Integrand size = 106, antiderivative size = 38 \[ \int \frac {128 e^8 x-16 e^8 x^2 (i \pi +\log (-\log (2 \log (\log (5)))))}{256 e^8-32 e^4 x^2+x^4+\left (-128 e^8 x+8 e^4 x^3\right ) (i \pi +\log (-\log (2 \log (\log (5)))))+16 e^8 x^2 (i \pi +\log (-\log (2 \log (\log (5)))))^2} \, dx=\frac {x}{-i \pi +\frac {16-\frac {x^2}{e^4}}{4 x}-\log (-\log (2 \log (\log (5))))} \] Output:
x/(1/4*(16-x^2/exp(1)^4)/x-ln(ln(2*ln(ln(5)))))
Time = 0.04 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.42 \[ \int \frac {128 e^8 x-16 e^8 x^2 (i \pi +\log (-\log (2 \log (\log (5)))))}{256 e^8-32 e^4 x^2+x^4+\left (-128 e^8 x+8 e^4 x^3\right ) (i \pi +\log (-\log (2 \log (\log (5)))))+16 e^8 x^2 (i \pi +\log (-\log (2 \log (\log (5)))))^2} \, dx=\frac {16 e^8 (-4+i \pi x+x \log (-\log (2 \log (\log (5)))))}{x^2+4 e^4 (-4+i \pi x+x \log (-\log (2 \log (\log (5)))))} \] Input:
Integrate[(128*E^8*x - 16*E^8*x^2*(I*Pi + Log[-Log[2*Log[Log[5]]]]))/(256* E^8 - 32*E^4*x^2 + x^4 + (-128*E^8*x + 8*E^4*x^3)*(I*Pi + Log[-Log[2*Log[L og[5]]]]) + 16*E^8*x^2*(I*Pi + Log[-Log[2*Log[Log[5]]]])^2),x]
Output:
(16*E^8*(-4 + I*Pi*x + x*Log[-Log[2*Log[Log[5]]]]))/(x^2 + 4*E^4*(-4 + I*P i*x + x*Log[-Log[2*Log[Log[5]]]]))
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(238\) vs. \(2(38)=76\).
Time = 0.74 (sec) , antiderivative size = 238, normalized size of antiderivative = 6.26, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.066, Rules used = {6, 2027, 2459, 1380, 27, 2345, 24}
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.
| \(\displaystyle \int \frac {128 e^8 x-16 e^8 x^2 (\log (-\log (2 \log (\log (5))))+i \pi )}{x^4+\left (8 e^4 x^3-128 e^8 x\right ) (\log (-\log (2 \log (\log (5))))+i \pi )-32 e^4 x^2+16 e^8 x^2 (\log (-\log (2 \log (\log (5))))+i \pi )^2+256 e^8} \, dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {128 e^8 x-16 e^8 x^2 (\log (-\log (2 \log (\log (5))))+i \pi )}{x^4+\left (8 e^4 x^3-128 e^8 x\right ) (\log (-\log (2 \log (\log (5))))+i \pi )+x^2 \left (-32 e^4+16 e^8 (\log (-\log (2 \log (\log (5))))+i \pi )^2\right )+256 e^8}dx\) |
| \(\Big \downarrow \) 2027 |
| \(\displaystyle \int \frac {x \left (128 e^8-16 e^8 x (\log (-\log (2 \log (\log (5))))+i \pi )\right )}{x^4+\left (8 e^4 x^3-128 e^8 x\right ) (\log (-\log (2 \log (\log (5))))+i \pi )+x^2 \left (-32 e^4+16 e^8 (\log (-\log (2 \log (\log (5))))+i \pi )^2\right )+256 e^8}dx\) |
| \(\Big \downarrow \) 2459 |
| \(\displaystyle \int \frac {-16 e^8 (\log (-\log (2 \log (\log (5))))+i \pi ) \left (x+\frac {1}{4} \left (8 e^4 \log (-\log (2 \log (\log (5))))+8 i e^4 \pi \right )\right )^2+64 e^8 \left (2-e^4 (\pi -i \log (-\log (2 \log (\log (5)))))^2\right ) \left (x+\frac {1}{4} \left (8 e^4 \log (-\log (2 \log (\log (5))))+8 i e^4 \pi \right )\right )-64 e^{12} (\pi -i \log (-\log (2 \log (\log (5))))) \left (2 i-i e^2 \pi -e^2 \log (-\log (2 \log (\log (5))))\right ) \left (2+e^2 \pi -i e^2 \log (-\log (2 \log (\log (5))))\right )}{\left (x+\frac {1}{4} \left (8 e^4 \log (-\log (2 \log (\log (5))))+8 i e^4 \pi \right )\right )^4-8 e^4 \left (4-e^4 (\pi -i \log (-\log (2 \log (\log (5)))))^2\right ) \left (x+\frac {1}{4} \left (8 e^4 \log (-\log (2 \log (\log (5))))+8 i e^4 \pi \right )\right )^2+16 e^8 \left (4-e^4 (\pi -i \log (-\log (2 \log (\log (5)))))^2\right )^2}d\left (x+\frac {1}{4} \left (8 e^4 \log (-\log (2 \log (\log (5))))+8 i e^4 \pi \right )\right )\) |
