\(\int \frac {(e^{\frac {x^3}{2+x}} (48 x^2+16 x^3) \log (\log (2))+(-128-128 x-32 x^2) \log ^2(\log (2))) \log (\log (4))}{e^{\frac {2 x^3}{2+x}} (4+4 x+x^2)+e^{\frac {x^3}{2+x}} (-32 x-32 x^2-8 x^3) \log (\log (2))+(64 x^2+64 x^3+16 x^4) \log ^2(\log (2))} \, dx\) [137]

Optimal result

Integrand size = 122, antiderivative size = 29 \[ \int \frac {\left (e^{\frac {x^3}{2+x}} \left (48 x^2+16 x^3\right ) \log (\log (2))+\left (-128-128 x-32 x^2\right ) \log ^2(\log (2))\right ) \log (\log (4))}{e^{\frac {2 x^3}{2+x}} \left (4+4 x+x^2\right )+e^{\frac {x^3}{2+x}} \left (-32 x-32 x^2-8 x^3\right ) \log (\log (2))+\left (64 x^2+64 x^3+16 x^4\right ) \log ^2(\log (2))} \, dx=\frac {2 \log (\log (4))}{x-\frac {e^{\frac {x^3}{2+x}}}{4 \log (\log (2))}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.025, Rules used = {12, 6820, 6818} \[ \int \frac {\left (e^{\frac {x^3}{2+x}} \left (48 x^2+16 x^3\right ) \log (\log (2))+\left (-128-128 x-32 x^2\right ) \log ^2(\log (2))\right ) \log (\log (4))}{e^{\frac {2 x^3}{2+x}} \left (4+4 x+x^2\right )+e^{\frac {x^3}{2+x}} \left (-32 x-32 x^2-8 x^3\right ) \log (\log (2))+\left (64 x^2+64 x^3+16 x^4\right ) \log ^2(\log (2))} \, dx=-\frac {8 \log (\log (2)) \log (\log (4))}{e^{\frac {x^3}{x+2}}-4 x \log (\log (2))} \]

[In]

Int[((E^(x^3/(2 + x))*(48*x^2 + 16*x^3)*Log[Log[2]] + (-128 - 128*x - 32*x^2)*Log[Log[2]]^2)*Log[Log[4]])/(E^(
(2*x^3)/(2 + x))*(4 + 4*x + x^2) + E^(x^3/(2 + x))*(-32*x - 32*x^2 - 8*x^3)*Log[Log[2]] + (64*x^2 + 64*x^3 + 1
6*x^4)*Log[Log[2]]^2),x]

[Out]

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

Rule 12

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

Rule 6818

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*(y^(m + 1)/(m + 1)), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6820

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

Rubi steps \begin{align*} \text {integral}& = \log (\log (4)) \int \frac {e^{\frac {x^3}{2+x}} \left (48 x^2+16 x^3\right ) \log (\log (2))+\left (-128-128 x-32 x^2\right ) \log ^2(\log (2))}{e^{\frac {2 x^3}{2+x}} \left (4+4 x+x^2\right )+e^{\frac {x^3}{2+x}} \left (-32 x-32 x^2-8 x^3\right ) \log (\log (2))+\left (64 x^2+64 x^3+16 x^4\right ) \log ^2(\log (2))} \, dx \\ & = \log (\log (4)) \int \frac {16 \log (\log (2)) \left (e^{\frac {x^3}{2+x}} x^2 (3+x)-2 (2+x)^2 \log (\log (2))\right )}{(2+x)^2 \left (e^{\frac {x^3}{2+x}}-4 x \log (\log (2))\right )^2} \, dx \\ & = (16 \log (\log (2)) \log (\log (4))) \int \frac {e^{\frac {x^3}{2+x}} x^2 (3+x)-2 (2+x)^2 \log (\log (2))}{(2+x)^2 \left (e^{\frac {x^3}{2+x}}-4 x \log (\log (2))\right )^2} \, dx \\ & = -\frac {8 \log (\log (2)) \log (\log (4))}{e^{\frac {x^3}{2+x}}-4 x \log (\log (2))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {\left (e^{\frac {x^3}{2+x}} \left (48 x^2+16 x^3\right ) \log (\log (2))+\left (-128-128 x-32 x^2\right ) \log ^2(\log (2))\right ) \log (\log (4))}{e^{\frac {2 x^3}{2+x}} \left (4+4 x+x^2\right )+e^{\frac {x^3}{2+x}} \left (-32 x-32 x^2-8 x^3\right ) \log (\log (2))+\left (64 x^2+64 x^3+16 x^4\right ) \log ^2(\log (2))} \, dx=-\frac {8 \log (\log (2)) \log (\log (4))}{e^{\frac {x^3}{2+x}}-4 x \log (\log (2))} \]

[In]

