Integrand size = 139, antiderivative size = 27 \[ \int \frac {e^{-e^{\frac {1}{3 \log (4+x)}}+x} \left (-8 e^{\frac {1}{3 \log (4+x)}} x+\left (96-72 x-24 x^2\right ) \log ^2(4+x)\right )}{e^{-2 e^{\frac {1}{3 \log (4+x)}}+2 x} (12+3 x) \log ^2(4+x)+e^{-e^{\frac {1}{3 \log (4+x)}}+x} \left (-24 x-6 x^2\right ) \log ^2(4+x)+\left (12 x^2+3 x^3\right ) \log ^2(4+x)} \, dx=\frac {8 x}{e^{-e^{\frac {1}{3 \log (4+x)}}+x}-x} \]
Time = 0.11 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {e^{-e^{\frac {1}{3 \log (4+x)}}+x} \left (-8 e^{\frac {1}{3 \log (4+x)}} x+\left (96-72 x-24 x^2\right ) \log ^2(4+x)\right )}{e^{-2 e^{\frac {1}{3 \log (4+x)}}+2 x} (12+3 x) \log ^2(4+x)+e^{-e^{\frac {1}{3 \log (4+x)}}+x} \left (-24 x-6 x^2\right ) \log ^2(4+x)+\left (12 x^2+3 x^3\right ) \log ^2(4+x)} \, dx=-\frac {8 e^x}{-e^x+e^{e^{\frac {1}{3 \log (4+x)}}} x} \]
Integrate[(E^(-E^(1/(3*Log[4 + x])) + x)*(-8*E^(1/(3*Log[4 + x]))*x + (96 - 72*x - 24*x^2)*Log[4 + x]^2))/(E^(-2*E^(1/(3*Log[4 + x])) + 2*x)*(12 + 3 *x)*Log[4 + x]^2 + E^(-E^(1/(3*Log[4 + x])) + x)*(-24*x - 6*x^2)*Log[4 + x ]^2 + (12*x^2 + 3*x^3)*Log[4 + x]^2),x]
Time = 2.29 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {7292, 27, 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.
\(\displaystyle \int \frac {e^{x-e^{\frac {1}{3 \log (x+4)}}} \left (\left (-24 x^2-72 x+96\right ) \log ^2(x+4)-8 x e^{\frac {1}{3 \log (x+4)}}\right )}{\left (-6 x^2-24 x\right ) e^{x-e^{\frac {1}{3 \log (x+4)}}} \log ^2(x+4)+\left (3 x^3+12 x^2\right ) \log ^2(x+4)+(3 x+12) e^{2 x-2 e^{\frac {1}{3 \log (x+4)}}} \log ^2(x+4)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{x+e^{\frac {1}{3 \log (x+4)}}} \left (\left (-24 x^2-72 x+96\right ) \log ^2(x+4)-8 x e^{\frac {1}{3 \log (x+4)}}\right )}{3 (x+4) \left (e^x-x e^{e^{\frac {1}{3 \log (x+4)}}}\right )^2 \log ^2(x+4)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \int -\frac {8 e^{x+e^{\frac {1}{3 \log (x+4)}}} \left (e^{\frac {1}{3 \log (x+4)}} x-3 \left (-x^2-3 x+4\right ) \log ^2(x+4)\right )}{(x+4) \left (e^x-e^{e^{\frac {1}{3 \log (x+4)}}} x\right )^2 \log ^2(x+4)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {8}{3} \int \frac {e^{x+e^{\frac {1}{3 \log (x+4)}}} \left (e^{\frac {1}{3 \log (x+4)}} x-3 \left (-x^2-3 x+4\right ) \log ^2(x+4)\right )}{(x+4) \left (e^x-e^{e^{\frac {1}{3 \log (x+4)}}} x\right )^2 \log ^2(x+4)}dx\) |
\(\Big \downarrow \) 7262 |
\(\displaystyle 8 \int \frac {1}{\left (1-\frac {e^{x-e^{\frac {1}{3 \log (x+4)}}}}{x}\right )^2}d\left (-\frac {e^{x-e^{\frac {1}{3 \log (x+4)}}}}{x}\right )\) |
\(\Big \downarrow \) 17 |
\(\displaystyle -\frac {8}{1-\frac {e^{x-e^{\frac {1}{3 \log (x+4)}}}}{x}}\) |
Int[(E^(-E^(1/(3*Log[4 + x])) + x)*(-8*E^(1/(3*Log[4 + x]))*x + (96 - 72*x - 24*x^2)*Log[4 + x]^2))/(E^(-2*E^(1/(3*Log[4 + x])) + 2*x)*(12 + 3*x)*Lo g[4 + x]^2 + E^(-E^(1/(3*Log[4 + x])) + x)*(-24*x - 6*x^2)*Log[4 + x]^2 + (12*x^2 + 3*x^3)*Log[4 + x]^2),x]
