\(\int \frac {e^{-8+4 x} (4 x^3-4 x^4)}{e^{-16+8 x}+x^8-2 x^8 \log (3)+x^8 \log ^2(3)+e^{-8+4 x} (2 x^4-2 x^4 \log (3))} \, dx\) [1091]

Optimal result

Integrand size = 68, antiderivative size = 19 \[ \int \frac {e^{-8+4 x} \left (4 x^3-4 x^4\right )}{e^{-16+8 x}+x^8-2 x^8 \log (3)+x^8 \log ^2(3)+e^{-8+4 x} \left (2 x^4-2 x^4 \log (3)\right )} \, dx=\frac {1}{1+\frac {e^{-8+4 x}}{x^4}-\log (3)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.37, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {6, 1607, 6820, 12, 6843, 32} \[ \int \frac {e^{-8+4 x} \left (4 x^3-4 x^4\right )}{e^{-16+8 x}+x^8-2 x^8 \log (3)+x^8 \log ^2(3)+e^{-8+4 x} \left (2 x^4-2 x^4 \log (3)\right )} \, dx=\frac {e^8}{\frac {e^{4 x}}{x^4}+e^8 (1-\log (3))} \]

[In]

Int[(E^(-8 + 4*x)*(4*x^3 - 4*x^4))/(E^(-16 + 8*x) + x^8 - 2*x^8*Log[3] + x^8*Log[3]^2 + E^(-8 + 4*x)*(2*x^4 -
2*x^4*Log[3])),x]

[Out]

E^8/(E^(4*x)/x^4 + E^8*(1 - Log[3]))

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6843

Int[(u_)*((a_.)*(v_)^(p_.) + (b_.)*(w_)^(q_.))^(m_.), x_Symbol] :> With[{c = Simplify[u/(p*w*D[v, x] - q*v*D[w
, x])]}, Dist[c*p, Subst[Int[(b + a*x^p)^m, x], x, v*w^(m*q + 1)], x] /; FreeQ[c, x]] /; FreeQ[{a, b, m, p, q}
, x] && EqQ[p + q*(m*p + 1), 0] && IntegerQ[p] && IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{-8+4 x} \left (4 x^3-4 x^4\right )}{e^{-16+8 x}+x^8 (1-2 \log (3))+x^8 \log ^2(3)+e^{-8+4 x} \left (2 x^4-2 x^4 \log (3)\right )} \, dx \\ & = \int \frac {e^{-8+4 x} \left (4 x^3-4 x^4\right )}{e^{-16+8 x}+e^{-8+4 x} \left (2 x^4-2 x^4 \log (3)\right )+x^8 \left (1-2 \log (3)+\log ^2(3)\right )} \, dx \\ & = \int \frac {e^{-8+4 x} (4-4 x) x^3}{e^{-16+8 x}+e^{-8+4 x} \left (2 x^4-2 x^4 \log (3)\right )+x^8 \left (1-2 \log (3)+\log ^2(3)\right )} \, dx \\ & = \int \frac {4 e^{8+4 x} (1-x) x^3}{\left (e^{4 x}-e^8 x^4 (-1+\log (3))\right )^2} \, dx \\ & = 4 \int \frac {e^{8+4 x} (1-x) x^3}{\left (e^{4 x}-e^8 x^4 (-1+\log (3))\right )^2} \, dx \\ & = -\left (e^8 \text {Subst}\left (\int \frac {1}{\left (x-e^8 (-1+\log (3))\right )^2} \, dx,x,\frac {e^{4 x}}{x^4}\right )\right ) \\ & = \frac {e^8}{\frac {e^{4 x}}{x^4}+e^8 (1-\log (3))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.58 \[ \int \frac {e^{-8+4 x} \left (4 x^3-4 x^4\right )}{e^{-16+8 x}+x^8-2 x^8 \log (3)+x^8 \log ^2(3)+e^{-8+4 x} \left (2 x^4-2 x^4 \log (3)\right )} \, dx=-\frac {4 e^8 x^4}{-4 e^{4 x}+4 e^8 x^4 (-1+\log (3))} \]

[In]

Integrate[(E^(-8 + 4*x)*(4*x^3 - 4*x^4))/(E^(-16 + 8*x) + x^8 - 2*x^8*Log[3] + x^8*Log[3]^2 + E^(-8 + 4*x)*(2*
x^4 - 2*x^4*Log[3])),x]

