3.12 \(\int \frac {A+B x^2}{(d+e x^2) (a+c x^4)^{3/2}} \, dx\)

Optimal. Leaf size=732 \[ \frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right ) \left (-\sqrt {a} \sqrt {c} (B d-A e)+a B e+A c d\right )}{4 a^{5/4} \sqrt [4]{c} \sqrt {a+c x^4} \left (a e^2+c d^2\right )}+\frac {\sqrt [4]{c} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} (B d-A e) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 a^{3/4} \sqrt {a+c x^4} \left (a e^2+c d^2\right )}+\frac {a^{3/4} e \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (\frac {\sqrt {c} d}{\sqrt {a}}+e\right )^2 (B d-A e) \Pi \left (-\frac {\left (\sqrt {c} d-\sqrt {a} e\right )^2}{4 \sqrt {a} \sqrt {c} d e};2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 \sqrt [4]{c} d \sqrt {a+c x^4} \left (c^2 d^4-a^2 e^4\right )}-\frac {\sqrt {c} x \sqrt {a+c x^4} (B d-A e)}{2 a \left (\sqrt {a}+\sqrt {c} x^2\right ) \left (a e^2+c d^2\right )}+\frac {x \left (a B e+c x^2 (B d-A e)+A c d\right )}{2 a \sqrt {a+c x^4} \left (a e^2+c d^2\right )}-\frac {\sqrt [4]{c} e \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} (B d-A e) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt {a+c x^4} \left (\sqrt {c} d-\sqrt {a} e\right ) \left (a e^2+c d^2\right )}-\frac {e^{3/2} (B d-A e) \tan ^{-1}\left (\frac {x \sqrt {a e^2+c d^2}}{\sqrt {d} \sqrt {e} \sqrt {a+c x^4}}\right )}{2 \sqrt {d} \left (a e^2+c d^2\right )^{3/2}} \]

[Out]

-1/2*e^(3/2)*(-A*e+B*d)*arctan(x*(a*e^2+c*d^2)^(1/2)/d^(1/2)/e^(1/2)/(c*x^4+a)^(1/2))/(a*e^2+c*d^2)^(3/2)/d^(1
/2)+1/2*x*(A*c*d+a*B*e+c*(-A*e+B*d)*x^2)/a/(a*e^2+c*d^2)/(c*x^4+a)^(1/2)-1/2*(-A*e+B*d)*x*c^(1/2)*(c*x^4+a)^(1
/2)/a/(a*e^2+c*d^2)/(a^(1/2)+x^2*c^(1/2))+1/2*c^(1/4)*(-A*e+B*d)*(cos(2*arctan(c^(1/4)*x/a^(1/4)))^2)^(1/2)/co
s(2*arctan(c^(1/4)*x/a^(1/4)))*EllipticE(sin(2*arctan(c^(1/4)*x/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+x^2*c^(1/2))*(
(c*x^4+a)/(a^(1/2)+x^2*c^(1/2))^2)^(1/2)/a^(3/4)/(a*e^2+c*d^2)/(c*x^4+a)^(1/2)-1/2*c^(1/4)*e*(-A*e+B*d)*(cos(2
*arctan(c^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x/a^(1/4)))*EllipticF(sin(2*arctan(c^(1/4)*x/a^(1/4)
)),1/2*2^(1/2))*(a^(1/2)+x^2*c^(1/2))*((c*x^4+a)/(a^(1/2)+x^2*c^(1/2))^2)^(1/2)/a^(1/4)/(a*e^2+c*d^2)/(-e*a^(1
/2)+d*c^(1/2))/(c*x^4+a)^(1/2)+1/4*a^(3/4)*e*(-A*e+B*d)*(cos(2*arctan(c^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arcta
n(c^(1/4)*x/a^(1/4)))*EllipticPi(sin(2*arctan(c^(1/4)*x/a^(1/4))),-1/4*(-e*a^(1/2)+d*c^(1/2))^2/d/e/a^(1/2)/c^
(1/2),1/2*2^(1/2))*(a^(1/2)+x^2*c^(1/2))*(e+d*c^(1/2)/a^(1/2))^2*((c*x^4+a)/(a^(1/2)+x^2*c^(1/2))^2)^(1/2)/c^(
1/4)/d/(-a^2*e^4+c^2*d^4)/(c*x^4+a)^(1/2)+1/4*(cos(2*arctan(c^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*
x/a^(1/4)))*EllipticF(sin(2*arctan(c^(1/4)*x/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+x^2*c^(1/2))*(A*c*d+a*B*e-(-A*e+B
*d)*a^(1/2)*c^(1/2))*((c*x^4+a)/(a^(1/2)+x^2*c^(1/2))^2)^(1/2)/a^(5/4)/c^(1/4)/(a*e^2+c*d^2)/(c*x^4+a)^(1/2)