| \(\Big \downarrow \) 1380 |
| \(\displaystyle \int \frac {16 \left (-e^8 (\log (-\log (2 \log (\log (5))))+i \pi ) \left (x+\frac {1}{4} \left (8 e^4 \log (-\log (2 \log (\log (5))))+8 i e^4 \pi \right )\right )^2+4 e^8 \left (2-e^4 (\pi -i \log (-\log (2 \log (\log (5)))))^2\right ) \left (x+\frac {1}{4} \left (8 e^4 \log (-\log (2 \log (\log (5))))+8 i e^4 \pi \right )\right )-4 e^{12} \left (2-e^2 (\pi -i \log (-\log (2 \log (\log (5)))))\right ) (\pi -i \log (-\log (2 \log (\log (5))))) \left (2 i+e^2 (\log (-\log (2 \log (\log (5))))+i \pi )\right )\right )}{\left (\left (x+\frac {1}{4} \left (8 e^4 \log (-\log (2 \log (\log (5))))+8 i e^4 \pi \right )\right )^2-4 e^4 \left (4-e^4 (\pi -i \log (-\log (2 \log (\log (5)))))^2\right )\right )^2}d\left (x+\frac {1}{4} \left (8 e^4 \log (-\log (2 \log (\log (5))))+8 i e^4 \pi \right )\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 16 \int \frac {-e^8 (i \pi +\log (-\log (2 \log (\log (5))))) \left (x+\frac {1}{4} \left (8 i e^4 \pi +8 e^4 \log (-\log (2 \log (\log (5))))\right )\right )^2+4 e^8 \left (2-e^4 (\pi -i \log (-\log (2 \log (\log (5)))))^2\right ) \left (x+\frac {1}{4} \left (8 i e^4 \pi +8 e^4 \log (-\log (2 \log (\log (5))))\right )\right )-4 e^{12} \left (2-e^2 (\pi -i \log (-\log (2 \log (\log (5)))))\right ) \left (2+e^2 (\pi -i \log (-\log (2 \log (\log (5)))))\right ) (i \pi +\log (-\log (2 \log (\log (5)))))}{\left (\left (x+\frac {1}{4} \left (8 i e^4 \pi +8 e^4 \log (-\log (2 \log (\log (5))))\right )\right )^2-4 e^4 \left (4-e^4 (\pi -i \log (-\log (2 \log (\log (5)))))^2\right )\right )^2}d\left (x+\frac {1}{4} \left (8 i e^4 \pi +8 e^4 \log (-\log (2 \log (\log (5))))\right )\right )\) |
| \(\Big \downarrow \) 2345 |
| \(\displaystyle 16 \left (\frac {\int 0d\left (x+\frac {1}{4} \left (8 i e^4 \pi +8 e^4 \log (-\log (2 \log (\log (5))))\right )\right )}{8 e^4 \left (4-e^4 (\pi -i \log (-\log (2 \log (\log (5)))))^2\right )}-\frac {e^8 \left (2 \left (2-e^4 (\pi -i \log (-\log (2 \log (\log (5)))))^2\right ) \left (4-e^4 (\pi -i \log (-\log (2 \log (\log (5)))))^2\right )-(\log (-\log (2 \log (\log (5))))+i \pi ) \left (2+e^2 \pi -i e^2 \log (-\log (2 \log (\log (5))))\right ) \left (2-e^2 \pi +i e^2 \log (-\log (2 \log (\log (5))))\right ) \left (x+\frac {1}{4} \left (8 e^4 \log (-\log (2 \log (\log (5))))+8 i e^4 \pi \right )\right )\right )}{\left (4-e^4 (\pi -i \log (-\log (2 \log (\log (5)))))^2\right ) \left (\left (x+\frac {1}{4} \left (8 e^4 \log (-\log (2 \log (\log (5))))+8 i e^4 \pi \right )\right )^2-4 e^4 \left (4-e^4 (\pi -i \log (-\log (2 \log (\log (5)))))^2\right )\right )}\right )\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle -\frac {16 e^8 \left (2 \left (2-e^4 (\pi -i \log (-\log (2 \log (\log (5)))))^2\right ) \left (4-e^4 (\pi -i \log (-\log (2 \log (\log (5)))))^2\right )-(\log (-\log (2 \log (\log (5))))+i \pi ) \left (2+e^2 \pi -i e^2 \log (-\log (2 \log (\log (5))))\right ) \left (2-e^2 \pi +i e^2 \log (-\log (2 \log (\log (5))))\right ) \left (x+\frac {1}{4} \left (8 e^4 \log (-\log (2 \log (\log (5))))+8 i e^4 \pi \right )\right )\right )}{\left (4-e^4 (\pi -i \log (-\log (2 \log (\log (5)))))^2\right ) \left (\left (x+\frac {1}{4} \left (8 e^4 \log (-\log (2 \log (\log (5))))+8 i e^4 \pi \right )\right )^2-4 e^4 \left (4-e^4 (\pi -i \log (-\log (2 \log (\log (5)))))^2\right )\right )}\) |
Input:
Int[(128*E^8*x - 16*E^8*x^2*(I*Pi + Log[-Log[2*Log[Log[5]]]]))/(256*E^8 - 32*E^4*x^2 + x^4 + (-128*E^8*x + 8*E^4*x^3)*(I*Pi + Log[-Log[2*Log[Log[5]] ]]) + 16*E^8*x^2*(I*Pi + Log[-Log[2*Log[Log[5]]]])^2),x]
Output:
(-16*E^8*(2*(2 - E^4*(Pi - I*Log[-Log[2*Log[Log[5]]]])^2)*(4 - E^4*(Pi - I *Log[-Log[2*Log[Log[5]]]])^2) - (I*Pi + Log[-Log[2*Log[Log[5]]]])*(2 + E^2 *Pi - I*E^2*Log[-Log[2*Log[Log[5]]]])*(2 - E^2*Pi + I*E^2*Log[-Log[2*Log[L og[5]]]])*(x + ((8*I)*E^4*Pi + 8*E^4*Log[-Log[2*Log[Log[5]]]])/4)))/((4 - E^4*(Pi - I*Log[-Log[2*Log[Log[5]]]])^2)*(-4*E^4*(4 - E^4*(Pi - I*Log[-Log [2*Log[Log[5]]]])^2) + (x + ((8*I)*E^4*Pi + 8*E^4*Log[-Log[2*Log[Log[5]]]] )/4)^2))
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[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> S imp[1/c^p Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Int[(Fx_.)*((a_.)*(x_)^(r_.) + (b_.)*(x_)^(s_.))^(p_.), x_Symbol] :> Int[x^ (p*r)*(a + b*x^(s - r))^p*Fx, x] /; FreeQ[{a, b, r, s}, x] && IntegerQ[p] & & PosQ[s - r] && !(EqQ[p, 1] && EqQ[u, 1])
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot ient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b *f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) In t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
Int[(Pn_)^(p_.)*(Qx_), x_Symbol] :> With[{S = Coeff[Pn, x, Expon[Pn, x] - 1 ]/(Expon[Pn, x]*Coeff[Pn, x, Expon[Pn, x]])}, Subst[Int[ExpandToSum[Pn /. x -> x - S, x]^p*ExpandToSum[Qx /. x -> x - S, x], x], x, x + S] /; Binomial Q[Pn /. x -> x - S, x] || (IntegerQ[Expon[Pn, x]/2] && TrinomialQ[Pn /. x - > x - S, x])] /; FreeQ[p, x] && PolyQ[Pn, x] && GtQ[Expon[Pn, x], 2] && NeQ [Coeff[Pn, x, Expon[Pn, x] - 1], 0] && PolyQ[Qx, x] && !(MonomialQ[Qx, x] && IGtQ[p, 0])
Time = 0.30 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.16
| method | result | size |
| gosper | \(\frac {16 \left (x \ln \left (\ln \left (2 \ln \left (\ln \left (5\right )\right )\right )\right )-4\right ) {\mathrm e}^{8}}{4 \ln \left (\ln \left (2 \ln \left (\ln \left (5\right )\right )\right )\right ) {\mathrm e}^{4} x -16 \,{\mathrm e}^{4}+x^{2}}\) | \(44\) |
| risch | \(\frac {4 x \,{\mathrm e}^{8} \ln \left (\ln \left (2\right )+\ln \left (\ln \left (\ln \left (5\right )\right )\right )\right )-16 \,{\mathrm e}^{8}}{\ln \left (\ln \left (2\right )+\ln \left (\ln \left (\ln \left (5\right )\right )\right )\right ) {\mathrm e}^{4} x -4 \,{\mathrm e}^{4}+\frac {x^{2}}{4}}\) | \(44\) |
