3.46.20 \(\int \frac {128 e^8 x-16 e^8 x^2 (i \pi +\log (-\log (2 \log (\log (5)))))}{256 e^8-32 e^4 x^2+x^4+(-128 e^8 x+8 e^4 x^3) (i \pi +\log (-\log (2 \log (\log (5)))))+16 e^8 x^2 (i \pi +\log (-\log (2 \log (\log (5)))))^2} \, dx\)

Optimal. Leaf size=38 \[ \frac {x}{-i \pi +\frac {16-\frac {x^2}{e^4}}{4 x}-\log (-\log (2 \log (\log (5))))} \]

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Rubi [A]  time = 0.24, antiderivative size = 57, normalized size of antiderivative = 1.50, number of steps used = 5, number of rules used = 5, integrand size = 106, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {6, 1593, 1680, 12, 776} \begin {gather*} -\frac {16 e^8 (4-x (\log (-\log (2 \log (\log (5))))+i \pi ))}{x^2+4 e^4 x (\log (-\log (2 \log (\log (5))))+i \pi )-16 e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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]

[Out]

(-16*E^8*(4 - x*(I*Pi + Log[-Log[2*Log[Log[5]]]])))/(-16*E^4 + x^2 + 4*E^4*x*(I*Pi + Log[-Log[2*Log[Log[5]]]])
)

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 776

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] /; FreeQ[{a, c, d, e, f, g, p}, x] &
& EqQ[a*e*g - c*d*f*(2*p + 3), 0] && NeQ[p, -1]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 1680

Int[(Pq_)*(Q4_)^(p_), x_Symbol] :> With[{a = Coeff[Q4, x, 0], b = Coeff[Q4, x, 1], c = Coeff[Q4, x, 2], d = Co
eff[Q4, x, 3], e = Coeff[Q4, x, 4]}, Subst[Int[SimplifyIntegrand[(Pq /. x -> -(d/(4*e)) + x)*(a + d^4/(256*e^3
) - (b*d)/(8*e) + (c - (3*d^2)/(8*e))*x^2 + e*x^4)^p, x], x], x, d/(4*e) + x] /; EqQ[d^3 - 4*c*d*e + 8*b*e^2,
0] && NeQ[d, 0]] /; FreeQ[p, x] && PolyQ[Pq, x] && PolyQ[Q4, x, 4] &&  !IGtQ[p, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {128 e^8 x-16 e^8 x^2 (i \pi +\log (-\log (2 \log (\log (5)))))}{256 e^8+x^4+\left (-128 e^8 x+8 e^4 x^3\right ) (i \pi +\log (-\log (2 \log (\log (5)))))+x^2 \left (-32 e^4+16 e^8 (i \pi +\log (-\log (2 \log (\log (5)))))^2\right )} \, dx\\ &=\int \frac {x \left (128 e^8-16 e^8 x (i \pi +\log (-\log (2 \log (\log (5)))))\right )}{256 e^8+x^4+\left (-128 e^8 x+8 e^4 x^3\right ) (i \pi +\log (-\log (2 \log (\log (5)))))+x^2 \left (-32 e^4+16 e^8 (i \pi +\log (-\log (2 \log (\log (5)))))^2\right )} \, dx\\ &=\operatorname {Subst}\left (\int \frac {16 e^8 \left (x-2 e^4 (i \pi +\log (-\log (2 \log (\log (5)))))\right ) \left (2 \left (4-e^4 (\pi -i \log (-\log (2 \log (\log (5)))))^2\right )-x (i \pi +\log (-\log (2 \log (\log (5)))))\right )}{\left (16 e^4-x^2-4 e^8 (\pi -i \log (-\log (2 \log (\log (5)))))^2\right )^2} \, dx,x,x+\frac {1}{4} \left (8 i e^4 \pi +8 e^4 \log (-\log (2 \log (\log (5))))\right )\right )\\ &=\left (16 e^8\right ) \operatorname {Subst}\left (\int \frac {\left (x-2 e^4 (i \pi +\log (-\log (2 \log (\log (5)))))\right ) \left (2 \left (4-e^4 (\pi -i \log (-\log (2 \log (\log (5)))))^2\right )-x (i \pi +\log (-\log (2 \log (\log (5)))))\right )}{\left (16 e^4-x^2-4 e^8 (\pi -i \log (-\log (2 \log (\log (5)))))^2\right )^2} \, dx,x,x+\frac {1}{4} \left (8 i e^4 \pi +8 e^4 \log (-\log (2 \log (\log (5))))\right )\right )\\ &=\frac {16 e^8 (4-x (i \pi +\log (-\log (2 \log (\log (5))))))}{16 e^4-x^2-4 e^4 x (i \pi +\log (-\log (2 \log (\log (5)))))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 54, normalized size = 1.42 \begin {gather*} \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)))))} \end {gather*}

Antiderivative was successfully verified.

