3.648 \(\int \frac{1}{(c x)^{5/2} (3 a-2 a x^2)^{3/2}} \, dx\)

Optimal. Leaf size=132 \[ \frac{5\ 2^{3/4} \sqrt{3-2 x^2} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt [4]{\frac{2}{3}} \sqrt{c x}}{\sqrt{c}}\right ),-1\right )}{27 \sqrt [4]{3} a c^{5/2} \sqrt{a \left (3-2 x^2\right )}}-\frac{5 \sqrt{3 a-2 a x^2}}{27 a^2 c (c x)^{3/2}}+\frac{1}{3 a c \sqrt{3 a-2 a x^2} (c x)^{3/2}} \]

[Out]

1/(3*a*c*(c*x)^(3/2)*Sqrt[3*a - 2*a*x^2]) - (5*Sqrt[3*a - 2*a*x^2])/(27*a^2*c*(c*x)^(3/2)) + (5*2^(3/4)*Sqrt[3
- 2*x^2]*EllipticF[ArcSin[((2/3)^(1/4)*Sqrt[c*x])/Sqrt[c]], -1])/(27*3^(1/4)*a*c^(5/2)*Sqrt[a*(3 - 2*x^2)])

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Rubi [A]  time = 0.0631519, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {290, 325, 329, 224, 221} \[ -\frac{5 \sqrt{3 a-2 a x^2}}{27 a^2 c (c x)^{3/2}}+\frac{5\ 2^{3/4} \sqrt{3-2 x^2} F\left (\left .\sin ^{-1}\left (\frac{\sqrt [4]{\frac{2}{3}} \sqrt{c x}}{\sqrt{c}}\right )\right |-1\right )}{27 \sqrt [4]{3} a c^{5/2} \sqrt{a \left (3-2 x^2\right )}}+\frac{1}{3 a c \sqrt{3 a-2 a x^2} (c x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((c*x)^(5/2)*(3*a - 2*a*x^2)^(3/2)),x]

[Out]

1/(3*a*c*(c*x)^(3/2)*Sqrt[3*a - 2*a*x^2]) - (5*Sqrt[3*a - 2*a*x^2])/(27*a^2*c*(c*x)^(3/2)) + (5*2^(3/4)*Sqrt[3
- 2*x^2]*EllipticF[ArcSin[((2/3)^(1/4)*Sqrt[c*x])/Sqrt[c]], -1])/(27*3^(1/4)*a*c^(5/2)*Sqrt[a*(3 - 2*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 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Dist[Sqrt[1 + (b*x^4)/a]/Sqrt[a + b*x^4], Int[1/Sqrt[1 + (b*x^4)
/a], x], x] /; FreeQ[{a, b}, x] && NegQ[b/a] &&  !GtQ[a, 0]

Rule 221

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

Rubi steps

\begin{align*} \int \frac{1}{(c x)^{5/2} \left (3 a-2 a x^2\right )^{3/2}} \, dx &=\frac{1}{3 a c (c x)^{3/2} \sqrt{3 a-2 a x^2}}+\frac{5 \int \frac{1}{(c x)^{5/2} \sqrt{3 a-2 a x^2}} \, dx}{6 a}\\ &=\frac{1}{3 a c (c x)^{3/2} \sqrt{3 a-2 a x^2}}-\frac{5 \sqrt{3 a-2 a x^2}}{27 a^2 c (c x)^{3/2}}+\frac{5 \int \frac{1}{\sqrt{c x} \sqrt{3 a-2 a x^2}} \, dx}{27 a c^2}\\ &=\frac{1}{3 a c (c x)^{3/2} \sqrt{3 a-2 a x^2}}-\frac{5 \sqrt{3 a-2 a x^2}}{27 a^2 c (c x)^{3/2}}+\frac{10 \operatorname{Subst}\left (\int \frac{1}{\sqrt{3 a-\frac{2 a x^4}{c^2}}} \, dx,x,\sqrt{c x}\right )}{27 a c^3}\\ &=\frac{1}{3 a c (c x)^{3/2} \sqrt{3 a-2 a x^2}}-\frac{5 \sqrt{3 a-2 a x^2}}{27 a^2 c (c x)^{3/2}}+\frac{\left (10 \sqrt{3-2 x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{2 x^4}{3 c^2}}} \, dx,x,\sqrt{c x}\right )}{27 \sqrt{3} a c^3 \sqrt{a \left (3-2 x^2\right )}}\\ &=\frac{1}{3 a c (c x)^{3/2} \sqrt{3 a-2 a x^2}}-\frac{5 \sqrt{3 a-2 a x^2}}{27 a^2 c (c x)^{3/2}}+\frac{5\ 2^{3/4} \sqrt{3-2 x^2} F\left (\left .\sin ^{-1}\left (\frac{\sqrt [4]{\frac{2}{3}} \sqrt{c x}}{\sqrt{c}}\right )\right |-1\right )}{27 \sqrt [4]{3} a c^{5/2} \sqrt{a \left (3-2 x^2\right )}}\\ \end{align*}

