3.834 \(\int \frac{x^8}{(a+b x^4)^2 \sqrt{c+d x^4}} \, dx\)

Optimal. Leaf size=996 \[ \text{result too large to display} \]

[Out]

(a*x*Sqrt[c + d*x^4])/(4*b*(b*c - a*d)*(a + b*x^4)) - ((-a)^(1/4)*(5*b*c - 3*a*d)*ArcTan[(Sqrt[b*c - a*d]*x)/(
(-a)^(1/4)*b^(1/4)*Sqrt[c + d*x^4])])/(16*b^(7/4)*(b*c - a*d)^(3/2)) + ((-a)^(1/4)*(5*b*c - 3*a*d)*ArcTan[(Sqr
t[-(b*c) + a*d]*x)/((-a)^(1/4)*b^(1/4)*Sqrt[c + d*x^4])])/(16*b^(7/4)*(-(b*c) + a*d)^(3/2)) + ((4*b*c - 3*a*d)
*(Sqrt[c] + Sqrt[d]*x^2)*Sqrt[(c + d*x^4)/(Sqrt[c] + Sqrt[d]*x^2)^2]*EllipticF[2*ArcTan[(d^(1/4)*x)/c^(1/4)],
1/2])/(8*b^2*c^(1/4)*d^(1/4)*(b*c - a*d)*Sqrt[c + d*x^4]) - (a*((Sqrt[b]*Sqrt[c])/Sqrt[-a] + Sqrt[d])*d^(1/4)*
(5*b*c - 3*a*d)*(Sqrt[c] + Sqrt[d]*x^2)*Sqrt[(c + d*x^4)/(Sqrt[c] + Sqrt[d]*x^2)^2]*EllipticF[2*ArcTan[(d^(1/4
)*x)/c^(1/4)], 1/2])/(16*b^2*c^(1/4)*(b*c - a*d)*(b*c + a*d)*Sqrt[c + d*x^4]) - (Sqrt[-a]*(Sqrt[b]*Sqrt[c] - S
qrt[-a]*Sqrt[d])*d^(1/4)*(5*b*c - 3*a*d)*(Sqrt[c] + Sqrt[d]*x^2)*Sqrt[(c + d*x^4)/(Sqrt[c] + Sqrt[d]*x^2)^2]*E
llipticF[2*ArcTan[(d^(1/4)*x)/c^(1/4)], 1/2])/(16*b^2*c^(1/4)*(b*c - a*d)*(b*c + a*d)*Sqrt[c + d*x^4]) - ((Sqr
t[b]*Sqrt[c] + Sqrt[-a]*Sqrt[d])^2*(5*b*c - 3*a*d)*(Sqrt[c] + Sqrt[d]*x^2)*Sqrt[(c + d*x^4)/(Sqrt[c] + Sqrt[d]
*x^2)^2]*EllipticPi[-(Sqrt[b]*Sqrt[c] - Sqrt[-a]*Sqrt[d])^2/(4*Sqrt[-a]*Sqrt[b]*Sqrt[c]*Sqrt[d]), 2*ArcTan[(d^
(1/4)*x)/c^(1/4)], 1/2])/(32*b^2*c^(1/4)*d^(1/4)*(b*c - a*d)*(b*c + a*d)*Sqrt[c + d*x^4]) - ((Sqrt[b]*Sqrt[c]
- Sqrt[-a]*Sqrt[d])^2*(5*b*c - 3*a*d)*(Sqrt[c] + Sqrt[d]*x^2)*Sqrt[(c + d*x^4)/(Sqrt[c] + Sqrt[d]*x^2)^2]*Elli
pticPi[(Sqrt[b]*Sqrt[c] + Sqrt[-a]*Sqrt[d])^2/(4*Sqrt[-a]*Sqrt[b]*Sqrt[c]*Sqrt[d]), 2*ArcTan[(d^(1/4)*x)/c^(1/
4)], 1/2])/(32*b^2*c^(1/4)*d^(1/4)*(b*c - a*d)*(b*c + a*d)*Sqrt[c + d*x^4])

