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

Optimal. Leaf size=912 \[ \text{result too large to display} \]

[Out]

(x*(A*c*d*(c*d^2 - 3*a*e^2) - a*B*e*(3*c*d^2 - a*e^2) + c*(B*c*d^3 + 3*A*c*d^2*e - 3*a*B*d*e^2 - a*A*e^3)*x^2)
)/(2*a*c^2*Sqrt[a + c*x^4]) + (B*e^3*x*Sqrt[a + c*x^4])/(3*c^2) + (e^2*(3*B*d + A*e)*x*Sqrt[a + c*x^4])/(c^(3/
2)*(Sqrt[a] + Sqrt[c]*x^2)) - ((B*c*d^3 + 3*A*c*d^2*e - 3*a*B*d*e^2 - a*A*e^3)*x*Sqrt[a + c*x^4])/(2*a*c^(3/2)
*(Sqrt[a] + Sqrt[c]*x^2)) - (a^(1/4)*e^2*(3*B*d + A*e)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqr
t[c]*x^2)^2]*EllipticE[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(c^(7/4)*Sqrt[a + c*x^4]) + ((B*c*d^3 + 3*A*c*d^2*
e - 3*a*B*d*e^2 - a*A*e^3)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticE[2*Arc
Tan[(c^(1/4)*x)/a^(1/4)], 1/2])/(2*a^(3/4)*c^(7/4)*Sqrt[a + c*x^4]) - (a^(3/4)*B*e^3*(Sqrt[a] + Sqrt[c]*x^2)*S
qrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(6*c^(9/4)*Sqrt[a +
c*x^4]) + (a^(1/4)*e^2*(3*B*d + A*e)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*Ellip
ticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(2*c^(7/4)*Sqrt[a + c*x^4]) + (e*(3*B*c*d^2 + 3*A*c*d*e - a*B*e^2)*(
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])/(2*a^(1/4)*c^(9/4)*Sqrt[a + c*x^4]) + ((A*c^2*d^3 + a^2*B*e^3 - 3*a*c*d*e*(B*d + A*e) + a^(3/2)*Sqrt[c]*e^
2*(3*B*d + A*e) - Sqrt[a]*c^(3/2)*d^2*(B*d + 3*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^(9/4)*Sqrt[a + c*x^4])

