3.837 \(\int \frac{1}{x^4 (a+b x^4)^2 \sqrt{c+d x^4}} \, dx\)

Optimal. Leaf size=1046 \[ \text{result too large to display} \]

[Out]

-((7*b*c - 4*a*d)*Sqrt[c + d*x^4])/(12*a^2*c*(b*c - a*d)*x^3) + (b*Sqrt[c + d*x^4])/(4*a*(b*c - a*d)*x^3*(a +
b*x^4)) + (b^(5/4)*(7*b*c - 9*a*d)*ArcTan[(Sqrt[b*c - a*d]*x)/((-a)^(1/4)*b^(1/4)*Sqrt[c + d*x^4])])/(16*(-a)^
(11/4)*(b*c - a*d)^(3/2)) - (b^(5/4)*(7*b*c - 9*a*d)*ArcTan[(Sqrt[-(b*c) + a*d]*x)/((-a)^(1/4)*b^(1/4)*Sqrt[c
+ d*x^4])])/(16*(-a)^(11/4)*(-(b*c) + a*d)^(3/2)) - (d^(3/4)*(7*b*c - 4*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])/(24*a^2*c^(5/4)*(b*c - a*d)*
Sqrt[c + d*x^4]) + (b*(Sqrt[b]*Sqrt[c] - Sqrt[-a]*Sqrt[d])*d^(1/4)*(7*b*c - 9*a*d)*(Sqrt[c] + Sqrt[d]*x^2)*Sqr
t[(c + d*x^4)/(Sqrt[c] + Sqrt[d]*x^2)^2]*EllipticF[2*ArcTan[(d^(1/4)*x)/c^(1/4)], 1/2])/(16*(-a)^(5/2)*c^(1/4)
*(b*c - a*d)*(b*c + a*d)*Sqrt[c + d*x^4]) - (b*(Sqrt[b]*Sqrt[c] + Sqrt[-a]*Sqrt[d])*d^(1/4)*(7*b*c - 9*a*d)*(S
qrt[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*(-a)^(5/2)*c^(1/4)*(b*c - a*d)*(b*c + a*d)*Sqrt[c + d*x^4]) - (b*(Sqrt[b]*Sqrt[c] + Sqrt[-a]*Sqrt[d])^2
*(7*b*c - 9*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*a^3*c
^(1/4)*d^(1/4)*(b*c - a*d)*(b*c + a*d)*Sqrt[c + d*x^4]) - (b*(Sqrt[b]*Sqrt[c] - Sqrt[-a]*Sqrt[d])^2*(7*b*c - 9
*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*a^3*c^(1/4)*d^(1/
4)*(b*c - a*d)*(b*c + a*d)*Sqrt[c + d*x^4])

________________________________________________________________________________________

Rubi [A]  time = 1.55713, antiderivative size = 1046, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.292, Rules used = {472, 583, 523, 220, 409, 1217, 1707} \[ -\frac{b (7 b c-9 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 a^3 \sqrt [4]{c} \sqrt [4]{d} (b c-a d) (b c+a d) \sqrt{d x^4+c}}+\frac{b \sqrt [4]{d} (7 b c-9 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 (-a)^{5/2} \sqrt [4]{c} (b c-a d) (b c+a d) \sqrt{d x^4+c}}+\frac{b^{5/4} (7 b c-9 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 (-a)^{11/4} (b c-a d)^{3/2}}-\frac{b^{5/4} (7 b c-9 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 (-a)^{11/4} (a d-b c)^{3/2}}-\frac{d^{3/4} (7 b c-4 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 )}{24 a^2 c^{5/4} (b c-a d) \sqrt{d x^4+c}}-\frac{b \left (\sqrt{b} \sqrt{c}+\sqrt{-a} \sqrt{d}\right ) \sqrt [4]{d} (7 b c-9 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 (-a)^{5/2} \sqrt [4]{c} (b c-a d) (b c+a d) \sqrt{d x^4+c}}-\frac{b \left (\sqrt{b} \sqrt{c}+\sqrt{-a} \sqrt{d}\right )^2 (7 b c-9 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 a^3 \sqrt [4]{c} \sqrt [4]{d} (b c-a d) (b c+a d) \sqrt{d x^4+c}}+\frac{b \sqrt{d x^4+c}}{4 a (b c-a d) x^3 \left (b x^4+a\right )}-\frac{(7 b c-4 a d) \sqrt{d x^4+c}}{12 a^2 c (b c-a d) x^3} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(a + b*x^4)^2*Sqrt[c + d*x^4]),x]