3.19.22.3.1 Defintions of rubi rules used
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[(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 = 3.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89
method | result | size |
risch | \(-\frac {8 x}{x -{\mathrm e}^{-{\mathrm e}^{\frac {1}{3 \ln \left (4+x \right )}}+x}}\) | \(24\) |
parallelrisch | \(-\frac {8 x}{x -{\mathrm e}^{-{\mathrm e}^{\frac {1}{3 \ln \left (4+x \right )}}+x}}\) | \(24\) |
int((-8*x*exp(1/3/ln(4+x))+(-24*x^2-72*x+96)*ln(4+x)^2)*exp(-exp(1/3/ln(4+ x))+x)/((3*x+12)*ln(4+x)^2*exp(-exp(1/3/ln(4+x))+x)^2+(-6*x^2-24*x)*ln(4+x )^2*exp(-exp(1/3/ln(4+x))+x)+(3*x^3+12*x^2)*ln(4+x)^2),x,method=_RETURNVER BOSE)
Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {e^{-e^{\frac {1}{3 \log (4+x)}}+x} \left (-8 e^{\frac {1}{3 \log (4+x)}} x+\left (96-72 x-24 x^2\right ) \log ^2(4+x)\right )}{e^{-2 e^{\frac {1}{3 \log (4+x)}}+2 x} (12+3 x) \log ^2(4+x)+e^{-e^{\frac {1}{3 \log (4+x)}}+x} \left (-24 x-6 x^2\right ) \log ^2(4+x)+\left (12 x^2+3 x^3\right ) \log ^2(4+x)} \, dx=-\frac {8 \, x}{x - e^{\left (x - e^{\left (\frac {1}{3 \, \log \left (x + 4\right )}\right )}\right )}} \]
integrate((-8*x*exp(1/3/log(4+x))+(-24*x^2-72*x+96)*log(4+x)^2)*exp(-exp(1 /3/log(4+x))+x)/((3*x+12)*log(4+x)^2*exp(-exp(1/3/log(4+x))+x)^2+(-6*x^2-2 4*x)*log(4+x)^2*exp(-exp(1/3/log(4+x))+x)+(3*x^3+12*x^2)*log(4+x)^2),x, al gorithm=\
Time = 0.18 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.63 \[ \int \frac {e^{-e^{\frac {1}{3 \log (4+x)}}+x} \left (-8 e^{\frac {1}{3 \log (4+x)}} x+\left (96-72 x-24 x^2\right ) \log ^2(4+x)\right )}{e^{-2 e^{\frac {1}{3 \log (4+x)}}+2 x} (12+3 x) \log ^2(4+x)+e^{-e^{\frac {1}{3 \log (4+x)}}+x} \left (-24 x-6 x^2\right ) \log ^2(4+x)+\left (12 x^2+3 x^3\right ) \log ^2(4+x)} \, dx=\frac {8 x}{- x + e^{x - e^{\frac {1}{3 \log {\left (x + 4 \right )}}}}} \]
integrate((-8*x*exp(1/3/ln(4+x))+(-24*x**2-72*x+96)*ln(4+x)**2)*exp(-exp(1 /3/ln(4+x))+x)/((3*x+12)*ln(4+x)**2*exp(-exp(1/3/ln(4+x))+x)**2+(-6*x**2-2 4*x)*ln(4+x)**2*exp(-exp(1/3/ln(4+x))+x)+(3*x**3+12*x**2)*ln(4+x)**2),x)
Time = 0.34 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {e^{-e^{\frac {1}{3 \log (4+x)}}+x} \left (-8 e^{\frac {1}{3 \log (4+x)}} x+\left (96-72 x-24 x^2\right ) \log ^2(4+x)\right )}{e^{-2 e^{\frac {1}{3 \log (4+x)}}+2 x} (12+3 x) \log ^2(4+x)+e^{-e^{\frac {1}{3 \log (4+x)}}+x} \left (-24 x-6 x^2\right ) \log ^2(4+x)+\left (12 x^2+3 x^3\right ) \log ^2(4+x)} \, dx=-\frac {8 \, e^{x}}{x e^{\left (e^{\left (\frac {1}{3 \, \log \left (x + 4\right )}\right )}\right )} - e^{x}} \]
integrate((-8*x*exp(1/3/log(4+x))+(-24*x^2-72*x+96)*log(4+x)^2)*exp(-exp(1 /3/log(4+x))+x)/((3*x+12)*log(4+x)^2*exp(-exp(1/3/log(4+x))+x)^2+(-6*x^2-2 4*x)*log(4+x)^2*exp(-exp(1/3/log(4+x))+x)+(3*x^3+12*x^2)*log(4+x)^2),x, al gorithm=\
