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

Optimal. Leaf size=125 \[ \frac{c^{3/2} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right ),\frac{1}{2}\right )}{2 \sqrt [4]{a} b^{5/4} \sqrt{a+b x^2}}-\frac{c \sqrt{c x}}{b \sqrt{a+b x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0707704, antiderivative size = 125, 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 = {288, 329, 220} \[ \frac{c^{3/2} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right )|\frac{1}{2}\right )}{2 \sqrt [4]{a} b^{5/4} \sqrt{a+b x^2}}-\frac{c \sqrt{c x}}{b \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(c*x)^(3/2)/(a + b*x^2)^(3/2),x]

[Out]

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

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 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{(c x)^{3/2}}{\left (a+b x^2\right )^{3/2}} \, dx &=-\frac{c \sqrt{c x}}{b \sqrt{a+b x^2}}+\frac{c^2 \int \frac{1}{\sqrt{c x} \sqrt{a+b x^2}} \, dx}{2 b}\\ &=-\frac{c \sqrt{c x}}{b \sqrt{a+b x^2}}+\frac{c \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+\frac{b x^4}{c^2}}} \, dx,x,\sqrt{c x}\right )}{b}\\ &=-\frac{c \sqrt{c x}}{b \sqrt{a+b x^2}}+\frac{c^{3/2} \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 )}{2 \sqrt [4]{a} b^{5/4} \sqrt{a+b x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0222413, size = 59, normalized size = 0.47 \[ \frac{c \sqrt{c x} \left (\sqrt{\frac{b x^2}{a}+1} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{b x^2}{a}\right )-1\right )}{b \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.012, size = 115, normalized size = 0.9 \begin{align*}{\frac{c}{2\,{b}^{2}x}\sqrt{cx} \left ({\it EllipticF} \left ( \sqrt{{ \left ( bx+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}}},{\frac{\sqrt{2}}{2}} \right ) \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}}}}}\sqrt{-ab}-2\,bx \right ){\frac{1}{\sqrt{b{x}^{2}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x)^(3/2)/(b*x^2 + a)^(3/2), 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} c x}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 2.94499, size = 44, normalized size = 0.35 \begin{align*} \frac{c^{\frac{3}{2}} x^{\frac{5}{2}} \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{5}{4}, \frac{3}{2} \\ \frac{9}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac{3}{2}} \Gamma \left (\frac{9}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c**(3/2)*x**(5/2)*gamma(5/4)*hyper((5/4, 3/2), (9/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(3/2)*gamma(9/4))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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