3.8.89 \(\int \frac {\sqrt {1-2 x^8} (-1+2 x^8) (1+2 x^8)}{x^7 (-1+x^4+2 x^8)} \, dx\)

Optimal. Leaf size=60 \[ \frac {\sqrt {1-2 x^8} \left (2 x^8-3 x^4-1\right )}{6 x^6}-\frac {1}{2} \tanh ^{-1}\left (\frac {x^2 \sqrt {1-2 x^8}}{2 x^8-1}\right ) \]

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Rubi [C]  time = 0.89, antiderivative size = 335, normalized size of antiderivative = 5.58, number of steps used = 47, number of rules used = 17, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.415, Rules used = {21, 6728, 275, 277, 195, 221, 279, 307, 1181, 424, 1491, 1209, 1177, 524, 248, 1213, 537} \begin {gather*} \frac {1}{15} \sqrt [4]{2} \left (3+5 \sqrt {2}\right ) F\left (\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )-\frac {\left (1+\sqrt {2}\right ) F\left (\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )}{2^{3/4}}+\frac {\left (1-\sqrt {2}\right ) F\left (\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )}{2 \sqrt [4]{2}}+\frac {1}{15} 2^{3/4} \left (5-3 \sqrt {2}\right ) F\left (\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )-\frac {2}{3} 2^{3/4} F\left (\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )+\frac {6}{5} \sqrt [4]{2} F\left (\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )+\frac {\Pi \left (-\frac {1}{\sqrt {2}};\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )}{2 \sqrt [4]{2}}+\frac {\Pi \left (\sqrt {2};\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )}{2 \sqrt [4]{2}}-\frac {6}{5} \sqrt {1-2 x^8} x^6-\frac {\left (1-2 x^8\right )^{3/2}}{6 x^6}-\frac {2}{3} \sqrt {1-2 x^8} x^2-\frac {\left (1-2 x^8\right )^{3/2}}{2 x^2}+\frac {1}{15} \left (5-3 x^4\right ) \sqrt {1-2 x^8} x^2+\frac {1}{15} \left (6 x^4+5\right ) \sqrt {1-2 x^8} x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sqrt[1 - 2*x^8]*(-1 + 2*x^8)*(1 + 2*x^8))/(x^7*(-1 + x^4 + 2*x^8)),x]

[Out]

(-2*x^2*Sqrt[1 - 2*x^8])/3 - (6*x^6*Sqrt[1 - 2*x^8])/5 + (x^2*(5 - 3*x^4)*Sqrt[1 - 2*x^8])/15 + (x^2*(5 + 6*x^
4)*Sqrt[1 - 2*x^8])/15 - (1 - 2*x^8)^(3/2)/(6*x^6) - (1 - 2*x^8)^(3/2)/(2*x^2) + (6*2^(1/4)*EllipticF[ArcSin[2
^(1/4)*x^2], -1])/5 - (2*2^(3/4)*EllipticF[ArcSin[2^(1/4)*x^2], -1])/3 + (2^(3/4)*(5 - 3*Sqrt[2])*EllipticF[Ar
cSin[2^(1/4)*x^2], -1])/15 + ((1 - Sqrt[2])*EllipticF[ArcSin[2^(1/4)*x^2], -1])/(2*2^(1/4)) - ((1 + Sqrt[2])*E
llipticF[ArcSin[2^(1/4)*x^2], -1])/2^(3/4) + (2^(1/4)*(3 + 5*Sqrt[2])*EllipticF[ArcSin[2^(1/4)*x^2], -1])/15 +
EllipticPi[-(1/Sqrt[2]), ArcSin[2^(1/4)*x^2], -1]/(2*2^(1/4)) + EllipticPi[Sqrt[2], ArcSin[2^(1/4)*x^2], -1]/
(2*2^(1/4))

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[(Rt[-b, 4]*x)/Rt[a, 4]], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rule 248

Int[((a1_.) + (b1_.)*(x_)^(n_))^(p_.)*((a2_.) + (b2_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[(a1*a2 + b1*b2*x^(2*
n))^p, x] /; FreeQ[{a1, b1, a2, b2, n, p}, x] && EqQ[a2*b1 + a1*b2, 0] && (IntegerQ[p] || (GtQ[a1, 0] && GtQ[a
2, 0]))

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
n*p + 1)), x] + Dist[(a*n*p)/(m + n*p + 1), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 307