\[ \int \frac {e^{-e^{\frac {1}{3 \log (4+x)}}+x} \left (-8 e^{\frac {1}{3 \log (4+x)}} x+\left (96-72 x-24 x^2\right ) \log ^2(4+x)\right )}{e^{-2 e^{\frac {1}{3 \log (4+x)}}+2 x} (12+3 x) \log ^2(4+x)+e^{-e^{\frac {1}{3 \log (4+x)}}+x} \left (-24 x-6 x^2\right ) \log ^2(4+x)+\left (12 x^2+3 x^3\right ) \log ^2(4+x)} \, dx=\int { -\frac {8 \, {\left (3 \, {\left (x^{2} + 3 \, x - 4\right )} \log \left (x + 4\right )^{2} + x e^{\left (\frac {1}{3 \, \log \left (x + 4\right )}\right )}\right )} e^{\left (x - e^{\left (\frac {1}{3 \, \log \left (x + 4\right )}\right )}\right )}}{3 \, {\left ({\left (x + 4\right )} e^{\left (2 \, x - 2 \, e^{\left (\frac {1}{3 \, \log \left (x + 4\right )}\right )}\right )} \log \left (x + 4\right )^{2} - 2 \, {\left (x^{2} + 4 \, x\right )} e^{\left (x - e^{\left (\frac {1}{3 \, \log \left (x + 4\right )}\right )}\right )} \log \left (x + 4\right )^{2} + {\left (x^{3} + 4 \, x^{2}\right )} \log \left (x + 4\right )^{2}\right )}} \,d x } \]
integrate((-8*x*exp(1/3/log(4+x))+(-24*x^2-72*x+96)*log(4+x)^2)*exp(-exp(1 /3/log(4+x))+x)/((3*x+12)*log(4+x)^2*exp(-exp(1/3/log(4+x))+x)^2+(-6*x^2-2 4*x)*log(4+x)^2*exp(-exp(1/3/log(4+x))+x)+(3*x^3+12*x^2)*log(4+x)^2),x, al gorithm=\
Time = 14.63 (sec) , antiderivative size = 172, normalized size of antiderivative = 6.37 \[ \int \frac {e^{-e^{\frac {1}{3 \log (4+x)}}+x} \left (-8 e^{\frac {1}{3 \log (4+x)}} x+\left (96-72 x-24 x^2\right ) \log ^2(4+x)\right )}{e^{-2 e^{\frac {1}{3 \log (4+x)}}+2 x} (12+3 x) \log ^2(4+x)+e^{-e^{\frac {1}{3 \log (4+x)}}+x} \left (-24 x-6 x^2\right ) \log ^2(4+x)+\left (12 x^2+3 x^3\right ) \log ^2(4+x)} \, dx=-\frac {8\,x\,{\left (x\,{\ln \left (x+4\right )}^2+4\,{\ln \left (x+4\right )}^2\right )}^2\,\left (x\,{\mathrm {e}}^{\frac {1}{3\,\ln \left (x+4\right )}}+9\,x\,{\ln \left (x+4\right )}^2-12\,{\ln \left (x+4\right )}^2+3\,x^2\,{\ln \left (x+4\right )}^2\right )}{{\ln \left (x+4\right )}^2\,\left (x-{\mathrm {e}}^{x-{\mathrm {e}}^{\frac {1}{3\,\ln \left (x+4\right )}}}\right )\,\left (x+4\right )\,\left (24\,x\,{\ln \left (x+4\right )}^4-48\,{\ln \left (x+4\right )}^4+21\,x^2\,{\ln \left (x+4\right )}^4+3\,x^3\,{\ln \left (x+4\right )}^4+x^2\,{\ln \left (x+4\right )}^2\,{\mathrm {e}}^{\frac {1}{3\,\ln \left (x+4\right )}}+4\,x\,{\ln \left (x+4\right )}^2\,{\mathrm {e}}^{\frac {1}{3\,\ln \left (x+4\right )}}\right )} \]
int(-(exp(x - exp(1/(3*log(x + 4))))*(8*x*exp(1/(3*log(x + 4))) + log(x + 4)^2*(72*x + 24*x^2 - 96)))/(log(x + 4)^2*(12*x^2 + 3*x^3) - log(x + 4)^2* exp(x - exp(1/(3*log(x + 4))))*(24*x + 6*x^2) + log(x + 4)^2*exp(2*x - 2*e xp(1/(3*log(x + 4))))*(3*x + 12)),x)
-(8*x*(x*log(x + 4)^2 + 4*log(x + 4)^2)^2*(x*exp(1/(3*log(x + 4))) + 9*x*l og(x + 4)^2 - 12*log(x + 4)^2 + 3*x^2*log(x + 4)^2))/(log(x + 4)^2*(x - ex p(x - exp(1/(3*log(x + 4)))))*(x + 4)*(24*x*log(x + 4)^4 - 48*log(x + 4)^4 + 21*x^2*log(x + 4)^4 + 3*x^3*log(x + 4)^4 + x^2*log(x + 4)^2*exp(1/(3*lo g(x + 4))) + 4*x*log(x + 4)^2*exp(1/(3*log(x + 4)))))