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

Optimal. Leaf size=367 \[ \frac{\left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac{1}{2}\right ) \left (-9 a^{3/2} B e^2+15 \sqrt{a} c d (2 A e+B d)-5 a \sqrt{c} e (A e+2 B d)+15 A c^{3/2} d^2\right )}{30 \sqrt [4]{a} c^{7/4} \sqrt{a+c x^4}}+\frac{x \sqrt{a+c x^4} \left (-3 a B e^2+10 A c d e+5 B c d^2\right )}{5 c^{3/2} \left (\sqrt{a}+\sqrt{c} x^2\right )}-\frac{\sqrt [4]{a} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right ) \left (-3 a B e^2+10 A c d e+5 B c d^2\right )}{5 c^{7/4} \sqrt{a+c x^4}}+\frac{e x \sqrt{a+c x^4} (A e+2 B d)}{3 c}+\frac{B e^2 x^3 \sqrt{a+c x^4}}{5 c} \]

[Out]

(e*(2*B*d + A*e)*x*Sqrt[a + c*x^4])/(3*c) + (B*e^2*x^3*Sqrt[a + c*x^4])/(5*c) + ((5*B*c*d^2 + 10*A*c*d*e - 3*a
*B*e^2)*x*Sqrt[a + c*x^4])/(5*c^(3/2)*(Sqrt[a] + Sqrt[c]*x^2)) - (a^(1/4)*(5*B*c*d^2 + 10*A*c*d*e - 3*a*B*e^2)
*(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]) + ((15*A*c^(3/2)*d^2 - 9*a^(3/2)*B*e^2 - 5*a*Sqrt[c]*e*(2*B*d + A*e) + 15*Sq
rt[a]*c*d*(B*d + 2*A*e))*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticF[2*ArcTa
n[(c^(1/4)*x)/a^(1/4)], 1/2])/(30*a^(1/4)*c^(7/4)*Sqrt[a + c*x^4])

