Integrand size = 122, antiderivative size = 29 \[ \int \frac {\left (e^{\frac {x^3}{2+x}} \left (48 x^2+16 x^3\right ) \log (\log (2))+\left (-128-128 x-32 x^2\right ) \log ^2(\log (2))\right ) \log (\log (4))}{e^{\frac {2 x^3}{2+x}} \left (4+4 x+x^2\right )+e^{\frac {x^3}{2+x}} \left (-32 x-32 x^2-8 x^3\right ) \log (\log (2))+\left (64 x^2+64 x^3+16 x^4\right ) \log ^2(\log (2))} \, dx=\frac {2 \log (\log (4))}{x-\frac {e^{\frac {x^3}{2+x}}}{4 \log (\log (2))}} \] Output:
2*ln(2*ln(2))/(-1/4*exp(x^3/(2+x))/ln(ln(2))+x)
Time = 0.05 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {\left (e^{\frac {x^3}{2+x}} \left (48 x^2+16 x^3\right ) \log (\log (2))+\left (-128-128 x-32 x^2\right ) \log ^2(\log (2))\right ) \log (\log (4))}{e^{\frac {2 x^3}{2+x}} \left (4+4 x+x^2\right )+e^{\frac {x^3}{2+x}} \left (-32 x-32 x^2-8 x^3\right ) \log (\log (2))+\left (64 x^2+64 x^3+16 x^4\right ) \log ^2(\log (2))} \, dx=-\frac {8 \log (\log (2)) \log (\log (4))}{e^{\frac {x^3}{2+x}}-4 x \log (\log (2))} \] Input:
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]
Output:
(-8*Log[Log[2]]*Log[Log[4]])/(E^(x^3/(2 + x)) - 4*x*Log[Log[2]])
Time = 1.07 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.041, Rules used = {27, 27, 7239, 27, 7237}
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.
\(\displaystyle \int \frac {\log (\log (4)) \left (\left (-32 x^2-128 x-128\right ) \log ^2(\log (2))+e^{\frac {x^3}{x+2}} \left (16 x^3+48 x^2\right ) \log (\log (2))\right )}{e^{\frac {2 x^3}{x+2}} \left (x^2+4 x+4\right )+e^{\frac {x^3}{x+2}} \left (-8 x^3-32 x^2-32 x\right ) \log (\log (2))+\left (16 x^4+64 x^3+64 x^2\right ) \log ^2(\log (2))} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \log (\log (4)) \int \frac {16 \left (e^{\frac {x^3}{x+2}} \left (x^3+3 x^2\right ) \log (\log (2))-2 \left (x^2+4 x+4\right ) \log ^2(\log (2))\right )}{e^{\frac {2 x^3}{x+2}} \left (x^2+4 x+4\right )+16 \left (x^4+4 x^3+4 x^2\right ) \log ^2(\log (2))-8 e^{\frac {x^3}{x+2}} \left (x^3+4 x^2+4 x\right ) \log (\log (2))}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 16 \log (\log (4)) \int \frac {e^{\frac {x^3}{x+2}} \left (x^3+3 x^2\right ) \log (\log (2))-2 \left (x^2+4 x+4\right ) \log ^2(\log (2))}{e^{\frac {2 x^3}{x+2}} \left (x^2+4 x+4\right )+16 \left (x^4+4 x^3+4 x^2\right ) \log ^2(\log (2))-8 e^{\frac {x^3}{x+2}} \left (x^3+4 x^2+4 x\right ) \log (\log (2))}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle 16 \log (\log (4)) \int \frac {\log (\log (2)) \left (e^{\frac {x^3}{x+2}} x^2 (x+3)-2 (x+2)^2 \log (\log (2))\right )}{(x+2)^2 \left (e^{\frac {x^3}{x+2}}-4 x \log (\log (2))\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 16 \log (\log (2)) \log (\log (4)) \int \frac {e^{\frac {x^3}{x+2}} x^2 (x+3)-2 (x+2)^2 \log (\log (2))}{(x+2)^2 \left (e^{\frac {x^3}{x+2}}-4 x \log (\log (2))\right )^2}dx\) |
\(\Big \downarrow \) 7237 |
\(\displaystyle -\frac {8 \log (\log (2)) \log (\log (4))}{e^{\frac {x^3}{x+2}}-4 x \log (\log (2))}\) |
Input:
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 + 16* x^4)*Log[Log[2]]^2),x]