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Rubi [A]  time = 0.78, antiderivative size = 732, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1721, 1179, 1198, 220, 1196, 1217, 1707} \[ \frac {a^{3/4} e \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (\frac {\sqrt {c} d}{\sqrt {a}}+e\right )^2 (B d-A e) \Pi \left (-\frac {\left (\sqrt {c} d-\sqrt {a} e\right )^2}{4 \sqrt {a} \sqrt {c} d e};2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 \sqrt [4]{c} d \sqrt {a+c x^4} \left (c^2 d^4-a^2 e^4\right )}+\frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right ) \left (-\sqrt {a} \sqrt {c} (B d-A e)+a B e+A c d\right )}{4 a^{5/4} \sqrt [4]{c} \sqrt {a+c x^4} \left (a e^2+c d^2\right )}+\frac {\sqrt [4]{c} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} (B d-A e) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 a^{3/4} \sqrt {a+c x^4} \left (a e^2+c d^2\right )}-\frac {\sqrt {c} x \sqrt {a+c x^4} (B d-A e)}{2 a \left (\sqrt {a}+\sqrt {c} x^2\right ) \left (a e^2+c d^2\right )}+\frac {x \left (a B e+c x^2 (B d-A e)+A c d\right )}{2 a \sqrt {a+c x^4} \left (a e^2+c d^2\right )}-\frac {e^{3/2} (B d-A e) \tan ^{-1}\left (\frac {x \sqrt {a e^2+c d^2}}{\sqrt {d} \sqrt {e} \sqrt {a+c x^4}}\right )}{2 \sqrt {d} \left (a e^2+c d^2\right )^{3/2}}-\frac {\sqrt [4]{c} e \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} (B d-A e) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt {a+c x^4} \left (\sqrt {c} d-\sqrt {a} e\right ) \left (a e^2+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/((d + e*x^2)*(a + c*x^4)^(3/2)),x]

[Out]

(x*(A*c*d + a*B*e + c*(B*d - A*e)*x^2))/(2*a*(c*d^2 + a*e^2)*Sqrt[a + c*x^4]) - (Sqrt[c]*(B*d - A*e)*x*Sqrt[a
+ c*x^4])/(2*a*(c*d^2 + a*e^2)*(Sqrt[a] + Sqrt[c]*x^2)) - (e^(3/2)*(B*d - A*e)*ArcTan[(Sqrt[c*d^2 + a*e^2]*x)/
(Sqrt[d]*Sqrt[e]*Sqrt[a + c*x^4])])/(2*Sqrt[d]*(c*d^2 + a*e^2)^(3/2)) + (c^(1/4)*(B*d - A*e)*(Sqrt[a] + Sqrt[c
]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticE[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(2*a^(3/4)*(
c*d^2 + a*e^2)*Sqrt[a + c*x^4]) - (c^(1/4)*e*(B*d - A*e)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + S
qrt[c]*x^2)^2]*EllipticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(2*a^(1/4)*(Sqrt[c]*d - Sqrt[a]*e)*(c*d^2 + a*e^
2)*Sqrt[a + c*x^4]) + ((A*c*d + a*B*e - Sqrt[a]*Sqrt[c]*(B*d - A*e))*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/
(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(4*a^(5/4)*c^(1/4)*(c*d^2 + a*e^2)*S
qrt[a + c*x^4]) + (a^(3/4)*e*((Sqrt[c]*d)/Sqrt[a] + e)^2*(B*d - A*e)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/
(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticPi[-(Sqrt[c]*d - Sqrt[a]*e)^2/(4*Sqrt[a]*Sqrt[c]*d*e), 2*ArcTan[(c^(1/4)*x)
/a^(1/4)], 1/2])/(4*c^(1/4)*d*(c^2*d^4 - a^2*e^4)*Sqrt[a + c*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 1179

Int[((d_) + (e_.)*(x_)^2)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> -Simp[(x*(d + e*x^2)*(a + c*x^4)^(p + 1))/(
4*a*(p + 1)), x] + Dist[1/(4*a*(p + 1)), Int[Simp[d*(4*p + 5) + e*(4*p + 7)*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] && LtQ[p, -1] && IntegerQ[2*p]