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Rubi [A]  time = 0.395883, antiderivative size = 706, normalized size of antiderivative = 1.92, number of steps used = 12, 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} \[ -\frac{a^{3/4} e \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} (A e+2 B d) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{6 c^{5/4} \sqrt{a+c x^4}}-\frac{3 a^{5/4} B e^2 \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{10 c^{7/4} \sqrt{a+c x^4}}+\frac{3 a^{5/4} B e^2 \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 c^{7/4} \sqrt{a+c x^4}}+\frac{\sqrt [4]{a} d \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} (2 A e+B d) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 c^{3/4} \sqrt{a+c x^4}}-\frac{\sqrt [4]{a} d \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} (2 A e+B d) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{c^{3/4} \sqrt{a+c x^4}}+\frac{e x \sqrt{a+c x^4} (A e+2 B d)}{3 c}+\frac{d x \sqrt{a+c x^4} (2 A e+B d)}{\sqrt{c} \left (\sqrt{a}+\sqrt{c} x^2\right )}+\frac{A d^2 \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt{a+c x^4}}-\frac{3 a B e^2 x \sqrt{a+c x^4}}{5 c^{3/2} \left (\sqrt{a}+\sqrt{c} x^2\right )}+\frac{B e^2 x^3 \sqrt{a+c x^4}}{5 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e*(2*B*d + A*e)*x*Sqrt[a + c*x^4])/(3*c) + (B*e^2*x^3*Sqrt[a + c*x^4])/(5*c) - (3*a*B*e^2*x*Sqrt[a + c*x^4])/
(5*c^(3/2)*(Sqrt[a] + Sqrt[c]*x^2)) + (d*(B*d + 2*A*e)*x*Sqrt[a + c*x^4])/(Sqrt[c]*(Sqrt[a] + Sqrt[c]*x^2)) +
(3*a^(5/4)*B*e^2*(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*(B*d + 2*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^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^(1/4)*Sqrt[a + c*x^4]) - (3*a^(5/4)*B*e^2*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(S
qrt[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^(3/4
)*e*(2*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])/(6*c^(5/4)*Sqrt[a + c*x^4]) + (a^(1/4)*d*(B*d + 2*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^(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 )^2}{\sqrt{a+c x^4}} \, dx &=\int \left (\frac{A d^2}{\sqrt{a+c x^4}}+\frac{d (B d+2 A e) x^2}{\sqrt{a+c x^4}}+\frac{e (2 B d+A e) x^4}{\sqrt{a+c x^4}}+\frac{B e^2 x^6}{\sqrt{a+c x^4}}\right ) \, dx\\ &=\left (A d^2\right ) \int \frac{1}{\sqrt{a+c x^4}} \, dx+\left (B e^2\right ) \int \frac{x^6}{\sqrt{a+c x^4}} \, dx+(e (2 B d+A e)) \int \frac{x^4}{\sqrt{a+c x^4}} \, dx+(d (B d+2 A e)) \int \frac{x^2}{\sqrt{a+c x^4}} \, dx\\ &=\frac{e (2 B d+A e) x \sqrt{a+c x^4}}{3 c}+\frac{B e^2 x^3 \sqrt{a+c x^4}}{5 c}+\frac{A d^2 \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 (3 a B e^2\right ) \int \frac{x^2}{\sqrt{a+c x^4}} \, dx}{5 c}-\frac{(a e (2 B d+A e)) \int \frac{1}{\sqrt{a+c x^4}} \, dx}{3 c}+\frac{\left (\sqrt{a} d (B d+2 A e)\right ) \int \frac{1}{\sqrt{a+c x^4}} \, dx}{\sqrt{c}}-\frac{\left (\sqrt{a} d (B d+2 A e)\right ) \int \frac{1-\frac{\sqrt{c} x^2}{\sqrt{a}}}{\sqrt{a+c x^4}} \, dx}{\sqrt{c}}\\ &=\frac{e (2 B d+A e) x \sqrt{a+c x^4}}{3 c}+\frac{B e^2 x^3 \sqrt{a+c x^4}}{5 c}+\frac{d (B d+2 A e) x \sqrt{a+c x^4}}{\sqrt{c} \left (\sqrt{a}+\sqrt{c} x^2\right )}-\frac{\sqrt [4]{a} d (B d+2 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^2 \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} e (2 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 )}{6 c^{5/4} \sqrt{a+c x^4}}+\frac{\sqrt [4]{a} d (B d+2 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 (3 a^{3/2} B e^2\right ) \int \frac{1}{\sqrt{a+c x^4}} \, dx}{5 c^{3/2}}+\frac{\left (3 a^{3/2} B e^2\right ) \int \frac{1-\frac{\sqrt{c} x^2}{\sqrt{a}}}{\sqrt{a+c x^4}} \, dx}{5 c^{3/2}}\\ &=\frac{e (2 B d+A e) x \sqrt{a+c x^4}}{3 c}+\frac{B e^2 x^3 \sqrt{a+c x^4}}{5 c}-\frac{3 a B e^2 x \sqrt{a+c x^4}}{5 c^{3/2} \left (\sqrt{a}+\sqrt{c} x^2\right )}+\frac{d (B d+2 A e) x \sqrt{a+c x^4}}{\sqrt{c} \left (\sqrt{a}+\sqrt{c} x^2\right )}+\frac{3 a^{5/4} B e^2 \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 (B d+2 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^2 \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{3 a^{5/4} B e^2 \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{a^{3/4} e (2 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 )}{6 c^{5/4} \sqrt{a+c x^4}}+\frac{\sqrt [4]{a} d (B d+2 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 = 0.170828, size = 159, normalized size = 0.43 \[ \frac{x^3 \sqrt{\frac{c x^4}{a}+1} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-\frac{c x^4}{a}\right ) \left (-3 a B e^2+10 A c d e+5 B c d^2\right )-5 x \sqrt{\frac{c x^4}{a}+1} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{c x^4}{a}\right ) \left (a A e^2+2 a B d e-3 A c d^2\right )+e x \left (a+c x^4\right ) \left (5 A e+10 B d+3 B e x^2\right )}{15 c \sqrt{a+c x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e*x*(10*B*d + 5*A*e + 3*B*e*x^2)*(a + c*x^4) - 5*(-3*A*c*d^2 + 2*a*B*d*e + a*A*e^2)*x*Sqrt[1 + (c*x^4)/a]*Hyp
ergeometric2F1[1/4, 1/2, 5/4, -((c*x^4)/a)] + (5*B*c*d^2 + 10*A*c*d*e - 3*a*B*e^2)*x^3*Sqrt[1 + (c*x^4)/a]*Hyp
ergeometric2F1[1/2, 3/4, 7/4, -((c*x^4)/a)])/(15*c*Sqrt[a + c*x^4])

