3.919 \(\int \frac{x}{(a+b x^8)^2 \sqrt{c+d x^8}} \, dx\)

Optimal. Leaf size=999 \[ \text{result too large to display} \]

[Out]

(b*x^2*Sqrt[c + d*x^8])/(8*a*(b*c - a*d)*(a + b*x^8)) + (b^(1/4)*(3*b*c - 5*a*d)*ArcTan[(Sqrt[b*c - a*d]*x^2)/
((-a)^(1/4)*b^(1/4)*Sqrt[c + d*x^8])])/(32*(-a)^(7/4)*(b*c - a*d)^(3/2)) - (b^(1/4)*(3*b*c - 5*a*d)*ArcTan[(Sq
rt[-(b*c) + a*d]*x^2)/((-a)^(1/4)*b^(1/4)*Sqrt[c + d*x^8])])/(32*(-a)^(7/4)*(-(b*c) + a*d)^(3/2)) + (d^(3/4)*(
Sqrt[c] + Sqrt[d]*x^4)*Sqrt[(c + d*x^8)/(Sqrt[c] + Sqrt[d]*x^4)^2]*EllipticF[2*ArcTan[(d^(1/4)*x^2)/c^(1/4)],
1/2])/(16*a*c^(1/4)*(b*c - a*d)*Sqrt[c + d*x^8]) + (((Sqrt[b]*Sqrt[c])/Sqrt[-a] + Sqrt[d])*d^(1/4)*(3*b*c - 5*
a*d)*(Sqrt[c] + Sqrt[d]*x^4)*Sqrt[(c + d*x^8)/(Sqrt[c] + Sqrt[d]*x^4)^2]*EllipticF[2*ArcTan[(d^(1/4)*x^2)/c^(1
/4)], 1/2])/(32*a*c^(1/4)*(b*c - a*d)*(b*c + a*d)*Sqrt[c + d*x^8]) + ((Sqrt[b]*Sqrt[c] - Sqrt[-a]*Sqrt[d])*d^(
1/4)*(3*b*c - 5*a*d)*(Sqrt[c] + Sqrt[d]*x^4)*Sqrt[(c + d*x^8)/(Sqrt[c] + Sqrt[d]*x^4)^2]*EllipticF[2*ArcTan[(d
^(1/4)*x^2)/c^(1/4)], 1/2])/(32*(-a)^(3/2)*c^(1/4)*(b*c - a*d)*(b*c + a*d)*Sqrt[c + d*x^8]) + ((Sqrt[b]*Sqrt[c
] + Sqrt[-a]*Sqrt[d])^2*(3*b*c - 5*a*d)*(Sqrt[c] + Sqrt[d]*x^4)*Sqrt[(c + d*x^8)/(Sqrt[c] + Sqrt[d]*x^4)^2]*El
lipticPi[-(Sqrt[b]*Sqrt[c] - Sqrt[-a]*Sqrt[d])^2/(4*Sqrt[-a]*Sqrt[b]*Sqrt[c]*Sqrt[d]), 2*ArcTan[(d^(1/4)*x^2)/
c^(1/4)], 1/2])/(64*a^2*c^(1/4)*d^(1/4)*(b*c - a*d)*(b*c + a*d)*Sqrt[c + d*x^8]) + ((Sqrt[b]*Sqrt[c] - Sqrt[-a
]*Sqrt[d])^2*(3*b*c - 5*a*d)*(Sqrt[c] + Sqrt[d]*x^4)*Sqrt[(c + d*x^8)/(Sqrt[c] + Sqrt[d]*x^4)^2]*EllipticPi[(S
qrt[b]*Sqrt[c] + Sqrt[-a]*Sqrt[d])^2/(4*Sqrt[-a]*Sqrt[b]*Sqrt[c]*Sqrt[d]), 2*ArcTan[(d^(1/4)*x^2)/c^(1/4)], 1/
2])/(64*a^2*c^(1/4)*d^(1/4)*(b*c - a*d)*(b*c + a*d)*Sqrt[c + d*x^8])

