Integrand size = 150, antiderivative size = 22 \[ \int \frac {-32+2 e^4+4 x-8 e^2 x+6 x^2+\left (16-e^4-x+2 e^2 x-x^2\right ) \log \left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right ) \log \left (\log ^2\left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right )\right )}{\left (-16 x^2+e^4 x^2+x^3-2 e^2 x^3+x^4\right ) \log \left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right )} \, dx=\frac {\log \left (\log ^2\left (x \left (-16+\left (e^2-x\right )^2+x\right )\right )\right )}{x} \]
Time = 0.06 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {-32+2 e^4+4 x-8 e^2 x+6 x^2+\left (16-e^4-x+2 e^2 x-x^2\right ) \log \left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right ) \log \left (\log ^2\left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right )\right )}{\left (-16 x^2+e^4 x^2+x^3-2 e^2 x^3+x^4\right ) \log \left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right )} \, dx=\frac {\log \left (\log ^2\left (x \left (-16+e^4+x-2 e^2 x+x^2\right )\right )\right )}{x} \]
Integrate[(-32 + 2*E^4 + 4*x - 8*E^2*x + 6*x^2 + (16 - E^4 - x + 2*E^2*x - x^2)*Log[-16*x + E^4*x + x^2 - 2*E^2*x^2 + x^3]*Log[Log[-16*x + E^4*x + x ^2 - 2*E^2*x^2 + x^3]^2])/((-16*x^2 + E^4*x^2 + x^3 - 2*E^2*x^3 + x^4)*Log [-16*x + E^4*x + x^2 - 2*E^2*x^2 + x^3]),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.
\(\displaystyle \int \frac {6 x^2+\left (-x^2+2 e^2 x-x-e^4+16\right ) \log \left (x^3-2 e^2 x^2+x^2+e^4 x-16 x\right ) \log \left (\log ^2\left (x^3-2 e^2 x^2+x^2+e^4 x-16 x\right )\right )-8 e^2 x+4 x+2 e^4-32}{\left (x^4-2 e^2 x^3+x^3+e^4 x^2-16 x^2\right ) \log \left (x^3-2 e^2 x^2+x^2+e^4 x-16 x\right )} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {6 x^2+\left (-x^2+2 e^2 x-x-e^4+16\right ) \log \left (x^3-2 e^2 x^2+x^2+e^4 x-16 x\right ) \log \left (\log ^2\left (x^3-2 e^2 x^2+x^2+e^4 x-16 x\right )\right )-8 e^2 x+4 x+2 e^4-32}{\left (x^4-2 e^2 x^3+x^3+\left (e^4-16\right ) x^2\right ) \log \left (x^3-2 e^2 x^2+x^2+e^4 x-16 x\right )}dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {6 x^2+\left (-x^2+2 e^2 x-x-e^4+16\right ) \log \left (x^3-2 e^2 x^2+x^2+e^4 x-16 x\right ) \log \left (\log ^2\left (x^3-2 e^2 x^2+x^2+e^4 x-16 x\right )\right )-8 e^2 x+4 x+2 e^4-32}{\left (x^4+\left (1-2 e^2\right ) x^3+\left (e^4-16\right ) x^2\right ) \log \left (x^3-2 e^2 x^2+x^2+e^4 x-16 x\right )}dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {6 x^2+\left (-x^2+2 e^2 x-x-e^4+16\right ) \log \left (x^3-2 e^2 x^2+x^2+e^4 x-16 x\right ) \log \left (\log ^2\left (x^3-2 e^2 x^2+x^2+e^4 x-16 x\right )\right )+\left (4-8 e^2\right ) x+2 e^4-32}{\left (x^4+\left (1-2 e^2\right ) x^3+\left (e^4-16\right ) x^2\right ) \log \left (x^3-2 e^2 x^2+x^2+e^4 x-16 x\right )}dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {6 x^2+\left (-x^2+2 e^2 x-x-e^4+16\right ) \log \left (x^3-2 e^2 x^2+x^2+e^4 x-16 x\right ) \log \left (\log ^2\left (x^3-2 e^2 x^2+x^2+e^4 x-16 x\right )\right )+\left (4-8 e^2\right ) x+2 e^4-32}{x^2 \left (x^2+\left (1-2 e^2\right ) x+e^4-16\right ) \log \left (x^3-2 e^2 x^2+x^2+e^4 x-16 x\right )}dx\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \int \left (\frac {2 \left (-3 x^2-2 \left (1-2 e^2\right ) x-e^4+16\right )}{x^2 \left (-x^2-\left (1-2 e^2\right ) x-e^4+16\right ) \log \left (x \left (x^2+\left (1-2 e^2\right ) x+e^4-16\right )\right )}-\frac {\log \left (\log ^2\left (x \left (x^2+\left (1-2 e^2\right ) x+e^4-16\right )\right )\right )}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\int \frac {\log \left (\log ^2\left (x \left (x^2+\left (1-2 e^2\right ) x+e^4-16\right )\right )\right )}{x^2}dx-\frac {2 \left (1-2 e^2\right ) \left (1+\frac {1-2 e^2}{\sqrt {65-4 e^2}}\right ) \int \frac {1}{\left (-2 x-\sqrt {65-4 e^2}+2 e^2-1\right ) \log \left (x \left (x^2+\left (1-2 e^2\right ) x+e^4-16\right )\right )}dx}{16-e^4}-\frac {2 \left (1-2 e^2\right ) \left (1-\frac {1-2 e^2}{\sqrt {65-4 e^2}}\right ) \int \frac {1}{\left (-2 x+\sqrt {65-4 e^2}+2 e^2-1\right ) \log \left (x \left (x^2+\left (1-2 e^2\right ) x+e^4-16\right )\right )}dx}{16-e^4}-\frac {4 \left (33-4 e^2+2 e^4\right ) \int \frac {1}{\left (-2 x+\sqrt {65-4 e^2}+2 e^2-1\right ) \log \left (x \left (x^2+\left (1-2 e^2\right ) x+e^4-16\right )\right )}dx}{\sqrt {65-4 e^2} \left (16-e^4\right )}+2 \int \frac {1}{x^2 \log \left (x \left (x^2+\left (1-2 e^2\right ) x+e^4-16\right )\right )}dx-\frac {2 \left (1-2 e^2\right ) \int \frac {1}{x \log \left (x \left (x^2+\left (1-2 e^2\right ) x+e^4-16\right )\right )}dx}{16-e^4}-\frac {4 \left (33-4 e^2+2 e^4\right ) \int \frac {1}{\left (2 x+\sqrt {65-4 e^2}-2 e^2+1\right ) \log \left (x \left (x^2+\left (1-2 e^2\right ) x+e^4-16\right )\right )}dx}{\sqrt {65-4 e^2} \left (16-e^4\right )}\) |
Int[(-32 + 2*E^4 + 4*x - 8*E^2*x + 6*x^2 + (16 - E^4 - x + 2*E^2*x - x^2)* Log[-16*x + E^4*x + x^2 - 2*E^2*x^2 + x^3]*Log[Log[-16*x + E^4*x + x^2 - 2 *E^2*x^2 + x^3]^2])/((-16*x^2 + E^4*x^2 + x^3 - 2*E^2*x^3 + x^4)*Log[-16*x + E^4*x + x^2 - 2*E^2*x^2 + x^3]),x]
3.10.15.3.1 Defintions of rubi rules used
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, 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_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
Time = 2.77 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27
method | result | size |
parallelrisch | \(\frac {\ln \left ({\ln \left (x \left (-16+{\mathrm e}^{4}-2 \,{\mathrm e}^{2} x +x^{2}+x \right )\right )}^{2}\right )}{x}\) | \(28\) |
int(((-exp(1)^4+2*x*exp(1)^2-x^2-x+16)*ln(x*exp(1)^4-2*x^2*exp(1)^2+x^3+x^ 2-16*x)*ln(ln(x*exp(1)^4-2*x^2*exp(1)^2+x^3+x^2-16*x)^2)+2*exp(1)^4-8*x*ex p(1)^2+6*x^2+4*x-32)/(x^2*exp(1)^4-2*x^3*exp(1)^2+x^4+x^3-16*x^2)/ln(x*exp (1)^4-2*x^2*exp(1)^2+x^3+x^2-16*x),x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {-32+2 e^4+4 x-8 e^2 x+6 x^2+\left (16-e^4-x+2 e^2 x-x^2\right ) \log \left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right ) \log \left (\log ^2\left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right )\right )}{\left (-16 x^2+e^4 x^2+x^3-2 e^2 x^3+x^4\right ) \log \left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right )} \, dx=\frac {\log \left (\log \left (x^{3} - 2 \, x^{2} e^{2} + x^{2} + x e^{4} - 16 \, x\right )^{2}\right )}{x} \]
