3.635 \(\int \frac{1}{(c x)^{5/2} (a+b x^2)^{5/2}} \, dx\)

Optimal. Leaf size=185 \[ -\frac{5 b^{3/4} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right ),\frac{1}{2}\right )}{4 a^{13/4} c^{5/2} \sqrt{a+b x^2}}-\frac{5 \sqrt{a+b x^2}}{2 a^3 c (c x)^{3/2}}+\frac{3}{2 a^2 c (c x)^{3/2} \sqrt{a+b x^2}}+\frac{1}{3 a c (c x)^{3/2} \left (a+b x^2\right )^{3/2}} \]

[Out]

1/(3*a*c*(c*x)^(3/2)*(a + b*x^2)^(3/2)) + 3/(2*a^2*c*(c*x)^(3/2)*Sqrt[a + b*x^2]) - (5*Sqrt[a + b*x^2])/(2*a^3
*c*(c*x)^(3/2)) - (5*b^(3/4)*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticF[2*ArcTa
n[(b^(1/4)*Sqrt[c*x])/(a^(1/4)*Sqrt[c])], 1/2])/(4*a^(13/4)*c^(5/2)*Sqrt[a + b*x^2])

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Rubi [A]  time = 0.110333, antiderivative size = 185, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {290, 325, 329, 220} \[ -\frac{5 b^{3/4} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right )|\frac{1}{2}\right )}{4 a^{13/4} c^{5/2} \sqrt{a+b x^2}}-\frac{5 \sqrt{a+b x^2}}{2 a^3 c (c x)^{3/2}}+\frac{3}{2 a^2 c (c x)^{3/2} \sqrt{a+b x^2}}+\frac{1}{3 a c (c x)^{3/2} \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((c*x)^(5/2)*(a + b*x^2)^(5/2)),x]

[Out]

1/(3*a*c*(c*x)^(3/2)*(a + b*x^2)^(3/2)) + 3/(2*a^2*c*(c*x)^(3/2)*Sqrt[a + b*x^2]) - (5*Sqrt[a + b*x^2])/(2*a^3
*c*(c*x)^(3/2)) - (5*b^(3/4)*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticF[2*ArcTa
n[(b^(1/4)*Sqrt[c*x])/(a^(1/4)*Sqrt[c])], 1/2])/(4*a^(13/4)*c^(5/2)*Sqrt[a + b*x^2])

Rule 290

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(
a*c*n*(p + 1)), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && 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{1}{(c x)^{5/2} \left (a+b x^2\right )^{5/2}} \, dx &=\frac{1}{3 a c (c x)^{3/2} \left (a+b x^2\right )^{3/2}}+\frac{3 \int \frac{1}{(c x)^{5/2} \left (a+b x^2\right )^{3/2}} \, dx}{2 a}\\ &=\frac{1}{3 a c (c x)^{3/2} \left (a+b x^2\right )^{3/2}}+\frac{3}{2 a^2 c (c x)^{3/2} \sqrt{a+b x^2}}+\frac{15 \int \frac{1}{(c x)^{5/2} \sqrt{a+b x^2}} \, dx}{4 a^2}\\ &=\frac{1}{3 a c (c x)^{3/2} \left (a+b x^2\right )^{3/2}}+\frac{3}{2 a^2 c (c x)^{3/2} \sqrt{a+b x^2}}-\frac{5 \sqrt{a+b x^2}}{2 a^3 c (c x)^{3/2}}-\frac{(5 b) \int \frac{1}{\sqrt{c x} \sqrt{a+b x^2}} \, dx}{4 a^3 c^2}\\ &=\frac{1}{3 a c (c x)^{3/2} \left (a+b x^2\right )^{3/2}}+\frac{3}{2 a^2 c (c x)^{3/2} \sqrt{a+b x^2}}-\frac{5 \sqrt{a+b x^2}}{2 a^3 c (c x)^{3/2}}-\frac{(5 b) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+\frac{b x^4}{c^2}}} \, dx,x,\sqrt{c x}\right )}{2 a^3 c^3}\\ &=\frac{1}{3 a c (c x)^{3/2} \left (a+b x^2\right )^{3/2}}+\frac{3}{2 a^2 c (c x)^{3/2} \sqrt{a+b x^2}}-\frac{5 \sqrt{a+b x^2}}{2 a^3 c (c x)^{3/2}}-\frac{5 b^{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 )}{4 a^{13/4} c^{5/2} \sqrt{a+b x^2}}\\ \end{align*}

Mathematica [C]  time = 0.013029, size = 59, normalized size = 0.32 \[ -\frac{2 x \sqrt{\frac{b x^2}{a}+1} \, _2F_1\left (-\frac{3}{4},\frac{5}{2};\frac{1}{4};-\frac{b x^2}{a}\right )}{3 a^2 (c x)^{5/2} \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((c*x)^(5/2)*(a + b*x^2)^(5/2)),x]

[Out]

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

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Maple [A]  time = 0.018, size = 227, normalized size = 1.2 \begin{align*} -{\frac{1}{12\,x{c}^{2}{a}^{3}} \left ( 15\,\sqrt{-ab}\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 ){x}^{3}b+15\,\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}xa+30\,{b}^{2}{x}^{4}+42\,ab{x}^{2}+8\,{a}^{2} \right ){\frac{1}{\sqrt{cx}}} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*x)^(5/2)/(b*x^2+a)^(5/2),x)

[Out]

-1/12*(15*(-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))*x^3*b+15*((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)*Ellip
ticF(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*(-a*b)^(1/2)*x*a+30*b^2*x^4+42*a*b*x^2+8*a^2)/x/c^2/
(c*x)^(1/2)/a^3/(b*x^2+a)^(3/2)

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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{5}{2}} \left (c x\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(5/2)/(b*x^2+a)^(5/2),x, algorithm="maxima")

[Out]

integrate(1/((b*x^2 + a)^(5/2)*(c*x)^(5/2)), 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}}{b^{3} c^{3} x^{9} + 3 \, a b^{2} c^{3} x^{7} + 3 \, a^{2} b c^{3} x^{5} + a^{3} c^{3} x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(5/2)/(b*x^2+a)^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*x^2 + a)*sqrt(c*x)/(b^3*c^3*x^9 + 3*a*b^2*c^3*x^7 + 3*a^2*b*c^3*x^5 + a^3*c^3*x^3), x)

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Sympy [C]  time = 85.2115, size = 48, normalized size = 0.26 \begin{align*} \frac{\Gamma \left (- \frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{4}, \frac{5}{2} \\ \frac{1}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac{5}{2}} c^{\frac{5}{2}} x^{\frac{3}{2}} \Gamma \left (\frac{1}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)**(5/2)/(b*x**2+a)**(5/2),x)

[Out]

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

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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{5}{2}} \left (c x\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(5/2)/(b*x^2+a)^(5/2),x, algorithm="giac")

[Out]

integrate(1/((b*x^2 + a)^(5/2)*(c*x)^(5/2)), x)