________________________________________________________________________________________

Rubi [A]  time = 1.30812, antiderivative size = 996, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {470, 523, 220, 409, 1217, 1707} \[ -\frac{(5 b c-3 a d) \left (\sqrt{d} x^2+\sqrt{c}\right ) \sqrt{\frac{d x^4+c}{\left (\sqrt{d} x^2+\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}{\sqrt [4]{c}}\right )|\frac{1}{2}\right ) \left (\sqrt{b} \sqrt{c}-\sqrt{-a} \sqrt{d}\right )^2}{32 b^2 \sqrt [4]{c} \sqrt [4]{d} (b c-a d) (b c+a d) \sqrt{d x^4+c}}-\frac{\sqrt{-a} \sqrt [4]{d} (5 b c-3 a d) \left (\sqrt{d} x^2+\sqrt{c}\right ) \sqrt{\frac{d x^4+c}{\left (\sqrt{d} x^2+\sqrt{c}\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} x}{\sqrt [4]{c}}\right )|\frac{1}{2}\right ) \left (\sqrt{b} \sqrt{c}-\sqrt{-a} \sqrt{d}\right )}{16 b^2 \sqrt [4]{c} (b c-a d) (b c+a d) \sqrt{d x^4+c}}-\frac{\sqrt [4]{-a} (5 b c-3 a d) \tan ^{-1}\left (\frac{\sqrt{b c-a d} x}{\sqrt [4]{-a} \sqrt [4]{b} \sqrt{d x^4+c}}\right )}{16 b^{7/4} (b c-a d)^{3/2}}+\frac{\sqrt [4]{-a} (5 b c-3 a d) \tan ^{-1}\left (\frac{\sqrt{a d-b c} x}{\sqrt [4]{-a} \sqrt [4]{b} \sqrt{d x^4+c}}\right )}{16 b^{7/4} (a d-b c)^{3/2}}+\frac{(4 b c-3 a d) \left (\sqrt{d} x^2+\sqrt{c}\right ) \sqrt{\frac{d x^4+c}{\left (\sqrt{d} x^2+\sqrt{c}\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} x}{\sqrt [4]{c}}\right )|\frac{1}{2}\right )}{8 b^2 \sqrt [4]{c} \sqrt [4]{d} (b c-a d) \sqrt{d x^4+c}}-\frac{a \left (\frac{\sqrt{b} \sqrt{c}}{\sqrt{-a}}+\sqrt{d}\right ) \sqrt [4]{d} (5 b c-3 a d) \left (\sqrt{d} x^2+\sqrt{c}\right ) \sqrt{\frac{d x^4+c}{\left (\sqrt{d} x^2+\sqrt{c}\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} x}{\sqrt [4]{c}}\right )|\frac{1}{2}\right )}{16 b^2 \sqrt [4]{c} (b c-a d) (b c+a d) \sqrt{d x^4+c}}-\frac{\left (\sqrt{b} \sqrt{c}+\sqrt{-a} \sqrt{d}\right )^2 (5 b c-3 a d) \left (\sqrt{d} x^2+\sqrt{c}\right ) \sqrt{\frac{d x^4+c}{\left (\sqrt{d} x^2+\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}{\sqrt [4]{c}}\right )|\frac{1}{2}\right )}{32 b^2 \sqrt [4]{c} \sqrt [4]{d} (b c-a d) (b c+a d) \sqrt{d x^4+c}}+\frac{a x \sqrt{d x^4+c}}{4 b (b c-a d) \left (b x^4+a\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^8/((a + b*x^4)^2*Sqrt[c + d*x^4]),x]

[Out]