________________________________________________________________________________________

Rubi [A]  time = 1.40638, antiderivative size = 999, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.318, Rules used = {465, 414, 523, 220, 409, 1217, 1707} \[ \frac{(3 b c-5 a d) \left (\sqrt{d} x^4+\sqrt{c}\right ) \sqrt{\frac{d x^8+c}{\left (\sqrt{d} x^4+\sqrt{c}\right )^2}} \Pi \left (\frac{\left (\sqrt{b} \sqrt{c}+\sqrt{-a} \sqrt{d}\right )^2}{4 \sqrt{-a} \sqrt{b} \sqrt{c} \sqrt{d}};2 \tan ^{-1}\left (\frac{\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac{1}{2}\right ) \left (\sqrt{b} \sqrt{c}-\sqrt{-a} \sqrt{d}\right )^2}{64 a^2 \sqrt [4]{c} \sqrt [4]{d} (b c-a d) (b c+a d) \sqrt{d x^8+c}}+\frac{\sqrt [4]{d} (3 b c-5 a d) \left (\sqrt{d} x^4+\sqrt{c}\right ) \sqrt{\frac{d x^8+c}{\left (\sqrt{d} x^4+\sqrt{c}\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac{1}{2}\right ) \left (\sqrt{b} \sqrt{c}-\sqrt{-a} \sqrt{d}\right )}{32 (-a)^{3/2} \sqrt [4]{c} (b c-a d) (b c+a d) \sqrt{d x^8+c}}+\frac{\sqrt [4]{b} (3 b c-5 a d) \tan ^{-1}\left (\frac{\sqrt{b c-a d} x^2}{\sqrt [4]{-a} \sqrt [4]{b} \sqrt{d x^8+c}}\right )}{32 (-a)^{7/4} (b c-a d)^{3/2}}-\frac{\sqrt [4]{b} (3 b c-5 a d) \tan ^{-1}\left (\frac{\sqrt{a d-b c} x^2}{\sqrt [4]{-a} \sqrt [4]{b} \sqrt{d x^8+c}}\right )}{32 (-a)^{7/4} (a d-b c)^{3/2}}+\frac{d^{3/4} \left (\sqrt{d} x^4+\sqrt{c}\right ) \sqrt{\frac{d x^8+c}{\left (\sqrt{d} x^4+\sqrt{c}\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac{1}{2}\right )}{16 a \sqrt [4]{c} (b c-a d) \sqrt{d x^8+c}}+\frac{\left (\frac{\sqrt{b} \sqrt{c}}{\sqrt{-a}}+\sqrt{d}\right ) \sqrt [4]{d} (3 b c-5 a d) \left (\sqrt{d} x^4+\sqrt{c}\right ) \sqrt{\frac{d x^8+c}{\left (\sqrt{d} x^4+\sqrt{c}\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac{1}{2}\right )}{32 a \sqrt [4]{c} (b c-a d) (b c+a d) \sqrt{d x^8+c}}+\frac{\left (\sqrt{b} \sqrt{c}+\sqrt{-a} \sqrt{d}\right )^2 (3 b c-5 a d) \left (\sqrt{d} x^4+\sqrt{c}\right ) \sqrt{\frac{d x^8+c}{\left (\sqrt{d} x^4+\sqrt{c}\right )^2}} \Pi \left (-\frac{\left (\sqrt{b} \sqrt{c}-\sqrt{-a} \sqrt{d}\right )^2}{4 \sqrt{-a} \sqrt{b} \sqrt{c} \sqrt{d}};2 \tan ^{-1}\left (\frac{\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac{1}{2}\right )}{64 a^2 \sqrt [4]{c} \sqrt [4]{d} (b c-a d) (b c+a d) \sqrt{d x^8+c}}+\frac{b x^2 \sqrt{d x^8+c}}{8 a (b c-a d) \left (b x^8+a\right )} \]

Antiderivative was successfully verified.