Mathematica [C]  time = 0.0187255, size = 58, normalized size = 0.44 \[ -\frac{2 x \left (3-2 x^2\right )^{3/2} \, _2F_1\left (-\frac{3}{4},\frac{3}{2};\frac{1}{4};\frac{2 x^2}{3}\right )}{9 \sqrt{3} \left (a \left (3-2 x^2\right )\right )^{3/2} (c x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((c*x)^(5/2)*(3*a - 2*a*x^2)^(3/2)),x]

[Out]

(-2*x*(3 - 2*x^2)^(3/2)*Hypergeometric2F1[-3/4, 3/2, 1/4, (2*x^2)/3])/(9*Sqrt[3]*(c*x)^(5/2)*(a*(3 - 2*x^2))^(
3/2))

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Maple [A]  time = 0.024, size = 133, normalized size = 1. \begin{align*} -{\frac{1}{162\,{a}^{2}x{c}^{2} \left ( 2\,{x}^{2}-3 \right ) }\sqrt{-a \left ( 2\,{x}^{2}-3 \right ) } \left ( 5\,\sqrt{ \left ( 2\,x+\sqrt{2}\sqrt{3} \right ) \sqrt{2}\sqrt{3}}\sqrt{ \left ( -2\,x+\sqrt{2}\sqrt{3} \right ) \sqrt{2}\sqrt{3}}\sqrt{-x\sqrt{2}\sqrt{3}}{\it EllipticF} \left ( 1/6\,\sqrt{3}\sqrt{2}\sqrt{ \left ( 2\,x+\sqrt{2}\sqrt{3} \right ) \sqrt{2}\sqrt{3}},1/2\,\sqrt{2} \right ) x+60\,{x}^{2}-36 \right ){\frac{1}{\sqrt{cx}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*x)^(5/2)/(-2*a*x^2+3*a)^(3/2),x)

[Out]

-1/162*(-a*(2*x^2-3))^(1/2)*(5*((2*x+2^(1/2)*3^(1/2))*2^(1/2)*3^(1/2))^(1/2)*((-2*x+2^(1/2)*3^(1/2))*2^(1/2)*3
^(1/2))^(1/2)*(-x*2^(1/2)*3^(1/2))^(1/2)*EllipticF(1/6*3^(1/2)*2^(1/2)*((2*x+2^(1/2)*3^(1/2))*2^(1/2)*3^(1/2))
^(1/2),1/2*2^(1/2))*x+60*x^2-36)/x/a^2/c^2/(c*x)^(1/2)/(2*x^2-3)

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

[Out]

integrate(1/((-2*a*x^2 + 3*a)^(3/2)*(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{-2 \, a x^{2} + 3 \, a} \sqrt{c x}}{4 \, a^{2} c^{3} x^{7} - 12 \, a^{2} c^{3} x^{5} + 9 \, a^{2} 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)/(-2*a*x^2+3*a)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(-2*a*x^2 + 3*a)*sqrt(c*x)/(4*a^2*c^3*x^7 - 12*a^2*c^3*x^5 + 9*a^2*c^3*x^3), x)

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Sympy [A]  time = 38.3393, size = 54, normalized size = 0.41 \begin{align*} \frac{\sqrt{3} \Gamma \left (- \frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{4}, \frac{3}{2} \\ \frac{1}{4} \end{matrix}\middle |{\frac{2 x^{2} e^{2 i \pi }}{3}} \right )}}{18 a^{\frac{3}{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)/(-2*a*x**2+3*a)**(3/2),x)

[Out]

sqrt(3)*gamma(-3/4)*hyper((-3/4, 3/2), (1/4,), 2*x**2*exp_polar(2*I*pi)/3)/(18*a**(3/2)*c**(5/2)*x**(3/2)*gamm
a(1/4))

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

[Out]

integrate(1/((-2*a*x^2 + 3*a)^(3/2)*(c*x)^(5/2)), x)