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Maple [C]  time = 0.008, size = 403, normalized size = 1.1 \begin{align*} B{e}^{2} \left ({\frac{{x}^{3}}{5\,c}\sqrt{c{x}^{4}+a}}-{{\frac{3\,i}{5}}{a}^{{\frac{3}{2}}}\sqrt{1-{i{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}} \left ({\it EllipticF} \left ( x\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}},i \right ) -{\it EllipticE} \left ( x\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}},i \right ) \right ){c}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{c{x}^{4}+a}}}} \right ) + \left ( A{e}^{2}+2\,Bde \right ) \left ({\frac{x}{3\,c}\sqrt{c{x}^{4}+a}}-{\frac{a}{3\,c}\sqrt{1-{i{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{c{x}^{4}+a}}}} \right ) +{i \left ( 2\,Ade+B{d}^{2} \right ) \sqrt{a}\sqrt{1-{i{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}} \left ({\it EllipticF} \left ( x\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}},i \right ) -{\it EllipticE} \left ( x\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}},i \right ) \right ){\frac{1}{\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{c{x}^{4}+a}}}{\frac{1}{\sqrt{c}}}}+{A{d}^{2}\sqrt{1-{i{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{c{x}^{4}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)*(e*x^2+d)^2/(c*x^4+a)^(1/2),x)

[Out]

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

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

[Out]

integrate((B*x^2 + A)*(e*x^2 + d)^2/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^{2} x^{6} +{\left (2 \, B d e + A e^{2}\right )} x^{4} + A d^{2} +{\left (B d^{2} + 2 \, A d 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)^2/(c*x^4+a)^(1/2),x, algorithm="fricas")

[Out]

integral((B*e^2*x^6 + (2*B*d*e + A*e^2)*x^4 + A*d^2 + (B*d^2 + 2*A*d*e)*x^2)/sqrt(c*x^4 + a), x)

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Sympy [C]  time = 4.88839, size = 262, normalized size = 0.71 \begin{align*} \frac{A d^{2} 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{A d 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 )}}{2 \sqrt{a} \Gamma \left (\frac{7}{4}\right )} + \frac{A 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{B d^{2} 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{B d 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 )}}{2 \sqrt{a} \Gamma \left (\frac{9}{4}\right )} + \frac{B 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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)*(e*x**2+d)**2/(c*x**4+a)**(1/2),x)

[Out]

A*d**2*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)) + A*d*e*x**3*ga
mma(3/4)*hyper((1/2, 3/4), (7/4,), c*x**4*exp_polar(I*pi)/a)/(2*sqrt(a)*gamma(7/4)) + A*e**2*x**5*gamma(5/4)*h
yper((1/2, 5/4), (9/4,), c*x**4*exp_polar(I*pi)/a)/(4*sqrt(a)*gamma(9/4)) + B*d**2*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)) + B*d*e*x**5*gamma(5/4)*hyper((1/2, 5/4), (9/4
,), c*x**4*exp_polar(I*pi)/a)/(2*sqrt(a)*gamma(9/4)) + B*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))

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

[Out]

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