\(\int \frac {e^{-e^{\frac {1}{3 \log (4+x)}}+x} (-8 e^{\frac {1}{3 \log (4+x)}} x+(96-72 x-24 x^2) \log ^2(4+x))}{e^{-2 e^{\frac {1}{3 \log (4+x)}}+2 x} (12+3 x) \log ^2(4+x)+e^{-e^{\frac {1}{3 \log (4+x)}}+x} (-24 x-6 x^2) \log ^2(4+x)+(12 x^2+3 x^3) \log ^2(4+x)} \, dx\) [7822]

Optimal result

Integrand size = 139, antiderivative size = 27 \[ \int \frac {e^{-e^{\frac {1}{3 \log (4+x)}}+x} \left (-8 e^{\frac {1}{3 \log (4+x)}} x+\left (96-72 x-24 x^2\right ) \log ^2(4+x)\right )}{e^{-2 e^{\frac {1}{3 \log (4+x)}}+2 x} (12+3 x) \log ^2(4+x)+e^{-e^{\frac {1}{3 \log (4+x)}}+x} \left (-24 x-6 x^2\right ) \log ^2(4+x)+\left (12 x^2+3 x^3\right ) \log ^2(4+x)} \, dx=\frac {8 x}{e^{-e^{\frac {1}{3 \log (4+x)}}+x}-x} \]

[Out]

x/(1/8*exp(-exp(1/3/ln(4+x))+x)-1/8*x)

Rubi [A] (verified)

Time = 1.51 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {6820, 12, 6843, 32} \[ \int \frac {e^{-e^{\frac {1}{3 \log (4+x)}}+x} \left (-8 e^{\frac {1}{3 \log (4+x)}} x+\left (96-72 x-24 x^2\right ) \log ^2(4+x)\right )}{e^{-2 e^{\frac {1}{3 \log (4+x)}}+2 x} (12+3 x) \log ^2(4+x)+e^{-e^{\frac {1}{3 \log (4+x)}}+x} \left (-24 x-6 x^2\right ) \log ^2(4+x)+\left (12 x^2+3 x^3\right ) \log ^2(4+x)} \, dx=-\frac {8}{1-\frac {e^{x-e^{\frac {1}{3 \log (x+4)}}}}{x}} \]

[In]

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)*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]

[Out]

-8/(1 - E^(-E^(1/(3*Log[4 + x])) + x)/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 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 {8 e^{e^{\frac {1}{3 \log (4+x)}}+x} \left (-e^{\frac {1}{3 \log (4+x)}} x-3 \left (-4+3 x+x^2\right ) \log ^2(4+x)\right )}{3 (4+x) \left (e^x-e^{e^{\frac {1}{3 \log (4+x)}}} x\right )^2 \log ^2(4+x)} \, dx \\ & = \frac {8}{3} \int \frac {e^{e^{\frac {1}{3 \log (4+x)}}+x} \left (-e^{\frac {1}{3 \log (4+x)}} x-3 \left (-4+3 x+x^2\right ) \log ^2(4+x)\right )}{(4+x) \left (e^x-e^{e^{\frac {1}{3 \log (4+x)}}} x\right )^2 \log ^2(4+x)} \, dx \\ & = 8 \text {Subst}\left (\int \frac {1}{(1+x)^2} \, dx,x,-\frac {e^{-e^{\frac {1}{3 \log (4+x)}}+x}}{x}\right ) \\ & = -\frac {8}{1-\frac {e^{-e^{\frac {1}{3 \log (4+x)}}+x}}{x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {e^{-e^{\frac {1}{3 \log (4+x)}}+x} \left (-8 e^{\frac {1}{3 \log (4+x)}} x+\left (96-72 x-24 x^2\right ) \log ^2(4+x)\right )}{e^{-2 e^{\frac {1}{3 \log (4+x)}}+2 x} (12+3 x) \log ^2(4+x)+e^{-e^{\frac {1}{3 \log (4+x)}}+x} \left (-24 x-6 x^2\right ) \log ^2(4+x)+\left (12 x^2+3 x^3\right ) \log ^2(4+x)} \, dx=-\frac {8 e^x}{-e^x+e^{e^{\frac {1}{3 \log (4+x)}}} x} \]