Integrate[((E^(x^3/(2 + x))*(48*x^2 + 16*x^3)*Log[Log[2]] + (-128 - 128*x - 32*x^2)*Log[Log[2]]^2)*Log[Log[4]]
)/(E^((2*x^3)/(2 + x))*(4 + 4*x + x^2) + E^(x^3/(2 + x))*(-32*x - 32*x^2 - 8*x^3)*Log[Log[2]] + (64*x^2 + 64*x
^3 + 16*x^4)*Log[Log[2]]^2),x]

[Out]

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

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10

method	result	size
parallelrisch	\(\frac {8 \ln \left (2 \ln \left (2\right )\right ) \ln \left (\ln \left (2\right )\right )}{4 x \ln \left (\ln \left (2\right )\right )-{\mathrm e}^{\frac {x^{3}}{2+x}}}\)	\(32\)
risch	\(\frac {8 \ln \left (\ln \left (2\right )\right )^{2}}{4 x \ln \left (\ln \left (2\right )\right )-{\mathrm e}^{\frac {x^{3}}{2+x}}}+\frac {8 \ln \left (\ln \left (2\right )\right ) \ln \left (2\right )}{4 x \ln \left (\ln \left (2\right )\right )-{\mathrm e}^{\frac {x^{3}}{2+x}}}\)	\(58\)
norman	\(\frac {\left (8 \ln \left (\ln \left (2\right )\right )^{2}+8 \ln \left (\ln \left (2\right )\right ) \ln \left (2\right )\right ) x +16 \ln \left (\ln \left (2\right )\right ) \ln \left (2\right )+16 \ln \left (\ln \left (2\right )\right )^{2}}{\left (2+x \right ) \left (4 x \ln \left (\ln \left (2\right )\right )-{\mathrm e}^{\frac {x^{3}}{2+x}}\right )}\)	\(60\)

[In]

int(((-32*x^2-128*x-128)*ln(ln(2))^2+(16*x^3+48*x^2)*exp(x^3/(2+x))*ln(ln(2)))*ln(2*ln(2))/((16*x^4+64*x^3+64*
x^2)*ln(ln(2))^2+(-8*x^3-32*x^2-32*x)*exp(x^3/(2+x))*ln(ln(2))+(x^2+4*x+4)*exp(x^3/(2+x))^2),x,method=_RETURNV
ERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21 \[ \int \frac {\left (e^{\frac {x^3}{2+x}} \left (48 x^2+16 x^3\right ) \log (\log (2))+\left (-128-128 x-32 x^2\right ) \log ^2(\log (2))\right ) \log (\log (4))}{e^{\frac {2 x^3}{2+x}} \left (4+4 x+x^2\right )+e^{\frac {x^3}{2+x}} \left (-32 x-32 x^2-8 x^3\right ) \log (\log (2))+\left (64 x^2+64 x^3+16 x^4\right ) \log ^2(\log (2))} \, dx=\frac {8 \, {\left (\log \left (2\right ) \log \left (\log \left (2\right )\right ) + \log \left (\log \left (2\right )\right )^{2}\right )}}{4 \, x \log \left (\log \left (2\right )\right ) - e^{\left (\frac {x^{3}}{x + 2}\right )}} \]

[In]

integrate(((-32*x^2-128*x-128)*log(log(2))^2+(16*x^3+48*x^2)*exp(x^3/(2+x))*log(log(2)))*log(2*log(2))/((16*x^
4+64*x^3+64*x^2)*log(log(2))^2+(-8*x^3-32*x^2-32*x)*exp(x^3/(2+x))*log(log(2))+(x^2+4*x+4)*exp(x^3/(2+x))^2),x
, algorithm="fricas")

[Out]

8*(log(2)*log(log(2)) + log(log(2))^2)/(4*x*log(log(2)) - e^(x^3/(x + 2)))

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {\left (e^{\frac {x^3}{2+x}} \left (48 x^2+16 x^3\right ) \log (\log (2))+\left (-128-128 x-32 x^2\right ) \log ^2(\log (2))\right ) \log (\log (4))}{e^{\frac {2 x^3}{2+x}} \left (4+4 x+x^2\right )+e^{\frac {x^3}{2+x}} \left (-32 x-32 x^2-8 x^3\right ) \log (\log (2))+\left (64 x^2+64 x^3+16 x^4\right ) \log ^2(\log (2))} \, dx=\frac {- 8 \log {\left (\log {\left (2 \right )} \right )}^{2} - 8 \log {\left (2 \right )} \log {\left (\log {\left (2 \right )} \right )}}{- 4 x \log {\left (\log {\left (2 \right )} \right )} + e^{\frac {x^{3}}{x + 2}}} \]

