3.966 \(\int \frac{1}{\sqrt{c x} (a+b x^2)^{3/4}} \, dx\)

Optimal. Leaf size=66 \[ -\frac{2 \sqrt{b} (c x)^{3/2} \left (\frac{a}{b x^2}+1\right )^{3/4} \text{EllipticF}\left (\frac{1}{2} \cot ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right ),2\right )}{\sqrt{a} c^2 \left (a+b x^2\right )^{3/4}} \]

[Out]

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

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Rubi [A]  time = 0.0629385, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {329, 237, 335, 275, 231} \[ -\frac{2 \sqrt{b} (c x)^{3/2} \left (\frac{a}{b x^2}+1\right )^{3/4} F\left (\left .\frac{1}{2} \cot ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{\sqrt{a} c^2 \left (a+b x^2\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 231

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

Rubi steps

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

Mathematica [C]  time = 0.0108494, size = 54, normalized size = 0.82 \[ \frac{2 x \left (\frac{b x^2}{a}+1\right )^{3/4} \, _2F_1\left (\frac{1}{4},\frac{3}{4};\frac{5}{4};-\frac{b x^2}{a}\right )}{\sqrt{c x} \left (a+b x^2\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.026, size = 0, normalized size = 0. \begin{align*} \int{{\frac{1}{\sqrt{cx}}} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*x)^(1/2)/(b*x^2+a)^(3/4),x)

[Out]

int(1/(c*x)^(1/2)/(b*x^2+a)^(3/4),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{2} + a\right )}^{\frac{3}{4}} \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)^(3/4),x, algorithm="maxima")

[Out]

integrate(1/((b*x^2 + a)^(3/4)*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}}{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)^(3/4),x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^(1/4)*sqrt(c*x)/(b*c*x^3 + a*c*x), x)

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Sympy [C]  time = 2.56298, size = 31, normalized size = 0.47 \begin{align*} - \frac{{{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{3}{4} \\ \frac{3}{2} \end{matrix}\middle |{\frac{a e^{i \pi }}{b x^{2}}} \right )}}{b^{\frac{3}{4}} \sqrt{c} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)**(1/2)/(b*x**2+a)**(3/4),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{2} + a\right )}^{\frac{3}{4}} \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)^(3/4),x, algorithm="giac")

[Out]

integrate(1/((b*x^2 + a)^(3/4)*sqrt(c*x)), x)