Integrand size = 92, antiderivative size = 21 \begin {dmath*} \int \frac {e^{2 e^{\frac {8}{-3+3 \log (x)}} x^4+\frac {8}{-3+3 \log (x)}} \left (8 x^4-48 x^4 \log (x)+24 x^4 \log ^2(x)+e^{-\frac {8}{-3+3 \log (x)}} \left (3-6 \log (x)+3 \log ^2(x)\right )\right )}{3-6 \log (x)+3 \log ^2(x)} \, dx=e^{2 e^{-\frac {8}{3-3 \log (x)}} x^4} x \end {dmath*}
Time = 1.88 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \begin {dmath*} \int \frac {e^{2 e^{\frac {8}{-3+3 \log (x)}} x^4+\frac {8}{-3+3 \log (x)}} \left (8 x^4-48 x^4 \log (x)+24 x^4 \log ^2(x)+e^{-\frac {8}{-3+3 \log (x)}} \left (3-6 \log (x)+3 \log ^2(x)\right )\right )}{3-6 \log (x)+3 \log ^2(x)} \, dx=e^{2 e^{\frac {8}{3 (-1+\log (x))}} x^4} x \end {dmath*}
Integrate[(E^(2*E^(8/(-3 + 3*Log[x]))*x^4 + 8/(-3 + 3*Log[x]))*(8*x^4 - 48 *x^4*Log[x] + 24*x^4*Log[x]^2 + (3 - 6*Log[x] + 3*Log[x]^2)/E^(8/(-3 + 3*L og[x]))))/(3 - 6*Log[x] + 3*Log[x]^2),x]
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.
\(\displaystyle \int \frac {e^{2 x^4 e^{\frac {8}{3 \log (x)-3}}+\frac {8}{3 \log (x)-3}} \left (8 x^4+24 x^4 \log ^2(x)-48 x^4 \log (x)+e^{-\frac {8}{3 \log (x)-3}} \left (3 \log ^2(x)-6 \log (x)+3\right )\right )}{3 \log ^2(x)-6 \log (x)+3} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{2 x^4 e^{\frac {8}{3 \log (x)-3}}+\frac {8}{3 \log (x)-3}} \left (8 x^4+24 x^4 \log ^2(x)-48 x^4 \log (x)+e^{-\frac {8}{3 \log (x)-3}} \left (3 \log ^2(x)-6 \log (x)+3\right )\right )}{3 (1-\log (x))^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \int \frac {\exp \left (2 e^{-\frac {8}{3 (1-\log (x))}} x^4-\frac {8}{3 (1-\log (x))}\right ) \left (24 \log ^2(x) x^4-48 \log (x) x^4+8 x^4+3 e^{\frac {8}{3 (1-\log (x))}} \left (\log ^2(x)-2 \log (x)+1\right )\right )}{(1-\log (x))^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{3} \int \left (\frac {8 \exp \left (2 e^{-\frac {8}{3 (1-\log (x))}} x^4-\frac {8}{3 (1-\log (x))}\right ) \left (3 \log ^2(x)-6 \log (x)+1\right ) x^4}{(\log (x)-1)^2}+3 \exp \left (2 e^{-\frac {8}{3 (1-\log (x))}} x^4-\frac {8}{3 (1-\log (x))}-\frac {8}{3 (\log (x)-1)}\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (24 \int \exp \left (2 e^{-\frac {8}{3 (1-\log (x))}} x^4-\frac {8}{3 (1-\log (x))}\right ) x^4dx-16 \int \frac {\exp \left (2 e^{-\frac {8}{3 (1-\log (x))}} x^4-\frac {8}{3 (1-\log (x))}\right ) x^4}{(\log (x)-1)^2}dx+3 \int e^{2 e^{\frac {8}{3 (\log (x)-1)}} x^4}dx\right )\) |
Int[(E^(2*E^(8/(-3 + 3*Log[x]))*x^4 + 8/(-3 + 3*Log[x]))*(8*x^4 - 48*x^4*L og[x] + 24*x^4*Log[x]^2 + (3 - 6*Log[x] + 3*Log[x]^2)/E^(8/(-3 + 3*Log[x]) )))/(3 - 6*Log[x] + 3*Log[x]^2),x]
3.10.24.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 10.19 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86
method | result | size |
risch | \(x \,{\mathrm e}^{2 x^{4} {\mathrm e}^{\frac {8}{3 \left (\ln \left (x \right )-1\right )}}}\) | \(18\) |
parallelrisch | \(-\frac {-30 \ln \left (x \right ) {\mathrm e}^{2 x^{4} {\mathrm e}^{\frac {8}{3 \left (\ln \left (x \right )-1\right )}}} x +30 x \,{\mathrm e}^{2 x^{4} {\mathrm e}^{\frac {8}{3 \left (\ln \left (x \right )-1\right )}}}}{30 \left (\ln \left (x \right )-1\right )}\) | \(52\) |
int(((3*ln(x)^2-6*ln(x)+3)*exp(-4/(3*ln(x)-3))^2+24*x^4*ln(x)^2-48*x^4*ln( x)+8*x^4)*exp(2*x^4/exp(-4/(3*ln(x)-3))^2)/(3*ln(x)^2-6*ln(x)+3)/exp(-4/(3 *ln(x)-3))^2,x,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (17) = 34\).
