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

Optimal. Leaf size=453 \[ \frac{\left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac{1}{2}\right ) \left (-63 a^{3/2} \sqrt{c} e^2 (A e+3 B d)+25 a^2 B e^3+105 \sqrt{a} c^{3/2} d^2 (3 A e+B d)-105 a c d e (A e+B d)+105 A c^2 d^3\right )}{210 \sqrt [4]{a} c^{9/4} \sqrt{a+c x^4}}+\frac{e x \sqrt{a+c x^4} \left (-5 a B e^2+21 A c d e+21 B c d^2\right )}{21 c^2}+\frac{x \sqrt{a+c x^4} \left (-3 a A e^3-9 a B d e^2+15 A c d^2 e+5 B c d^3\right )}{5 c^{3/2} \left (\sqrt{a}+\sqrt{c} x^2\right )}-\frac{\sqrt [4]{a} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right ) \left (-3 a A e^3-9 a B d e^2+15 A c d^2 e+5 B c d^3\right )}{5 c^{7/4} \sqrt{a+c x^4}}+\frac{e^2 x^3 \sqrt{a+c x^4} (A e+3 B d)}{5 c}+\frac{B e^3 x^5 \sqrt{a+c x^4}}{7 c} \]

[Out]

(e*(21*B*c*d^2 + 21*A*c*d*e - 5*a*B*e^2)*x*Sqrt[a + c*x^4])/(21*c^2) + (e^2*(3*B*d + A*e)*x^3*Sqrt[a + c*x^4])
/(5*c) + (B*e^3*x^5*Sqrt[a + c*x^4])/(7*c) + ((5*B*c*d^3 + 15*A*c*d^2*e - 9*a*B*d*e^2 - 3*a*A*e^3)*x*Sqrt[a +
c*x^4])/(5*c^(3/2)*(Sqrt[a] + Sqrt[c]*x^2)) - (a^(1/4)*(5*B*c*d^3 + 15*A*c*d^2*e - 9*a*B*d*e^2 - 3*a*A*e^3)*(S
qrt[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
])/(5*c^(7/4)*Sqrt[a + c*x^4]) + ((105*A*c^2*d^3 + 25*a^2*B*e^3 - 105*a*c*d*e*(B*d + A*e) - 63*a^(3/2)*Sqrt[c]
*e^2*(3*B*d + A*e) + 105*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])/(210*a^(1/4)*c^(9/4)*Sqrt[a + c*x^4])