(a*x*Sqrt[c + d*x^4])/(4*b*(b*c - a*d)*(a + b*x^4)) - ((-a)^(1/4)*(5*b*c - 3*a*d)*ArcTan[(Sqrt[b*c - a*d]*x)/(
(-a)^(1/4)*b^(1/4)*Sqrt[c + d*x^4])])/(16*b^(7/4)*(b*c - a*d)^(3/2)) + ((-a)^(1/4)*(5*b*c - 3*a*d)*ArcTan[(Sqr
t[-(b*c) + a*d]*x)/((-a)^(1/4)*b^(1/4)*Sqrt[c + d*x^4])])/(16*b^(7/4)*(-(b*c) + a*d)^(3/2)) + ((4*b*c - 3*a*d)
*(Sqrt[c] + Sqrt[d]*x^2)*Sqrt[(c + d*x^4)/(Sqrt[c] + Sqrt[d]*x^2)^2]*EllipticF[2*ArcTan[(d^(1/4)*x)/c^(1/4)],
1/2])/(8*b^2*c^(1/4)*d^(1/4)*(b*c - a*d)*Sqrt[c + d*x^4]) - (a*((Sqrt[b]*Sqrt[c])/Sqrt[-a] + Sqrt[d])*d^(1/4)*
(5*b*c - 3*a*d)*(Sqrt[c] + Sqrt[d]*x^2)*Sqrt[(c + d*x^4)/(Sqrt[c] + Sqrt[d]*x^2)^2]*EllipticF[2*ArcTan[(d^(1/4
)*x)/c^(1/4)], 1/2])/(16*b^2*c^(1/4)*(b*c - a*d)*(b*c + a*d)*Sqrt[c + d*x^4]) - (Sqrt[-a]*(Sqrt[b]*Sqrt[c] - S
qrt[-a]*Sqrt[d])*d^(1/4)*(5*b*c - 3*a*d)*(Sqrt[c] + Sqrt[d]*x^2)*Sqrt[(c + d*x^4)/(Sqrt[c] + Sqrt[d]*x^2)^2]*E
llipticF[2*ArcTan[(d^(1/4)*x)/c^(1/4)], 1/2])/(16*b^2*c^(1/4)*(b*c - a*d)*(b*c + a*d)*Sqrt[c + d*x^4]) - ((Sqr
t[b]*Sqrt[c] + Sqrt[-a]*Sqrt[d])^2*(5*b*c - 3*a*d)*(Sqrt[c] + Sqrt[d]*x^2)*Sqrt[(c + d*x^4)/(Sqrt[c] + Sqrt[d]
*x^2)^2]*EllipticPi[-(Sqrt[b]*Sqrt[c] - Sqrt[-a]*Sqrt[d])^2/(4*Sqrt[-a]*Sqrt[b]*Sqrt[c]*Sqrt[d]), 2*ArcTan[(d^
(1/4)*x)/c^(1/4)], 1/2])/(32*b^2*c^(1/4)*d^(1/4)*(b*c - a*d)*(b*c + a*d)*Sqrt[c + d*x^4]) - ((Sqrt[b]*Sqrt[c]
- Sqrt[-a]*Sqrt[d])^2*(5*b*c - 3*a*d)*(Sqrt[c] + Sqrt[d]*x^2)*Sqrt[(c + d*x^4)/(Sqrt[c] + Sqrt[d]*x^2)^2]*Elli
pticPi[(Sqrt[b]*Sqrt[c] + Sqrt[-a]*Sqrt[d])^2/(4*Sqrt[-a]*Sqrt[b]*Sqrt[c]*Sqrt[d]), 2*ArcTan[(d^(1/4)*x)/c^(1/
4)], 1/2])/(32*b^2*c^(1/4)*d^(1/4)*(b*c - a*d)*(b*c + a*d)*Sqrt[c + d*x^4])

