3.158 \(\int \frac{(d+e x^2)^2}{\sqrt{a-c x^4}} \, dx\)

Optimal. Leaf size=162 \[ \frac{\sqrt [4]{a} \sqrt{1-\frac{c x^4}{a}} \left (-6 \sqrt{a} \sqrt{c} d e+a e^2+3 c d^2\right ) \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),-1\right )}{3 c^{5/4} \sqrt{a-c x^4}}+\frac{2 a^{3/4} d e \sqrt{1-\frac{c x^4}{a}} E\left (\left .\sin ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )\right |-1\right )}{c^{3/4} \sqrt{a-c x^4}}-\frac{e^2 x \sqrt{a-c x^4}}{3 c} \]

[Out]

-(e^2*x*Sqrt[a - c*x^4])/(3*c) + (2*a^(3/4)*d*e*Sqrt[1 - (c*x^4)/a]*EllipticE[ArcSin[(c^(1/4)*x)/a^(1/4)], -1]
)/(c^(3/4)*Sqrt[a - c*x^4]) + (a^(1/4)*(3*c*d^2 - 6*Sqrt[a]*Sqrt[c]*d*e + a*e^2)*Sqrt[1 - (c*x^4)/a]*EllipticF
[ArcSin[(c^(1/4)*x)/a^(1/4)], -1])/(3*c^(5/4)*Sqrt[a - c*x^4])

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Rubi [A]  time = 0.144698, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.318, Rules used = {1207, 1201, 224, 221, 1200, 1199, 424} \[ \frac{2 a^{3/4} d e \sqrt{1-\frac{c x^4}{a}} E\left (\left .\sin ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )\right |-1\right )}{c^{3/4} \sqrt{a-c x^4}}+\frac{\sqrt [4]{a} \sqrt{1-\frac{c x^4}{a}} \left (-6 \sqrt{a} \sqrt{c} d e+a e^2+3 c d^2\right ) F\left (\left .\sin ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )\right |-1\right )}{3 c^{5/4} \sqrt{a-c x^4}}-\frac{e^2 x \sqrt{a-c x^4}}{3 c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x^2)^2/Sqrt[a - c*x^4],x]

[Out]

-(e^2*x*Sqrt[a - c*x^4])/(3*c) + (2*a^(3/4)*d*e*Sqrt[1 - (c*x^4)/a]*EllipticE[ArcSin[(c^(1/4)*x)/a^(1/4)], -1]
)/(c^(3/4)*Sqrt[a - c*x^4]) + (a^(1/4)*(3*c*d^2 - 6*Sqrt[a]*Sqrt[c]*d*e + a*e^2)*Sqrt[1 - (c*x^4)/a]*EllipticF
[ArcSin[(c^(1/4)*x)/a^(1/4)], -1])/(3*c^(5/4)*Sqrt[a - c*x^4])

Rule 1207

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(e^q*x^(2*q - 3)*(a + c*x^4)^(p +
1))/(c*(4*p + 2*q + 1)), x] + Dist[1/(c*(4*p + 2*q + 1)), Int[(a + c*x^4)^p*ExpandToSum[c*(4*p + 2*q + 1)*(d
+ e*x^2)^q - a*(2*q - 3)*e^q*x^(2*q - 4) - c*(4*p + 2*q + 1)*e^q*x^(2*q), x], x], x] /; FreeQ[{a, c, d, e, p},
x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[q, 1]

Rule 1201

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

Rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Dist[Sqrt[1 + (b*x^4)/a]/Sqrt[a + b*x^4], Int[1/Sqrt[1 + (b*x^4)
/a], x], x] /; FreeQ[{a, b}, x] && NegQ[b/a] &&  !GtQ[a, 0]

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[(Rt[-b, 4]*x)/Rt[a, 4]], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rule 1200

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> Dist[Sqrt[1 + (c*x^4)/a]/Sqrt[a + c*x^4], In
t[(d + e*x^2)/Sqrt[1 + (c*x^4)/a], x], x] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && EqQ[c*d^2 + a*e^2, 0] &&
!GtQ[a, 0]

Rule 1199

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

Rule 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rubi steps

\begin{align*} \int \frac{\left (d+e x^2\right )^2}{\sqrt{a-c x^4}} \, dx &=-\frac{e^2 x \sqrt{a-c x^4}}{3 c}-\frac{\int \frac{-3 c d^2-a e^2-6 c d e x^2}{\sqrt{a-c x^4}} \, dx}{3 c}\\ &=-\frac{e^2 x \sqrt{a-c x^4}}{3 c}+\frac{\left (2 \sqrt{a} d e\right ) \int \frac{1+\frac{\sqrt{c} x^2}{\sqrt{a}}}{\sqrt{a-c x^4}} \, dx}{\sqrt{c}}-\frac{\left (-3 c d^2+6 \sqrt{a} \sqrt{c} d e-a e^2\right ) \int \frac{1}{\sqrt{a-c x^4}} \, dx}{3 c}\\ &=-\frac{e^2 x \sqrt{a-c x^4}}{3 c}+\frac{\left (2 \sqrt{a} d e \sqrt{1-\frac{c x^4}{a}}\right ) \int \frac{1+\frac{\sqrt{c} x^2}{\sqrt{a}}}{\sqrt{1-\frac{c x^4}{a}}} \, dx}{\sqrt{c} \sqrt{a-c x^4}}-\frac{\left (\left (-3 c d^2+6 \sqrt{a} \sqrt{c} d e-a e^2\right ) \sqrt{1-\frac{c x^4}{a}}\right ) \int \frac{1}{\sqrt{1-\frac{c x^4}{a}}} \, dx}{3 c \sqrt{a-c x^4}}\\ &=-\frac{e^2 x \sqrt{a-c x^4}}{3 c}+\frac{\sqrt [4]{a} \left (3 c d^2-6 \sqrt{a} \sqrt{c} d e+a e^2\right ) \sqrt{1-\frac{c x^4}{a}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )\right |-1\right )}{3 c^{5/4} \sqrt{a-c x^4}}+\frac{\left (2 \sqrt{a} d e \sqrt{1-\frac{c x^4}{a}}\right ) \int \frac{\sqrt{1+\frac{\sqrt{c} x^2}{\sqrt{a}}}}{\sqrt{1-\frac{\sqrt{c} x^2}{\sqrt{a}}}} \, dx}{\sqrt{c} \sqrt{a-c x^4}}\\ &=-\frac{e^2 x \sqrt{a-c x^4}}{3 c}+\frac{2 a^{3/4} d e \sqrt{1-\frac{c x^4}{a}} E\left (\left .\sin ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )\right |-1\right )}{c^{3/4} \sqrt{a-c x^4}}+\frac{\sqrt [4]{a} \left (3 c d^2-6 \sqrt{a} \sqrt{c} d e+a e^2\right ) \sqrt{1-\frac{c x^4}{a}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )\right |-1\right )}{3 c^{5/4} \sqrt{a-c x^4}}\\ \end{align*}