[In]

Int[x/((a + b*x^8)^2*Sqrt[c + d*x^8]),x]

[Out]

(b*x^2*Sqrt[c + d*x^8])/(8*a*(b*c - a*d)*(a + b*x^8)) + (b^(1/4)*(3*b*c - 5*a*d)*ArcTan[(Sqrt[b*c - a*d]*x^2)/
((-a)^(1/4)*b^(1/4)*Sqrt[c + d*x^8])])/(32*(-a)^(7/4)*(b*c - a*d)^(3/2)) - (b^(1/4)*(3*b*c - 5*a*d)*ArcTan[(Sq
rt[-(b*c) + a*d]*x^2)/((-a)^(1/4)*b^(1/4)*Sqrt[c + d*x^8])])/(32*(-a)^(7/4)*(-(b*c) + a*d)^(3/2)) + (d^(3/4)*(
Sqrt[c] + Sqrt[d]*x^4)*Sqrt[(c + d*x^8)/(Sqrt[c] + Sqrt[d]*x^4)^2]*EllipticF[2*ArcTan[(d^(1/4)*x^2)/c^(1/4)],
1/2])/(16*a*c^(1/4)*(b*c - a*d)*Sqrt[c + d*x^8]) + (((Sqrt[b]*Sqrt[c])/Sqrt[-a] + Sqrt[d])*d^(1/4)*(3*b*c - 5*
a*d)*(Sqrt[c] + Sqrt[d]*x^4)*Sqrt[(c + d*x^8)/(Sqrt[c] + Sqrt[d]*x^4)^2]*EllipticF[2*ArcTan[(d^(1/4)*x^2)/c^(1
/4)], 1/2])/(32*a*c^(1/4)*(b*c - a*d)*(b*c + a*d)*Sqrt[c + d*x^8]) + ((Sqrt[b]*Sqrt[c] - Sqrt[-a]*Sqrt[d])*d^(
1/4)*(3*b*c - 5*a*d)*(Sqrt[c] + Sqrt[d]*x^4)*Sqrt[(c + d*x^8)/(Sqrt[c] + Sqrt[d]*x^4)^2]*EllipticF[2*ArcTan[(d
^(1/4)*x^2)/c^(1/4)], 1/2])/(32*(-a)^(3/2)*c^(1/4)*(b*c - a*d)*(b*c + a*d)*Sqrt[c + d*x^8]) + ((Sqrt[b]*Sqrt[c
] + Sqrt[-a]*Sqrt[d])^2*(3*b*c - 5*a*d)*(Sqrt[c] + Sqrt[d]*x^4)*Sqrt[(c + d*x^8)/(Sqrt[c] + Sqrt[d]*x^4)^2]*El
lipticPi[-(Sqrt[b]*Sqrt[c] - Sqrt[-a]*Sqrt[d])^2/(4*Sqrt[-a]*Sqrt[b]*Sqrt[c]*Sqrt[d]), 2*ArcTan[(d^(1/4)*x^2)/
c^(1/4)], 1/2])/(64*a^2*c^(1/4)*d^(1/4)*(b*c - a*d)*(b*c + a*d)*Sqrt[c + d*x^8]) + ((Sqrt[b]*Sqrt[c] - Sqrt[-a
]*Sqrt[d])^2*(3*b*c - 5*a*d)*(Sqrt[c] + Sqrt[d]*x^4)*Sqrt[(c + d*x^8)/(Sqrt[c] + Sqrt[d]*x^4)^2]*EllipticPi[(S
qrt[b]*Sqrt[c] + Sqrt[-a]*Sqrt[d])^2/(4*Sqrt[-a]*Sqrt[b]*Sqrt[c]*Sqrt[d]), 2*ArcTan[(d^(1/4)*x^2)/c^(1/4)], 1/
2])/(64*a^2*c^(1/4)*d^(1/4)*(b*c - a*d)*(b*c + a*d)*Sqrt[c + d*x^8])

Rule 465

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

Rule 414

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
 n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 523

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
 c, d, e, f, n}, x]

