Integrand size = 84, antiderivative size = 20 \[ \int \frac {e^4 (-8+2 x)+\left (16 x-8 x^2+x^3+e^4 \left (-16 x+8 x^2-x^3\right )\right ) \log \left (x^2\right )-e^4 x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{e^4 \left (16 x-8 x^2+x^3\right ) \log \left (x^2\right )} \, dx=-x+\frac {x}{e^4}+\frac {\log \left (\log \left (x^2\right )\right )}{-4+x} \]
Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {e^4 (-8+2 x)+\left (16 x-8 x^2+x^3+e^4 \left (-16 x+8 x^2-x^3\right )\right ) \log \left (x^2\right )-e^4 x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{e^4 \left (16 x-8 x^2+x^3\right ) \log \left (x^2\right )} \, dx=\left (-1+\frac {1}{e^4}\right ) x+\frac {\log \left (\log \left (x^2\right )\right )}{-4+x} \]
Integrate[(E^4*(-8 + 2*x) + (16*x - 8*x^2 + x^3 + E^4*(-16*x + 8*x^2 - x^3 ))*Log[x^2] - E^4*x*Log[x^2]*Log[Log[x^2]])/(E^4*(16*x - 8*x^2 + 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.
\(\displaystyle \int \frac {-e^4 x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )+\left (x^3-8 x^2+e^4 \left (-x^3+8 x^2-16 x\right )+16 x\right ) \log \left (x^2\right )+e^4 (2 x-8)}{e^4 \left (x^3-8 x^2+16 x\right ) \log \left (x^2\right )} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int -\frac {2 e^4 (4-x)-\left (x^3-8 x^2+16 x-e^4 \left (x^3-8 x^2+16 x\right )\right ) \log \left (x^2\right )+e^4 x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{\left (x^3-8 x^2+16 x\right ) \log \left (x^2\right )}dx}{e^4}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {2 e^4 (4-x)-\left (x^3-8 x^2+16 x-e^4 \left (x^3-8 x^2+16 x\right )\right ) \log \left (x^2\right )+e^4 x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{\left (x^3-8 x^2+16 x\right ) \log \left (x^2\right )}dx}{e^4}\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle -\frac {\int \frac {2 e^4 (4-x)-\left (x^3-8 x^2+16 x-e^4 \left (x^3-8 x^2+16 x\right )\right ) \log \left (x^2\right )+e^4 x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{x \left (x^2-8 x+16\right ) \log \left (x^2\right )}dx}{e^4}\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle -\frac {\int \left (\frac {e^4 \log \left (\log \left (x^2\right )\right )}{(x-4)^2}-\frac {2 e^4}{(x-4) x \log \left (x^2\right )}+e^4-1\right )dx}{e^4}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-2 e^4 \int \frac {1}{(x-4) x \log \left (x^2\right )}dx+e^4 \int \frac {\log \left (\log \left (x^2\right )\right )}{(x-4)^2}dx-\left (\left (1-e^4\right ) x\right )}{e^4}\) |
Int[(E^4*(-8 + 2*x) + (16*x - 8*x^2 + x^3 + E^4*(-16*x + 8*x^2 - x^3))*Log [x^2] - E^4*x*Log[x^2]*Log[Log[x^2]])/(E^4*(16*x - 8*x^2 + x^3)*Log[x^2]), x]
3.14.14.3.1 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]]
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.80 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.95
method | result | size |
parallelrisch | \(-\frac {{\mathrm e}^{-4} \left (2 x^{2} {\mathrm e}^{4}+32-2 \ln \left (\ln \left (x^{2}\right )\right ) {\mathrm e}^{4}-2 x^{2}-32 \,{\mathrm e}^{4}\right )}{2 \left (x -4\right )}\) | \(39\) |