Rule 1196

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

Rule 1198

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

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]

Rule 1721

Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Int[ExpandIntegrand[1/Sqrt[a +
c*x^4], Px*(d + e*x^2)^q*(a + c*x^4)^(p + 1/2), x], x] /; FreeQ[{a, c, d, e}, x] && PolyQ[Px, x^2] && NeQ[c*d^
2 + a*e^2, 0] && IntegerQ[p + 1/2] && IntegerQ[q]

Rubi steps

\begin {align*} \int \frac {A+B x^2}{\left (d+e x^2\right ) \left (a+c x^4\right )^{3/2}} \, dx &=\int \left (\frac {A c d+a B e+c (B d-A e) x^2}{\left (c d^2+a e^2\right ) \left (a+c x^4\right )^{3/2}}+\frac {e (-B d+A e)}{\left (c d^2+a e^2\right ) \left (d+e x^2\right ) \sqrt {a+c x^4}}\right ) \, dx\\ &=\frac {\int \frac {A c d+a B e+c (B d-A e) x^2}{\left (a+c x^4\right )^{3/2}} \, dx}{c d^2+a e^2}-\frac {(e (B d-A e)) \int \frac {1}{\left (d+e x^2\right ) \sqrt {a+c x^4}} \, dx}{c d^2+a e^2}\\ &=\frac {x \left (A c d+a B e+c (B d-A e) x^2\right )}{2 a \left (c d^2+a e^2\right ) \sqrt {a+c x^4}}-\frac {\int \frac {-A c d-a B e+c (B d-A e) x^2}{\sqrt {a+c x^4}} \, dx}{2 a \left (c d^2+a e^2\right )}-\frac {\left (\sqrt {c} e (B d-A e)\right ) \int \frac {1}{\sqrt {a+c x^4}} \, dx}{\left (\sqrt {c} d-\sqrt {a} e\right ) \left (c d^2+a e^2\right )}+\frac {\left (\sqrt {a} e^2 (B d-A e)\right ) \int \frac {1+\frac {\sqrt {c} x^2}{\sqrt {a}}}{\left (d+e x^2\right ) \sqrt {a+c x^4}} \, dx}{\left (\sqrt {c} d-\sqrt {a} e\right ) \left (c d^2+a e^2\right )}\\ &=\frac {x \left (A c d+a B e+c (B d-A e) x^2\right )}{2 a \left (c d^2+a e^2\right ) \sqrt {a+c x^4}}-\frac {e^{3/2} (B d-A e) \tan ^{-1}\left (\frac {\sqrt {c d^2+a e^2} x}{\sqrt {d} \sqrt {e} \sqrt {a+c x^4}}\right )}{2 \sqrt {d} \left (c d^2+a e^2\right )^{3/2}}-\frac {\sqrt [4]{c} e (B d-A e) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 \sqrt [4]{a} \left (\sqrt {c} d-\sqrt {a} e\right ) \left (c d^2+a e^2\right ) \sqrt {a+c x^4}}+\frac {\sqrt [4]{a} e \left (\frac {\sqrt {c} d}{\sqrt {a}}+e\right ) (B d-A e) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \Pi \left (-\frac {\left (\sqrt {c} d-\sqrt {a} e\right )^2}{4 \sqrt {a} \sqrt {c} d e};2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 \sqrt [4]{c} d \left (\sqrt {c} d-\sqrt {a} e\right ) \left (c d^2+a e^2\right ) \sqrt {a+c x^4}}+\frac {\left (\sqrt {c} (B d-A e)\right ) \int \frac {1-\frac {\sqrt {c} x^2}{\sqrt {a}}}{\sqrt {a+c x^4}} \, dx}{2 \sqrt {a} \left (c d^2+a e^2\right )}+\frac {\left (A c d+a B e-\sqrt {a} \sqrt {c} (B d-A e)\right ) \int \frac {1}{\sqrt {a+c x^4}} \, dx}{2 a \left (c d^2+a e^2\right )}\\ &=\frac {x \left (A c d+a B e+c (B d-A e) x^2\right )}{2 a \left (c d^2+a e^2\right ) \sqrt {a+c x^4}}-\frac {\sqrt {c} (B d-A e) x \sqrt {a+c x^4}}{2 a \left (c d^2+a e^2\right ) \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {e^{3/2} (B d-A e) \tan ^{-1}\left (\frac {\sqrt {c d^2+a e^2} x}{\sqrt {d} \sqrt {e} \sqrt {a+c x^4}}\right )}{2 \sqrt {d} \left (c d^2+a e^2\right )^{3/2}}+\frac {\sqrt [4]{c} (B d-A e) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 a^{3/4} \left (c d^2+a e^2\right ) \sqrt {a+c x^4}}-\frac {\sqrt [4]{c} e (B d-A e) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 \sqrt [4]{a} \left (\sqrt {c} d-\sqrt {a} e\right ) \left (c d^2+a e^2\right ) \sqrt {a+c x^4}}+\frac {\left (A c d+a B e-\sqrt {a} \sqrt {c} (B d-A e)\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 a^{5/4} \sqrt [4]{c} \left (c d^2+a e^2\right ) \sqrt {a+c x^4}}+\frac {\sqrt [4]{a} e \left (\frac {\sqrt {c} d}{\sqrt {a}}+e\right ) (B d-A e) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \Pi \left (-\frac {\left (\sqrt {c} d-\sqrt {a} e\right )^2}{4 \sqrt {a} \sqrt {c} d e};2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 \sqrt [4]{c} d \left (\sqrt {c} d-\sqrt {a} e\right ) \left (c d^2+a e^2\right ) \sqrt {a+c x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.79, size = 432, normalized size = 0.59 \[ \frac {d \sqrt {\frac {c x^4}{a}+1} \left (\sqrt {a} B-i A \sqrt {c}\right ) \left (\sqrt {c} d-i \sqrt {a} e\right ) F\left (\left .i \sinh ^{-1}\left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}} x\right )\right |-1\right )-\sqrt {a} \sqrt {c} d \sqrt {\frac {c x^4}{a}+1} (B d-A e) E\left (\left .i \sinh ^{-1}\left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}} x\right )\right |-1\right )+A c d^2 x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}-2 i a A e^2 \sqrt {\frac {c x^4}{a}+1} \Pi \left (-\frac {i \sqrt {a} e}{\sqrt {c} d};\left .i \sinh ^{-1}\left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}} x\right )\right |-1\right )-A c d e x^3 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}+B c d^2 x^3 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}+2 i a B d e \sqrt {\frac {c x^4}{a}+1} \Pi \left (-\frac {i \sqrt {a} e}{\sqrt {c} d};\left .i \sinh ^{-1}\left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}} x\right )\right |-1\right )+a B d e x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}}{2 a d \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}} \sqrt {a+c x^4} \left (a e^2+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^2)/((d + e*x^2)*(a + c*x^4)^(3/2)),x]

