3.182 \(\int \frac {1}{(a+b x^4)^{3/2} (c+d x^4)} \, dx\)

Optimal. Leaf size=913 \[ -\frac {d \left (\sqrt {b} x^2+\sqrt {a}\right ) \sqrt {\frac {b x^4+a}{\left (\sqrt {b} x^2+\sqrt {a}\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]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right ) \left (\sqrt {b} \sqrt {-c}-\sqrt {a} \sqrt {d}\right )^2}{8 \sqrt [4]{a} \sqrt [4]{b} c (b c-a d) (b c+a d) \sqrt {b x^4+a}}+\frac {d^{5/4} \tan ^{-1}\left (\frac {\sqrt {b c-a d} x}{\sqrt [4]{-c} \sqrt [4]{d} \sqrt {b x^4+a}}\right )}{4 (-c)^{3/4} (b c-a d)^{3/2}}-\frac {d^{5/4} \tan ^{-1}\left (\frac {\sqrt {a d-b c} x}{\sqrt [4]{-c} \sqrt [4]{d} \sqrt {b x^4+a}}\right )}{4 (-c)^{3/4} (a d-b c)^{3/2}}+\frac {b^{3/4} \left (\sqrt {b} x^2+\sqrt {a}\right ) \sqrt {\frac {b x^4+a}{\left (\sqrt {b} x^2+\sqrt {a}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 a^{5/4} (b c-a d) \sqrt {b x^4+a}}-\frac {\sqrt [4]{b} \left (\sqrt {b}+\frac {\sqrt {a} \sqrt {d}}{\sqrt {-c}}\right ) d \left (\sqrt {b} x^2+\sqrt {a}\right ) \sqrt {\frac {b x^4+a}{\left (\sqrt {b} x^2+\sqrt {a}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 \sqrt [4]{a} (b c-a d) (b c+a d) \sqrt {b x^4+a}}-\frac {\sqrt [4]{b} \left (\sqrt {b} c+\sqrt {a} \sqrt {-c} \sqrt {d}\right ) d \left (\sqrt {b} x^2+\sqrt {a}\right ) \sqrt {\frac {b x^4+a}{\left (\sqrt {b} x^2+\sqrt {a}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 \sqrt [4]{a} c \left (b^2 c^2-a^2 d^2\right ) \sqrt {b x^4+a}}-\frac {\left (\sqrt {b} \sqrt {-c}+\sqrt {a} \sqrt {d}\right )^2 d \left (\sqrt {b} x^2+\sqrt {a}\right ) \sqrt {\frac {b x^4+a}{\left (\sqrt {b} x^2+\sqrt {a}\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]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{8 \sqrt [4]{a} \sqrt [4]{b} c (b c-a d) (b c+a d) \sqrt {b x^4+a}}+\frac {b x}{2 a (b c-a d) \sqrt {b x^4+a}} \]

[Out]