[Out]

-((7*b*c - 4*a*d)*Sqrt[c + d*x^4])/(12*a^2*c*(b*c - a*d)*x^3) + (b*Sqrt[c + d*x^4])/(4*a*(b*c - a*d)*x^3*(a +
b*x^4)) + (b^(5/4)*(7*b*c - 9*a*d)*ArcTan[(Sqrt[b*c - a*d]*x)/((-a)^(1/4)*b^(1/4)*Sqrt[c + d*x^4])])/(16*(-a)^
(11/4)*(b*c - a*d)^(3/2)) - (b^(5/4)*(7*b*c - 9*a*d)*ArcTan[(Sqrt[-(b*c) + a*d]*x)/((-a)^(1/4)*b^(1/4)*Sqrt[c
+ d*x^4])])/(16*(-a)^(11/4)*(-(b*c) + a*d)^(3/2)) - (d^(3/4)*(7*b*c - 4*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])/(24*a^2*c^(5/4)*(b*c - a*d)*
Sqrt[c + d*x^4]) + (b*(Sqrt[b]*Sqrt[c] - Sqrt[-a]*Sqrt[d])*d^(1/4)*(7*b*c - 9*a*d)*(Sqrt[c] + Sqrt[d]*x^2)*Sqr
t[(c + d*x^4)/(Sqrt[c] + Sqrt[d]*x^2)^2]*EllipticF[2*ArcTan[(d^(1/4)*x)/c^(1/4)], 1/2])/(16*(-a)^(5/2)*c^(1/4)
*(b*c - a*d)*(b*c + a*d)*Sqrt[c + d*x^4]) - (b*(Sqrt[b]*Sqrt[c] + Sqrt[-a]*Sqrt[d])*d^(1/4)*(7*b*c - 9*a*d)*(S
qrt[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*(-a)^(5/2)*c^(1/4)*(b*c - a*d)*(b*c + a*d)*Sqrt[c + d*x^4]) - (b*(Sqrt[b]*Sqrt[c] + Sqrt[-a]*Sqrt[d])^2
*(7*b*c - 9*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*a^3*c
^(1/4)*d^(1/4)*(b*c - a*d)*(b*c + a*d)*Sqrt[c + d*x^4]) - (b*(Sqrt[b]*Sqrt[c] - Sqrt[-a]*Sqrt[d])^2*(7*b*c - 9
*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*a^3*c^(1/4)*d^(1/
4)*(b*c - a*d)*(b*c + a*d)*Sqrt[c + d*x^4])

Rule 472

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

Rule 583

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*g*(m + 1)), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
 IGtQ[n, 0] && LtQ[m, -1]