Rule 220

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

Rule 409

Int[1/(Sqrt[(a_) + (b_.)*(x_)^4]*((c_) + (d_.)*(x_)^4)), x_Symbol] :> Dist[1/(2*c), Int[1/(Sqrt[a + b*x^4]*(1
- Rt[-(d/c), 2]*x^2)), x], x] + Dist[1/(2*c), Int[1/(Sqrt[a + b*x^4]*(1 + Rt[-(d/c), 2]*x^2)), x], x] /; FreeQ
[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 1217

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

Rule 1707

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[B/A, 2]
}, -Simp[((B*d - A*e)*ArcTan[(Rt[(c*d)/e + (a*e)/d, 2]*x)/Sqrt[a + c*x^4]])/(2*d*e*Rt[(c*d)/e + (a*e)/d, 2]),
x] + Simp[((B*d + A*e)*(A + B*x^2)*Sqrt[(A^2*(a + c*x^4))/(a*(A + B*x^2)^2)]*EllipticPi[Cancel[-((B*d - A*e)^2
/(4*d*e*A*B))], 2*ArcTan[q*x], 1/2])/(4*d*e*A*q*Sqrt[a + c*x^4]), x]] /; FreeQ[{a, c, d, e, A, B}, x] && NeQ[c
*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[c/a] && EqQ[c*A^2 - a*B^2, 0]