1/4*d^(5/4)*arctan(x*(-a*d+b*c)^(1/2)/(-c)^(1/4)/d^(1/4)/(b*x^4+a)^(1/2))/(-c)^(3/4)/(-a*d+b*c)^(3/2)-1/4*d^(5
/4)*arctan(x*(a*d-b*c)^(1/2)/(-c)^(1/4)/d^(1/4)/(b*x^4+a)^(1/2))/(-c)^(3/4)/(a*d-b*c)^(3/2)+1/2*b*x/a/(-a*d+b*
c)/(b*x^4+a)^(1/2)+1/4*b^(3/4)*(cos(2*arctan(b^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1/4)*x/a^(1/4)))*Ell
ipticF(sin(2*arctan(b^(1/4)*x/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+x^2*b^(1/2))*((b*x^4+a)/(a^(1/2)+x^2*b^(1/2))^2)
^(1/2)/a^(5/4)/(-a*d+b*c)/(b*x^4+a)^(1/2)-1/8*d*(cos(2*arctan(b^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1/4
)*x/a^(1/4)))*EllipticPi(sin(2*arctan(b^(1/4)*x/a^(1/4))),1/4*(b^(1/2)*(-c)^(1/2)+a^(1/2)*d^(1/2))^2/a^(1/2)/b
^(1/2)/(-c)^(1/2)/d^(1/2),1/2*2^(1/2))*(a^(1/2)+x^2*b^(1/2))*(b^(1/2)*(-c)^(1/2)-a^(1/2)*d^(1/2))^2*((b*x^4+a)
/(a^(1/2)+x^2*b^(1/2))^2)^(1/2)/a^(1/4)/b^(1/4)/c/(-a^2*d^2+b^2*c^2)/(b*x^4+a)^(1/2)-1/8*d*(cos(2*arctan(b^(1/
4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1/4)*x/a^(1/4)))*EllipticPi(sin(2*arctan(b^(1/4)*x/a^(1/4))),-1/4*(b^(
1/2)*(-c)^(1/2)-a^(1/2)*d^(1/2))^2/a^(1/2)/b^(1/2)/(-c)^(1/2)/d^(1/2),1/2*2^(1/2))*(a^(1/2)+x^2*b^(1/2))*(b^(1
/2)*(-c)^(1/2)+a^(1/2)*d^(1/2))^2*((b*x^4+a)/(a^(1/2)+x^2*b^(1/2))^2)^(1/2)/a^(1/4)/b^(1/4)/c/(-a^2*d^2+b^2*c^
2)/(b*x^4+a)^(1/2)-1/4*b^(1/4)*d*(cos(2*arctan(b^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1/4)*x/a^(1/4)))*E
llipticF(sin(2*arctan(b^(1/4)*x/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+x^2*b^(1/2))*(b^(1/2)+a^(1/2)*d^(1/2)/(-c)^(1/
2))*((b*x^4+a)/(a^(1/2)+x^2*b^(1/2))^2)^(1/2)/a^(1/4)/(-a*d+b*c)/(a*d+b*c)/(b*x^4+a)^(1/2)-1/4*b^(1/4)*d*(cos(
2*arctan(b^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1/4)*x/a^(1/4)))*EllipticF(sin(2*arctan(b^(1/4)*x/a^(1/4
))),1/2*2^(1/2))*(a^(1/2)+x^2*b^(1/2))*(c*b^(1/2)+a^(1/2)*(-c)^(1/2)*d^(1/2))*((b*x^4+a)/(a^(1/2)+x^2*b^(1/2))
^2)^(1/2)/a^(1/4)/c/(-a^2*d^2+b^2*c^2)/(b*x^4+a)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 1.13, antiderivative size = 913, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {414, 523, 220, 409, 1217, 1707} \[ -\frac {d \left (\sqrt {b} x^2+\sqrt {a}\right ) \sqrt {\frac {b x^4+a}{\left (\sqrt {b} x^2+\sqrt {a}\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]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right ) \left (\sqrt {b} \sqrt {-c}-\sqrt {a} \sqrt {d}\right )^2}{8 \sqrt [4]{a} \sqrt [4]{b} c (b c-a d) (b c+a d) \sqrt {b x^4+a}}+\frac {d^{5/4} \tan ^{-1}\left (\frac {\sqrt {b c-a d} x}{\sqrt [4]{-c} \sqrt [4]{d} \sqrt {b x^4+a}}\right )}{4 (-c)^{3/4} (b c-a d)^{3/2}}-\frac {d^{5/4} \tan ^{-1}\left (\frac {\sqrt {a d-b c} x}{\sqrt [4]{-c} \sqrt [4]{d} \sqrt {b x^4+a}}\right )}{4 (-c)^{3/4} (a d-b c)^{3/2}}+\frac {b^{3/4} \left (\sqrt {b} x^2+\sqrt {a}\right ) \sqrt {\frac {b x^4+a}{\left (\sqrt {b} x^2+\sqrt {a}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 a^{5/4} (b c-a d) \sqrt {b x^4+a}}-\frac {\sqrt [4]{b} \left (\sqrt {b}+\frac {\sqrt {a} \sqrt {d}}{\sqrt {-c}}\right ) d \left (\sqrt {b} x^2+\sqrt {a}\right ) \sqrt {\frac {b x^4+a}{\left (\sqrt {b} x^2+\sqrt {a}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 \sqrt [4]{a} (b c-a d) (b c+a d) \sqrt {b x^4+a}}-\frac {\sqrt [4]{b} \left (\sqrt {b} c+\sqrt {a} \sqrt {-c} \sqrt {d}\right ) d \left (\sqrt {b} x^2+\sqrt {a}\right ) \sqrt {\frac {b x^4+a}{\left (\sqrt {b} x^2+\sqrt {a}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 \sqrt [4]{a} c \left (b^2 c^2-a^2 d^2\right ) \sqrt {b x^4+a}}-\frac {\left (\sqrt {b} \sqrt {-c}+\sqrt {a} \sqrt {d}\right )^2 d \left (\sqrt {b} x^2+\sqrt {a}\right ) \sqrt {\frac {b x^4+a}{\left (\sqrt {b} x^2+\sqrt {a}\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]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{8 \sqrt [4]{a} \sqrt [4]{b} c (b c-a d) (b c+a d) \sqrt {b x^4+a}}+\frac {b x}{2 a (b c-a d) \sqrt {b x^4+a}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x^4)^(3/2)*(c + d*x^4)),x]