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[-(b/a), 2]}, -Dist[q^(-1), Int[1/Sqrt[a + b*x^
4], x], x] + Dist[1/q, Int[(1 + q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && NegQ[b/a]

Rule 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rule 524

Int[((e_) + (f_.)*(x_)^(n_))/(Sqrt[(a_) + (b_.)*(x_)^(n_)]*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/
b, Int[Sqrt[a + b*x^n]/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/(Sqrt[a + b*x^n]*Sqrt[c + d*x^n]),
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&  !(EqQ[n, 2] && ((PosQ[b/a] && PosQ[d/c]) || (NegQ[b/a] && (PosQ[
d/c] || (GtQ[a, 0] && ( !GtQ[c, 0] || SimplerSqrtQ[-(b/a), -(d/c)]))))))

Rule 537

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(1*Ellipt
icPi[(b*c)/(a*d), ArcSin[Rt[-(d/c), 2]*x], (c*f)/(d*e)])/(a*Sqrt[c]*Sqrt[e]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b,
c, d, e, f}, x] &&  !GtQ[d/c, 0] && GtQ[c, 0] && GtQ[e, 0] &&  !( !GtQ[f/e, 0] && SimplerSqrtQ[-(f/e), -(d/c)
])

Rule 1177

Int[((d_) + (e_.)*(x_)^2)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(x*(d*(4*p + 3) + e*(4*p + 1)*x^2)*(a
+ c*x^4)^p)/((4*p + 1)*(4*p + 3)), x] + Dist[(2*p)/((4*p + 1)*(4*p + 3)), Int[Simp[2*a*d*(4*p + 3) + (2*a*e*(4
*p + 1))*x^2, x]*(a + c*x^4)^(p - 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] &
& FractionQ[p] && IntegerQ[2*p]

Rule 1181

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[Sqrt[-c], Int
[(d + e*x^2)/(Sqrt[q + c*x^2]*Sqrt[q - c*x^2]), x], x]] /; FreeQ[{a, c, d, e}, x] && GtQ[a, 0] && LtQ[c, 0]

Rule 1209

Int[((a_) + (c_.)*(x_)^4)^(p_)/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Dist[(e^2)^(-1), Int[(c*d - c*e*x^2)*(a +
c*x^4)^(p - 1), x], x] + Dist[(c*d^2 + a*e^2)/e^2, Int[(a + c*x^4)^(p - 1)/(d + e*x^2), x], x] /; FreeQ[{a, c,
d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p + 1/2, 0]

Rule 1213

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[Sqrt[-c],
Int[1/((d + e*x^2)*Sqrt[q + c*x^2]*Sqrt[q - c*x^2]), x], x]] /; FreeQ[{a, c, d, e}, x] && GtQ[a, 0] && LtQ[c,
0]

Rule 1491

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{k = GCD[m +
1, n]}, Dist[1/k, Subst[Int[x^((m + 1)/k - 1)*(d + e*x^(n/k))^q*(a + c*x^((2*n)/k))^p, x], x, x^k], x] /; k !=
1] /; FreeQ[{a, c, d, e, p, q}, x] && EqQ[n2, 2*n] && IGtQ[n, 0] && IntegerQ[m]

