3.167 \(\int \frac{1+\sqrt{\frac{c}{a}} x^2}{\sqrt{-a+c x^4}} \, dx\)

Optimal. Leaf size=52 \[ \frac{\sqrt{1-\frac{c x^4}{a}} E\left (\left .\sin ^{-1}\left (\sqrt [4]{\frac{c}{a}} x\right )\right |-1\right )}{\sqrt [4]{\frac{c}{a}} \sqrt{c x^4-a}} \]

[Out]

(Sqrt[1 - (c*x^4)/a]*EllipticE[ArcSin[(c/a)^(1/4)*x], -1])/((c/a)^(1/4)*Sqrt[-a + c*x^4])

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Rubi [A]  time = 0.0463449, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {1200, 1199, 424} \[ \frac{\sqrt{1-\frac{c x^4}{a}} E\left (\left .\sin ^{-1}\left (\sqrt [4]{\frac{c}{a}} x\right )\right |-1\right )}{\sqrt [4]{\frac{c}{a}} \sqrt{c x^4-a}} \]

Antiderivative was successfully verified.

[In]

Int[(1 + Sqrt[c/a]*x^2)/Sqrt[-a + c*x^4],x]

[Out]

(Sqrt[1 - (c*x^4)/a]*EllipticE[ArcSin[(c/a)^(1/4)*x], -1])/((c/a)^(1/4)*Sqrt[-a + c*x^4])

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

Mathematica [C]  time = 0.0305537, size = 85, normalized size = 1.63 \[ \frac{\sqrt{1-\frac{c x^4}{a}} \left (x^3 \sqrt{\frac{c}{a}} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};\frac{c x^4}{a}\right )+3 x \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};\frac{c x^4}{a}\right )\right )}{3 \sqrt{c x^4-a}} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + Sqrt[c/a]*x^2)/Sqrt[-a + c*x^4],x]

[Out]

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

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Maple [B]  time = 0.071, size = 165, normalized size = 3.2 \begin{align*}{\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}}}}+{\sqrt{{\frac{c}{a}}}\sqrt{a}\sqrt{1+{{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}\sqrt{1-{{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}} \left ({\it EllipticF} \left ( x\sqrt{-{\sqrt{c}{\frac{1}{\sqrt{a}}}}},i \right ) -{\it EllipticE} \left ( x\sqrt{-{\sqrt{c}{\frac{1}{\sqrt{a}}}}},i \right ) \right ){\frac{1}{\sqrt{-{\sqrt{c}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{c{x}^{4}-a}}}{\frac{1}{\sqrt{c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+x^2*(c/a)^(1/2))/(c*x^4-a)^(1/2),x)

[Out]

1/(-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)*E
llipticF(x*(-1/a^(1/2)*c^(1/2))^(1/2),I)+(c/a)^(1/2)*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)-E
llipticE(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{x^{2} \sqrt{\frac{c}{a}} + 1}{\sqrt{c x^{4} - a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((x^2*sqrt(c/a) + 1)/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{x^{2} \sqrt{\frac{c}{a}} + 1}{\sqrt{c x^{4} - a}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((x^2*sqrt(c/a) + 1)/sqrt(c*x^4 - a), x)

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Sympy [B]  time = 1.75264, size = 76, normalized size = 1.46 \begin{align*} - \frac{i x^{3} \sqrt{\frac{c}{a}} \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}}{a}} \right )}}{4 \sqrt{a} \Gamma \left (\frac{7}{4}\right )} - \frac{i 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}}{a}} \right )}}{4 \sqrt{a} \Gamma \left (\frac{5}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*x**3*sqrt(c/a)*gamma(3/4)*hyper((1/2, 3/4), (7/4,), c*x**4/a)/(4*sqrt(a)*gamma(7/4)) - I*x*gamma(1/4)*hyper
((1/4, 1/2), (5/4,), c*x**4/a)/(4*sqrt(a)*gamma(5/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2} \sqrt{\frac{c}{a}} + 1}{\sqrt{c x^{4} - a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((x^2*sqrt(c/a) + 1)/sqrt(c*x^4 - a), x)