Time = 0.28 (sec) , antiderivative size = 45, normalized size of antiderivative = 2.14 \begin {dmath*} \int \frac {e^{2 e^{\frac {8}{-3+3 \log (x)}} x^4+\frac {8}{-3+3 \log (x)}} \left (8 x^4-48 x^4 \log (x)+24 x^4 \log ^2(x)+e^{-\frac {8}{-3+3 \log (x)}} \left (3-6 \log (x)+3 \log ^2(x)\right )\right )}{3-6 \log (x)+3 \log ^2(x)} \, dx=x e^{\left (\frac {2 \, {\left (3 \, {\left (x^{4} \log \left (x\right ) - x^{4}\right )} e^{\left (\frac {8}{3 \, {\left (\log \left (x\right ) - 1\right )}}\right )} + 4\right )}}{3 \, {\left (\log \left (x\right ) - 1\right )}} - \frac {8}{3 \, {\left (\log \left (x\right ) - 1\right )}}\right )} \end {dmath*}
integrate(((3*log(x)^2-6*log(x)+3)*exp(-4/(3*log(x)-3))^2+24*x^4*log(x)^2- 48*x^4*log(x)+8*x^4)*exp(2*x^4/exp(-4/(3*log(x)-3))^2)/(3*log(x)^2-6*log(x )+3)/exp(-4/(3*log(x)-3))^2,x, algorithm=\
Time = 32.89 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \begin {dmath*} \int \frac {e^{2 e^{\frac {8}{-3+3 \log (x)}} x^4+\frac {8}{-3+3 \log (x)}} \left (8 x^4-48 x^4 \log (x)+24 x^4 \log ^2(x)+e^{-\frac {8}{-3+3 \log (x)}} \left (3-6 \log (x)+3 \log ^2(x)\right )\right )}{3-6 \log (x)+3 \log ^2(x)} \, dx=x e^{2 x^{4} e^{\frac {8}{3 \log {\left (x \right )} - 3}}} \end {dmath*}
integrate(((3*ln(x)**2-6*ln(x)+3)*exp(-4/(3*ln(x)-3))**2+24*x**4*ln(x)**2- 48*x**4*ln(x)+8*x**4)*exp(2*x**4/exp(-4/(3*ln(x)-3))**2)/(3*ln(x)**2-6*ln( x)+3)/exp(-4/(3*ln(x)-3))**2,x)
Time = 0.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \begin {dmath*} \int \frac {e^{2 e^{\frac {8}{-3+3 \log (x)}} x^4+\frac {8}{-3+3 \log (x)}} \left (8 x^4-48 x^4 \log (x)+24 x^4 \log ^2(x)+e^{-\frac {8}{-3+3 \log (x)}} \left (3-6 \log (x)+3 \log ^2(x)\right )\right )}{3-6 \log (x)+3 \log ^2(x)} \, dx=x e^{\left (2 \, x^{4} e^{\left (\frac {8}{3 \, {\left (\log \left (x\right ) - 1\right )}}\right )}\right )} \end {dmath*}
integrate(((3*log(x)^2-6*log(x)+3)*exp(-4/(3*log(x)-3))^2+24*x^4*log(x)^2- 48*x^4*log(x)+8*x^4)*exp(2*x^4/exp(-4/(3*log(x)-3))^2)/(3*log(x)^2-6*log(x )+3)/exp(-4/(3*log(x)-3))^2,x, algorithm=\
\begin {dmath*} \int \frac {e^{2 e^{\frac {8}{-3+3 \log (x)}} x^4+\frac {8}{-3+3 \log (x)}} \left (8 x^4-48 x^4 \log (x)+24 x^4 \log ^2(x)+e^{-\frac {8}{-3+3 \log (x)}} \left (3-6 \log (x)+3 \log ^2(x)\right )\right )}{3-6 \log (x)+3 \log ^2(x)} \, dx=\int { \frac {{\left (24 \, x^{4} \log \left (x\right )^{2} - 48 \, x^{4} \log \left (x\right ) + 8 \, x^{4} + 3 \, {\left (\log \left (x\right )^{2} - 2 \, \log \left (x\right ) + 1\right )} e^{\left (-\frac {8}{3 \, {\left (\log \left (x\right ) - 1\right )}}\right )}\right )} e^{\left (2 \, x^{4} e^{\left (\frac {8}{3 \, {\left (\log \left (x\right ) - 1\right )}}\right )} + \frac {8}{3 \, {\left (\log \left (x\right ) - 1\right )}}\right )}}{3 \, {\left (\log \left (x\right )^{2} - 2 \, \log \left (x\right ) + 1\right )}} \,d x } \end {dmath*}
integrate(((3*log(x)^2-6*log(x)+3)*exp(-4/(3*log(x)-3))^2+24*x^4*log(x)^2- 48*x^4*log(x)+8*x^4)*exp(2*x^4/exp(-4/(3*log(x)-3))^2)/(3*log(x)^2-6*log(x )+3)/exp(-4/(3*log(x)-3))^2,x, algorithm=\
integrate(1/3*(24*x^4*log(x)^2 - 48*x^4*log(x) + 8*x^4 + 3*(log(x)^2 - 2*l og(x) + 1)*e^(-8/3/(log(x) - 1)))*e^(2*x^4*e^(8/3/(log(x) - 1)) + 8/3/(log (x) - 1))/(log(x)^2 - 2*log(x) + 1), x)
Time = 14.61 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \begin {dmath*} \int \frac {e^{2 e^{\frac {8}{-3+3 \log (x)}} x^4+\frac {8}{-3+3 \log (x)}} \left (8 x^4-48 x^4 \log (x)+24 x^4 \log ^2(x)+e^{-\frac {8}{-3+3 \log (x)}} \left (3-6 \log (x)+3 \log ^2(x)\right )\right )}{3-6 \log (x)+3 \log ^2(x)} \, dx=x\,{\mathrm {e}}^{2\,x^4\,{\mathrm {e}}^{\frac {8}{3\,\ln \left (x\right )-3}}} \end {dmath*}