Rule 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(a*e^(2
*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c - a*d)*(p + 1)), x] + Dist[e^(2
*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) +
(a*d*(m - n + n*q + 1) + b*c*n*(p + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, 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^8}{\left (a+b x^4\right )^2 \sqrt{c+d x^4}} \, dx &=\frac{a x \sqrt{c+d x^4}}{4 b (b c-a d) \left (a+b x^4\right )}-\frac{\int \frac{a c+(-4 b c+3 a d) x^4}{\left (a+b x^4\right ) \sqrt{c+d x^4}} \, dx}{4 b (b c-a d)}\\ &=\frac{a x \sqrt{c+d x^4}}{4 b (b c-a d) \left (a+b x^4\right )}+\frac{(4 b c-3 a d) \int \frac{1}{\sqrt{c+d x^4}} \, dx}{4 b^2 (b c-a d)}-\frac{(a (5 b c-3 a d)) \int \frac{1}{\left (a+b x^4\right ) \sqrt{c+d x^4}} \, dx}{4 b^2 (b c-a d)}\\ &=\frac{a x \sqrt{c+d x^4}}{4 b (b c-a d) \left (a+b x^4\right )}+\frac{(4 b c-3 a d) \left (\sqrt{c}+\sqrt{d} x^2\right ) \sqrt{\frac{c+d x^4}{\left (\sqrt{c}+\sqrt{d} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} x}{\sqrt [4]{c}}\right )|\frac{1}{2}\right )}{8 b^2 \sqrt [4]{c} \sqrt [4]{d} (b c-a d) \sqrt{c+d x^4}}-\frac{(5 b c-3 a d) \int \frac{1}{\left (1-\frac{\sqrt{b} x^2}{\sqrt{-a}}\right ) \sqrt{c+d x^4}} \, dx}{8 b^2 (b c-a d)}-\frac{(5 b c-3 a d) \int \frac{1}{\left (1+\frac{\sqrt{b} x^2}{\sqrt{-a}}\right ) \sqrt{c+d x^4}} \, dx}{8 b^2 (b c-a d)}\\ &=\frac{a x \sqrt{c+d x^4}}{4 b (b c-a d) \left (a+b x^4\right )}+\frac{(4 b c-3 a d) \left (\sqrt{c}+\sqrt{d} x^2\right ) \sqrt{\frac{c+d x^4}{\left (\sqrt{c}+\sqrt{d} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} x}{\sqrt [4]{c}}\right )|\frac{1}{2}\right )}{8 b^2 \sqrt [4]{c} \sqrt [4]{d} (b c-a d) \sqrt{c+d x^4}}-\frac{\left (\sqrt{c} \left (\sqrt{b} \sqrt{c}-\sqrt{-a} \sqrt{d}\right ) (5 b c-3 a d)\right ) \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}{8 b^{3/2} (b c-a d) (b c+a d)}-\frac{\left (\sqrt{c} \left (\sqrt{b} \sqrt{c}+\sqrt{-a} \sqrt{d}\right ) (5 b c-3 a d)\right ) \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}{8 b^{3/2} (b c-a d) (b c+a d)}-\frac{\left (a \left (\frac{\sqrt{b} \sqrt{c}}{\sqrt{-a}}+\sqrt{d}\right ) \sqrt{d} (5 b c-3 a d)\right ) \int \frac{1}{\sqrt{c+d x^4}} \, dx}{8 b^2 (b c-a d) (b c+a d)}-\frac{\left (\sqrt{-a} \left (\sqrt{b} \sqrt{c}-\sqrt{-a} \sqrt{d}\right ) \sqrt{d} (5 b c-3 a d)\right ) \int \frac{1}{\sqrt{c+d x^4}} \, dx}{8 b^2 (b c-a d) (b c+a d)}\\ &=\frac{a x \sqrt{c+d x^4}}{4 b (b c-a d) \left (a+b x^4\right )}-\frac{\sqrt [4]{-a} (5 b c-3 a d) \tan ^{-1}\left (\frac{\sqrt{b c-a d} x}{\sqrt [4]{-a} \sqrt [4]{b} \sqrt{c+d x^4}}\right )}{16 b^{7/4} (b c-a d)^{3/2}}+\frac{\sqrt [4]{-a} (5 b c-3 a d) \tan ^{-1}\left (\frac{\sqrt{-b c+a d} x}{\sqrt [4]{-a} \sqrt [4]{b} \sqrt{c+d x^4}}\right )}{16 b^{7/4} (-b c+a d)^{3/2}}+\frac{(4 b c-3 a d) \left (\sqrt{c}+\sqrt{d} x^2\right ) \sqrt{\frac{c+d x^4}{\left (\sqrt{c}+\sqrt{d} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} x}{\sqrt [4]{c}}\right )|\frac{1}{2}\right )}{8 b^2 \sqrt [4]{c} \sqrt [4]{d} (b c-a d) \sqrt{c+d x^4}}-\frac{a \left (\frac{\sqrt{b} \sqrt{c}}{\sqrt{-a}}+\sqrt{d}\right ) \sqrt [4]{d} (5 b c-3 a d) \left (\sqrt{c}+\sqrt{d} x^2\right ) \sqrt{\frac{c+d x^4}{\left (\sqrt{c}+\sqrt{d} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} x}{\sqrt [4]{c}}\right )|\frac{1}{2}\right )}{16 b^2 \sqrt [4]{c} (b c-a d) (b c+a d) \sqrt{c+d x^4}}-\frac{\sqrt{-a} \left (\sqrt{b} \sqrt{c}-\sqrt{-a} \sqrt{d}\right ) \sqrt [4]{d} (5 b c-3 a d) \left (\sqrt{c}+\sqrt{d} x^2\right ) \sqrt{\frac{c+d x^4}{\left (\sqrt{c}+\sqrt{d} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} x}{\sqrt [4]{c}}\right )|\frac{1}{2}\right )}{16 b^2 \sqrt [4]{c} (b c-a d) (b c+a d) \sqrt{c+d x^4}}-\frac{\left (\sqrt{b} \sqrt{c}+\sqrt{-a} \sqrt{d}\right )^2 (5 b c-3 a d) \left (\sqrt{c}+\sqrt{d} x^2\right ) \sqrt{\frac{c+d x^4}{\left (\sqrt{c}+\sqrt{d} x^2\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}{\sqrt [4]{c}}\right )|\frac{1}{2}\right )}{32 b^2 \sqrt [4]{c} \sqrt [4]{d} (b c-a d) (b c+a d) \sqrt{c+d x^4}}-\frac{\left (\sqrt{b} \sqrt{c}-\sqrt{-a} \sqrt{d}\right )^2 (5 b c-3 a d) \left (\sqrt{c}+\sqrt{d} x^2\right ) \sqrt{\frac{c+d x^4}{\left (\sqrt{c}+\sqrt{d} x^2\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}{\sqrt [4]{c}}\right )|\frac{1}{2}\right )}{32 b^2 \sqrt [4]{c} \sqrt [4]{d} (b c-a d) (b c+a d) \sqrt{c+d x^4}}\\ \end{align*}