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{1}{x^4 \left (a+b x^4\right )^2 \sqrt{c+d x^4}} \, dx &=\frac{b \sqrt{c+d x^4}}{4 a (b c-a d) x^3 \left (a+b x^4\right )}-\frac{\int \frac{-7 b c+4 a d-5 b d x^4}{x^4 \left (a+b x^4\right ) \sqrt{c+d x^4}} \, dx}{4 a (b c-a d)}\\ &=-\frac{(7 b c-4 a d) \sqrt{c+d x^4}}{12 a^2 c (b c-a d) x^3}+\frac{b \sqrt{c+d x^4}}{4 a (b c-a d) x^3 \left (a+b x^4\right )}+\frac{\int \frac{-21 b^2 c^2+20 a b c d+4 a^2 d^2-b d (7 b c-4 a d) x^4}{\left (a+b x^4\right ) \sqrt{c+d x^4}} \, dx}{12 a^2 c (b c-a d)}\\ &=-\frac{(7 b c-4 a d) \sqrt{c+d x^4}}{12 a^2 c (b c-a d) x^3}+\frac{b \sqrt{c+d x^4}}{4 a (b c-a d) x^3 \left (a+b x^4\right )}-\frac{(b (7 b c-9 a d)) \int \frac{1}{\left (a+b x^4\right ) \sqrt{c+d x^4}} \, dx}{4 a^2 (b c-a d)}-\frac{(d (7 b c-4 a d)) \int \frac{1}{\sqrt{c+d x^4}} \, dx}{12 a^2 c (b c-a d)}\\ &=-\frac{(7 b c-4 a d) \sqrt{c+d x^4}}{12 a^2 c (b c-a d) x^3}+\frac{b \sqrt{c+d x^4}}{4 a (b c-a d) x^3 \left (a+b x^4\right )}-\frac{d^{3/4} (7 b c-4 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 )}{24 a^2 c^{5/4} (b c-a d) \sqrt{c+d x^4}}-\frac{(b (7 b c-9 a d)) \int \frac{1}{\left (1-\frac{\sqrt{b} x^2}{\sqrt{-a}}\right ) \sqrt{c+d x^4}} \, dx}{8 a^3 (b c-a d)}-\frac{(b (7 b c-9 a d)) \int \frac{1}{\left (1+\frac{\sqrt{b} x^2}{\sqrt{-a}}\right ) \sqrt{c+d x^4}} \, dx}{8 a^3 (b c-a d)}\\ &=-\frac{(7 b c-4 a d) \sqrt{c+d x^4}}{12 a^2 c (b c-a d) x^3}+\frac{b \sqrt{c+d x^4}}{4 a (b c-a d) x^3 \left (a+b x^4\right )}-\frac{d^{3/4} (7 b c-4 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 )}{24 a^2 c^{5/4} (b c-a d) \sqrt{c+d x^4}}-\frac{\left (b^{3/2} \sqrt{c} \left (\sqrt{b} \sqrt{c}-\sqrt{-a} \sqrt{d}\right ) (7 b c-9 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 a^3 (b c-a d) (b c+a d)}-\frac{\left (b^{3/2} \sqrt{c} \left (\sqrt{b} \sqrt{c}+\sqrt{-a} \sqrt{d}\right ) (7 b c-9 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 a^3 (b c-a d) (b c+a d)}-\frac{\left (b \left (\frac{\sqrt{b} \sqrt{c}}{\sqrt{-a}}+\sqrt{d}\right ) \sqrt{d} (7 b c-9 a d)\right ) \int \frac{1}{\sqrt{c+d x^4}} \, dx}{8 a^2 (b c-a d) (b c+a d)}+\frac{\left (b \left (\sqrt{b} \sqrt{c}-\sqrt{-a} \sqrt{d}\right ) \sqrt{d} (7 b c-9 a d)\right ) \int \frac{1}{\sqrt{c+d x^4}} \, dx}{8 (-a)^{5/2} (b c-a d) (b c+a d)}\\ &=-\frac{(7 b c-4 a d) \sqrt{c+d x^4}}{12 a^2 c (b c-a d) x^3}+\frac{b \sqrt{c+d x^4}}{4 a (b c-a d) x^3 \left (a+b x^4\right )}+\frac{b^{5/4} (7 b c-9 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 (-a)^{11/4} (b c-a d)^{3/2}}-\frac{b^{5/4} (7 b c-9 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 (-a)^{11/4} (-b c+a d)^{3/2}}-\frac{d^{3/4} (7 b c-4 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 )}{24 a^2 c^{5/4} (b c-a d) \sqrt{c+d x^4}}-\frac{b \left (\frac{\sqrt{b} \sqrt{c}}{\sqrt{-a}}+\sqrt{d}\right ) \sqrt [4]{d} (7 b c-9 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 a^2 \sqrt [4]{c} (b c-a d) (b c+a d) \sqrt{c+d x^4}}+\frac{b \left (\sqrt{b} \sqrt{c}-\sqrt{-a} \sqrt{d}\right ) \sqrt [4]{d} (7 b c-9 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 (-a)^{5/2} \sqrt [4]{c} (b c-a d) (b c+a d) \sqrt{c+d x^4}}-\frac{b \left (\sqrt{b} \sqrt{c}+\sqrt{-a} \sqrt{d}\right )^2 (7 b c-9 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 a^3 \sqrt [4]{c} \sqrt [4]{d} (b c-a d) (b c+a d) \sqrt{c+d x^4}}-\frac{b \left (\sqrt{b} \sqrt{c}-\sqrt{-a} \sqrt{d}\right )^2 (7 b c-9 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 a^3 \sqrt [4]{c} \sqrt [4]{d} (b c-a d) (b c+a d) \sqrt{c+d x^4}}\\ \end{align*}