Rule 6728

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] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\sqrt {1-2 x^8} \left (-1+2 x^8\right ) \left (1+2 x^8\right )}{x^7 \left (-1+x^4+2 x^8\right )} \, dx &=-\int \frac {\left (1-2 x^8\right )^{3/2} \left (1+2 x^8\right )}{x^7 \left (-1+x^4+2 x^8\right )} \, dx\\ &=-\int \left (-\frac {\left (1-2 x^8\right )^{3/2}}{x^7}-\frac {\left (1-2 x^8\right )^{3/2}}{x^3}-\frac {x \left (1-2 x^8\right )^{3/2}}{1+x^4}+\frac {4 x \left (1-2 x^8\right )^{3/2}}{-1+2 x^4}\right ) \, dx\\ &=-\left (4 \int \frac {x \left (1-2 x^8\right )^{3/2}}{-1+2 x^4} \, dx\right )+\int \frac {\left (1-2 x^8\right )^{3/2}}{x^7} \, dx+\int \frac {\left (1-2 x^8\right )^{3/2}}{x^3} \, dx+\int \frac {x \left (1-2 x^8\right )^{3/2}}{1+x^4} \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {\left (1-2 x^4\right )^{3/2}}{x^4} \, dx,x,x^2\right )+\frac {1}{2} \operatorname {Subst}\left (\int \frac {\left (1-2 x^4\right )^{3/2}}{x^2} \, dx,x,x^2\right )+\frac {1}{2} \operatorname {Subst}\left (\int \frac {\left (1-2 x^4\right )^{3/2}}{1+x^2} \, dx,x,x^2\right )-2 \operatorname {Subst}\left (\int \frac {\left (1-2 x^4\right )^{3/2}}{-1+2 x^2} \, dx,x,x^2\right )\\ &=-\frac {\left (1-2 x^8\right )^{3/2}}{6 x^6}-\frac {\left (1-2 x^8\right )^{3/2}}{2 x^2}-\frac {1}{2} \operatorname {Subst}\left (\int \frac {\sqrt {1-2 x^4}}{1+x^2} \, dx,x,x^2\right )-\frac {1}{2} \operatorname {Subst}\left (\int \left (-2+2 x^2\right ) \sqrt {1-2 x^4} \, dx,x,x^2\right )+\frac {1}{2} \operatorname {Subst}\left (\int \left (2+4 x^2\right ) \sqrt {1-2 x^4} \, dx,x,x^2\right )-2 \operatorname {Subst}\left (\int \sqrt {1-2 x^4} \, dx,x,x^2\right )-6 \operatorname {Subst}\left (\int x^2 \sqrt {1-2 x^4} \, dx,x,x^2\right )-\operatorname {Subst}\left (\int \frac {\sqrt {1-2 x^4}}{-1+2 x^2} \, dx,x,x^2\right )\\ &=-\frac {2}{3} x^2 \sqrt {1-2 x^8}-\frac {6}{5} x^6 \sqrt {1-2 x^8}+\frac {1}{15} x^2 \left (5-3 x^4\right ) \sqrt {1-2 x^8}+\frac {1}{15} x^2 \left (5+6 x^4\right ) \sqrt {1-2 x^8}-\frac {\left (1-2 x^8\right )^{3/2}}{6 x^6}-\frac {\left (1-2 x^8\right )^{3/2}}{2 x^2}-\frac {1}{30} \operatorname {Subst}\left (\int \frac {-20+12 x^2}{\sqrt {1-2 x^4}} \, dx,x,x^2\right )+\frac {1}{30} \operatorname {Subst}\left (\int \frac {20+24 x^2}{\sqrt {1-2 x^4}} \, dx,x,x^2\right )+\frac {1}{4} \operatorname {Subst}\left (\int \frac {2+4 x^2}{\sqrt {1-2 x^4}} \, dx,x,x^2\right )+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \sqrt {1-2 x^4}} \, dx,x,x^2\right )+\frac {1}{2} \operatorname {Subst}\left (\int \frac {-2+2 x^2}{\sqrt {1-2 x^4}} \, dx,x,x^2\right )-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{\left (-1+2 x^2\right ) \sqrt {1-2 x^4}} \, dx,x,x^2\right )-\frac {4}{3} \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-2 x^4}} \, dx,x,x^2\right )-\frac {12}{5} \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1-2 x^4}} \, dx,x,x^2\right )\\ &=-\frac {2}{3} x^2 \sqrt {1-2 x^8}-\frac {6}{5} x^6 \sqrt {1-2 x^8}+\frac {1}{15} x^2 \left (5-3 x^4\right ) \sqrt {1-2 x^8}+\frac {1}{15} x^2 \left (5+6 x^4\right ) \sqrt {1-2 x^8}-\frac {\left (1-2 x^8\right )^{3/2}}{6 x^6}-\frac {\left (1-2 x^8\right )^{3/2}}{2 x^2}-\frac {2}{3} 2^{3/4} F\left (\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )-\frac {\operatorname {Subst}\left (\int \frac {-20+12 x^2}{\sqrt {\sqrt {2}-2 x^2} \sqrt {\sqrt {2}+2 x^2}} \, dx,x,x^2\right )}{15 \sqrt {2}}+\frac {\operatorname {Subst}\left (\int \frac {20+24 x^2}{\sqrt {\sqrt {2}-2 x^2} \sqrt {\sqrt {2}+2 x^2}} \, dx,x,x^2\right )}{15 \sqrt {2}}+\frac {\operatorname {Subst}\left (\int \frac {2+4 x^2}{\sqrt {\sqrt {2}-2 x^2} \sqrt {\sqrt {2}+2 x^2}} \, dx,x,x^2\right )}{2 \sqrt {2}}+\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt {\sqrt {2}-2 x^2} \left (1+x^2\right ) \sqrt {\sqrt {2}+2 x^2}} \, dx,x,x^2\right )}{\sqrt {2}}+\frac {\operatorname {Subst}\left (\int \frac {-2+2 x^2}{\sqrt {\sqrt {2}-2 x^2} \sqrt {\sqrt {2}+2 x^2}} \, dx,x,x^2\right )}{\sqrt {2}}-\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt {\sqrt {2}-2 x^2} \left (-1+2 x^2\right ) \sqrt {\sqrt {2}+2 x^2}} \, dx,x,x^2\right )}{\sqrt {2}}+\frac {1}{5} \left (6 \sqrt {2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-2 x^4}} \, dx,x,x^2\right )-\frac {1}{5} \left (6 \sqrt {2}\right ) \operatorname {Subst}\left (\int \frac {1+\sqrt {2} x^2}{\sqrt {1-2 x^4}} \, dx,x,x^2\right )\\ &=-\frac {2}{3} x^2 \sqrt {1-2 x^8}-\frac {6}{5} x^6 \sqrt {1-2 x^8}+\frac {1}{15} x^2 \left (5-3 x^4\right ) \sqrt {1-2 x^8}+\frac {1}{15} x^2 \left (5+6 x^4\right ) \sqrt {1-2 x^8}-\frac {\left (1-2 x^8\right )^{3/2}}{6 x^6}-\frac {\left (1-2 x^8\right )^{3/2}}{2 x^2}+\frac {6}{5} \sqrt [4]{2} F\left (\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )-\frac {2}{3} 2^{3/4} F\left (\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )+\frac {\Pi \left (-\frac {1}{\sqrt {2}};\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )}{2 \sqrt [4]{2}}+\frac {\Pi \left (\sqrt {2};\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )}{2 \sqrt [4]{2}}-\frac {12}{5} \operatorname {Subst}\left (\int \frac {1+\sqrt {2} x^2}{\sqrt {\sqrt {2}-2 x^2} \sqrt {\sqrt {2}+2 x^2}} \, dx,x,x^2\right )+2 \frac {\operatorname {Subst}\left (\int \frac {\sqrt {\sqrt {2}+2 x^2}}{\sqrt {\sqrt {2}-2 x^2}} \, dx,x,x^2\right )}{\sqrt {2}}-\frac {1}{5} \sqrt {2} \operatorname {Subst}\left (\int \frac {\sqrt {\sqrt {2}+2 x^2}}{\sqrt {\sqrt {2}-2 x^2}} \, dx,x,x^2\right )+\frac {1}{5} \left (2 \sqrt {2}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {\sqrt {2}+2 x^2}}{\sqrt {\sqrt {2}-2 x^2}} \, dx,x,x^2\right )-\frac {1}{15} \left (2 \left (6-5 \sqrt {2}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {\sqrt {2}-2 x^2} \sqrt {\sqrt {2}+2 x^2}} \, dx,x,x^2\right )+\frac {1}{2} \left (-2+\sqrt {2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {\sqrt {2}-2 x^2} \sqrt {\sqrt {2}+2 x^2}} \, dx,x,x^2\right )-\left (1+\sqrt {2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {\sqrt {2}-2 x^2} \sqrt {\sqrt {2}+2 x^2}} \, dx,x,x^2\right )+\frac {1}{15} \left (2 \left (3+5 \sqrt {2}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {\sqrt {2}-2 x^2} \sqrt {\sqrt {2}+2 x^2}} \, dx,x,x^2\right )\\ &=-\frac {2}{3} x^2 \sqrt {1-2 x^8}-\frac {6}{5} x^6 \sqrt {1-2 x^8}+\frac {1}{15} x^2 \left (5-3 x^4\right ) \sqrt {1-2 x^8}+\frac {1}{15} x^2 \left (5+6 x^4\right ) \sqrt {1-2 x^8}-\frac {\left (1-2 x^8\right )^{3/2}}{6 x^6}-\frac {\left (1-2 x^8\right )^{3/2}}{2 x^2}+\frac {6}{5} \sqrt [4]{2} E\left (\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )+\frac {6}{5} \sqrt [4]{2} F\left (\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )-\frac {2}{3} 2^{3/4} F\left (\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )+\frac {\Pi \left (-\frac {1}{\sqrt {2}};\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )}{2 \sqrt [4]{2}}+\frac {\Pi \left (\sqrt {2};\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )}{2 \sqrt [4]{2}}-\frac {1}{5} \left (6 \sqrt {2}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {\sqrt {2}+2 x^2}}{\sqrt {\sqrt {2}-2 x^2}} \, dx,x,x^2\right )-\frac {1}{15} \left (2 \left (6-5 \sqrt {2}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {2-4 x^4}} \, dx,x,x^2\right )+\frac {1}{2} \left (-2+\sqrt {2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {2-4 x^4}} \, dx,x,x^2\right )-\left (1+\sqrt {2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {2-4 x^4}} \, dx,x,x^2\right )+\frac {1}{15} \left (2 \left (3+5 \sqrt {2}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {2-4 x^4}} \, dx,x,x^2\right )\\ &=-\frac {2}{3} x^2 \sqrt {1-2 x^8}-\frac {6}{5} x^6 \sqrt {1-2 x^8}+\frac {1}{15} x^2 \left (5-3 x^4\right ) \sqrt {1-2 x^8}+\frac {1}{15} x^2 \left (5+6 x^4\right ) \sqrt {1-2 x^8}-\frac {\left (1-2 x^8\right )^{3/2}}{6 x^6}-\frac {\left (1-2 x^8\right )^{3/2}}{2 x^2}+\frac {6}{5} \sqrt [4]{2} F\left (\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )-\frac {2}{3} 2^{3/4} F\left (\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )+\frac {1}{15} 2^{3/4} \left (5-3 \sqrt {2}\right ) F\left (\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )+\frac {\left (1-\sqrt {2}\right ) F\left (\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )}{2 \sqrt [4]{2}}-\frac {\left (1+\sqrt {2}\right ) F\left (\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )}{2^{3/4}}+\frac {1}{15} \sqrt [4]{2} \left (3+5 \sqrt {2}\right ) F\left (\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )+\frac {\Pi \left (-\frac {1}{\sqrt {2}};\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )}{2 \sqrt [4]{2}}+\frac {\Pi \left (\sqrt {2};\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )}{2 \sqrt [4]{2}}\\ \end {align*}