[Out]

(b*x)/(2*a*(b*c - a*d)*Sqrt[a + b*x^4]) + (d^(5/4)*ArcTan[(Sqrt[b*c - a*d]*x)/((-c)^(1/4)*d^(1/4)*Sqrt[a + b*x
^4])])/(4*(-c)^(3/4)*(b*c - a*d)^(3/2)) - (d^(5/4)*ArcTan[(Sqrt[-(b*c) + a*d]*x)/((-c)^(1/4)*d^(1/4)*Sqrt[a +
b*x^4])])/(4*(-c)^(3/4)*(-(b*c) + a*d)^(3/2)) + (b^(3/4)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + S
qrt[b]*x^2)^2]*EllipticF[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(4*a^(5/4)*(b*c - a*d)*Sqrt[a + b*x^4]) - (b^(1/
4)*(Sqrt[b] + (Sqrt[a]*Sqrt[d])/Sqrt[-c])*d*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2
]*EllipticF[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(4*a^(1/4)*(b*c - a*d)*(b*c + a*d)*Sqrt[a + b*x^4]) - (b^(1/4
)*(Sqrt[b]*c + Sqrt[a]*Sqrt[-c]*Sqrt[d])*d*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]
*EllipticF[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(4*a^(1/4)*c*(b^2*c^2 - a^2*d^2)*Sqrt[a + b*x^4]) - ((Sqrt[b]*
Sqrt[-c] + Sqrt[a]*Sqrt[d])^2*d*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticPi
[-(Sqrt[b]*Sqrt[-c] - Sqrt[a]*Sqrt[d])^2/(4*Sqrt[a]*Sqrt[b]*Sqrt[-c]*Sqrt[d]), 2*ArcTan[(b^(1/4)*x)/a^(1/4)],
1/2])/(8*a^(1/4)*b^(1/4)*c*(b*c - a*d)*(b*c + a*d)*Sqrt[a + b*x^4]) - ((Sqrt[b]*Sqrt[-c] - Sqrt[a]*Sqrt[d])^2*
d*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticPi[(Sqrt[b]*Sqrt[-c] + Sqrt[a]*S
qrt[d])^2/(4*Sqrt[a]*Sqrt[b]*Sqrt[-c]*Sqrt[d]), 2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(8*a^(1/4)*b^(1/4)*c*(b*c
- a*d)*(b*c + a*d)*Sqrt[a + b*x^4])

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

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Mathematica [C]  time = 0.28, size = 331, normalized size = 0.36 \[ \frac {x \left (\frac {5 \left (2 b x^4 \left (c+d x^4\right ) \left (2 a d F_1\left (\frac {5}{4};\frac {1}{2},2;\frac {9}{4};-\frac {b x^4}{a},-\frac {d x^4}{c}\right )+b c F_1\left (\frac {5}{4};\frac {3}{2},1;\frac {9}{4};-\frac {b x^4}{a},-\frac {d x^4}{c}\right )\right )+5 a c \left (2 a d-b \left (2 c+d x^4\right )\right ) F_1\left (\frac {1}{4};\frac {1}{2},1;\frac {5}{4};-\frac {b x^4}{a},-\frac {d x^4}{c}\right )\right )}{\left (c+d x^4\right ) \left (5 a c F_1\left (\frac {1}{4};\frac {1}{2},1;\frac {5}{4};-\frac {b x^4}{a},-\frac {d x^4}{c}\right )-2 x^4 \left (2 a d F_1\left (\frac {5}{4};\frac {1}{2},2;\frac {9}{4};-\frac {b x^4}{a},-\frac {d x^4}{c}\right )+b c F_1\left (\frac {5}{4};\frac {3}{2},1;\frac {9}{4};-\frac {b x^4}{a},-\frac {d x^4}{c}\right )\right )\right )}-\frac {b d x^4 \sqrt {\frac {b x^4}{a}+1} F_1\left (\frac {5}{4};\frac {1}{2},1;\frac {9}{4};-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{c}\right )}{10 a \sqrt {a+b x^4} (a d-b c)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((a + b*x^4)^(3/2)*(c + d*x^4)),x]

[Out]