Mathematica [C]  time = 0.562522, size = 408, normalized size = 0.39 \[ \frac{\frac{a \left (25 a c \left (4 a^2 d \left (c+2 d x^4\right )+4 a b \left (-c^2+5 c d x^4+d^2 x^8\right )-7 b^2 c x^4 \left (4 c+d x^4\right )\right ) F_1\left (\frac{1}{4};\frac{1}{2},1;\frac{5}{4};-\frac{d x^4}{c},-\frac{b x^4}{a}\right )+10 x^4 \left (c+d x^4\right ) \left (-4 a^2 d+4 a b \left (c-d x^4\right )+7 b^2 c x^4\right ) \left (2 b c F_1\left (\frac{5}{4};\frac{1}{2},2;\frac{9}{4};-\frac{d x^4}{c},-\frac{b x^4}{a}\right )+a d F_1\left (\frac{5}{4};\frac{3}{2},1;\frac{9}{4};-\frac{d x^4}{c},-\frac{b x^4}{a}\right )\right )\right )}{\left (a+b x^4\right ) \left (2 x^4 \left (2 b c F_1\left (\frac{5}{4};\frac{1}{2},2;\frac{9}{4};-\frac{d x^4}{c},-\frac{b x^4}{a}\right )+a d F_1\left (\frac{5}{4};\frac{3}{2},1;\frac{9}{4};-\frac{d x^4}{c},-\frac{b x^4}{a}\right )\right )-5 a c F_1\left (\frac{1}{4};\frac{1}{2},1;\frac{5}{4};-\frac{d x^4}{c},-\frac{b x^4}{a}\right )\right )}+b d x^8 \sqrt{\frac{d x^4}{c}+1} (7 b c-4 a d) F_1\left (\frac{5}{4};\frac{1}{2},1;\frac{9}{4};-\frac{d x^4}{c},-\frac{b x^4}{a}\right )}{60 a^3 c x^3 \sqrt{c+d x^4} (a d-b c)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(x^4*(a + b*x^4)^2*Sqrt[c + d*x^4]),x]

[Out]

(b*d*(7*b*c - 4*a*d)*x^8*Sqrt[1 + (d*x^4)/c]*AppellF1[5/4, 1/2, 1, 9/4, -((d*x^4)/c), -((b*x^4)/a)] + (a*(25*a
*c*(-7*b^2*c*x^4*(4*c + d*x^4) + 4*a^2*d*(c + 2*d*x^4) + 4*a*b*(-c^2 + 5*c*d*x^4 + d^2*x^8))*AppellF1[1/4, 1/2
, 1, 5/4, -((d*x^4)/c), -((b*x^4)/a)] + 10*x^4*(c + d*x^4)*(-4*a^2*d + 7*b^2*c*x^4 + 4*a*b*(c - d*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)])))/((a + b*x^4)*(-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)]))))
/(60*a^3*c*(-(b*c) + a*d)*x^3*Sqrt[c + d*x^4])

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Maple [C]  time = 0.013, size = 626, 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(1/x^4/(b*x^4+a)^2/(d*x^4+c)^(1/2),x)

[Out]

-1/8/a^2*sum(1/_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))-b/a*(-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)*Elli
pticPi(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)))+1/a^2*(-1/3/c*(d*x^4+c)^(1/2)/x^3-1/3*d/c/(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))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{4} + a\right )}^{2} \sqrt{d x^{4} + c} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{4} + a\right )}^{2} \sqrt{d x^{4} + c} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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