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Mathematica [C]  time = 0.30, size = 149, normalized size = 2.48 \begin {gather*} -\frac {6 \sqrt {1-2 x^8} x^8 \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};2 x^8\right )+8 x^{16}-12 x^{12}-8 x^8+6 x^4-3\ 2^{3/4} \sqrt {1-2 x^8} x^6 \Pi \left (-\frac {1}{\sqrt {2}};\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )-3\ 2^{3/4} \sqrt {1-2 x^8} x^6 \Pi \left (\sqrt {2};\left .\sin ^{-1}\left (\sqrt [4]{2} x^2\right )\right |-1\right )+2}{12 x^6 \sqrt {1-2 x^8}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[1 - 2*x^8]*(-1 + 2*x^8)*(1 + 2*x^8))/(x^7*(-1 + x^4 + 2*x^8)),x]

[Out]

-1/12*(2 + 6*x^4 - 8*x^8 - 12*x^12 + 8*x^16 - 3*2^(3/4)*x^6*Sqrt[1 - 2*x^8]*EllipticPi[-(1/Sqrt[2]), ArcSin[2^
(1/4)*x^2], -1] - 3*2^(3/4)*x^6*Sqrt[1 - 2*x^8]*EllipticPi[Sqrt[2], ArcSin[2^(1/4)*x^2], -1] + 6*x^8*Sqrt[1 -
2*x^8]*Hypergeometric2F1[1/4, 1/2, 5/4, 2*x^8])/(x^6*Sqrt[1 - 2*x^8])

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IntegrateAlgebraic [A]  time = 22.22, size = 60, normalized size = 1.00 \begin {gather*} \frac {\sqrt {1-2 x^8} \left (-1-3 x^4+2 x^8\right )}{6 x^6}-\frac {1}{2} \tanh ^{-1}\left (\frac {x^2 \sqrt {1-2 x^8}}{-1+2 x^8}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(Sqrt[1 - 2*x^8]*(-1 + 2*x^8)*(1 + 2*x^8))/(x^7*(-1 + x^4 + 2*x^8)),x]

[Out]

(Sqrt[1 - 2*x^8]*(-1 - 3*x^4 + 2*x^8))/(6*x^6) - ArcTanh[(x^2*Sqrt[1 - 2*x^8])/(-1 + 2*x^8)]/2

