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

Optimal. Leaf size=54 \[ \frac {a^{3/4} \sqrt {1-\frac {c x^4}{a}} E\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )\right |-1\right )}{\sqrt [4]{c} \sqrt {c x^4-a}} \]

[Out]

a^(3/4)*EllipticE(c^(1/4)*x/a^(1/4),I)*(1-c*x^4/a)^(1/2)/c^(1/4)/(c*x^4-a)^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 54, normalized size of antiderivative = 1.00, 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} \[ \frac {a^{3/4} \sqrt {1-\frac {c x^4}{a}} E\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )\right |-1\right )}{\sqrt [4]{c} \sqrt {c x^4-a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^(3/4)*Sqrt[1 - (c*x^4)/a]*EllipticE[ArcSin[(c^(1/4)*x)/a^(1/4)], -1])/(c^(1/4)*Sqrt[-a + c*x^4])

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]

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 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]

Rubi steps

\begin {align*} \int \frac {\sqrt {a}+\sqrt {c} x^2}{\sqrt {-a+c x^4}} \, dx &=\frac {\sqrt {1-\frac {c x^4}{a}} \int \frac {\sqrt {a}+\sqrt {c} x^2}{\sqrt {1-\frac {c x^4}{a}}} \, dx}{\sqrt {-a+c x^4}}\\ &=\frac {\left (\sqrt {a} \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 {-a+c x^4}}\\ &=\frac {a^{3/4} \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 )}{\sqrt [4]{c} \sqrt {-a+c x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 86, normalized size = 1.59 \[ \frac {\sqrt {1-\frac {c x^4}{a}} \left (3 \sqrt {a} x \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\frac {c x^4}{a}\right )+\sqrt {c} x^3 \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};\frac {c x^4}{a}\right )\right )}{3 \sqrt {c x^4-a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c} x^{2} + \sqrt {a}}{\sqrt {c x^{4} - a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c} x^{2} + \sqrt {a}}{\sqrt {c x^{4} - a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.05, size = 158, normalized size = 2.93 \[ \frac {\sqrt {\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {-\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {a}\, \EllipticF \left (\sqrt {-\frac {\sqrt {c}}{\sqrt {a}}}\, x , i\right )}{\sqrt {-\frac {\sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}-a}}+\frac {\sqrt {\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {-\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \left (-\EllipticE \left (\sqrt {-\frac {\sqrt {c}}{\sqrt {a}}}\, x , i\right )+\EllipticF \left (\sqrt {-\frac {\sqrt {c}}{\sqrt {a}}}\, x , i\right )\right ) \sqrt {a}}{\sqrt {-\frac {\sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}-a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^(1/2)/(-1/a^(1/2)*c^(1/2))^(1/2)*(1/a^(1/2)*c^(1/2)*x^2+1)^(1/2)*(-1/a^(1/2)*c^(1/2)*x^2+1)^(1/2)/(c*x^4-a)^
(1/2)*(EllipticF((-1/a^(1/2)*c^(1/2))^(1/2)*x,I)-EllipticE((-1/a^(1/2)*c^(1/2))^(1/2)*x,I))+a^(1/2)/(-1/a^(1/2
)*c^(1/2))^(1/2)*(1/a^(1/2)*c^(1/2)*x^2+1)^(1/2)*(-1/a^(1/2)*c^(1/2)*x^2+1)^(1/2)/(c*x^4-a)^(1/2)*EllipticF((-
1/a^(1/2)*c^(1/2))^(1/2)*x,I)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c} x^{2} + \sqrt {a}}{\sqrt {c x^{4} - a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\sqrt {a}+\sqrt {c}\,x^2}{\sqrt {c\,x^4-a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 2.41, size = 70, normalized size = 1.30 \[ - \frac {i x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {c x^{4}}{a}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} - \frac {i \sqrt {c} x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {c x^{4}}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {7}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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