3.933 \(\int \frac{\sqrt [4]{a-b x^2}}{\sqrt{c x}} \, dx\)

Optimal. Leaf size=92 \[ \frac{\sqrt{c x} \sqrt [4]{a-b x^2}}{c}-\frac{\sqrt{a} \sqrt{b} (c x)^{3/2} \left (1-\frac{a}{b x^2}\right )^{3/4} \text{EllipticF}\left (\frac{1}{2} \csc ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right ),2\right )}{c^2 \left (a-b x^2\right )^{3/4}} \]

[Out]

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

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Rubi [A]  time = 0.0791617, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {279, 329, 237, 335, 275, 232} \[ \frac{\sqrt{c x} \sqrt [4]{a-b x^2}}{c}-\frac{\sqrt{a} \sqrt{b} (c x)^{3/2} \left (1-\frac{a}{b x^2}\right )^{3/4} F\left (\left .\frac{1}{2} \csc ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{c^2 \left (a-b x^2\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Int[(a - b*x^2)^(1/4)/Sqrt[c*x],x]

[Out]

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

Rule 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
n*p + 1)), x] + Dist[(a*n*p)/(m + n*p + 1), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]

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]

Rule 237

Int[((a_) + (b_.)*(x_)^4)^(-3/4), x_Symbol] :> Dist[(x^3*(1 + a/(b*x^4))^(3/4))/(a + b*x^4)^(3/4), Int[1/(x^3*
(1 + a/(b*x^4))^(3/4)), x], x] /; FreeQ[{a, b}, x]

Rule 335

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rule 275

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

Rule 232

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2*EllipticF[(1*ArcSin[Rt[-(b/a), 2]*x])/2, 2])/(a^(3/4)*R
t[-(b/a), 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b/a]

Rubi steps

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

Mathematica [C]  time = 0.0135242, size = 55, normalized size = 0.6 \[ \frac{2 x \sqrt [4]{a-b x^2} \, _2F_1\left (-\frac{1}{4},\frac{1}{4};\frac{5}{4};\frac{b x^2}{a}\right )}{\sqrt{c x} \sqrt [4]{1-\frac{b x^2}{a}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.02, size = 0, normalized size = 0. \begin{align*} \int{\sqrt [4]{-b{x}^{2}+a}{\frac{1}{\sqrt{cx}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (-b x^{2} + a\right )}^{\frac{1}{4}} \sqrt{c x}}{c x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 1.44166, size = 39, normalized size = 0.42 \begin{align*} - \frac{i \sqrt [4]{b} x e^{\frac{3 i \pi }{4}}{{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, - \frac{1}{4} \\ \frac{1}{2} \end{matrix}\middle |{\frac{a}{b x^{2}}} \right )}}{\sqrt{c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*b**(1/4)*x*exp(3*I*pi/4)*hyper((-1/2, -1/4), (1/2,), a/(b*x**2))/sqrt(c)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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