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fricas [A]  time = 0.50, size = 75, normalized size = 1.25 \begin {gather*} \frac {3 \, x^{6} \log \left (-\frac {2 \, x^{8} - x^{4} - 2 \, \sqrt {-2 \, x^{8} + 1} x^{2} - 1}{2 \, x^{8} + x^{4} - 1}\right ) + 2 \, {\left (2 \, x^{8} - 3 \, x^{4} - 1\right )} \sqrt {-2 \, x^{8} + 1}}{12 \, x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x^8+1)^(1/2)*(2*x^8-1)*(2*x^8+1)/x^7/(2*x^8+x^4-1),x, algorithm="fricas")

[Out]

1/12*(3*x^6*log(-(2*x^8 - x^4 - 2*sqrt(-2*x^8 + 1)*x^2 - 1)/(2*x^8 + x^4 - 1)) + 2*(2*x^8 - 3*x^4 - 1)*sqrt(-2
*x^8 + 1))/x^6

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (2 \, x^{8} + 1\right )} {\left (2 \, x^{8} - 1\right )} \sqrt {-2 \, x^{8} + 1}}{{\left (2 \, x^{8} + x^{4} - 1\right )} x^{7}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x^8+1)^(1/2)*(2*x^8-1)*(2*x^8+1)/x^7/(2*x^8+x^4-1),x, algorithm="giac")

[Out]

integrate((2*x^8 + 1)*(2*x^8 - 1)*sqrt(-2*x^8 + 1)/((2*x^8 + x^4 - 1)*x^7), x)

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maple [A]  time = 0.66, size = 72, normalized size = 1.20

method	result	size
trager	\(\frac {\sqrt {-2 x^{8}+1}\, \left (2 x^{8}-3 x^{4}-1\right )}{6 x^{6}}+\frac {\ln \left (\frac {-2 x^{8}+x^{4}+2 \sqrt {-2 x^{8}+1}\, x^{2}+1}{\left (x^{4}+1\right ) \left (2 x^{4}-1\right )}\right )}{4}\)	\(72\)
risch	\(-\frac {4 x^{16}-6 x^{12}-4 x^{8}+3 x^{4}+1}{6 x^{6} \sqrt {-2 x^{8}+1}}+\frac {\ln \left (-\frac {-2 x^{8}+x^{4}+2 \sqrt {-2 x^{8}+1}\, x^{2}+1}{\left (x^{4}+1\right ) \left (2 x^{4}-1\right )}\right )}{4}\)	\(83\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*x^8+1)^(1/2)*(2*x^8-1)*(2*x^8+1)/x^7/(2*x^8+x^4-1),x,method=_RETURNVERBOSE)

[Out]

1/6*(-2*x^8+1)^(1/2)*(2*x^8-3*x^4-1)/x^6+1/4*ln((-2*x^8+x^4+2*(-2*x^8+1)^(1/2)*x^2+1)/(x^4+1)/(2*x^4-1))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (2 \, x^{8} + 1\right )} {\left (2 \, x^{8} - 1\right )} \sqrt {-2 \, x^{8} + 1}}{{\left (2 \, x^{8} + x^{4} - 1\right )} x^{7}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x^8+1)^(1/2)*(2*x^8-1)*(2*x^8+1)/x^7/(2*x^8+x^4-1),x, algorithm="maxima")

[Out]

integrate((2*x^8 + 1)*(2*x^8 - 1)*sqrt(-2*x^8 + 1)/((2*x^8 + x^4 - 1)*x^7), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int -\frac {{\left (1-2\,x^8\right )}^{3/2}\,\left (2\,x^8+1\right )}{x^7\,\left (2\,x^8+x^4-1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((1 - 2*x^8)^(3/2)*(2*x^8 + 1))/(x^7*(x^4 + 2*x^8 - 1)),x)

[Out]

int(-((1 - 2*x^8)^(3/2)*(2*x^8 + 1))/(x^7*(x^4 + 2*x^8 - 1)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {1 - 2 x^{8}} \left (2 x^{8} - 1\right ) \left (2 x^{8} + 1\right )}{x^{7} \left (x^{4} + 1\right ) \left (2 x^{4} - 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x**8+1)**(1/2)*(2*x**8-1)*(2*x**8+1)/x**7/(2*x**8+x**4-1),x)

[Out]

Integral(sqrt(1 - 2*x**8)*(2*x**8 - 1)*(2*x**8 + 1)/(x**7*(x**4 + 1)*(2*x**4 - 1)), x)

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