[Out]

(-4*E^8*x^4)/(-4*E^(4*x) + 4*E^8*x^4*(-1 + Log[3]))

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.47

method	result	size
risch	\(-\frac {x^{4}}{x^{4} \ln \left (3\right )-{\mathrm e}^{4 x -8}-x^{4}}\)	\(28\)
parallelrisch	\(-\frac {x^{4}}{x^{4} \ln \left (3\right )-{\mathrm e}^{4 x -8}-x^{4}}\)	\(28\)

[In]

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

[Out]

-x^4/(x^4*ln(3)-exp(4*x-8)-x^4)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.42 \[ \int \frac {e^{-8+4 x} \left (4 x^3-4 x^4\right )}{e^{-16+8 x}+x^8-2 x^8 \log (3)+x^8 \log ^2(3)+e^{-8+4 x} \left (2 x^4-2 x^4 \log (3)\right )} \, dx=-\frac {x^{4}}{x^{4} \log \left (3\right ) - x^{4} - e^{\left (4 \, x - 8\right )}} \]

[In]

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

[Out]

-x^4/(x^4*log(3) - x^4 - e^(4*x - 8))

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-8+4 x} \left (4 x^3-4 x^4\right )}{e^{-16+8 x}+x^8-2 x^8 \log (3)+x^8 \log ^2(3)+e^{-8+4 x} \left (2 x^4-2 x^4 \log (3)\right )} \, dx=\frac {x^{4}}{- x^{4} \log {\left (3 \right )} + x^{4} + e^{4 x - 8}} \]

[In]

integrate((-4*x**4+4*x**3)*exp(-2+x)**4/(exp(-2+x)**8+(-2*x**4*ln(3)+2*x**4)*exp(-2+x)**4+x**8*ln(3)**2-2*x**8
*ln(3)+x**8),x)

[Out]

x**4/(-x**4*log(3) + x**4 + exp(4*x - 8))

Maxima [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.37 \[ \int \frac {e^{-8+4 x} \left (4 x^3-4 x^4\right )}{e^{-16+8 x}+x^8-2 x^8 \log (3)+x^8 \log ^2(3)+e^{-8+4 x} \left (2 x^4-2 x^4 \log (3)\right )} \, dx=-\frac {x^{4} e^{8}}{x^{4} {\left (\log \left (3\right ) - 1\right )} e^{8} - e^{\left (4 \, x\right )}} \]

[In]

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

[Out]

-x^4*e^8/(x^4*(log(3) - 1)*e^8 - e^(4*x))

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.63 \[ \int \frac {e^{-8+4 x} \left (4 x^3-4 x^4\right )}{e^{-16+8 x}+x^8-2 x^8 \log (3)+x^8 \log ^2(3)+e^{-8+4 x} \left (2 x^4-2 x^4 \log (3)\right )} \, dx=-\frac {x^{4} e^{8}}{x^{4} e^{8} \log \left (3\right ) - x^{4} e^{8} - e^{\left (4 \, x\right )}} \]

[In]

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

[Out]

-x^4*e^8/(x^4*e^8*log(3) - x^4*e^8 - e^(4*x))

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{-8+4 x} \left (4 x^3-4 x^4\right )}{e^{-16+8 x}+x^8-2 x^8 \log (3)+x^8 \log ^2(3)+e^{-8+4 x} \left (2 x^4-2 x^4 \log (3)\right )} \, dx=\int \frac {{\mathrm {e}}^{4\,x-8}\,\left (4\,x^3-4\,x^4\right )}{{\mathrm {e}}^{8\,x-16}+x^8\,{\ln \left (3\right )}^2-{\mathrm {e}}^{4\,x-8}\,\left (2\,x^4\,\ln \left (3\right )-2\,x^4\right )-2\,x^8\,\ln \left (3\right )+x^8} \,d x \]

[In]

int((exp(4*x - 8)*(4*x^3 - 4*x^4))/(exp(8*x - 16) + x^8*log(3)^2 - exp(4*x - 8)*(2*x^4*log(3) - 2*x^4) - 2*x^8
*log(3) + x^8),x)

[Out]

int((exp(4*x - 8)*(4*x^3 - 4*x^4))/(exp(8*x - 16) + x^8*log(3)^2 - exp(4*x - 8)*(2*x^4*log(3) - 2*x^4) - 2*x^8
*log(3) + x^8), x)