Rubi steps

\begin{align*} \int \frac{x}{\left (a+b x^8\right )^2 \sqrt{c+d x^8}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{\left (a+b x^4\right )^2 \sqrt{c+d x^4}} \, dx,x,x^2\right )\\ &=\frac{b x^2 \sqrt{c+d x^8}}{8 a (b c-a d) \left (a+b x^8\right )}-\frac{\operatorname{Subst}\left (\int \frac{-3 b c+4 a d-b d x^4}{\left (a+b x^4\right ) \sqrt{c+d x^4}} \, dx,x,x^2\right )}{8 a (b c-a d)}\\ &=\frac{b x^2 \sqrt{c+d x^8}}{8 a (b c-a d) \left (a+b x^8\right )}+\frac{d \operatorname{Subst}\left (\int \frac{1}{\sqrt{c+d x^4}} \, dx,x,x^2\right )}{8 a (b c-a d)}+\frac{(3 b c-5 a d) \operatorname{Subst}\left (\int \frac{1}{\left (a+b x^4\right ) \sqrt{c+d x^4}} \, dx,x,x^2\right )}{8 a (b c-a d)}\\ &=\frac{b x^2 \sqrt{c+d x^8}}{8 a (b c-a d) \left (a+b x^8\right )}+\frac{d^{3/4} \left (\sqrt{c}+\sqrt{d} x^4\right ) \sqrt{\frac{c+d x^8}{\left (\sqrt{c}+\sqrt{d} x^4\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac{1}{2}\right )}{16 a \sqrt [4]{c} (b c-a d) \sqrt{c+d x^8}}+\frac{(3 b c-5 a d) \operatorname{Subst}\left (\int \frac{1}{\left (1-\frac{\sqrt{b} x^2}{\sqrt{-a}}\right ) \sqrt{c+d x^4}} \, dx,x,x^2\right )}{16 a^2 (b c-a d)}+\frac{(3 b c-5 a d) \operatorname{Subst}\left (\int \frac{1}{\left (1+\frac{\sqrt{b} x^2}{\sqrt{-a}}\right ) \sqrt{c+d x^4}} \, dx,x,x^2\right )}{16 a^2 (b c-a d)}\\ &=\frac{b x^2 \sqrt{c+d x^8}}{8 a (b c-a d) \left (a+b x^8\right )}+\frac{d^{3/4} \left (\sqrt{c}+\sqrt{d} x^4\right ) \sqrt{\frac{c+d x^8}{\left (\sqrt{c}+\sqrt{d} x^4\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac{1}{2}\right )}{16 a \sqrt [4]{c} (b c-a d) \sqrt{c+d x^8}}+\frac{\left (\sqrt{b} \sqrt{c} \left (\sqrt{b} \sqrt{c}-\sqrt{-a} \sqrt{d}\right ) (3 b c-5 a d)\right ) \operatorname{Subst}\left (\int \frac{1+\frac{\sqrt{d} x^2}{\sqrt{c}}}{\left (1-\frac{\sqrt{b} x^2}{\sqrt{-a}}\right ) \sqrt{c+d x^4}} \, dx,x,x^2\right )}{16 a^2 (b c-a d) (b c+a d)}+\frac{\left (\sqrt{b} \sqrt{c} \left (\sqrt{b} \sqrt{c}+\sqrt{-a} \sqrt{d}\right ) (3 b c-5 a d)\right ) \operatorname{Subst}\left (\int \frac{1+\frac{\sqrt{d} x^2}{\sqrt{c}}}{\left (1+\frac{\sqrt{b} x^2}{\sqrt{-a}}\right ) \sqrt{c+d x^4}} \, dx,x,x^2\right )}{16 a^2 (b c-a d) (b c+a d)}+\frac{\left (\left (\frac{\sqrt{b} \sqrt{c}}{\sqrt{-a}}+\sqrt{d}\right ) \sqrt{d} (3 b c-5 a d)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c+d x^4}} \, dx,x,x^2\right )}{16 a (b c-a d) (b c+a d)}+\frac{\left (\left (\sqrt{b} \sqrt{c}-\sqrt{-a} \sqrt{d}\right ) \sqrt{d} (3 b c-5 a d)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c+d x^4}} \, dx,x,x^2\right )}{16 (-a)^{3/2} (b c-a d) (b c+a d)}\\ &=\frac{b x^2 \sqrt{c+d x^8}}{8 a (b c-a d) \left (a+b x^8\right )}+\frac{\sqrt [4]{b} (3 b c-5 a d) \tan ^{-1}\left (\frac{\sqrt{b c-a d} x^2}{\sqrt [4]{-a} \sqrt [4]{b} \sqrt{c+d x^8}}\right )}{32 (-a)^{7/4} (b c-a d)^{3/2}}-\frac{\sqrt [4]{b} (3 b c-5 a d) \tan ^{-1}\left (\frac{\sqrt{-b c+a d} x^2}{\sqrt [4]{-a} \sqrt [4]{b} \sqrt{c+d x^8}}\right )}{32 (-a)^{7/4} (-b c+a d)^{3/2}}+\frac{d^{3/4} \left (\sqrt{c}+\sqrt{d} x^4\right ) \sqrt{\frac{c+d x^8}{\left (\sqrt{c}+\sqrt{d} x^4\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac{1}{2}\right )}{16 a \sqrt [4]{c} (b c-a d) \sqrt{c+d x^8}}+\frac{\left (\frac{\sqrt{b} \sqrt{c}}{\sqrt{-a}}+\sqrt{d}\right ) \sqrt [4]{d} (3 b c-5 a d) \left (\sqrt{c}+\sqrt{d} x^4\right ) \sqrt{\frac{c+d x^8}{\left (\sqrt{c}+\sqrt{d} x^4\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac{1}{2}\right )}{32 a \sqrt [4]{c} (b c-a d) (b c+a d) \sqrt{c+d x^8}}+\frac{\left (\sqrt{b} \sqrt{c}-\sqrt{-a} \sqrt{d}\right ) \sqrt [4]{d} (3 b c-5 a d) \left (\sqrt{c}+\sqrt{d} x^4\right ) \sqrt{\frac{c+d x^8}{\left (\sqrt{c}+\sqrt{d} x^4\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac{1}{2}\right )}{32 (-a)^{3/2} \sqrt [4]{c} (b c-a d) (b c+a d) \sqrt{c+d x^8}}+\frac{\left (\sqrt{b} \sqrt{c}+\sqrt{-a} \sqrt{d}\right )^2 (3 b c-5 a d) \left (\sqrt{c}+\sqrt{d} x^4\right ) \sqrt{\frac{c+d x^8}{\left (\sqrt{c}+\sqrt{d} x^4\right )^2}} \Pi \left (-\frac{\left (\sqrt{b} \sqrt{c}-\sqrt{-a} \sqrt{d}\right )^2}{4 \sqrt{-a} \sqrt{b} \sqrt{c} \sqrt{d}};2 \tan ^{-1}\left (\frac{\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac{1}{2}\right )}{64 a^2 \sqrt [4]{c} \sqrt [4]{d} (b c-a d) (b c+a d) \sqrt{c+d x^8}}+\frac{\left (\sqrt{b} \sqrt{c}-\sqrt{-a} \sqrt{d}\right )^2 (3 b c-5 a d) \left (\sqrt{c}+\sqrt{d} x^4\right ) \sqrt{\frac{c+d x^8}{\left (\sqrt{c}+\sqrt{d} x^4\right )^2}} \Pi \left (\frac{\left (\sqrt{b} \sqrt{c}+\sqrt{-a} \sqrt{d}\right )^2}{4 \sqrt{-a} \sqrt{b} \sqrt{c} \sqrt{d}};2 \tan ^{-1}\left (\frac{\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac{1}{2}\right )}{64 a^2 \sqrt [4]{c} \sqrt [4]{d} (b c-a d) (b c+a d) \sqrt{c+d x^8}}\\ \end{align*}