Mathematica [C]  time = 0.279361, size = 253, normalized size = 0.25 \[ \frac{x \left (\frac{5 a \left (\frac{5 a c^2 F_1\left (\frac{1}{4};\frac{1}{2},1;\frac{5}{4};-\frac{d x^4}{c},-\frac{b x^4}{a}\right )}{2 x^4 \left (2 b c F_1\left (\frac{5}{4};\frac{1}{2},2;\frac{9}{4};-\frac{d x^4}{c},-\frac{b x^4}{a}\right )+a d F_1\left (\frac{5}{4};\frac{3}{2},1;\frac{9}{4};-\frac{d x^4}{c},-\frac{b x^4}{a}\right )\right )-5 a c F_1\left (\frac{1}{4};\frac{1}{2},1;\frac{5}{4};-\frac{d x^4}{c},-\frac{b x^4}{a}\right )}+c+d x^4\right )}{b \left (a+b x^4\right )}+\frac{x^4 \sqrt{\frac{d x^4}{c}+1} (4 b c-3 a d) F_1\left (\frac{5}{4};\frac{1}{2},1;\frac{9}{4};-\frac{d x^4}{c},-\frac{b x^4}{a}\right )}{a b}\right )}{20 \sqrt{c+d x^4} (b c-a d)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^8/((a + b*x^4)^2*Sqrt[c + d*x^4]),x]

[Out]

(x*(((4*b*c - 3*a*d)*x^4*Sqrt[1 + (d*x^4)/c]*AppellF1[5/4, 1/2, 1, 9/4, -((d*x^4)/c), -((b*x^4)/a)])/(a*b) + (
5*a*(c + d*x^4 + (5*a*c^2*AppellF1[1/4, 1/2, 1, 5/4, -((d*x^4)/c), -((b*x^4)/a)])/(-5*a*c*AppellF1[1/4, 1/2, 1
, 5/4, -((d*x^4)/c), -((b*x^4)/a)] + 2*x^4*(2*b*c*AppellF1[5/4, 1/2, 2, 9/4, -((d*x^4)/c), -((b*x^4)/a)] + a*d
*AppellF1[5/4, 3/2, 1, 9/4, -((d*x^4)/c), -((b*x^4)/a)]))))/(b*(a + b*x^4))))/(20*(b*c - a*d)*Sqrt[c + d*x^4])