[In]

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]

[Out]

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

Maple [A] (verified)

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	\(-\frac {8 x}{x -{\mathrm e}^{-{\mathrm e}^{\frac {1}{3 \ln \left (4+x \right )}}+x}}\)	\(24\)

[In]

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=_RETURNV
ERBOSE)

[Out]

-8*x/(x-exp(-exp(1/3/ln(4+x))+x))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {e^{-e^{\frac {1}{3 \log (4+x)}}+x} \left (-8 e^{\frac {1}{3 \log (4+x)}} x+\left (96-72 x-24 x^2\right ) \log ^2(4+x)\right )}{e^{-2 e^{\frac {1}{3 \log (4+x)}}+2 x} (12+3 x) \log ^2(4+x)+e^{-e^{\frac {1}{3 \log (4+x)}}+x} \left (-24 x-6 x^2\right ) \log ^2(4+x)+\left (12 x^2+3 x^3\right ) \log ^2(4+x)} \, dx=-\frac {8 \, x}{x - e^{\left (x - e^{\left (\frac {1}{3 \, \log \left (x + 4\right )}\right )}\right )}} \]

[In]

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^2-24*x)*log(4+x)^2*exp(-exp(1/3/log(4+x))+x)+(3*x^3+12*x^2)*log(4+x)^2),x,
algorithm="fricas")

[Out]

-8*x/(x - e^(x - e^(1/3/log(x + 4))))

Sympy [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.63 \[ \int \frac {e^{-e^{\frac {1}{3 \log (4+x)}}+x} \left (-8 e^{\frac {1}{3 \log (4+x)}} x+\left (96-72 x-24 x^2\right ) \log ^2(4+x)\right )}{e^{-2 e^{\frac {1}{3 \log (4+x)}}+2 x} (12+3 x) \log ^2(4+x)+e^{-e^{\frac {1}{3 \log (4+x)}}+x} \left (-24 x-6 x^2\right ) \log ^2(4+x)+\left (12 x^2+3 x^3\right ) \log ^2(4+x)} \, dx=\frac {8 x}{- x + e^{x - e^{\frac {1}{3 \log {\left (x + 4 \right )}}}}} \]

[In]

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**2-24*x)*ln(4+x)**2*exp(-exp(1/3/ln(4+x))+x)+(3*x**3+12*x**2)*ln(4+x)**2),x)

[Out]

8*x/(-x + exp(x - exp(1/(3*log(x + 4)))))

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {e^{-e^{\frac {1}{3 \log (4+x)}}+x} \left (-8 e^{\frac {1}{3 \log (4+x)}} x+\left (96-72 x-24 x^2\right ) \log ^2(4+x)\right )}{e^{-2 e^{\frac {1}{3 \log (4+x)}}+2 x} (12+3 x) \log ^2(4+x)+e^{-e^{\frac {1}{3 \log (4+x)}}+x} \left (-24 x-6 x^2\right ) \log ^2(4+x)+\left (12 x^2+3 x^3\right ) \log ^2(4+x)} \, dx=-\frac {8 \, e^{x}}{x e^{\left (e^{\left (\frac {1}{3 \, \log \left (x + 4\right )}\right )}\right )} - e^{x}} \]

[In]