Output:
(-8*Log[Log[2]]*Log[Log[4]])/(E^(x^3/(2 + x)) - 4*x*Log[Log[2]])
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Si mp[q*(y^(m + 1)/(m + 1)), x] /; !FalseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.54 (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 | \(\frac {8 \ln \left (\ln \left (2\right )\right ) \ln \left (2\right )}{4 x \ln \left (\ln \left (2\right )\right )-{\mathrm e}^{\frac {x^{3}}{2+x}}}+\frac {8 \ln \left (\ln \left (2\right )\right )^{2}}{4 x \ln \left (\ln \left (2\right )\right )-{\mathrm e}^{\frac {x^{3}}{2+x}}}\) | \(58\) |
norman | \(\frac {\left (8 \ln \left (\ln \left (2\right )\right ) \ln \left (2\right )+8 \ln \left (\ln \left (2\right )\right )^{2}\right ) x +16 \ln \left (\ln \left (2\right )\right )^{2}+16 \ln \left (\ln \left (2\right )\right ) \ln \left (2\right )}{\left (2+x \right ) \left (4 x \ln \left (\ln \left (2\right )\right )-{\mathrm e}^{\frac {x^{3}}{2+x}}\right )}\) | \(60\) |
Input:
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=_RETURNVER BOSE)
Output:
8*ln(2*ln(2))*ln(ln(2))/(4*x*ln(ln(2))-exp(x^3/(2+x)))
Time = 0.07 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21 \[ \int \frac {\left (e^{\frac {x^3}{2+x}} \left (48 x^2+16 x^3\right ) \log (\log (2))+\left (-128-128 x-32 x^2\right ) \log ^2(\log (2))\right ) \log (\log (4))}{e^{\frac {2 x^3}{2+x}} \left (4+4 x+x^2\right )+e^{\frac {x^3}{2+x}} \left (-32 x-32 x^2-8 x^3\right ) \log (\log (2))+\left (64 x^2+64 x^3+16 x^4\right ) \log ^2(\log (2))} \, dx=\frac {8 \, {\left (\log \left (2\right ) \log \left (\log \left (2\right )\right ) + \log \left (\log \left (2\right )\right )^{2}\right )}}{4 \, x \log \left (\log \left (2\right )\right ) - e^{\left (\frac {x^{3}}{x + 2}\right )}} \] Input:
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")
Output:
8*(log(2)*log(log(2)) + log(log(2))^2)/(4*x*log(log(2)) - e^(x^3/(x + 2)))
Time = 0.11 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {\left (e^{\frac {x^3}{2+x}} \left (48 x^2+16 x^3\right ) \log (\log (2))+\left (-128-128 x-32 x^2\right ) \log ^2(\log (2))\right ) \log (\log (4))}{e^{\frac {2 x^3}{2+x}} \left (4+4 x+x^2\right )+e^{\frac {x^3}{2+x}} \left (-32 x-32 x^2-8 x^3\right ) \log (\log (2))+\left (64 x^2+64 x^3+16 x^4\right ) \log ^2(\log (2))} \, dx=\frac {- 8 \log {\left (\log {\left (2 \right )} \right )}^{2} - 8 \log {\left (2 \right )} \log {\left (\log {\left (2 \right )} \right )}}{- 4 x \log {\left (\log {\left (2 \right )} \right )} + e^{\frac {x^{3}}{x + 2}}} \] Input:
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)
Output:
(-8*log(log(2))**2 - 8*log(2)*log(log(2)))/(-4*x*log(log(2)) + exp(x**3/(x + 2)))
Time = 0.19 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.76 \[ \int \frac {\left (e^{\frac {x^3}{2+x}} \left (48 x^2+16 x^3\right ) \log (\log (2))+\left (-128-128 x-32 x^2\right ) \log ^2(\log (2))\right ) \log (\log (4))}{e^{\frac {2 x^3}{2+x}} \left (4+4 x+x^2\right )+e^{\frac {x^3}{2+x}} \left (-32 x-32 x^2-8 x^3\right ) \log (\log (2))+\left (64 x^2+64 x^3+16 x^4\right ) \log ^2(\log (2))} \, dx=\frac {8 \, e^{\left (2 \, x + \frac {8}{x + 2}\right )} \log \left (2 \, \log \left (2\right )\right ) \log \left (\log \left (2\right )\right )}{4 \, x e^{\left (2 \, x + \frac {8}{x + 2}\right )} \log \left (\log \left (2\right )\right ) - e^{\left (x^{2} + 4\right )}} \] Input:
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")
Output:
8*e^(2*x + 8/(x + 2))*log(2*log(2))*log(log(2))/(4*x*e^(2*x + 8/(x + 2))*l og(log(2)) - e^(x^2 + 4))
Time = 0.13 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {\left (e^{\frac {x^3}{2+x}} \left (48 x^2+16 x^3\right ) \log (\log (2))+\left (-128-128 x-32 x^2\right ) \log ^2(\log (2))\right ) \log (\log (4))}{e^{\frac {2 x^3}{2+x}} \left (4+4 x+x^2\right )+e^{\frac {x^3}{2+x}} \left (-32 x-32 x^2-8 x^3\right ) \log (\log (2))+\left (64 x^2+64 x^3+16 x^4\right ) \log ^2(\log (2))} \, dx=\frac {8 \, \log \left (2 \, \log \left (2\right )\right ) \log \left (\log \left (2\right )\right )}{4 \, x \log \left (\log \left (2\right )\right ) - e^{\left (\frac {x^{3}}{x + 2}\right )}} \] Input:
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")
Output:
8*log(2*log(2))*log(log(2))/(4*x*log(log(2)) - e^(x^3/(x + 2)))
Time = 0.44 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {\left (e^{\frac {x^3}{2+x}} \left (48 x^2+16 x^3\right ) \log (\log (2))+\left (-128-128 x-32 x^2\right ) \log ^2(\log (2))\right ) \log (\log (4))}{e^{\frac {2 x^3}{2+x}} \left (4+4 x+x^2\right )+e^{\frac {x^3}{2+x}} \left (-32 x-32 x^2-8 x^3\right ) \log (\log (2))+\left (64 x^2+64 x^3+16 x^4\right ) \log ^2(\log (2))} \, dx=-\frac {8\,\ln \left (2\,\ln \left (2\right )\right )\,\ln \left (\ln \left (2\right )\right )}{{\mathrm {e}}^{\frac {x^3}{x+2}}-4\,x\,\ln \left (\ln \left (2\right )\right )} \] Input:
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)
Output:
-(8*log(2*log(2))*log(log(2)))/(exp(x^3/(x + 2)) - 4*x*log(log(2)))
Time = 0.39 (sec) , antiderivative size = 100, normalized size of antiderivative = 3.45 \[ \int \frac {\left (e^{\frac {x^3}{2+x}} \left (48 x^2+16 x^3\right ) \log (\log (2))+\left (-128-128 x-32 x^2\right ) \log ^2(\log (2))\right ) \log (\log (4))}{e^{\frac {2 x^3}{2+x}} \left (4+4 x+x^2\right )+e^{\frac {x^3}{2+x}} \left (-32 x-32 x^2-8 x^3\right ) \log (\log (2))+\left (64 x^2+64 x^3+16 x^4\right ) \log ^2(\log (2))} \, dx=\frac {\mathrm {log}\left (2 \,\mathrm {log}\left (2\right )\right ) \left (-e^{x^{2}} e^{4}+4 e^{\frac {2 x^{2}+4 x +8}{x +2}} \mathrm {log}\left (\mathrm {log}\left (2\right )\right ) x -8 e^{\frac {2 x^{2}+4 x +8}{x +2}} \mathrm {log}\left (\mathrm {log}\left (2\right )\right )\right )}{e^{x^{2}} e^{4}-4 e^{\frac {2 x^{2}+4 x +8}{x +2}} \mathrm {log}\left (\mathrm {log}\left (2\right )\right ) x} \] Input:
int(((-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)
Output:
(log(2*log(2))*( - e**(x**2)*e**4 + 4*e**((2*x**2 + 4*x + 8)/(x + 2))*log( log(2))*x - 8*e**((2*x**2 + 4*x + 8)/(x + 2))*log(log(2))))/(e**(x**2)*e** 4 - 4*e**((2*x**2 + 4*x + 8)/(x + 2))*log(log(2))*x)