\(\int \frac {-10-20 e^{2 x} x^3-10 e^{4 x} x^6+e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} (-12 x^2-8 x^3)}{5+10 e^{2 x} x^3+5 e^{4 x} x^6} \, dx\) [5836]

Optimal result

Integrand size = 81, antiderivative size = 24 \[ \int \frac {-10-20 e^{2 x} x^3-10 e^{4 x} x^6+e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} \left (-12 x^2-8 x^3\right )}{5+10 e^{2 x} x^3+5 e^{4 x} x^6} \, dx=e^{\frac {4 x}{5 \left (x+e^{2 x} x^4\right )}}-2 x \]

[Out]

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

Rubi [F]

\[ \int \frac {-10-20 e^{2 x} x^3-10 e^{4 x} x^6+e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} \left (-12 x^2-8 x^3\right )}{5+10 e^{2 x} x^3+5 e^{4 x} x^6} \, dx=\int \frac {-10-20 e^{2 x} x^3-10 e^{4 x} x^6+e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} \left (-12 x^2-8 x^3\right )}{5+10 e^{2 x} x^3+5 e^{4 x} x^6} \, dx \]

[In]

Int[(-10 - 20*E^(2*x)*x^3 - 10*E^(4*x)*x^6 + E^(2*x + 4/(5 + 5*E^(2*x)*x^3))*(-12*x^2 - 8*x^3))/(5 + 10*E^(2*x
)*x^3 + 5*E^(4*x)*x^6),x]

[Out]

-2*x - (12*Defer[Int][(E^(2*x + 4/(5 + 5*E^(2*x)*x^3))*x^2)/(1 + E^(2*x)*x^3)^2, x])/5 - (8*Defer[Int][(E^(2*x
+ 4/(5 + 5*E^(2*x)*x^3))*x^3)/(1 + E^(2*x)*x^3)^2, x])/5

Rubi steps \begin{align*} \text {integral}& = \int \frac {-10-20 e^{2 x} x^3-10 e^{4 x} x^6+e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} \left (-12 x^2-8 x^3\right )}{5 \left (1+e^{2 x} x^3\right )^2} \, dx \\ & = \frac {1}{5} \int \frac {-10-20 e^{2 x} x^3-10 e^{4 x} x^6+e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} \left (-12 x^2-8 x^3\right )}{\left (1+e^{2 x} x^3\right )^2} \, dx \\ & = \frac {1}{5} \int \left (-10-\frac {4 e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} x^2 (3+2 x)}{\left (1+e^{2 x} x^3\right )^2}\right ) \, dx \\ & = -2 x-\frac {4}{5} \int \frac {e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} x^2 (3+2 x)}{\left (1+e^{2 x} x^3\right )^2} \, dx \\ & = -2 x-\frac {4}{5} \int \left (\frac {3 e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} x^2}{\left (1+e^{2 x} x^3\right )^2}+\frac {2 e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} x^3}{\left (1+e^{2 x} x^3\right )^2}\right ) \, dx \\ & = -2 x-\frac {8}{5} \int \frac {e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} x^3}{\left (1+e^{2 x} x^3\right )^2} \, dx-\frac {12}{5} \int \frac {e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} x^2}{\left (1+e^{2 x} x^3\right )^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.85 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {-10-20 e^{2 x} x^3-10 e^{4 x} x^6+e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} \left (-12 x^2-8 x^3\right )}{5+10 e^{2 x} x^3+5 e^{4 x} x^6} \, dx=-\frac {2}{5} \left (-\frac {5}{2} e^{\frac {4}{5 \left (1+e^{2 x} x^3\right )}}+5 x\right ) \]

[In]

Integrate[(-10 - 20*E^(2*x)*x^3 - 10*E^(4*x)*x^6 + E^(2*x + 4/(5 + 5*E^(2*x)*x^3))*(-12*x^2 - 8*x^3))/(5 + 10*
E^(2*x)*x^3 + 5*E^(4*x)*x^6),x]

[Out]

(-2*((-5*E^(4/(5*(1 + E^(2*x)*x^3))))/2 + 5*x))/5

Maple [A] (verified)

Time = 7.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83

method	result	size
risch	\({\mathrm e}^{\frac {4}{5 \left ({\mathrm e}^{2 x} x^{3}+1\right )}}-2 x\)	\(20\)

[In]

int(((-8*x^3-12*x^2)*exp(x)^2*exp(2/(5*exp(x)^2*x^3+5))^2-10*x^6*exp(x)^4-20*exp(x)^2*x^3-10)/(5*x^6*exp(x)^4+
10*exp(x)^2*x^3+5),x,method=_RETURNVERBOSE)

[Out]

exp(4/5/(exp(2*x)*x^3+1))-2*x

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (20) = 40\).