[Out]

(A*Sqrt[(I*Sqrt[c])/Sqrt[a]]*c*d^2*x + a*B*Sqrt[(I*Sqrt[c])/Sqrt[a]]*d*e*x + B*Sqrt[(I*Sqrt[c])/Sqrt[a]]*c*d^2
*x^3 - A*Sqrt[(I*Sqrt[c])/Sqrt[a]]*c*d*e*x^3 - Sqrt[a]*Sqrt[c]*d*(B*d - A*e)*Sqrt[1 + (c*x^4)/a]*EllipticE[I*A
rcSinh[Sqrt[(I*Sqrt[c])/Sqrt[a]]*x], -1] + (Sqrt[a]*B - I*A*Sqrt[c])*d*(Sqrt[c]*d - I*Sqrt[a]*e)*Sqrt[1 + (c*x
^4)/a]*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[c])/Sqrt[a]]*x], -1] + (2*I)*a*B*d*e*Sqrt[1 + (c*x^4)/a]*EllipticPi[((
-I)*Sqrt[a]*e)/(Sqrt[c]*d), I*ArcSinh[Sqrt[(I*Sqrt[c])/Sqrt[a]]*x], -1] - (2*I)*a*A*e^2*Sqrt[1 + (c*x^4)/a]*El
lipticPi[((-I)*Sqrt[a]*e)/(Sqrt[c]*d), I*ArcSinh[Sqrt[(I*Sqrt[c])/Sqrt[a]]*x], -1])/(2*a*Sqrt[(I*Sqrt[c])/Sqrt
[a]]*d*(c*d^2 + a*e^2)*Sqrt[a + c*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((B*x^2+A)/(e*x^2+d)/(c*x^4+a)^(3/2),x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/(e*x^2+d)/(c*x^4+a)^(3/2),x, algorithm="giac")

[Out]

integrate((B*x^2 + A)/((c*x^4 + a)^(3/2)*(e*x^2 + d)), x)