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Rubi [A]  time = 0.509473, antiderivative size = 878, normalized size of antiderivative = 1.94, number of steps used = 15, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {1721, 220, 305, 1196, 321} \[ \frac{B e^3 \sqrt{c x^4+a} x^5}{7 c}+\frac{e^2 (3 B d+A e) \sqrt{c x^4+a} x^3}{5 c}-\frac{5 a B e^3 \sqrt{c x^4+a} x}{21 c^2}+\frac{d e (B d+A e) \sqrt{c x^4+a} x}{c}-\frac{3 a e^2 (3 B d+A e) \sqrt{c x^4+a} x}{5 c^{3/2} \left (\sqrt{c} x^2+\sqrt{a}\right )}+\frac{d^2 (B d+3 A e) \sqrt{c x^4+a} x}{\sqrt{c} \left (\sqrt{c} x^2+\sqrt{a}\right )}+\frac{3 a^{5/4} e^2 (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 )}{5 c^{7/4} \sqrt{c x^4+a}}-\frac{\sqrt [4]{a} d^2 (B d+3 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 )}{c^{3/4} \sqrt{c x^4+a}}+\frac{A d^3 \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 )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt{c x^4+a}}+\frac{5 a^{7/4} B e^3 \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 )}{42 c^{9/4} \sqrt{c x^4+a}}-\frac{a^{3/4} d e (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 )}{2 c^{5/4} \sqrt{c x^4+a}}-\frac{3 a^{5/4} e^2 (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 )}{10 c^{7/4} \sqrt{c x^4+a}}+\frac{\sqrt [4]{a} d^2 (B d+3 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 )}{2 c^{3/4} \sqrt{c x^4+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*a*B*e^3*x*Sqrt[a + c*x^4])/(21*c^2) + (d*e*(B*d + A*e)*x*Sqrt[a + c*x^4])/c + (e^2*(3*B*d + A*e)*x^3*Sqrt[
a + c*x^4])/(5*c) + (B*e^3*x^5*Sqrt[a + c*x^4])/(7*c) - (3*a*e^2*(3*B*d + A*e)*x*Sqrt[a + c*x^4])/(5*c^(3/2)*(
Sqrt[a] + Sqrt[c]*x^2)) + (d^2*(B*d + 3*A*e)*x*Sqrt[a + c*x^4])/(Sqrt[c]*(Sqrt[a] + Sqrt[c]*x^2)) + (3*a^(5/4)
*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]*EllipticE[2*ArcTan[(c^(
1/4)*x)/a^(1/4)], 1/2])/(5*c^(7/4)*Sqrt[a + c*x^4]) - (a^(1/4)*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]*EllipticE[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(c^(3/4)*Sqrt[a + c*x^4]
) + (A*d^3*(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^(1/4)*Sqrt[a + c*x^4]) + (5*a^(7/4)*B*e^3*(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])/(42*c^(9/4)*Sqrt[a + c*x^4]) - (a^
(3/4)*d*e*(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])/(2*c^(5/4)*Sqrt[a + c*x^4]) - (3*a^(5/4)*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]*EllipticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(10*c^(7/4)*Sqrt[a
+ c*x^4]) + (a^(1/4)*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]*Ell
ipticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(2*c^(3/4)*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 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 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 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 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}{\sqrt{a+c x^4}} \, dx &=\int \left (\frac{A d^3}{\sqrt{a+c x^4}}+\frac{d^2 (B d+3 A e) x^2}{\sqrt{a+c x^4}}+\frac{3 d e (B d+A e) x^4}{\sqrt{a+c x^4}}+\frac{e^2 (3 B d+A e) x^6}{\sqrt{a+c x^4}}+\frac{B e^3 x^8}{\sqrt{a+c x^4}}\right ) \, dx\\ &=\left (A d^3\right ) \int \frac{1}{\sqrt{a+c x^4}} \, dx+\left (B e^3\right ) \int \frac{x^8}{\sqrt{a+c x^4}} \, dx+(3 d e (B d+A e)) \int \frac{x^4}{\sqrt{a+c x^4}} \, dx+\left (e^2 (3 B d+A e)\right ) \int \frac{x^6}{\sqrt{a+c x^4}} \, dx+\left (d^2 (B d+3 A e)\right ) \int \frac{x^2}{\sqrt{a+c x^4}} \, dx\\ &=\frac{d e (B d+A e) x \sqrt{a+c x^4}}{c}+\frac{e^2 (3 B d+A e) x^3 \sqrt{a+c x^4}}{5 c}+\frac{B e^3 x^5 \sqrt{a+c x^4}}{7 c}+\frac{A d^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 )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt{a+c x^4}}-\frac{\left (5 a B e^3\right ) \int \frac{x^4}{\sqrt{a+c x^4}} \, dx}{7 c}-\frac{(a d e (B d+A e)) \int \frac{1}{\sqrt{a+c x^4}} \, dx}{c}-\frac{\left (3 a e^2 (3 B d+A e)\right ) \int \frac{x^2}{\sqrt{a+c x^4}} \, dx}{5 c}+\frac{\left (\sqrt{a} d^2 (B d+3 A e)\right ) \int \frac{1}{\sqrt{a+c x^4}} \, dx}{\sqrt{c}}-\frac{\left (\sqrt{a} d^2 (B d+3 A e)\right ) \int \frac{1-\frac{\sqrt{c} x^2}{\sqrt{a}}}{\sqrt{a+c x^4}} \, dx}{\sqrt{c}}\\ &=-\frac{5 a B e^3 x \sqrt{a+c x^4}}{21 c^2}+\frac{d e (B d+A e) x \sqrt{a+c x^4}}{c}+\frac{e^2 (3 B d+A e) x^3 \sqrt{a+c x^4}}{5 c}+\frac{B e^3 x^5 \sqrt{a+c x^4}}{7 c}+\frac{d^2 (B d+3 A e) x \sqrt{a+c x^4}}{\sqrt{c} \left (\sqrt{a}+\sqrt{c} x^2\right )}-\frac{\sqrt [4]{a} d^2 (B d+3 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^{3/4} \sqrt{a+c x^4}}+\frac{A d^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 )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt{a+c x^4}}-\frac{a^{3/4} d 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 c^{5/4} \sqrt{a+c x^4}}+\frac{\sqrt [4]{a} d^2 (B d+3 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^{3/4} \sqrt{a+c x^4}}+\frac{\left (5 a^2 B e^3\right ) \int \frac{1}{\sqrt{a+c x^4}} \, dx}{21 c^2}-\frac{\left (3 a^{3/2} e^2 (3 B d+A e)\right ) \int \frac{1}{\sqrt{a+c x^4}} \, dx}{5 c^{3/2}}+\frac{\left (3 a^{3/2} e^2 (3 B d+A e)\right ) \int \frac{1-\frac{\sqrt{c} x^2}{\sqrt{a}}}{\sqrt{a+c x^4}} \, dx}{5 c^{3/2}}\\ &=-\frac{5 a B e^3 x \sqrt{a+c x^4}}{21 c^2}+\frac{d e (B d+A e) x \sqrt{a+c x^4}}{c}+\frac{e^2 (3 B d+A e) x^3 \sqrt{a+c x^4}}{5 c}+\frac{B e^3 x^5 \sqrt{a+c x^4}}{7 c}-\frac{3 a e^2 (3 B d+A e) x \sqrt{a+c x^4}}{5 c^{3/2} \left (\sqrt{a}+\sqrt{c} x^2\right )}+\frac{d^2 (B d+3 A e) x \sqrt{a+c x^4}}{\sqrt{c} \left (\sqrt{a}+\sqrt{c} x^2\right )}+\frac{3 a^{5/4} 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 )}{5 c^{7/4} \sqrt{a+c x^4}}-\frac{\sqrt [4]{a} d^2 (B d+3 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^{3/4} \sqrt{a+c x^4}}+\frac{A d^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 )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt{a+c x^4}}+\frac{5 a^{7/4} 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 )}{42 c^{9/4} \sqrt{a+c x^4}}-\frac{a^{3/4} d 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 c^{5/4} \sqrt{a+c x^4}}-\frac{3 a^{5/4} 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 )}{10 c^{7/4} \sqrt{a+c x^4}}+\frac{\sqrt [4]{a} d^2 (B d+3 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^{3/4} \sqrt{a+c x^4}}\\ \end{align*}

