{4}\right ) x^2-3 e^{10} x^3}+\frac {3 e^{10} x^3}{4+e^{10+x} x^2+4 e^{10} \left (1-\frac {e^5}{4}\right ) x^2-3 e^{10} x^3}+\frac {4}{-4-e^{10+x} x^2-4 e^{10} \left (1-\frac {e^5}{4}\right ) x^2+3 e^{10} x^3}+\frac {8}{x \left (-4-e^{10+x} x^2-4 e^{10} \left (1-\frac {e^5}{4}\right ) x^2+3 e^{10} x^3\right )}\right ) \, dx \\ & = x+4 \int \frac {1}{-4-e^{10+x} x^2-4 e^{10} \left (1-\frac {e^5}{4}\right ) x^2+3 e^{10} x^3} \, dx+8 \int \frac {1}{x \left (-4-e^{10+x} x^2-4 e^{10} \left (1-\frac {e^5}{4}\right ) x^2+3 e^{10} x^3\right )} \, dx+\left (3 e^{10}\right ) \int \frac {x^3}{4+e^{10+x} x^2+4 e^{10} \left (1-\frac {e^5}{4}\right ) x^2-3 e^{10} x^3} \, dx-\left (e^{10} \left (7-e^5\right )\right ) \int \frac {x^2}{4+e^{10+x} x^2+4 e^{10} \left (1-\frac {e^5}{4}\right ) x^2-3 e^{10} x^3} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 5.12 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.68 \[ \int \frac {-8-3 e^{10} x^3+e^{10+x} x^3}{4 x-e^{15} x^3+e^{10+x} x^3+e^{10} \left (4 x^3-3 x^4\right )} \, dx=-2 \log (x)+\log \left (4-e^{15} x^2+e^{10+x} x^2+e^{10} (4-3 x) x^2\right ) \]

[In]

Integrate[(-8 - 3*E^10*x^3 + E^(10 + x)*x^3)/(4*x - E^15*x^3 + E^(10 + x)*x^3 + E^10*(4*x^3 - 3*x^4)),x]

[Out]

-2*Log[x] + Log[4 - E^15*x^2 + E^(10 + x)*x^2 + E^10*(4 - 3*x)*x^2]

Maple [A] (verified)

Time = 0.92 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.55

method	result	size
risch	\(\ln \left (\frac {-3 x^{3}-x^{2} {\mathrm e}^{5}+4 x^{2}+{\mathrm e}^{x} x^{2}+4 \,{\mathrm e}^{-10}}{x^{2}}\right )\)	\(34\)
norman	\(-2 \ln \left (x \right )+\ln \left (x^{2} {\mathrm e}^{15}-{\mathrm e}^{x} {\mathrm e}^{10} x^{2}+3 x^{3} {\mathrm e}^{10}-4 x^{2} {\mathrm e}^{10}-4\right )\)	\(46\)
parallelrisch	\(-2 \ln \left (x \right )+\ln \left (\frac {\left (x^{2} {\mathrm e}^{15}-{\mathrm e}^{x} {\mathrm e}^{10} x^{2}+3 x^{3} {\mathrm e}^{10}-4 x^{2} {\mathrm e}^{10}-4\right ) {\mathrm e}^{-10}}{3}\right )\)	\(52\)

[In]

int((x^3*exp(5)^2*exp(x)-3*x^3*exp(5)^2-8)/(x^3*exp(5)^2*exp(x)-x^3*exp(5)^3+(-3*x^4+4*x^3)*exp(5)^2+4*x),x,me
thod=_RETURNVERBOSE)

[Out]

ln((-3*x^3-x^2*exp(5)+4*x^2+exp(x)*x^2+4*exp(-10))/x^2)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.68 \[ \int \frac {-8-3 e^{10} x^3+e^{10+x} x^3}{4 x-e^{15} x^3+e^{10+x} x^3+e^{10} \left (4 x^3-3 x^4\right )} \, dx=\log \left (-\frac {x^{2} e^{15} - x^{2} e^{\left (x + 10\right )} + {\left (3 \, x^{3} - 4 \, x^{2}\right )} e^{10} - 4}{x^{2}}\right ) \]

[In]

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

[Out]

log(-(x^2*e^15 - x^2*e^(x + 10) + (3*x^3 - 4*x^2)*e^10 - 4)/x^2)

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.64 \[ \int \frac {-8-3 e^{10} x^3+e^{10+x} x^3}{4 x-e^{15} x^3+e^{10+x} x^3+e^{10} \left (4 x^3-3 x^4\right )} \, dx=\log {\left (e^{x} + \frac {- 3 x^{3} e^{10} - x^{2} e^{15} + 4 x^{2} e^{10} + 4}{x^{2} e^{10}} \right )} \]

[In]

integrate((x**3*exp(5)**2*exp(x)-3*x**3*exp(5)**2-8)/(x**3*exp(5)**2*exp(x)-x**3*exp(5)**3+(-3*x**4+4*x**3)*ex
p(5)**2+4*x),x)

[Out]

log(exp(x) + (-3*x**3*exp(10) - x**2*exp(15) + 4*x**2*exp(10) + 4)*exp(-10)/x**2)

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.68 \[ \int \frac {-8-3 e^{10} x^3+e^{10+x} x^3}{4 x-e^{15} x^3+e^{10+x} x^3+e^{10} \left (4 x^3-3 x^4\right )} \, dx=\log \left (-\frac {{\left (3 \, x^{3} e^{10} + x^{2} {\left (e^{15} - 4 \, e^{10}\right )} - x^{2} e^{\left (x + 10\right )} - 4\right )} e^{\left (-10\right )}}{x^{2}}\right ) \]

[In]

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

[Out]

log(-(3*x^3*e^10 + x^2*(e^15 - 4*e^10) - x^2*e^(x + 10) - 4)*e^(-10)/x^2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.68 \[ \int \frac {-8-3 e^{10} x^3+e^{10+x} x^3}{4 x-e^{15} x^3+e^{10+x} x^3+e^{10} \left (4 x^3-3 x^4\right )} \, dx=\log \left (-3 \, x^{3} e^{10} - x^{2} e^{15} + 4 \, x^{2} e^{10} + x^{2} e^{\left (x + 10\right )} + 4\right ) - 2 \, \log \left (x\right ) \]

[In]

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

[Out]

log(-3*x^3*e^10 - x^2*e^15 + 4*x^2*e^10 + x^2*e^(x + 10) + 4) - 2*log(x)

Mupad [B] (verification not implemented)

Time = 15.03 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.50 \[ \int \frac {-8-3 e^{10} x^3+e^{10+x} x^3}{4 x-e^{15} x^3+e^{10+x} x^3+e^{10} \left (4 x^3-3 x^4\right )} \, dx=\ln \left (\frac {x^2\,{\mathrm {e}}^5}{3}-\frac {x^2\,{\mathrm {e}}^x}{3}-\frac {4\,{\mathrm {e}}^{-10}}{3}-\frac {4\,x^2}{3}+x^3\right )-2\,\ln \left (x\right ) \]

[In]

int(-(3*x^3*exp(10) - x^3*exp(10)*exp(x) + 8)/(4*x + exp(10)*(4*x^3 - 3*x^4) - x^3*exp(15) + x^3*exp(10)*exp(x
)),x)

[Out]

log((x^2*exp(5))/3 - (x^2*exp(x))/3 - (4*exp(-10))/3 - (4*x^2)/3 + x^3) - 2*log(x)