[In]

integrate(((-32*x**2-128*x-128)*ln(ln(2))**2+(16*x**3+48*x**2)*exp(x**3/(2+x))*ln(ln(2)))*ln(2*ln(2))/((16*x**
4+64*x**3+64*x**2)*ln(ln(2))**2+(-8*x**3-32*x**2-32*x)*exp(x**3/(2+x))*ln(ln(2))+(x**2+4*x+4)*exp(x**3/(2+x))*
*2),x)

[Out]

(-8*log(log(2))**2 - 8*log(2)*log(log(2)))/(-4*x*log(log(2)) + exp(x**3/(x + 2)))

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.76 \[ \int \frac {\left (e^{\frac {x^3}{2+x}} \left (48 x^2+16 x^3\right ) \log (\log (2))+\left (-128-128 x-32 x^2\right ) \log ^2(\log (2))\right ) \log (\log (4))}{e^{\frac {2 x^3}{2+x}} \left (4+4 x+x^2\right )+e^{\frac {x^3}{2+x}} \left (-32 x-32 x^2-8 x^3\right ) \log (\log (2))+\left (64 x^2+64 x^3+16 x^4\right ) \log ^2(\log (2))} \, dx=\frac {8 \, e^{\left (2 \, x + \frac {8}{x + 2}\right )} \log \left (2 \, \log \left (2\right )\right ) \log \left (\log \left (2\right )\right )}{4 \, x e^{\left (2 \, x + \frac {8}{x + 2}\right )} \log \left (\log \left (2\right )\right ) - e^{\left (x^{2} + 4\right )}} \]

[In]

integrate(((-32*x^2-128*x-128)*log(log(2))^2+(16*x^3+48*x^2)*exp(x^3/(2+x))*log(log(2)))*log(2*log(2))/((16*x^
4+64*x^3+64*x^2)*log(log(2))^2+(-8*x^3-32*x^2-32*x)*exp(x^3/(2+x))*log(log(2))+(x^2+4*x+4)*exp(x^3/(2+x))^2),x
, algorithm="maxima")

[Out]

8*e^(2*x + 8/(x + 2))*log(2*log(2))*log(log(2))/(4*x*e^(2*x + 8/(x + 2))*log(log(2)) - e^(x^2 + 4))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {\left (e^{\frac {x^3}{2+x}} \left (48 x^2+16 x^3\right ) \log (\log (2))+\left (-128-128 x-32 x^2\right ) \log ^2(\log (2))\right ) \log (\log (4))}{e^{\frac {2 x^3}{2+x}} \left (4+4 x+x^2\right )+e^{\frac {x^3}{2+x}} \left (-32 x-32 x^2-8 x^3\right ) \log (\log (2))+\left (64 x^2+64 x^3+16 x^4\right ) \log ^2(\log (2))} \, dx=\frac {8 \, \log \left (2 \, \log \left (2\right )\right ) \log \left (\log \left (2\right )\right )}{4 \, x \log \left (\log \left (2\right )\right ) - e^{\left (\frac {x^{3}}{x + 2}\right )}} \]

[In]

integrate(((-32*x^2-128*x-128)*log(log(2))^2+(16*x^3+48*x^2)*exp(x^3/(2+x))*log(log(2)))*log(2*log(2))/((16*x^
4+64*x^3+64*x^2)*log(log(2))^2+(-8*x^3-32*x^2-32*x)*exp(x^3/(2+x))*log(log(2))+(x^2+4*x+4)*exp(x^3/(2+x))^2),x
, algorithm="giac")

[Out]

8*log(2*log(2))*log(log(2))/(4*x*log(log(2)) - e^(x^3/(x + 2)))

Mupad [B] (verification not implemented)

Time = 8.39 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {\left (e^{\frac {x^3}{2+x}} \left (48 x^2+16 x^3\right ) \log (\log (2))+\left (-128-128 x-32 x^2\right ) \log ^2(\log (2))\right ) \log (\log (4))}{e^{\frac {2 x^3}{2+x}} \left (4+4 x+x^2\right )+e^{\frac {x^3}{2+x}} \left (-32 x-32 x^2-8 x^3\right ) \log (\log (2))+\left (64 x^2+64 x^3+16 x^4\right ) \log ^2(\log (2))} \, dx=-\frac {8\,\ln \left (2\,\ln \left (2\right )\right )\,\ln \left (\ln \left (2\right )\right )}{{\mathrm {e}}^{\frac {x^3}{x+2}}-4\,x\,\ln \left (\ln \left (2\right )\right )} \]

[In]

int(-(log(2*log(2))*(log(log(2))^2*(128*x + 32*x^2 + 128) - exp(x^3/(x + 2))*log(log(2))*(48*x^2 + 16*x^3)))/(
exp((2*x^3)/(x + 2))*(4*x + x^2 + 4) + log(log(2))^2*(64*x^2 + 64*x^3 + 16*x^4) - exp(x^3/(x + 2))*log(log(2))
*(32*x + 32*x^2 + 8*x^3)),x)

[Out]

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