3.640 \(\int \frac {1}{\sqrt {c x} \sqrt {3 a-2 a x^2}} \, dx\)

Optimal. Leaf size=63 \[ \frac {2^{3/4} \sqrt {3-2 x^2} \operatorname {EllipticF}\left (\sin ^{-1}\left (\frac {\sqrt [4]{\frac {2}{3}} \sqrt {c x}}{\sqrt {c}}\right ),-1\right )}{\sqrt [4]{3} \sqrt {c} \sqrt {a \left (3-2 x^2\right )}} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {329, 224, 221} \[ \frac {2^{3/4} \sqrt {3-2 x^2} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{\frac {2}{3}} \sqrt {c x}}{\sqrt {c}}\right )\right |-1\right )}{\sqrt [4]{3} \sqrt {c} \sqrt {a \left (3-2 x^2\right )}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[c*x]*Sqrt[3*a - 2*a*x^2]),x]

[Out]

(2^(3/4)*Sqrt[3 - 2*x^2]*EllipticF[ArcSin[((2/3)^(1/4)*Sqrt[c*x])/Sqrt[c]], -1])/(3^(1/4)*Sqrt[c]*Sqrt[a*(3 -
2*x^2)])

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 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 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rubi steps

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

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Mathematica [C]  time = 0.02, size = 56, normalized size = 0.89 \[ \frac {2 x \sqrt {3-2 x^2} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\frac {2 x^2}{3}\right )}{\sqrt {3} \sqrt {a \left (3-2 x^2\right )} \sqrt {c x}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[c*x]*Sqrt[3*a - 2*a*x^2]),x]

[Out]

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

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fricas [F]  time = 0.84, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-2 \, a x^{2} + 3 \, a} \sqrt {c x}}{2 \, a c x^{3} - 3 \, a c x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(1/2)/(-2*a*x^2+3*a)^(1/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-2*a*x^2 + 3*a)*sqrt(c*x)/(2*a*c*x^3 - 3*a*c*x), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {-2 \, a x^{2} + 3 \, a} \sqrt {c x}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(1/2)/(-2*a*x^2+3*a)^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(-2*a*x^2 + 3*a)*sqrt(c*x)), x)

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maple [B]  time = 0.02, size = 117, normalized size = 1.86 \[ -\frac {\sqrt {-\left (2 x^{2}-3\right ) a}\, \sqrt {\left (2 x +\sqrt {2}\, \sqrt {3}\right ) \sqrt {2}\, \sqrt {3}}\, \sqrt {\left (-2 x +\sqrt {2}\, \sqrt {3}\right ) \sqrt {2}\, \sqrt {3}}\, \sqrt {-\sqrt {2}\, \sqrt {3}\, x}\, \EllipticF \left (\frac {\sqrt {3}\, \sqrt {2}\, \sqrt {\left (2 x +\sqrt {2}\, \sqrt {3}\right ) \sqrt {2}\, \sqrt {3}}}{6}, \frac {\sqrt {2}}{2}\right )}{6 \sqrt {c x}\, \left (2 x^{2}-3\right ) a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*x)^(1/2)/(-2*a*x^2+3*a)^(1/2),x)

[Out]

-1/6*(-(2*x^2-3)*a)^(1/2)*((2*x+2^(1/2)*3^(1/2))*2^(1/2)*3^(1/2))^(1/2)*((-2*x+2^(1/2)*3^(1/2))*2^(1/2)*3^(1/2
))^(1/2)*(-2^(1/2)*3^(1/2)*x)^(1/2)*EllipticF(1/6*3^(1/2)*2^(1/2)*((2*x+2^(1/2)*3^(1/2))*2^(1/2)*3^(1/2))^(1/2
),1/2*2^(1/2))/(c*x)^(1/2)/a/(2*x^2-3)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {-2 \, a x^{2} + 3 \, a} \sqrt {c x}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(1/2)/(-2*a*x^2+3*a)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(-2*a*x^2 + 3*a)*sqrt(c*x)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{\sqrt {c\,x}\,\sqrt {3\,a-2\,a\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((c*x)^(1/2)*(3*a - 2*a*x^2)^(1/2)),x)

[Out]

int(1/((c*x)^(1/2)*(3*a - 2*a*x^2)^(1/2)), x)

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sympy [A]  time = 1.04, size = 51, normalized size = 0.81 \[ \frac {\sqrt {3} \sqrt {x} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {2 x^{2} e^{2 i \pi }}{3}} \right )}}{6 \sqrt {a} \sqrt {c} \Gamma \left (\frac {5}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)**(1/2)/(-2*a*x**2+3*a)**(1/2),x)

[Out]

sqrt(3)*sqrt(x)*gamma(1/4)*hyper((1/4, 1/2), (5/4,), 2*x**2*exp_polar(2*I*pi)/3)/(6*sqrt(a)*sqrt(c)*gamma(5/4)
)

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