(x*(-((b*d*x^4*Sqrt[1 + (b*x^4)/a]*AppellF1[5/4, 1/2, 1, 9/4, -((b*x^4)/a), -((d*x^4)/c)])/c) + (5*(5*a*c*(2*a
*d - b*(2*c + d*x^4))*AppellF1[1/4, 1/2, 1, 5/4, -((b*x^4)/a), -((d*x^4)/c)] + 2*b*x^4*(c + d*x^4)*(2*a*d*Appe
llF1[5/4, 1/2, 2, 9/4, -((b*x^4)/a), -((d*x^4)/c)] + b*c*AppellF1[5/4, 3/2, 1, 9/4, -((b*x^4)/a), -((d*x^4)/c)
])))/((c + d*x^4)*(5*a*c*AppellF1[1/4, 1/2, 1, 5/4, -((b*x^4)/a), -((d*x^4)/c)] - 2*x^4*(2*a*d*AppellF1[5/4, 1
/2, 2, 9/4, -((b*x^4)/a), -((d*x^4)/c)] + b*c*AppellF1[5/4, 3/2, 1, 9/4, -((b*x^4)/a), -((d*x^4)/c)])))))/(10*
a*(-(b*c) + a*d)*Sqrt[a + b*x^4])

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^4+a)^(3/2)/(d*x^4+c),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{4} + a\right )}^{\frac {3}{2}} {\left (d x^{4} + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^4+a)^(3/2)/(d*x^4+c),x, algorithm="giac")

[Out]

integrate(1/((b*x^4 + a)^(3/2)*(d*x^4 + c)), x)

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maple [C]  time = 0.29, size = 313, normalized size = 0.34 \[ -\frac {b x}{2 \left (a d -b c \right ) \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}\, a}-\frac {\sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, b \EllipticF \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{2 \left (a d -b c \right ) \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, a}+\frac {\frac {2 \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \RootOf \left (d \,\textit {\_Z}^{4}+c \right )^{3} d \EllipticPi \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , \frac {i \RootOf \left (d \,\textit {\_Z}^{4}+c \right )^{2} \sqrt {a}\, d}{\sqrt {b}\, c}, \frac {\sqrt {-\frac {i \sqrt {b}}{\sqrt {a}}}}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}}\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, c}-\frac {\arctanh \left (\frac {2 \RootOf \left (d \,\textit {\_Z}^{4}+c \right )^{2} b \,x^{2}+2 a}{2 \sqrt {\frac {a d -b c}{d}}\, \sqrt {b \,x^{4}+a}}\right )}{\sqrt {\frac {a d -b c}{d}}}}{8 \left (a d -b c \right ) \RootOf \left (d \,\textit {\_Z}^{4}+c \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x^4+a)^(3/2)/(d*x^4+c),x)

[Out]

-1/2*b/a*x/(a*d-b*c)/((x^4+a/b)*b)^(1/2)-1/2/a*b/(a*d-b*c)/(I/a^(1/2)*b^(1/2))^(1/2)*(-I/a^(1/2)*b^(1/2)*x^2+1
)^(1/2)*(I/a^(1/2)*b^(1/2)*x^2+1)^(1/2)/(b*x^4+a)^(1/2)*EllipticF((I/a^(1/2)*b^(1/2))^(1/2)*x,I)+1/8*sum(1/(a*
d-b*c)/_alpha^3*(-1/((a*d-b*c)/d)^(1/2)*arctanh(1/2*(2*_alpha^2*b*x^2+2*a)/((a*d-b*c)/d)^(1/2)/(b*x^4+a)^(1/2)
)+2/(I/a^(1/2)*b^(1/2))^(1/2)*_alpha^3*d/c*(-I/a^(1/2)*b^(1/2)*x^2+1)^(1/2)*(I/a^(1/2)*b^(1/2)*x^2+1)^(1/2)/(b
*x^4+a)^(1/2)*EllipticPi((I/a^(1/2)*b^(1/2))^(1/2)*x,I*_alpha^2*a^(1/2)/b^(1/2)/c*d,(-I/a^(1/2)*b^(1/2))^(1/2)
/(I/a^(1/2)*b^(1/2))^(1/2))),_alpha=RootOf(_Z^4*d+c))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{4} + a\right )}^{\frac {3}{2}} {\left (d x^{4} + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^4+a)^(3/2)/(d*x^4+c),x, algorithm="maxima")

[Out]

integrate(1/((b*x^4 + a)^(3/2)*(d*x^4 + c)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\left (b\,x^4+a\right )}^{3/2}\,\left (d\,x^4+c\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b*x^4)^(3/2)*(c + d*x^4)),x)

[Out]

int(1/((a + b*x^4)^(3/2)*(c + d*x^4)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b x^{4}\right )^{\frac {3}{2}} \left (c + d x^{4}\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x**4+a)**(3/2)/(d*x**4+c),x)

[Out]

Integral(1/((a + b*x**4)**(3/2)*(c + d*x**4)), x)

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