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Rubi [A]  time = 0.72556, antiderivative size = 912, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 7, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {1721, 1179, 1198, 220, 1196, 305, 321} \[ -\frac{a^{3/4} B \left (\sqrt{c} x^2+\sqrt{a}\right ) \sqrt{\frac{c x^4+a}{\left (\sqrt{c} x^2+\sqrt{a}\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right ) e^3}{6 c^{9/4} \sqrt{c x^4+a}}+\frac{B x \sqrt{c x^4+a} e^3}{3 c^2}-\frac{\sqrt [4]{a} (3 B d+A e) \left (\sqrt{c} x^2+\sqrt{a}\right ) \sqrt{\frac{c x^4+a}{\left (\sqrt{c} x^2+\sqrt{a}\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right ) e^2}{c^{7/4} \sqrt{c x^4+a}}+\frac{\sqrt [4]{a} (3 B d+A e) \left (\sqrt{c} x^2+\sqrt{a}\right ) \sqrt{\frac{c x^4+a}{\left (\sqrt{c} x^2+\sqrt{a}\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right ) e^2}{2 c^{7/4} \sqrt{c x^4+a}}+\frac{(3 B d+A e) x \sqrt{c x^4+a} e^2}{c^{3/2} \left (\sqrt{c} x^2+\sqrt{a}\right )}+\frac{\left (3 B c d^2+3 A c e d-a B e^2\right ) \left (\sqrt{c} x^2+\sqrt{a}\right ) \sqrt{\frac{c x^4+a}{\left (\sqrt{c} x^2+\sqrt{a}\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right ) e}{2 \sqrt [4]{a} c^{9/4} \sqrt{c x^4+a}}+\frac{\left (B c d^3+3 A c e d^2-3 a B e^2 d-a A e^3\right ) \left (\sqrt{c} x^2+\sqrt{a}\right ) \sqrt{\frac{c x^4+a}{\left (\sqrt{c} x^2+\sqrt{a}\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} c^{7/4} \sqrt{c x^4+a}}+\frac{\left (A c^2 d^3-\sqrt{a} c^{3/2} (B d+3 A e) d^2-3 a c e (B d+A e) d+a^2 B e^3+a^{3/2} \sqrt{c} e^2 (3 B d+A e)\right ) \left (\sqrt{c} x^2+\sqrt{a}\right ) \sqrt{\frac{c x^4+a}{\left (\sqrt{c} x^2+\sqrt{a}\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} c^{9/4} \sqrt{c x^4+a}}-\frac{\left (B c d^3+3 A c e d^2-3 a B e^2 d-a A e^3\right ) x \sqrt{c x^4+a}}{2 a c^{3/2} \left (\sqrt{c} x^2+\sqrt{a}\right )}+\frac{x \left (c \left (B c d^3+3 A c e d^2-3 a B e^2 d-a A e^3\right ) x^2+A c d \left (c d^2-3 a e^2\right )-a B e \left (3 c d^2-a e^2\right )\right )}{2 a c^2 \sqrt{c x^4+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(A*c*d*(c*d^2 - 3*a*e^2) - a*B*e*(3*c*d^2 - a*e^2) + c*(B*c*d^3 + 3*A*c*d^2*e - 3*a*B*d*e^2 - a*A*e^3)*x^2)
)/(2*a*c^2*Sqrt[a + c*x^4]) + (B*e^3*x*Sqrt[a + c*x^4])/(3*c^2) + (e^2*(3*B*d + A*e)*x*Sqrt[a + c*x^4])/(c^(3/
2)*(Sqrt[a] + Sqrt[c]*x^2)) - ((B*c*d^3 + 3*A*c*d^2*e - 3*a*B*d*e^2 - a*A*e^3)*x*Sqrt[a + c*x^4])/(2*a*c^(3/2)
*(Sqrt[a] + Sqrt[c]*x^2)) - (a^(1/4)*e^2*(3*B*d + A*e)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqr
t[c]*x^2)^2]*EllipticE[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(c^(7/4)*Sqrt[a + c*x^4]) + ((B*c*d^3 + 3*A*c*d^2*
e - 3*a*B*d*e^2 - a*A*e^3)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticE[2*Arc
Tan[(c^(1/4)*x)/a^(1/4)], 1/2])/(2*a^(3/4)*c^(7/4)*Sqrt[a + c*x^4]) - (a^(3/4)*B*e^3*(Sqrt[a] + Sqrt[c]*x^2)*S
qrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(6*c^(9/4)*Sqrt[a +
c*x^4]) + (a^(1/4)*e^2*(3*B*d + A*e)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*Ellip
ticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(2*c^(7/4)*Sqrt[a + c*x^4]) + (e*(3*B*c*d^2 + 3*A*c*d*e - a*B*e^2)*(
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])/(2*a^(1/4)*c^(9/4)*Sqrt[a + c*x^4]) + ((A*c^2*d^3 + a^2*B*e^3 - 3*a*c*d*e*(B*d + A*e) + a^(3/2)*Sqrt[c]*e^
2*(3*B*d + A*e) - Sqrt[a]*c^(3/2)*d^2*(B*d + 3*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^(9/4)*Sqrt[a + c*x^4])

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]

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 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 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 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 305

