Integrand size = 68, antiderivative size = 19 \[ \int \frac {e^{-8+4 x} \left (4 x^3-4 x^4\right )}{e^{-16+8 x}+x^8-2 x^8 \log (3)+x^8 \log ^2(3)+e^{-8+4 x} \left (2 x^4-2 x^4 \log (3)\right )} \, dx=\frac {1}{1+\frac {e^{-8+4 x}}{x^4}-\log (3)} \] Output:
1/(exp(-2+x)^4/x^4-ln(3)+1)
Time = 0.33 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.58 \[ \int \frac {e^{-8+4 x} \left (4 x^3-4 x^4\right )}{e^{-16+8 x}+x^8-2 x^8 \log (3)+x^8 \log ^2(3)+e^{-8+4 x} \left (2 x^4-2 x^4 \log (3)\right )} \, dx=-\frac {4 e^8 x^4}{-4 e^{4 x}+4 e^8 x^4 (-1+\log (3))} \] Input:
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]
Output:
(-4*E^8*x^4)/(-4*E^(4*x) + 4*E^8*x^4*(-1 + Log[3]))
Time = 1.39 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.37, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {6, 6, 2027, 7239, 27, 7262, 17}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {e^{4 x-8} \left (4 x^3-4 x^4\right )}{x^8+x^8 \log ^2(3)-2 x^8 \log (3)+e^{4 x-8} \left (2 x^4-2 x^4 \log (3)\right )+e^{8 x-16}} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {e^{4 x-8} \left (4 x^3-4 x^4\right )}{x^8 \log ^2(3)+x^8 (1-2 \log (3))+e^{4 x-8} \left (2 x^4-2 x^4 \log (3)\right )+e^{8 x-16}}dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {e^{4 x-8} \left (4 x^3-4 x^4\right )}{x^8 \left (1+\log ^2(3)-2 \log (3)\right )+e^{4 x-8} \left (2 x^4-2 x^4 \log (3)\right )+e^{8 x-16}}dx\) |
\(\Big \downarrow \) 2027 |
\(\displaystyle \int \frac {e^{4 x-8} (4-4 x) x^3}{x^8 \left (1+\log ^2(3)-2 \log (3)\right )+e^{4 x-8} \left (2 x^4-2 x^4 \log (3)\right )+e^{8 x-16}}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {4 e^{4 x+8} (1-x) x^3}{\left (e^{4 x}-e^8 x^4 (\log (3)-1)\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 4 \int \frac {e^{4 x+8} (1-x) x^3}{\left (e^8 (1-\log (3)) x^4+e^{4 x}\right )^2}dx\) |
\(\Big \downarrow \) 7262 |
\(\displaystyle -e^8 \int \frac {1}{\left (e^8 (1-\log (3))+\frac {e^{4 x}}{x^4}\right )^2}d\frac {e^{4 x}}{x^4}\) |
\(\Big \downarrow \) 17 |
\(\displaystyle \frac {e^8}{\frac {e^{4 x}}{x^4}+e^8 (1-\log (3))}\) |
Input:
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]
Output:
E^8/(E^(4*x)/x^4 + E^8*(1 - Log[3]))
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 )/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(Fx_.)*((a_.)*(x_)^(r_.) + (b_.)*(x_)^(s_.))^(p_.), x_Symbol] :> Int[x^ (p*r)*(a + b*x^(s - r))^p*Fx, x] /; FreeQ[{a, b, r, s}, x] && IntegerQ[p] & & PosQ[s - r] && !(EqQ[p, 1] && EqQ[u, 1])
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
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])]}, Simp[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]
Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.47
method | result | size |
risch | \(-\frac {x^{4}}{x^{4} \ln \left (3\right )-x^{4}-{\mathrm e}^{4 x -8}}\) | \(28\) |
parallelrisch | \(-\frac {x^{4}}{x^{4} \ln \left (3\right )-x^{4}-{\mathrm e}^{4 x -8}}\) | \(28\) |
Input:
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,method=_RETURNVERBOSE)
Output:
-x^4/(x^4*ln(3)-x^4-exp(4*x-8))
Time = 0.07 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.42 \[ \int \frac {e^{-8+4 x} \left (4 x^3-4 x^4\right )}{e^{-16+8 x}+x^8-2 x^8 \log (3)+x^8 \log ^2(3)+e^{-8+4 x} \left (2 x^4-2 x^4 \log (3)\right )} \, dx=-\frac {x^{4}}{x^{4} \log \left (3\right ) - x^{4} - e^{\left (4 \, x - 8\right )}} \] Input:
integrate((-4*x^4+4*x^3)*exp(-2+x)^4/(exp(-2+x)^8+(-2*x^4*log(3)+2*x^4)*ex p(-2+x)^4+x^8*log(3)^2-2*x^8*log(3)+x^8),x, algorithm="fricas")
Output:
-x^4/(x^4*log(3) - x^4 - e^(4*x - 8))
Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-8+4 x} \left (4 x^3-4 x^4\right )}{e^{-16+8 x}+x^8-2 x^8 \log (3)+x^8 \log ^2(3)+e^{-8+4 x} \left (2 x^4-2 x^4 \log (3)\right )} \, dx=\frac {x^{4}}{- x^{4} \log {\left (3 \right )} + x^{4} + e^{4 x - 8}} \] Input:
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)
Output:
x**4/(-x**4*log(3) + x**4 + exp(4*x - 8))
Time = 0.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.37 \[ \int \frac {e^{-8+4 x} \left (4 x^3-4 x^4\right )}{e^{-16+8 x}+x^8-2 x^8 \log (3)+x^8 \log ^2(3)+e^{-8+4 x} \left (2 x^4-2 x^4 \log (3)\right )} \, dx=-\frac {x^{4} e^{8}}{x^{4} {\left (\log \left (3\right ) - 1\right )} e^{8} - e^{\left (4 \, x\right )}} \] Input:
integrate((-4*x^4+4*x^3)*exp(-2+x)^4/(exp(-2+x)^8+(-2*x^4*log(3)+2*x^4)*ex p(-2+x)^4+x^8*log(3)^2-2*x^8*log(3)+x^8),x, algorithm="maxima")
Output:
-x^4*e^8/(x^4*(log(3) - 1)*e^8 - e^(4*x))
Time = 0.12 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.63 \[ \int \frac {e^{-8+4 x} \left (4 x^3-4 x^4\right )}{e^{-16+8 x}+x^8-2 x^8 \log (3)+x^8 \log ^2(3)+e^{-8+4 x} \left (2 x^4-2 x^4 \log (3)\right )} \, dx=-\frac {x^{4} e^{8}}{x^{4} e^{8} \log \left (3\right ) - x^{4} e^{8} - e^{\left (4 \, x\right )}} \] Input:
integrate((-4*x^4+4*x^3)*exp(-2+x)^4/(exp(-2+x)^8+(-2*x^4*log(3)+2*x^4)*ex p(-2+x)^4+x^8*log(3)^2-2*x^8*log(3)+x^8),x, algorithm="giac")
Output:
-x^4*e^8/(x^4*e^8*log(3) - x^4*e^8 - e^(4*x))
Timed out. \[ \int \frac {e^{-8+4 x} \left (4 x^3-4 x^4\right )}{e^{-16+8 x}+x^8-2 x^8 \log (3)+x^8 \log ^2(3)+e^{-8+4 x} \left (2 x^4-2 x^4 \log (3)\right )} \, dx=\int \frac {{\mathrm {e}}^{4\,x-8}\,\left (4\,x^3-4\,x^4\right )}{{\mathrm {e}}^{8\,x-16}+x^8\,{\ln \left (3\right )}^2-{\mathrm {e}}^{4\,x-8}\,\left (2\,x^4\,\ln \left (3\right )-2\,x^4\right )-2\,x^8\,\ln \left (3\right )+x^8} \,d x \] Input:
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)
Output:
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)
\[ \int \frac {e^{-8+4 x} \left (4 x^3-4 x^4\right )}{e^{-16+8 x}+x^8-2 x^8 \log (3)+x^8 \log ^2(3)+e^{-8+4 x} \left (2 x^4-2 x^4 \log (3)\right )} \, dx=4 e^{8} \left (-\left (\int \frac {e^{4 x} x^{4}}{e^{8 x}-2 e^{4 x} \mathrm {log}\left (3\right ) e^{8} x^{4}+2 e^{4 x} e^{8} x^{4}+\mathrm {log}\left (3\right )^{2} e^{16} x^{8}-2 \,\mathrm {log}\left (3\right ) e^{16} x^{8}+e^{16} x^{8}}d x \right )+\int \frac {e^{4 x} x^{3}}{e^{8 x}-2 e^{4 x} \mathrm {log}\left (3\right ) e^{8} x^{4}+2 e^{4 x} e^{8} x^{4}+\mathrm {log}\left (3\right )^{2} e^{16} x^{8}-2 \,\mathrm {log}\left (3\right ) e^{16} x^{8}+e^{16} x^{8}}d x \right ) \] Input:
int((-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)
Output:
4*e**8*( - int((e**(4*x)*x**4)/(e**(8*x) - 2*e**(4*x)*log(3)*e**8*x**4 + 2 *e**(4*x)*e**8*x**4 + log(3)**2*e**16*x**8 - 2*log(3)*e**16*x**8 + e**16*x **8),x) + int((e**(4*x)*x**3)/(e**(8*x) - 2*e**(4*x)*log(3)*e**8*x**4 + 2* e**(4*x)*e**8*x**4 + log(3)**2*e**16*x**8 - 2*log(3)*e**16*x**8 + e**16*x* *8),x))