________________________________________________________________________________________

Maple [C]  time = 0.03, size = 604, normalized size = 0.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/(b*x^4+a)^2/(d*x^4+c)^(1/2),x)

[Out]

1/b^2/(I/c^(1/2)*d^(1/2))^(1/2)*(1-I/c^(1/2)*d^(1/2)*x^2)^(1/2)*(1+I/c^(1/2)*d^(1/2)*x^2)^(1/2)/(d*x^4+c)^(1/2
)*EllipticF(x*(I/c^(1/2)*d^(1/2))^(1/2),I)-1/4*a/b^3*sum(1/_alpha^3*(-1/((-a*d+b*c)/b)^(1/2)*arctanh(1/2*(2*_a
lpha^2*d*x^2+2*c)/((-a*d+b*c)/b)^(1/2)/(d*x^4+c)^(1/2))+2/(I/c^(1/2)*d^(1/2))^(1/2)*_alpha^3*b/a*(1-I/c^(1/2)*
d^(1/2)*x^2)^(1/2)*(1+I/c^(1/2)*d^(1/2)*x^2)^(1/2)/(d*x^4+c)^(1/2)*EllipticPi(x*(I/c^(1/2)*d^(1/2))^(1/2),I*c^
(1/2)/d^(1/2)*_alpha^2/a*b,(-I/c^(1/2)*d^(1/2))^(1/2)/(I/c^(1/2)*d^(1/2))^(1/2))),_alpha=RootOf(_Z^4*b+a))+a^2
/b^2*(-1/4*b/(a*d-b*c)/a*x*(d*x^4+c)^(1/2)/(b*x^4+a)-1/4*d/(a*d-b*c)/a/(I/c^(1/2)*d^(1/2))^(1/2)*(1-I/c^(1/2)*
d^(1/2)*x^2)^(1/2)*(1+I/c^(1/2)*d^(1/2)*x^2)^(1/2)/(d*x^4+c)^(1/2)*EllipticF(x*(I/c^(1/2)*d^(1/2))^(1/2),I)-1/
32/b/a*sum((-5*a*d+3*b*c)/(a*d-b*c)/_alpha^3*(-1/((-a*d+b*c)/b)^(1/2)*arctanh(1/2*(2*_alpha^2*d*x^2+2*c)/((-a*
d+b*c)/b)^(1/2)/(d*x^4+c)^(1/2))+2/(I/c^(1/2)*d^(1/2))^(1/2)*_alpha^3*b/a*(1-I/c^(1/2)*d^(1/2)*x^2)^(1/2)*(1+I
/c^(1/2)*d^(1/2)*x^2)^(1/2)/(d*x^4+c)^(1/2)*EllipticPi(x*(I/c^(1/2)*d^(1/2))^(1/2),I*c^(1/2)/d^(1/2)*_alpha^2/
a*b,(-I/c^(1/2)*d^(1/2))^(1/2)/(I/c^(1/2)*d^(1/2))^(1/2))),_alpha=RootOf(_Z^4*b+a)))

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{8}}{{\left (b x^{4} + a\right )}^{2} \sqrt{d x^{4} + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(b*x^4+a)^2/(d*x^4+c)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^8/((b*x^4 + a)^2*sqrt(d*x^4 + 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^8/(b*x^4+a)^2/(d*x^4+c)^(1/2),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8/(b*x**4+a)**2/(d*x**4+c)**(1/2),x)

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{8}}{{\left (b x^{4} + a\right )}^{2} \sqrt{d x^{4} + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(b*x^4+a)^2/(d*x^4+c)^(1/2),x, algorithm="giac")

[Out]

integrate(x^8/((b*x^4 + a)^2*sqrt(d*x^4 + c)), x)