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

Rule 321

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

Rubi steps

\begin{align*} \int \frac{\left (A+B x^2\right ) \left (d+e x^2\right )^3}{\left (a+c x^4\right )^{3/2}} \, dx &=\int \left (\frac{A c d \left (c d^2-3 a e^2\right )-a B e \left (3 c d^2-a e^2\right )+c \left (B c d^3+3 A c d^2 e-3 a B d e^2-a A e^3\right ) x^2}{c^2 \left (a+c x^4\right )^{3/2}}+\frac{e \left (3 B c d^2+3 A c d e-a B e^2\right )}{c^2 \sqrt{a+c x^4}}+\frac{e^2 (3 B d+A e) x^2}{c \sqrt{a+c x^4}}+\frac{B e^3 x^4}{c \sqrt{a+c x^4}}\right ) \, dx\\ &=\frac{\int \frac{A c d \left (c d^2-3 a e^2\right )-a B e \left (3 c d^2-a e^2\right )+c \left (B c d^3+3 A c d^2 e-3 a B d e^2-a A e^3\right ) x^2}{\left (a+c x^4\right )^{3/2}} \, dx}{c^2}+\frac{\left (B e^3\right ) \int \frac{x^4}{\sqrt{a+c x^4}} \, dx}{c}+\frac{\left (e^2 (3 B d+A e)\right ) \int \frac{x^2}{\sqrt{a+c x^4}} \, dx}{c}+\frac{\left (e \left (3 B c d^2+3 A c d e-a B e^2\right )\right ) \int \frac{1}{\sqrt{a+c x^4}} \, dx}{c^2}\\ &=\frac{x \left (A c d \left (c d^2-3 a e^2\right )-a B e \left (3 c d^2-a e^2\right )+c \left (B c d^3+3 A c d^2 e-3 a B d e^2-a A e^3\right ) x^2\right )}{2 a c^2 \sqrt{a+c x^4}}+\frac{B e^3 x \sqrt{a+c x^4}}{3 c^2}+\frac{e \left (3 B c d^2+3 A c d e-a B e^2\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 )}{2 \sqrt [4]{a} c^{9/4} \sqrt{a+c x^4}}-\frac{\int \frac{-A c d \left (c d^2-3 a e^2\right )+a B e \left (3 c d^2-a e^2\right )+c \left (B c d^3+3 A c d^2 e-3 a B d e^2-a A e^3\right ) x^2}{\sqrt{a+c x^4}} \, dx}{2 a c^2}-\frac{\left (a B e^3\right ) \int \frac{1}{\sqrt{a+c x^4}} \, dx}{3 c^2}+\frac{\left (\sqrt{a} e^2 (3 B d+A e)\right ) \int \frac{1}{\sqrt{a+c x^4}} \, dx}{c^{3/2}}-\frac{\left (\sqrt{a} e^2 (3 B d+A e)\right ) \int \frac{1-\frac{\sqrt{c} x^2}{\sqrt{a}}}{\sqrt{a+c x^4}} \, dx}{c^{3/2}}\\ &=\frac{x \left (A c d \left (c d^2-3 a e^2\right )-a B e \left (3 c d^2-a e^2\right )+c \left (B c d^3+3 A c d^2 e-3 a B d e^2-a A e^3\right ) x^2\right )}{2 a c^2 \sqrt{a+c x^4}}+\frac{B e^3 x \sqrt{a+c x^4}}{3 c^2}+\frac{e^2 (3 B d+A e) x \sqrt{a+c x^4}}{c^{3/2} \left (\sqrt{a}+\sqrt{c} x^2\right )}-\frac{\sqrt [4]{a} e^2 (3 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 )}{c^{7/4} \sqrt{a+c x^4}}-\frac{a^{3/4} B e^3 \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 )}{6 c^{9/4} \sqrt{a+c x^4}}+\frac{\sqrt [4]{a} e^2 (3 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 c^{7/4} \sqrt{a+c x^4}}+\frac{e \left (3 B c d^2+3 A c d e-a B e^2\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 )}{2 \sqrt [4]{a} c^{9/4} \sqrt{a+c x^4}}+\frac{\left (B c d^3+3 A c d^2 e-3 a B d e^2-a A e^3\right ) \int \frac{1-\frac{\sqrt{c} x^2}{\sqrt{a}}}{\sqrt{a+c x^4}} \, dx}{2 \sqrt{a} c^{3/2}}+\frac{\left (A c^2 d^3+a^2 B e^3-3 a c d e (B d+A e)+a^{3/2} \sqrt{c} e^2 (3 B d+A e)-\sqrt{a} c^{3/2} d^2 (B d+3 A e)\right ) \int \frac{1}{\sqrt{a+c x^4}} \, dx}{2 a c^2}\\ &=\frac{x \left (A c d \left (c d^2-3 a e^2\right )-a B e \left (3 c d^2-a e^2\right )+c \left (B c d^3+3 A c d^2 e-3 a B d e^2-a A e^3\right ) x^2\right )}{2 a c^2 \sqrt{a+c x^4}}+\frac{B e^3 x \sqrt{a+c x^4}}{3 c^2}+\frac{e^2 (3 B d+A e) x \sqrt{a+c x^4}}{c^{3/2} \left (\sqrt{a}+\sqrt{c} x^2\right )}-\frac{\left (B c d^3+3 A c d^2 e-3 a B d e^2-a A e^3\right ) x \sqrt{a+c x^4}}{2 a c^{3/2} \left (\sqrt{a}+\sqrt{c} x^2\right )}-\frac{\sqrt [4]{a} e^2 (3 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 )}{c^{7/4} \sqrt{a+c x^4}}+\frac{\left (B c d^3+3 A c d^2 e-3 a B d e^2-a A e^3\right ) \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} c^{7/4} \sqrt{a+c x^4}}-\frac{a^{3/4} B e^3 \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 )}{6 c^{9/4} \sqrt{a+c x^4}}+\frac{\sqrt [4]{a} e^2 (3 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 c^{7/4} \sqrt{a+c x^4}}+\frac{e \left (3 B c d^2+3 A c d e-a B e^2\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 )}{2 \sqrt [4]{a} c^{9/4} \sqrt{a+c x^4}}+\frac{\left (A c^2 d^3+a^2 B e^3-3 a c d e (B d+A e)+a^{3/2} \sqrt{c} e^2 (3 B d+A e)-\sqrt{a} c^{3/2} d^2 (B d+3 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} c^{9/4} \sqrt{a+c x^4}}\\ \end{align*}