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^2-24*x)*log(4+x)^2*exp(-exp(1/3/log(4+x))+x)+(3*x^3+12*x^2)*log(4+x)^2),x,
algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {e^{-e^{\frac {1}{3 \log (4+x)}}+x} \left (-8 e^{\frac {1}{3 \log (4+x)}} x+\left (96-72 x-24 x^2\right ) \log ^2(4+x)\right )}{e^{-2 e^{\frac {1}{3 \log (4+x)}}+2 x} (12+3 x) \log ^2(4+x)+e^{-e^{\frac {1}{3 \log (4+x)}}+x} \left (-24 x-6 x^2\right ) \log ^2(4+x)+\left (12 x^2+3 x^3\right ) \log ^2(4+x)} \, dx=\int { -\frac {8 \, {\left (3 \, {\left (x^{2} + 3 \, x - 4\right )} \log \left (x + 4\right )^{2} + x e^{\left (\frac {1}{3 \, \log \left (x + 4\right )}\right )}\right )} e^{\left (x - e^{\left (\frac {1}{3 \, \log \left (x + 4\right )}\right )}\right )}}{3 \, {\left ({\left (x + 4\right )} e^{\left (2 \, x - 2 \, e^{\left (\frac {1}{3 \, \log \left (x + 4\right )}\right )}\right )} \log \left (x + 4\right )^{2} - 2 \, {\left (x^{2} + 4 \, x\right )} e^{\left (x - e^{\left (\frac {1}{3 \, \log \left (x + 4\right )}\right )}\right )} \log \left (x + 4\right )^{2} + {\left (x^{3} + 4 \, x^{2}\right )} \log \left (x + 4\right )^{2}\right )}} \,d x } \]

[In]

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^2-24*x)*log(4+x)^2*exp(-exp(1/3/log(4+x))+x)+(3*x^3+12*x^2)*log(4+x)^2),x,
algorithm="giac")

[Out]

undef

Mupad [B] (verification not implemented)

Time = 14.63 (sec) , antiderivative size = 172, normalized size of antiderivative = 6.37 \[ \int \frac {e^{-e^{\frac {1}{3 \log (4+x)}}+x} \left (-8 e^{\frac {1}{3 \log (4+x)}} x+\left (96-72 x-24 x^2\right ) \log ^2(4+x)\right )}{e^{-2 e^{\frac {1}{3 \log (4+x)}}+2 x} (12+3 x) \log ^2(4+x)+e^{-e^{\frac {1}{3 \log (4+x)}}+x} \left (-24 x-6 x^2\right ) \log ^2(4+x)+\left (12 x^2+3 x^3\right ) \log ^2(4+x)} \, dx=-\frac {8\,x\,{\left (x\,{\ln \left (x+4\right )}^2+4\,{\ln \left (x+4\right )}^2\right )}^2\,\left (x\,{\mathrm {e}}^{\frac {1}{3\,\ln \left (x+4\right )}}+9\,x\,{\ln \left (x+4\right )}^2-12\,{\ln \left (x+4\right )}^2+3\,x^2\,{\ln \left (x+4\right )}^2\right )}{{\ln \left (x+4\right )}^2\,\left (x-{\mathrm {e}}^{x-{\mathrm {e}}^{\frac {1}{3\,\ln \left (x+4\right )}}}\right )\,\left (x+4\right )\,\left (24\,x\,{\ln \left (x+4\right )}^4-48\,{\ln \left (x+4\right )}^4+21\,x^2\,{\ln \left (x+4\right )}^4+3\,x^3\,{\ln \left (x+4\right )}^4+x^2\,{\ln \left (x+4\right )}^2\,{\mathrm {e}}^{\frac {1}{3\,\ln \left (x+4\right )}}+4\,x\,{\ln \left (x+4\right )}^2\,{\mathrm {e}}^{\frac {1}{3\,\ln \left (x+4\right )}}\right )} \]

[In]

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
*exp(1/(3*log(x + 4))))*(3*x + 12)),x)

[Out]

-(8*x*(x*log(x + 4)^2 + 4*log(x + 4)^2)^2*(x*exp(1/(3*log(x + 4))) + 9*x*log(x + 4)^2 - 12*log(x + 4)^2 + 3*x^
2*log(x + 4)^2))/(log(x + 4)^2*(x - exp(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*log(x + 4))) + 4*x*log(x + 4)^2*exp(
1/(3*log(x + 4)))))