| parallelrisch | \(\frac {16 x \,{\mathrm e}^{8} \ln \left (\ln \left (2 \ln \left (\ln \left (5\right )\right )\right )\right )-64 \,{\mathrm e}^{8}}{4 \ln \left (\ln \left (2 \ln \left (\ln \left (5\right )\right )\right )\right ) {\mathrm e}^{4} x -16 \,{\mathrm e}^{4}+x^{2}}\) | \(49\) |
| norman | \(\frac {16 x \,{\mathrm e}^{8} \ln \left (\ln \left (2\right )+\ln \left (\ln \left (\ln \left (5\right )\right )\right )\right )-64 \,{\mathrm e}^{8}}{4 \ln \left (\ln \left (2 \ln \left (\ln \left (5\right )\right )\right )\right ) {\mathrm e}^{4} x -16 \,{\mathrm e}^{4}+x^{2}}\) | \(50\) |
| default | \(-4 \,{\mathrm e}^{8} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}+8 \ln \left (\ln \left (2 \ln \left (\ln \left (5\right )\right )\right )\right ) {\mathrm e}^{4} \textit {\_Z}^{3}+\left (16 \,{\mathrm e}^{8} \ln \left (\ln \left (2 \ln \left (\ln \left (5\right )\right )\right )\right )^{2}-32 \,{\mathrm e}^{4}\right ) \textit {\_Z}^{2}-128 \textit {\_Z} \,{\mathrm e}^{8} \ln \left (\ln \left (2 \ln \left (\ln \left (5\right )\right )\right )\right )+256 \,{\mathrm e}^{8}\right )}{\sum }\frac {\left (\ln \left (\ln \left (2 \ln \left (\ln \left (5\right )\right )\right )\right ) \textit {\_R}^{2}-8 \textit {\_R} \right ) \ln \left (x -\textit {\_R} \right )}{8 \textit {\_R} \,{\mathrm e}^{8} \ln \left (\ln \left (2 \ln \left (\ln \left (5\right )\right )\right )\right )^{2}-32 \,{\mathrm e}^{8} \ln \left (\ln \left (2 \ln \left (\ln \left (5\right )\right )\right )\right )+6 \ln \left (\ln \left (2 \ln \left (\ln \left (5\right )\right )\right )\right ) {\mathrm e}^{4} \textit {\_R}^{2}-16 \textit {\_R} \,{\mathrm e}^{4}+\textit {\_R}^{3}}\right )\) | \(139\) |
Input:
int((-16*x^2*exp(1)^8*ln(ln(2*ln(ln(5))))+128*x*exp(1)^8)/(16*x^2*exp(1)^8 *ln(ln(2*ln(ln(5))))^2+(-128*x*exp(1)^8+8*x^3*exp(1)^4)*ln(ln(2*ln(ln(5))) )+256*exp(1)^8-32*x^2*exp(1)^4+x^4),x,method=_RETURNVERBOSE)
Output:
16*(x*ln(ln(2*ln(ln(5))))-4)*exp(1)^8/(4*ln(ln(2*ln(ln(5))))*exp(1)^4*x-16 *exp(1)^4+x^2)
Time = 0.08 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.05 \[ \int \frac {128 e^8 x-16 e^8 x^2 (i \pi +\log (-\log (2 \log (\log (5)))))}{256 e^8-32 e^4 x^2+x^4+\left (-128 e^8 x+8 e^4 x^3\right ) (i \pi +\log (-\log (2 \log (\log (5)))))+16 e^8 x^2 (i \pi +\log (-\log (2 \log (\log (5)))))^2} \, dx=\frac {16 \, {\left (x e^{8} \log \left (\log \left (2 \, \log \left (\log \left (5\right )\right )\right )\right ) - 4 \, e^{8}\right )}}{4 \, x e^{4} \log \left (\log \left (2 \, \log \left (\log \left (5\right )\right )\right )\right ) + x^{2} - 16 \, e^{4}} \] Input:
integrate((-16*x^2*exp(1)^8*log(log(2*log(log(5))))+128*x*exp(1)^8)/(16*x^ 2*exp(1)^8*log(log(2*log(log(5))))^2+(-128*x*exp(1)^8+8*x^3*exp(1)^4)*log( log(2*log(log(5))))+256*exp(1)^8-32*x^2*exp(1)^4+x^4),x, algorithm="fricas ")
Output:
16*(x*e^8*log(log(2*log(log(5)))) - 4*e^8)/(4*x*e^4*log(log(2*log(log(5))) ) + x^2 - 16*e^4)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (27) = 54\).