[In]

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[Log[5]]]]) + 16*E^8*x^2*(I*Pi + Log[-Log[2*Log[Log[5]]]])^2),x]

[Out]

(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]]]]))

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fricas [A]  time = 0.51, size = 40, normalized size = 1.05 \begin {gather*} \frac {16 \, {\left (x e^{8} \log \left (\log \left (2 \, \log \left (\log \relax (5)\right )\right )\right ) - 4 \, e^{8}\right )}}{4 \, x e^{4} \log \left (\log \left (2 \, \log \left (\log \relax (5)\right )\right )\right ) + x^{2} - 16 \, e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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="fric
as")

[Out]

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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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
")

[Out]

undef

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maple [A]  time = 0.34, size = 44, normalized size = 1.16

method	result	size
gosper	\(\frac {16 \left (\ln \left (\ln \left (2 \ln \left (\ln \relax (5)\right )\right )\right ) x -4\right ) {\mathrm e}^{8}}{4 \,{\mathrm e}^{4} \ln \left (\ln \left (2 \ln \left (\ln \relax (5)\right )\right )\right ) x -16 \,{\mathrm e}^{4}+x^{2}}\)	\(44\)
risch	\(\frac {4 x \,{\mathrm e}^{8} \ln \left (\ln \relax (2)+\ln \left (\ln \left (\ln \relax (5)\right )\right )\right )-16 \,{\mathrm e}^{8}}{{\mathrm e}^{4} \ln \left (\ln \relax (2)+\ln \left (\ln \left (\ln \relax (5)\right )\right )\right ) x -4 \,{\mathrm e}^{4}+\frac {x^{2}}{4}}\)	\(44\)
norman	\(\frac {16 x \,{\mathrm e}^{8} \ln \left (\ln \relax (2)+\ln \left (\ln \left (\ln \relax (5)\right )\right )\right )-64 \,{\mathrm e}^{8}}{4 \,{\mathrm e}^{4} \ln \left (\ln \left (2 \ln \left (\ln \relax (5)\right )\right )\right ) x -16 \,{\mathrm e}^{4}+x^{2}}\)	\(50\)
default	\(-4 \,{\mathrm e}^{8} \left (\munderset {\textit {\_R} =\RootOf \left (\textit {\_Z}^{4}+8 \,{\mathrm e}^{4} \ln \left (\ln \left (2 \ln \left (\ln \relax (5)\right )\right )\right ) \textit {\_Z}^{3}+\left (16 \,{\mathrm e}^{8} \ln \left (\ln \left (2 \ln \left (\ln \relax (5)\right )\right )\right )^{2}-32 \,{\mathrm e}^{4}\right ) \textit {\_Z}^{2}-128 \textit {\_Z} \,{\mathrm e}^{8} \ln \left (\ln \left (2 \ln \left (\ln \relax (5)\right )\right )\right )+256 \,{\mathrm e}^{8}\right )}{\sum }\frac {\left (\ln \left (\ln \left (2 \ln \left (\ln \relax (5)\right )\right )\right ) \textit {\_R}^{2}-8 \textit {\_R} \right ) \ln \left (x -\textit {\_R} \right )}{8 \textit {\_R} \,{\mathrm e}^{8} \ln \left (\ln \left (2 \ln \left (\ln \relax (5)\right )\right )\right )^{2}-32 \,{\mathrm e}^{8} \ln \left (\ln \left (2 \ln \left (\ln \relax (5)\right )\right )\right )+6 \,{\mathrm e}^{4} \ln \left (\ln \left (2 \ln \left (\ln \relax (5)\right )\right )\right ) \textit {\_R}^{2}-16 \textit {\_R} \,{\mathrm e}^{4}+\textit {\_R}^{3}}\right )\)	\(139\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)

[Out]

16*(ln(ln(2*ln(ln(5))))*x-4)*exp(1)^8/(4*exp(1)^4*ln(ln(2*ln(ln(5))))*x-16*exp(1)^4+x^2)

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maxima [A]  time = 0.37, size = 40, normalized size = 1.05 \begin {gather*} \frac {16 \, {\left (x e^{8} \log \left (\log \left (2 \, \log \left (\log \relax (5)\right )\right )\right ) - 4 \, e^{8}\right )}}{4 \, x e^{4} \log \left (\log \left (2 \, \log \left (\log \relax (5)\right )\right )\right ) + x^{2} - 16 \, e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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="maxi
ma")

[Out]

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)

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mupad [B]  time = 3.85, size = 58, normalized size = 1.53 \begin {gather*} -\frac {64\,{\mathrm {e}}^8-16\,x\,{\mathrm {e}}^8\,\left (\ln \left (-\ln \left (2\,\ln \left (\ln \relax (5)\right )\right )\right )+\pi \,1{}\mathrm {i}\right )}{x^2+\left (4\,{\mathrm {e}}^4\,\ln \left (-\ln \left (2\,\ln \left (\ln \relax (5)\right )\right )\right )+\pi \,{\mathrm {e}}^4\,4{}\mathrm {i}\right )\,x-16\,{\mathrm {e}}^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((128*x*exp(8) - 16*x^2*log(log(2*log(log(5))))*exp(8))/(256*exp(8) - 32*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)

[Out]

-(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)

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sympy [B]  time = 1.46, size = 73, normalized size = 1.92 \begin {gather*} - \frac {x \left (16 e^{8} \log {\left (- \log {\relax (2 )} - \log {\left (\log {\left (\log {\relax (5 )} \right )} \right )} \right )} + 16 i \pi e^{8}\right ) - 64 e^{8}}{- x^{2} + x \left (- 4 e^{4} \log {\left (- \log {\relax (2 )} - \log {\left (\log {\left (\log {\relax (5 )} \right )} \right )} \right )} - 4 i \pi e^{4}\right ) + 16 e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)

[Out]

-(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))

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