Mathematica [C]  time = 0.286017, size = 222, normalized size = 0.24 \[ \frac{2 c x^3 \sqrt{\frac{c x^4}{a}+1} \, _2F_1\left (\frac{3}{4},\frac{3}{2};\frac{7}{4};-\frac{c x^4}{a}\right ) \left (-3 a A e^3-9 a B d e^2+3 A c d^2 e+B c d^3\right )+x \sqrt{\frac{c x^4}{a}+1} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{c x^4}{a}\right ) \left (3 A c d \left (3 a e^2+c d^2\right )+a B e \left (9 c d^2-5 a e^2\right )\right )+3 A c x \left (a e^2 \left (2 e x^2-3 d\right )+c d^3\right )+a B e x \left (5 a e^2+c \left (-9 d^2+18 d e x^2+2 e^2 x^4\right )\right )}{6 a c^2 \sqrt{a+c x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*A*c*x*(c*d^3 + a*e^2*(-3*d + 2*e*x^2)) + a*B*e*x*(5*a*e^2 + c*(-9*d^2 + 18*d*e*x^2 + 2*e^2*x^4)) + (a*B*e*(
9*c*d^2 - 5*a*e^2) + 3*A*c*d*(c*d^2 + 3*a*e^2))*x*Sqrt[1 + (c*x^4)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((c*x^
4)/a)] + 2*c*(B*c*d^3 + 3*A*c*d^2*e - 9*a*B*d*e^2 - 3*a*A*e^3)*x^3*Sqrt[1 + (c*x^4)/a]*Hypergeometric2F1[3/4,
3/2, 7/4, -((c*x^4)/a)])/(6*a*c^2*Sqrt[a + c*x^4])