integrate(((-exp(1)^4+2*x*exp(1)^2-x^2-x+16)*log(x*exp(1)^4-2*x^2*exp(1)^2 +x^3+x^2-16*x)*log(log(x*exp(1)^4-2*x^2*exp(1)^2+x^3+x^2-16*x)^2)+2*exp(1) ^4-8*x*exp(1)^2+6*x^2+4*x-32)/(x^2*exp(1)^4-2*x^3*exp(1)^2+x^4+x^3-16*x^2) /log(x*exp(1)^4-2*x^2*exp(1)^2+x^3+x^2-16*x),x, algorithm=\
Time = 0.20 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {-32+2 e^4+4 x-8 e^2 x+6 x^2+\left (16-e^4-x+2 e^2 x-x^2\right ) \log \left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right ) \log \left (\log ^2\left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right )\right )}{\left (-16 x^2+e^4 x^2+x^3-2 e^2 x^3+x^4\right ) \log \left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right )} \, dx=\frac {\log {\left (\log {\left (x^{3} - 2 x^{2} e^{2} + x^{2} - 16 x + x e^{4} \right )}^{2} \right )}}{x} \]
integrate(((-exp(1)**4+2*x*exp(1)**2-x**2-x+16)*ln(x*exp(1)**4-2*x**2*exp( 1)**2+x**3+x**2-16*x)*ln(ln(x*exp(1)**4-2*x**2*exp(1)**2+x**3+x**2-16*x)** 2)+2*exp(1)**4-8*x*exp(1)**2+6*x**2+4*x-32)/(x**2*exp(1)**4-2*x**3*exp(1)* *2+x**4+x**3-16*x**2)/ln(x*exp(1)**4-2*x**2*exp(1)**2+x**3+x**2-16*x),x)
Time = 0.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {-32+2 e^4+4 x-8 e^2 x+6 x^2+\left (16-e^4-x+2 e^2 x-x^2\right ) \log \left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right ) \log \left (\log ^2\left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right )\right )}{\left (-16 x^2+e^4 x^2+x^3-2 e^2 x^3+x^4\right ) \log \left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right )} \, dx=\frac {2 \, \log \left (\log \left (x^{2} - x {\left (2 \, e^{2} - 1\right )} + e^{4} - 16\right ) + \log \left (x\right )\right )}{x} \]
integrate(((-exp(1)^4+2*x*exp(1)^2-x^2-x+16)*log(x*exp(1)^4-2*x^2*exp(1)^2 +x^3+x^2-16*x)*log(log(x*exp(1)^4-2*x^2*exp(1)^2+x^3+x^2-16*x)^2)+2*exp(1) ^4-8*x*exp(1)^2+6*x^2+4*x-32)/(x^2*exp(1)^4-2*x^3*exp(1)^2+x^4+x^3-16*x^2) /log(x*exp(1)^4-2*x^2*exp(1)^2+x^3+x^2-16*x),x, algorithm=\
Time = 14.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {-32+2 e^4+4 x-8 e^2 x+6 x^2+\left (16-e^4-x+2 e^2 x-x^2\right ) \log \left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right ) \log \left (\log ^2\left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right )\right )}{\left (-16 x^2+e^4 x^2+x^3-2 e^2 x^3+x^4\right ) \log \left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right )} \, dx=\frac {\log \left (\log \left (x^{3} - 2 \, x^{2} e^{2} + x^{2} + x e^{4} - 16 \, x\right )^{2}\right )}{x} \]
integrate(((-exp(1)^4+2*x*exp(1)^2-x^2-x+16)*log(x*exp(1)^4-2*x^2*exp(1)^2 +x^3+x^2-16*x)*log(log(x*exp(1)^4-2*x^2*exp(1)^2+x^3+x^2-16*x)^2)+2*exp(1) ^4-8*x*exp(1)^2+6*x^2+4*x-32)/(x^2*exp(1)^4-2*x^3*exp(1)^2+x^4+x^3-16*x^2) /log(x*exp(1)^4-2*x^2*exp(1)^2+x^3+x^2-16*x),x, algorithm=\
Time = 9.97 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {-32+2 e^4+4 x-8 e^2 x+6 x^2+\left (16-e^4-x+2 e^2 x-x^2\right ) \log \left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right ) \log \left (\log ^2\left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right )\right )}{\left (-16 x^2+e^4 x^2+x^3-2 e^2 x^3+x^4\right ) \log \left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right )} \, dx=\frac {\ln \left ({\ln \left (x\,{\mathrm {e}}^4-16\,x-2\,x^2\,{\mathrm {e}}^2+x^2+x^3\right )}^2\right )}{x} \]