Mathematica [C]  time = 0.104588, size = 121, normalized size = 0.75 \[ \frac{x \sqrt{1-\frac{c x^4}{a}} \left (a e^2+3 c d^2\right ) \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};\frac{c x^4}{a}\right )+e x \left (2 c d x^2 \sqrt{1-\frac{c x^4}{a}} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};\frac{c x^4}{a}\right )-a e+c e x^4\right )}{3 c \sqrt{a-c x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.055, size = 246, normalized size = 1.5 \begin{align*}{e}^{2} \left ( -{\frac{x}{3\,c}\sqrt{-c{x}^{4}+a}}+{\frac{a}{3\,c}\sqrt{1-{{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{\sqrt{c}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{\sqrt{c}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{-c{x}^{4}+a}}}} \right ) -2\,{\frac{de\sqrt{a}}{\sqrt{-c{x}^{4}+a}\sqrt{c}}\sqrt{1-{\frac{{x}^{2}\sqrt{c}}{\sqrt{a}}}}\sqrt{1+{\frac{{x}^{2}\sqrt{c}}{\sqrt{a}}}} \left ({\it EllipticF} \left ( x\sqrt{{\frac{\sqrt{c}}{\sqrt{a}}}},i \right ) -{\it EllipticE} \left ( x\sqrt{{\frac{\sqrt{c}}{\sqrt{a}}}},i \right ) \right ){\frac{1}{\sqrt{{\frac{\sqrt{c}}{\sqrt{a}}}}}}}+{{d}^{2}\sqrt{1-{{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{\sqrt{c}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{\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((e*x^2+d)^2/(-c*x^4+a)^(1/2),x)

[Out]

e^2*(-1/3/c*x*(-c*x^4+a)^(1/2)+1/3*a/c/(1/a^(1/2)*c^(1/2))^(1/2)*(1-1/a^(1/2)*c^(1/2)*x^2)^(1/2)*(1+1/a^(1/2)*
c^(1/2)*x^2)^(1/2)/(-c*x^4+a)^(1/2)*EllipticF(x*(1/a^(1/2)*c^(1/2))^(1/2),I))-2*d*e*a^(1/2)/(1/a^(1/2)*c^(1/2)
)^(1/2)*(1-1/a^(1/2)*c^(1/2)*x^2)^(1/2)*(1+1/a^(1/2)*c^(1/2)*x^2)^(1/2)/(-c*x^4+a)^(1/2)/c^(1/2)*(EllipticF(x*
(1/a^(1/2)*c^(1/2))^(1/2),I)-EllipticE(x*(1/a^(1/2)*c^(1/2))^(1/2),I))+d^2/(1/a^(1/2)*c^(1/2))^(1/2)*(1-1/a^(1
/2)*c^(1/2)*x^2)^(1/2)*(1+1/a^(1/2)*c^(1/2)*x^2)^(1/2)/(-c*x^4+a)^(1/2)*EllipticF(x*(1/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 (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((e*x^2+d)^2/(-c*x^4+a)^(1/2),x, algorithm="maxima")

[Out]

integrate((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{{\left (e^{2} x^{4} + 2 \, d e x^{2} + d^{2}\right )} \sqrt{-c x^{4} + a}}{c x^{4} - a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 2.81924, size = 129, normalized size = 0.8 \begin{align*} \frac{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^{2 i \pi }}{a}} \right )}}{4 \sqrt{a} \Gamma \left (\frac{5}{4}\right )} + \frac{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^{2 i \pi }}{a}} \right )}}{2 \sqrt{a} \Gamma \left (\frac{7}{4}\right )} + \frac{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^{2 i \pi }}{a}} \right )}}{4 \sqrt{a} \Gamma \left (\frac{9}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d**2*x*gamma(1/4)*hyper((1/4, 1/2), (5/4,), c*x**4*exp_polar(2*I*pi)/a)/(4*sqrt(a)*gamma(5/4)) + d*e*x**3*gamm
a(3/4)*hyper((1/2, 3/4), (7/4,), c*x**4*exp_polar(2*I*pi)/a)/(2*sqrt(a)*gamma(7/4)) + e**2*x**5*gamma(5/4)*hyp
er((1/2, 5/4), (9/4,), c*x**4*exp_polar(2*I*pi)/a)/(4*sqrt(a)*gamma(9/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\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((e*x^2+d)^2/(-c*x^4+a)^(1/2),x, algorithm="giac")

[Out]

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