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Maple [C]  time = 0.036, size = 588, 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((B*x^2+A)*(e*x^2+d)^3/(c*x^4+a)^(3/2),x)

[Out]

B*e^3*(1/2/c^2*a*x/((x^4+a/c)*c)^(1/2)+1/3/c^2*x*(c*x^4+a)^(1/2)-5/6*a/c^2/(I/a^(1/2)*c^(1/2))^(1/2)*(1-I/a^(1
/2)*c^(1/2)*x^2)^(1/2)*(1+I/a^(1/2)*c^(1/2)*x^2)^(1/2)/(c*x^4+a)^(1/2)*EllipticF(x*(I/a^(1/2)*c^(1/2))^(1/2),I
))+(A*e^3+3*B*d*e^2)*(-1/2/c*x^3/((x^4+a/c)*c)^(1/2)+3/2*I/c^(3/2)*a^(1/2)/(I/a^(1/2)*c^(1/2))^(1/2)*(1-I/a^(1
/2)*c^(1/2)*x^2)^(1/2)*(1+I/a^(1/2)*c^(1/2)*x^2)^(1/2)/(c*x^4+a)^(1/2)*(EllipticF(x*(I/a^(1/2)*c^(1/2))^(1/2),
I)-EllipticE(x*(I/a^(1/2)*c^(1/2))^(1/2),I)))+(3*A*d*e^2+3*B*d^2*e)*(-1/2/c*x/((x^4+a/c)*c)^(1/2)+1/2/c/(I/a^(
1/2)*c^(1/2))^(1/2)*(1-I/a^(1/2)*c^(1/2)*x^2)^(1/2)*(1+I/a^(1/2)*c^(1/2)*x^2)^(1/2)/(c*x^4+a)^(1/2)*EllipticF(
x*(I/a^(1/2)*c^(1/2))^(1/2),I))+(3*A*d^2*e+B*d^3)*(1/2/a*x^3/((x^4+a/c)*c)^(1/2)-1/2*I/a^(1/2)/(I/a^(1/2)*c^(1
/2))^(1/2)*(1-I/a^(1/2)*c^(1/2)*x^2)^(1/2)*(1+I/a^(1/2)*c^(1/2)*x^2)^(1/2)/(c*x^4+a)^(1/2)/c^(1/2)*(EllipticF(
x*(I/a^(1/2)*c^(1/2))^(1/2),I)-EllipticE(x*(I/a^(1/2)*c^(1/2))^(1/2),I)))+A*d^3*(1/2/a*x/((x^4+a/c)*c)^(1/2)+1
/2/a/(I/a^(1/2)*c^(1/2))^(1/2)*(1-I/a^(1/2)*c^(1/2)*x^2)^(1/2)*(1+I/a^(1/2)*c^(1/2)*x^2)^(1/2)/(c*x^4+a)^(1/2)
*EllipticF(x*(I/a^(1/2)*c^(1/2))^(1/2),I))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x^{2} + A\right )}{\left (e x^{2} + d\right )}^{3}}{{\left (c x^{4} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B e^{3} x^{8} +{\left (3 \, B d e^{2} + A e^{3}\right )} x^{6} + 3 \,{\left (B d^{2} e + A d e^{2}\right )} x^{4} + A d^{3} +{\left (B d^{3} + 3 \, A d^{2} e\right )} x^{2}\right )} \sqrt{c x^{4} + a}}{c^{2} x^{8} + 2 \, a c x^{4} + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*e^3*x^8 + (3*B*d*e^2 + A*e^3)*x^6 + 3*(B*d^2*e + A*d*e^2)*x^4 + A*d^3 + (B*d^3 + 3*A*d^2*e)*x^2)*s
qrt(c*x^4 + a)/(c^2*x^8 + 2*a*c*x^4 + a^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B x^{2}\right ) \left (d + e x^{2}\right )^{3}}{\left (a + c x^{4}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x^{2} + A\right )}{\left (e x^{2} + d\right )}^{3}}{{\left (c x^{4} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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