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

Optimal. Leaf size=152 \[ \frac{4 a^{7/4} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right ),\frac{1}{2}\right )}{7 \sqrt [4]{b} \sqrt{c} \sqrt{a+b x^2}}+\frac{4 a \sqrt{c x} \sqrt{a+b x^2}}{7 c}+\frac{2 \sqrt{c x} \left (a+b x^2\right )^{3/2}}{7 c} \]

[Out]

(4*a*Sqrt[c*x]*Sqrt[a + b*x^2])/(7*c) + (2*Sqrt[c*x]*(a + b*x^2)^(3/2))/(7*c) + (4*a^(7/4)*(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])/
(7*b^(1/4)*Sqrt[c]*Sqrt[a + b*x^2])

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Rubi [A]  time = 0.0904508, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {279, 329, 220} \[ \frac{4 a^{7/4} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right )|\frac{1}{2}\right )}{7 \sqrt [4]{b} \sqrt{c} \sqrt{a+b x^2}}+\frac{4 a \sqrt{c x} \sqrt{a+b x^2}}{7 c}+\frac{2 \sqrt{c x} \left (a+b x^2\right )^{3/2}}{7 c} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(3/2)/Sqrt[c*x],x]

[Out]

(4*a*Sqrt[c*x]*Sqrt[a + b*x^2])/(7*c) + (2*Sqrt[c*x]*(a + b*x^2)^(3/2))/(7*c) + (4*a^(7/4)*(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])/
(7*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{\left (a+b x^2\right )^{3/2}}{\sqrt{c x}} \, dx &=\frac{2 \sqrt{c x} \left (a+b x^2\right )^{3/2}}{7 c}+\frac{1}{7} (6 a) \int \frac{\sqrt{a+b x^2}}{\sqrt{c x}} \, dx\\ &=\frac{4 a \sqrt{c x} \sqrt{a+b x^2}}{7 c}+\frac{2 \sqrt{c x} \left (a+b x^2\right )^{3/2}}{7 c}+\frac{1}{7} \left (4 a^2\right ) \int \frac{1}{\sqrt{c x} \sqrt{a+b x^2}} \, dx\\ &=\frac{4 a \sqrt{c x} \sqrt{a+b x^2}}{7 c}+\frac{2 \sqrt{c x} \left (a+b x^2\right )^{3/2}}{7 c}+\frac{\left (8 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+\frac{b x^4}{c^2}}} \, dx,x,\sqrt{c x}\right )}{7 c}\\ &=\frac{4 a \sqrt{c x} \sqrt{a+b x^2}}{7 c}+\frac{2 \sqrt{c x} \left (a+b x^2\right )^{3/2}}{7 c}+\frac{4 a^{7/4} \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 )}{7 \sqrt [4]{b} \sqrt{c} \sqrt{a+b x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0104361, size = 55, normalized size = 0.36 \[ \frac{2 a x \sqrt{a+b x^2} \, _2F_1\left (-\frac{3}{2},\frac{1}{4};\frac{5}{4};-\frac{b x^2}{a}\right )}{\sqrt{c x} \sqrt{\frac{b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.01, size = 134, normalized size = 0.9 \begin{align*}{\frac{2}{7\,b} \left ( 2\,\sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{2}\sqrt{{\frac{-bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{bx}{\sqrt{-ab}}}}{\it EllipticF} \left ( \sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ) \sqrt{-ab}{a}^{2}+{b}^{3}{x}^{5}+4\,a{b}^{2}{x}^{3}+3\,{a}^{2}bx \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)^(3/2)/(c*x)^(1/2),x)

[Out]

2/7/(b*x^2+a)^(1/2)*(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^2+b^
3*x^5+4*a*b^2*x^3+3*a^2*b*x)/b/(c*x)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Sympy [C]  time = 2.83888, size = 46, normalized size = 0.3 \begin{align*} \frac{a^{\frac{3}{2}} \sqrt{x} \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{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)**(3/2)/(c*x)**(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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