int((-x*exp(4)*ln(x^2)*ln(ln(x^2))+((-x^3+8*x^2-16*x)*exp(4)+x^3-8*x^2+16* x)*ln(x^2)+(2*x-8)*exp(4))/(x^3-8*x^2+16*x)/exp(4)/ln(x^2),x,method=_RETUR NVERBOSE)
Time = 0.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.70 \[ \int \frac {e^4 (-8+2 x)+\left (16 x-8 x^2+x^3+e^4 \left (-16 x+8 x^2-x^3\right )\right ) \log \left (x^2\right )-e^4 x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{e^4 \left (16 x-8 x^2+x^3\right ) \log \left (x^2\right )} \, dx=\frac {{\left (x^{2} - {\left (x^{2} - 4 \, x\right )} e^{4} + e^{4} \log \left (\log \left (x^{2}\right )\right ) - 4 \, x\right )} e^{\left (-4\right )}}{x - 4} \]
integrate((-x*exp(4)*log(x^2)*log(log(x^2))+((-x^3+8*x^2-16*x)*exp(4)+x^3- 8*x^2+16*x)*log(x^2)+(2*x-8)*exp(4))/(x^3-8*x^2+16*x)/exp(4)/log(x^2),x, a lgorithm=\
Time = 0.13 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {e^4 (-8+2 x)+\left (16 x-8 x^2+x^3+e^4 \left (-16 x+8 x^2-x^3\right )\right ) \log \left (x^2\right )-e^4 x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{e^4 \left (16 x-8 x^2+x^3\right ) \log \left (x^2\right )} \, dx=\frac {x \left (1 - e^{4}\right )}{e^{4}} + \frac {\log {\left (\log {\left (x^{2} \right )} \right )}}{x - 4} \]
integrate((-x*exp(4)*ln(x**2)*ln(ln(x**2))+((-x**3+8*x**2-16*x)*exp(4)+x** 3-8*x**2+16*x)*ln(x**2)+(2*x-8)*exp(4))/(x**3-8*x**2+16*x)/exp(4)/ln(x**2) ,x)
Time = 0.33 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.90 \[ \int \frac {e^4 (-8+2 x)+\left (16 x-8 x^2+x^3+e^4 \left (-16 x+8 x^2-x^3\right )\right ) \log \left (x^2\right )-e^4 x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{e^4 \left (16 x-8 x^2+x^3\right ) \log \left (x^2\right )} \, dx=-\frac {{\left (x^{2} {\left (e^{4} - 1\right )} - 4 \, x {\left (e^{4} - 1\right )} - e^{4} \log \left (2\right ) - e^{4} \log \left (\log \left (x\right )\right )\right )} e^{\left (-4\right )}}{x - 4} \]
integrate((-x*exp(4)*log(x^2)*log(log(x^2))+((-x^3+8*x^2-16*x)*exp(4)+x^3- 8*x^2+16*x)*log(x^2)+(2*x-8)*exp(4))/(x^3-8*x^2+16*x)/exp(4)/log(x^2),x, a lgorithm=\
Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.90 \[ \int \frac {e^4 (-8+2 x)+\left (16 x-8 x^2+x^3+e^4 \left (-16 x+8 x^2-x^3\right )\right ) \log \left (x^2\right )-e^4 x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{e^4 \left (16 x-8 x^2+x^3\right ) \log \left (x^2\right )} \, dx=-\frac {{\left (x^{2} e^{4} - x^{2} - 4 \, x e^{4} - e^{4} \log \left (\log \left (x^{2}\right )\right ) + 4 \, x\right )} e^{\left (-4\right )}}{x - 4} \]
integrate((-x*exp(4)*log(x^2)*log(log(x^2))+((-x^3+8*x^2-16*x)*exp(4)+x^3- 8*x^2+16*x)*log(x^2)+(2*x-8)*exp(4))/(x^3-8*x^2+16*x)/exp(4)/log(x^2),x, a lgorithm=\
Time = 10.84 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {e^4 (-8+2 x)+\left (16 x-8 x^2+x^3+e^4 \left (-16 x+8 x^2-x^3\right )\right ) \log \left (x^2\right )-e^4 x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{e^4 \left (16 x-8 x^2+x^3\right ) \log \left (x^2\right )} \, dx=\frac {\ln \left (\ln \left (x^2\right )\right )}{x-4}+x\,\left ({\mathrm {e}}^{-4}-1\right ) \]