Mathematica [C]  time = 0.157496, size = 169, normalized size = 0.17 \[ \frac{x^2 \left (b d x^8 \left (a+b x^8\right ) \sqrt{\frac{d x^8}{c}+1} F_1\left (\frac{5}{4};\frac{1}{2},1;\frac{9}{4};-\frac{d x^8}{c},-\frac{b x^8}{a}\right )+5 \left (a+b x^8\right ) \sqrt{\frac{d x^8}{c}+1} (3 b c-4 a d) F_1\left (\frac{1}{4};\frac{1}{2},1;\frac{5}{4};-\frac{d x^8}{c},-\frac{b x^8}{a}\right )+5 a b \left (c+d x^8\right )\right )}{40 a^2 \left (a+b x^8\right ) \sqrt{c+d x^8} (b c-a d)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x/((a + b*x^8)^2*Sqrt[c + d*x^8]),x]

[Out]

(x^2*(5*a*b*(c + d*x^8) + 5*(3*b*c - 4*a*d)*(a + b*x^8)*Sqrt[1 + (d*x^8)/c]*AppellF1[1/4, 1/2, 1, 5/4, -((d*x^
8)/c), -((b*x^8)/a)] + b*d*x^8*(a + b*x^8)*Sqrt[1 + (d*x^8)/c]*AppellF1[5/4, 1/2, 1, 9/4, -((d*x^8)/c), -((b*x
^8)/a)]))/(40*a^2*(b*c - a*d)*(a + b*x^8)*Sqrt[c + d*x^8])

________________________________________________________________________________________

Maple [F]  time = 0.035, size = 0, normalized size = 0. \begin{align*} \int{\frac{x}{ \left ( b{x}^{8}+a \right ) ^{2}}{\frac{1}{\sqrt{d{x}^{8}+c}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(b*x^8+a)^2/(d*x^8+c)^(1/2),x)

[Out]

int(x/(b*x^8+a)^2/(d*x^8+c)^(1/2),x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{{\left (b x^{8} + a\right )}^{2} \sqrt{d x^{8} + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(b*x^8+a)^2/(d*x^8+c)^(1/2),x, algorithm="maxima")

[Out]

integrate(x/((b*x^8 + a)^2*sqrt(d*x^8 + c)), x)

________________________________________________________________________________________

Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(b*x^8+a)^2/(d*x^8+c)^(1/2),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(b*x**8+a)**2/(d*x**8+c)**(1/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{{\left (b x^{8} + a\right )}^{2} \sqrt{d x^{8} + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(b*x^8+a)^2/(d*x^8+c)^(1/2),x, algorithm="giac")

[Out]

integrate(x/((b*x^8 + a)^2*sqrt(d*x^8 + c)), x)