Time = 1.01 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.92 \[ \int \frac {128 e^8 x-16 e^8 x^2 (i \pi +\log (-\log (2 \log (\log (5)))))}{256 e^8-32 e^4 x^2+x^4+\left (-128 e^8 x+8 e^4 x^3\right ) (i \pi +\log (-\log (2 \log (\log (5)))))+16 e^8 x^2 (i \pi +\log (-\log (2 \log (\log (5)))))^2} \, dx=- \frac {x \left (- 16 e^{8} \log {\left (- \log {\left (2 \right )} - \log {\left (\log {\left (\log {\left (5 \right )} \right )} \right )} \right )} - 16 i \pi e^{8}\right ) + 64 e^{8}}{x^{2} + x \left (4 e^{4} \log {\left (- \log {\left (2 \right )} - \log {\left (\log {\left (\log {\left (5 \right )} \right )} \right )} \right )} + 4 i \pi e^{4}\right ) - 16 e^{4}} \] Input:
integrate((-16*x**2*exp(1)**8*ln(ln(2*ln(ln(5))))+128*x*exp(1)**8)/(16*x** 2*exp(1)**8*ln(ln(2*ln(ln(5))))**2+(-128*x*exp(1)**8+8*x**3*exp(1)**4)*ln( ln(2*ln(ln(5))))+256*exp(1)**8-32*x**2*exp(1)**4+x**4),x)
Output:
-(x*(-16*exp(8)*log(-log(2) - log(log(log(5)))) - 16*I*pi*exp(8)) + 64*exp (8))/(x**2 + x*(4*exp(4)*log(-log(2) - log(log(log(5)))) + 4*I*pi*exp(4)) - 16*exp(4))
Time = 0.04 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.05 \[ \int \frac {128 e^8 x-16 e^8 x^2 (i \pi +\log (-\log (2 \log (\log (5)))))}{256 e^8-32 e^4 x^2+x^4+\left (-128 e^8 x+8 e^4 x^3\right ) (i \pi +\log (-\log (2 \log (\log (5)))))+16 e^8 x^2 (i \pi +\log (-\log (2 \log (\log (5)))))^2} \, dx=\frac {16 \, {\left (x e^{8} \log \left (\log \left (2 \, \log \left (\log \left (5\right )\right )\right )\right ) - 4 \, e^{8}\right )}}{4 \, x e^{4} \log \left (\log \left (2 \, \log \left (\log \left (5\right )\right )\right )\right ) + x^{2} - 16 \, e^{4}} \] Input:
integrate((-16*x^2*exp(1)^8*log(log(2*log(log(5))))+128*x*exp(1)^8)/(16*x^ 2*exp(1)^8*log(log(2*log(log(5))))^2+(-128*x*exp(1)^8+8*x^3*exp(1)^4)*log( log(2*log(log(5))))+256*exp(1)^8-32*x^2*exp(1)^4+x^4),x, algorithm="maxima ")
Output:
16*(x*e^8*log(log(2*log(log(5)))) - 4*e^8)/(4*x*e^4*log(log(2*log(log(5))) ) + x^2 - 16*e^4)
\[ \int \frac {128 e^8 x-16 e^8 x^2 (i \pi +\log (-\log (2 \log (\log (5)))))}{256 e^8-32 e^4 x^2+x^4+\left (-128 e^8 x+8 e^4 x^3\right ) (i \pi +\log (-\log (2 \log (\log (5)))))+16 e^8 x^2 (i \pi +\log (-\log (2 \log (\log (5)))))^2} \, dx=\int { -\frac {16 \, {\left (x^{2} e^{8} \log \left (\log \left (2 \, \log \left (\log \left (5\right )\right )\right )\right ) - 8 \, x e^{8}\right )}}{16 \, x^{2} e^{8} \log \left (\log \left (2 \, \log \left (\log \left (5\right )\right )\right )\right )^{2} + x^{4} - 32 \, x^{2} e^{4} + 8 \, {\left (x^{3} e^{4} - 16 \, x e^{8}\right )} \log \left (\log \left (2 \, \log \left (\log \left (5\right )\right )\right )\right ) + 256 \, e^{8}} \,d x } \] Input:
integrate((-16*x^2*exp(1)^8*log(log(2*log(log(5))))+128*x*exp(1)^8)/(16*x^ 2*exp(1)^8*log(log(2*log(log(5))))^2+(-128*x*exp(1)^8+8*x^3*exp(1)^4)*log( log(2*log(log(5))))+256*exp(1)^8-32*x^2*exp(1)^4+x^4),x, algorithm="giac")
Output:
sage0*x
Time = 4.15 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.53 \[ \int \frac {128 e^8 x-16 e^8 x^2 (i \pi +\log (-\log (2 \log (\log (5)))))}{256 e^8-32 e^4 x^2+x^4+\left (-128 e^8 x+8 e^4 x^3\right ) (i \pi +\log (-\log (2 \log (\log (5)))))+16 e^8 x^2 (i \pi +\log (-\log (2 \log (\log (5)))))^2} \, dx=-\frac {64\,{\mathrm {e}}^8-16\,x\,{\mathrm {e}}^8\,\left (\ln \left (-\ln \left (2\,\ln \left (\ln \left (5\right )\right )\right )\right )+\pi \,1{}\mathrm {i}\right )}{x^2+\left (4\,{\mathrm {e}}^4\,\ln \left (-\ln \left (2\,\ln \left (\ln \left (5\right )\right )\right )\right )+\pi \,{\mathrm {e}}^4\,4{}\mathrm {i}\right )\,x-16\,{\mathrm {e}}^4} \] Input:
int((128*x*exp(8) - 16*x^2*log(log(2*log(log(5))))*exp(8))/(256*exp(8) - 3 2*x^2*exp(4) - log(log(2*log(log(5))))*(128*x*exp(8) - 8*x^3*exp(4)) + x^4 + 16*x^2*log(log(2*log(log(5))))^2*exp(8)),x)
Output:
-(64*exp(8) - 16*x*exp(8)*(pi*1i + log(-log(2*log(log(5))))))/(x*(pi*exp(4 )*4i + 4*exp(4)*log(-log(2*log(log(5))))) - 16*exp(4) + x^2)
Time = 0.18 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.84 \[ \int \frac {128 e^8 x-16 e^8 x^2 (i \pi +\log (-\log (2 \log (\log (5)))))}{256 e^8-32 e^4 x^2+x^4+\left (-128 e^8 x+8 e^4 x^3\right ) (i \pi +\log (-\log (2 \log (\log (5)))))+16 e^8 x^2 (i \pi +\log (-\log (2 \log (\log (5)))))^2} \, dx=-\frac {4 e^{4} x^{2}}{4 \,\mathrm {log}\left (\mathrm {log}\left (2 \,\mathrm {log}\left (\mathrm {log}\left (5\right )\right )\right )\right ) e^{4} x -16 e^{4}+x^{2}} \] Input:
int((-16*x^2*exp(1)^8*log(log(2*log(log(5))))+128*x*exp(1)^8)/(16*x^2*exp( 1)^8*log(log(2*log(log(5))))^2+(-128*x*exp(1)^8+8*x^3*exp(1)^4)*log(log(2* log(log(5))))+256*exp(1)^8-32*x^2*exp(1)^4+x^4),x)
Output:
( - 4*e**4*x**2)/(4*log(log(2*log(log(5))))*e**4*x - 16*e**4 + x**2)