Time = 0.27 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.88 \[ \int \frac {-10-20 e^{2 x} x^3-10 e^{4 x} x^6+e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} \left (-12 x^2-8 x^3\right )}{5+10 e^{2 x} x^3+5 e^{4 x} x^6} \, dx=-{\left (2 \, x e^{\left (2 \, x\right )} - e^{\left (\frac {2 \, {\left (5 \, x^{4} e^{\left (2 \, x\right )} + 5 \, x + 2\right )}}{5 \, {\left (x^{3} e^{\left (2 \, x\right )} + 1\right )}}\right )}\right )} e^{\left (-2 \, x\right )} \]

[In]

integrate(((-8*x^3-12*x^2)*exp(x)^2*exp(2/(5*exp(x)^2*x^3+5))^2-10*x^6*exp(x)^4-20*exp(x)^2*x^3-10)/(5*x^6*exp
(x)^4+10*exp(x)^2*x^3+5),x, algorithm="fricas")

[Out]

-(2*x*e^(2*x) - e^(2/5*(5*x^4*e^(2*x) + 5*x + 2)/(x^3*e^(2*x) + 1)))*e^(-2*x)

Sympy [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {-10-20 e^{2 x} x^3-10 e^{4 x} x^6+e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} \left (-12 x^2-8 x^3\right )}{5+10 e^{2 x} x^3+5 e^{4 x} x^6} \, dx=- 2 x + e^{\frac {4}{5 x^{3} e^{2 x} + 5}} \]

[In]

integrate(((-8*x**3-12*x**2)*exp(x)**2*exp(2/(5*exp(x)**2*x**3+5))**2-10*x**6*exp(x)**4-20*exp(x)**2*x**3-10)/
(5*x**6*exp(x)**4+10*exp(x)**2*x**3+5),x)

[Out]

-2*x + exp(4/(5*x**3*exp(2*x) + 5))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {-10-20 e^{2 x} x^3-10 e^{4 x} x^6+e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} \left (-12 x^2-8 x^3\right )}{5+10 e^{2 x} x^3+5 e^{4 x} x^6} \, dx=-2 \, x + e^{\left (\frac {4}{5 \, {\left (x^{3} e^{\left (2 \, x\right )} + 1\right )}}\right )} \]

[In]

integrate(((-8*x^3-12*x^2)*exp(x)^2*exp(2/(5*exp(x)^2*x^3+5))^2-10*x^6*exp(x)^4-20*exp(x)^2*x^3-10)/(5*x^6*exp
(x)^4+10*exp(x)^2*x^3+5),x, algorithm="maxima")

[Out]

-2*x + e^(4/5/(x^3*e^(2*x) + 1))

Giac [F]

\[ \int \frac {-10-20 e^{2 x} x^3-10 e^{4 x} x^6+e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} \left (-12 x^2-8 x^3\right )}{5+10 e^{2 x} x^3+5 e^{4 x} x^6} \, dx=\int { -\frac {2 \, {\left (5 \, x^{6} e^{\left (4 \, x\right )} + 10 \, x^{3} e^{\left (2 \, x\right )} + 2 \, {\left (2 \, x^{3} + 3 \, x^{2}\right )} e^{\left (2 \, x + \frac {4}{5 \, {\left (x^{3} e^{\left (2 \, x\right )} + 1\right )}}\right )} + 5\right )}}{5 \, {\left (x^{6} e^{\left (4 \, x\right )} + 2 \, x^{3} e^{\left (2 \, x\right )} + 1\right )}} \,d x } \]

[In]

integrate(((-8*x^3-12*x^2)*exp(x)^2*exp(2/(5*exp(x)^2*x^3+5))^2-10*x^6*exp(x)^4-20*exp(x)^2*x^3-10)/(5*x^6*exp
(x)^4+10*exp(x)^2*x^3+5),x, algorithm="giac")

[Out]

integrate(-2/5*(5*x^6*e^(4*x) + 10*x^3*e^(2*x) + 2*(2*x^3 + 3*x^2)*e^(2*x + 4/5/(x^3*e^(2*x) + 1)) + 5)/(x^6*e
^(4*x) + 2*x^3*e^(2*x) + 1), x)

Mupad [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {-10-20 e^{2 x} x^3-10 e^{4 x} x^6+e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} \left (-12 x^2-8 x^3\right )}{5+10 e^{2 x} x^3+5 e^{4 x} x^6} \, dx={\mathrm {e}}^{\frac {4}{5\,x^3\,{\mathrm {e}}^{2\,x}+5}}-2\,x \]

[In]

int(-(20*x^3*exp(2*x) + 10*x^6*exp(4*x) + exp(2*x)*exp(4/(5*x^3*exp(2*x) + 5))*(12*x^2 + 8*x^3) + 10)/(10*x^3*
exp(2*x) + 5*x^6*exp(4*x) + 5),x)

[Out]

exp(4/(5*x^3*exp(2*x) + 5)) - 2*x