3.593 \(\int \frac{\sqrt{a+b x^2}}{\sqrt{c x}} \, dx\)

Optimal. Leaf size=126 \[ \frac{2 a^{3/4} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right ),\frac{1}{2}\right )}{3 \sqrt [4]{b} \sqrt{c} \sqrt{a+b x^2}}+\frac{2 \sqrt{c x} \sqrt{a+b x^2}}{3 c} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0743437, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {279, 329, 220} \[ \frac{2 a^{3/4} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right )|\frac{1}{2}\right )}{3 \sqrt [4]{b} \sqrt{c} \sqrt{a+b x^2}}+\frac{2 \sqrt{c x} \sqrt{a+b x^2}}{3 c} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x^2]/Sqrt[c*x],x]

[Out]

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

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 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rubi steps

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

Mathematica [C]  time = 0.0122697, size = 54, normalized size = 0.43 \[ \frac{2 x \sqrt{a+b x^2} \, _2F_1\left (-\frac{1}{2},\frac{1}{4};\frac{5}{4};-\frac{b x^2}{a}\right )}{\sqrt{c x} \sqrt{\frac{b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*x^2]/Sqrt[c*x],x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.018, size = 119, normalized size = 0.9 \begin{align*}{\frac{2}{3\,b} \left ( \sqrt{{ \left ( bx+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}}}\sqrt{2}\sqrt{{ \left ( -bx+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}}}\sqrt{-{bx{\frac{1}{\sqrt{-ab}}}}}{\it EllipticF} \left ( \sqrt{{ \left ( bx+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}}},{\frac{\sqrt{2}}{2}} \right ) \sqrt{-ab}a+{b}^{2}{x}^{3}+abx \right ){\frac{1}{\sqrt{cx}}}{\frac{1}{\sqrt{b{x}^{2}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Sympy [C]  time = 0.843216, size = 46, normalized size = 0.37 \begin{align*} \frac{\sqrt{a} \sqrt{x} \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{1}{4} \\ \frac{5}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt{c} \Gamma \left (\frac{5}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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