Mathematica [C]  time = 0.281571, size = 217, normalized size = 0.48 \[ \frac{7 c x^3 \sqrt{\frac{c x^4}{a}+1} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-\frac{c x^4}{a}\right ) \left (-3 a A e^3-9 a B d e^2+15 A c d^2 e+5 B c d^3\right )+5 x \sqrt{\frac{c x^4}{a}+1} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{c x^4}{a}\right ) \left (21 A c d \left (c d^2-a e^2\right )+a B e \left (5 a e^2-21 c d^2\right )\right )+e x \left (a+c x^4\right ) \left (-25 a B e^2+21 A c e \left (5 d+e x^2\right )+3 B c \left (35 d^2+21 d e x^2+5 e^2 x^4\right )\right )}{105 c^2 \sqrt{a+c x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e*x*(a + c*x^4)*(-25*a*B*e^2 + 21*A*c*e*(5*d + e*x^2) + 3*B*c*(35*d^2 + 21*d*e*x^2 + 5*e^2*x^4)) + 5*(21*A*c*
d*(c*d^2 - a*e^2) + a*B*e*(-21*c*d^2 + 5*a*e^2))*x*Sqrt[1 + (c*x^4)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((c*x
^4)/a)] + 7*c*(5*B*c*d^3 + 15*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[1
/2, 3/4, 7/4, -((c*x^4)/a)])/(105*c^2*Sqrt[a + c*x^4])

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Maple [C]  time = 0.054, size = 533, normalized size = 1.2 \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)^(1/2),x)

[Out]