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maple [C]  time = 0.03, size = 564, normalized size = 0.77 \[ \frac {\left (\frac {x}{2 \sqrt {\left (x^{4}+\frac {a}{c}\right ) c}\, a}+\frac {\sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \EllipticF \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )}{2 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, a}\right ) B}{e}+\frac {\left (A e -B d \right ) \left (-\frac {i \sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {c}\, e \EllipticE \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )}{2 \left (a \,e^{2}+c \,d^{2}\right ) \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, \sqrt {a}}+\frac {i \sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {c}\, e \EllipticF \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )}{2 \left (a \,e^{2}+c \,d^{2}\right ) \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, \sqrt {a}}+\frac {\sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, c d \EllipticF \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )}{2 \left (a \,e^{2}+c \,d^{2}\right ) \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, a}+\frac {\sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, e^{2} \EllipticPi \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , \frac {i \sqrt {a}\, e}{\sqrt {c}\, d}, \frac {\sqrt {-\frac {i \sqrt {c}}{\sqrt {a}}}}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}}\right )}{\left (a \,e^{2}+c \,d^{2}\right ) \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, d}-\frac {2 \left (\frac {e \,x^{3}}{4 \left (a \,e^{2}+c \,d^{2}\right ) a}-\frac {d x}{4 \left (a \,e^{2}+c \,d^{2}\right ) a}\right ) c}{\sqrt {\left (x^{4}+\frac {a}{c}\right ) c}}\right )}{e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)/(e*x^2+d)/(c*x^4+a)^(3/2),x)

[Out]

B/e*(1/2/((x^4+a/c)*c)^(1/2)/a*x+1/2/a/(I/a^(1/2)*c^(1/2))^(1/2)*(-I/a^(1/2)*c^(1/2)*x^2+1)^(1/2)*(I/a^(1/2)*c
^(1/2)*x^2+1)^(1/2)/(c*x^4+a)^(1/2)*EllipticF((I/a^(1/2)*c^(1/2))^(1/2)*x,I))+(A*e-B*d)/e*(-2*c*(1/4/a*e/(a*e^
2+c*d^2)*x^3-1/4/a*d/(a*e^2+c*d^2)*x)/((x^4+a/c)*c)^(1/2)+1/2/a*c*d/(a*e^2+c*d^2)/(I/a^(1/2)*c^(1/2))^(1/2)*(-
I/a^(1/2)*c^(1/2)*x^2+1)^(1/2)*(I/a^(1/2)*c^(1/2)*x^2+1)^(1/2)/(c*x^4+a)^(1/2)*EllipticF((I/a^(1/2)*c^(1/2))^(
1/2)*x,I)+1/2*I/a^(1/2)*c^(1/2)*e/(a*e^2+c*d^2)/(I/a^(1/2)*c^(1/2))^(1/2)*(-I/a^(1/2)*c^(1/2)*x^2+1)^(1/2)*(I/
a^(1/2)*c^(1/2)*x^2+1)^(1/2)/(c*x^4+a)^(1/2)*EllipticF((I/a^(1/2)*c^(1/2))^(1/2)*x,I)-1/2*I/a^(1/2)*c^(1/2)*e/
(a*e^2+c*d^2)/(I/a^(1/2)*c^(1/2))^(1/2)*(-I/a^(1/2)*c^(1/2)*x^2+1)^(1/2)*(I/a^(1/2)*c^(1/2)*x^2+1)^(1/2)/(c*x^
4+a)^(1/2)*EllipticE((I/a^(1/2)*c^(1/2))^(1/2)*x,I)+1/(a*e^2+c*d^2)*e^2/d/(I/a^(1/2)*c^(1/2))^(1/2)*(-I/a^(1/2
)*c^(1/2)*x^2+1)^(1/2)*(I/a^(1/2)*c^(1/2)*x^2+1)^(1/2)/(c*x^4+a)^(1/2)*EllipticPi((I/a^(1/2)*c^(1/2))^(1/2)*x,
I*a^(1/2)/c^(1/2)/d*e,(-I/a^(1/2)*c^(1/2))^(1/2)/(I/a^(1/2)*c^(1/2))^(1/2)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/(e*x^2+d)/(c*x^4+a)^(3/2),x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)/((c*x^4 + a)^(3/2)*(e*x^2 + d)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x^2)/((a + c*x^4)^(3/2)*(d + e*x^2)),x)

[Out]

int((A + B*x^2)/((a + c*x^4)^(3/2)*(d + e*x^2)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)/(e*x**2+d)/(c*x**4+a)**(3/2),x)

[Out]

Integral((A + B*x**2)/((a + c*x**4)**(3/2)*(d + e*x**2)), x)

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