3.617 \(\int \frac{1}{\sqrt{c x} \sqrt{a+b x^2}} \, dx\)

Optimal. Leaf size=97 \[ \frac{\left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right ),\frac{1}{2}\right )}{\sqrt [4]{a} \sqrt [4]{b} \sqrt{c} \sqrt{a+b x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0577287, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {329, 220} \[ \frac{\left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right )|\frac{1}{2}\right )}{\sqrt [4]{a} \sqrt [4]{b} \sqrt{c} \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[c*x]*Sqrt[a + b*x^2]),x]

[Out]

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

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

Mathematica [C]  time = 0.012867, size = 54, normalized size = 0.56 \[ \frac{2 x \sqrt{\frac{b x^2}{a}+1} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{b x^2}{a}\right )}{\sqrt{c x} \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.013, size = 104, normalized size = 1.1 \begin{align*}{\frac{\sqrt{2}}{b}\sqrt{-ab}\sqrt{{ \left ( bx+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}}}\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 ){\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(1/(c*x)^(1/2)/(b*x^2+a)^(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b x^{2} + a} \sqrt{c x}}{b c x^{3} + a c x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 0.988725, size = 44, normalized size = 0.45 \begin{align*} \frac{\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{b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt{a} \sqrt{c} \Gamma \left (\frac{5}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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