B*e^3*(1/7/c*x^5*(c*x^4+a)^(1/2)-5/21*a/c^2*x*(c*x^4+a)^(1/2)+5/21*a^2/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)*(1/5/c*x^3*(c*x^4+a)^(1/2)-3/5*I*a^(3/2)/c^(3/2)/(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)-El
lipticE(x*(I/a^(1/2)*c^(1/2))^(1/2),I)))+(3*A*d*e^2+3*B*d^2*e)*(1/3/c*x*(c*x^4+a)^(1/2)-1/3*a/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))+I*(3*A*d^2*e+B*d^3)*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/(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}}{\sqrt{c x^{4} + a}}\,{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)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{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}}{\sqrt{c x^{4} + a}}, 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)^(1/2),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), x)

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Sympy [C]  time = 6.84416, size = 364, normalized size = 0.8 \begin{align*} \frac{A d^{3} x \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{4}, \frac{1}{2} \\ \frac{5}{4} \end{matrix}\middle |{\frac{c x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt{a} \Gamma \left (\frac{5}{4}\right )} + \frac{3 A d^{2} e x^{3} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{3}{4} \\ \frac{7}{4} \end{matrix}\middle |{\frac{c x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt{a} \Gamma \left (\frac{7}{4}\right )} + \frac{3 A d e^{2} x^{5} \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{5}{4} \\ \frac{9}{4} \end{matrix}\middle |{\frac{c x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt{a} \Gamma \left (\frac{9}{4}\right )} + \frac{A e^{3} x^{7} \Gamma \left (\frac{7}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{7}{4} \\ \frac{11}{4} \end{matrix}\middle |{\frac{c x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt{a} \Gamma \left (\frac{11}{4}\right )} + \frac{B d^{3} x^{3} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{3}{4} \\ \frac{7}{4} \end{matrix}\middle |{\frac{c x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt{a} \Gamma \left (\frac{7}{4}\right )} + \frac{3 B d^{2} e x^{5} \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{5}{4} \\ \frac{9}{4} \end{matrix}\middle |{\frac{c x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt{a} \Gamma \left (\frac{9}{4}\right )} + \frac{3 B d e^{2} x^{7} \Gamma \left (\frac{7}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{7}{4} \\ \frac{11}{4} \end{matrix}\middle |{\frac{c x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt{a} \Gamma \left (\frac{11}{4}\right )} + \frac{B e^{3} x^{9} \Gamma \left (\frac{9}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{9}{4} \\ \frac{13}{4} \end{matrix}\middle |{\frac{c x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt{a} \Gamma \left (\frac{13}{4}\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)**(1/2),x)

[Out]

A*d**3*x*gamma(1/4)*hyper((1/4, 1/2), (5/4,), c*x**4*exp_polar(I*pi)/a)/(4*sqrt(a)*gamma(5/4)) + 3*A*d**2*e*x*
*3*gamma(3/4)*hyper((1/2, 3/4), (7/4,), c*x**4*exp_polar(I*pi)/a)/(4*sqrt(a)*gamma(7/4)) + 3*A*d*e**2*x**5*gam
ma(5/4)*hyper((1/2, 5/4), (9/4,), c*x**4*exp_polar(I*pi)/a)/(4*sqrt(a)*gamma(9/4)) + A*e**3*x**7*gamma(7/4)*hy
per((1/2, 7/4), (11/4,), c*x**4*exp_polar(I*pi)/a)/(4*sqrt(a)*gamma(11/4)) + B*d**3*x**3*gamma(3/4)*hyper((1/2
, 3/4), (7/4,), c*x**4*exp_polar(I*pi)/a)/(4*sqrt(a)*gamma(7/4)) + 3*B*d**2*e*x**5*gamma(5/4)*hyper((1/2, 5/4)
, (9/4,), c*x**4*exp_polar(I*pi)/a)/(4*sqrt(a)*gamma(9/4)) + 3*B*d*e**2*x**7*gamma(7/4)*hyper((1/2, 7/4), (11/
4,), c*x**4*exp_polar(I*pi)/a)/(4*sqrt(a)*gamma(11/4)) + B*e**3*x**9*gamma(9/4)*hyper((1/2, 9/4), (13/4,), c*x
**4*exp_polar(I*pi)/a)/(4*sqrt(a)*gamma(13/4))

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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}}{\sqrt{c x^{4} + a}}\,{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)^(1/2),x, algorithm="giac")

[Out]

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