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

Optimal. Leaf size=107 \[ -\frac {9 \sqrt [4]{3} c^2 \sqrt {3-2 x^2} \sqrt {c x} E\left (\left .\sin ^{-1}\left (\frac {\sqrt {3-\sqrt {6} x}}{\sqrt {6}}\right )\right |2\right )}{5\ 2^{3/4} \sqrt {x} \sqrt {3 a-2 a x^2}}-\frac {c \sqrt {3 a-2 a x^2} (c x)^{3/2}}{5 a} \]

[Out]

-9/10*3^(1/4)*c^2*EllipticE(1/6*(3-x*6^(1/2))^(1/2)*6^(1/2),2^(1/2))*(c*x)^(1/2)*(-2*x^2+3)^(1/2)*2^(1/4)/x^(1
/2)/(-2*a*x^2+3*a)^(1/2)-1/5*c*(c*x)^(3/2)*(-2*a*x^2+3*a)^(1/2)/a

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Rubi [A]  time = 0.04, antiderivative size = 107, normalized size of antiderivative = 1.00, 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 = {321, 320, 319, 318, 424} \[ -\frac {9 \sqrt [4]{3} c^2 \sqrt {3-2 x^2} \sqrt {c x} E\left (\left .\sin ^{-1}\left (\frac {\sqrt {3-\sqrt {6} x}}{\sqrt {6}}\right )\right |2\right )}{5\ 2^{3/4} \sqrt {x} \sqrt {3 a-2 a x^2}}-\frac {c \sqrt {3 a-2 a x^2} (c x)^{3/2}}{5 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(c*(c*x)^(3/2)*Sqrt[3*a - 2*a*x^2])/(5*a) - (9*3^(1/4)*c^2*Sqrt[c*x]*Sqrt[3 - 2*x^2]*EllipticE[ArcSin[Sqrt[3
- Sqrt[6]*x]/Sqrt[6]], 2])/(5*2^(3/4)*Sqrt[x]*Sqrt[3*a - 2*a*x^2])

Rule 318

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

Rule 319

Int[Sqrt[x_]/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[1 + (b*x^2)/a]/Sqrt[a + b*x^2], Int[Sqrt[x]/Sqr
t[1 + (b*x^2)/a], x], x] /; FreeQ[{a, b}, x] && GtQ[-(b/a), 0] &&  !GtQ[a, 0]

Rule 320

Int[Sqrt[(c_)*(x_)]/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[c*x]/Sqrt[x], Int[Sqrt[x]/Sqrt[a + b*x^2
], x], x] /; FreeQ[{a, b, c}, x] && GtQ[-(b/a), 0]

Rule 321

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*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rubi steps

\begin {align*} \int \frac {(c x)^{5/2}}{\sqrt {3 a-2 a x^2}} \, dx &=-\frac {c (c x)^{3/2} \sqrt {3 a-2 a x^2}}{5 a}+\frac {1}{10} \left (9 c^2\right ) \int \frac {\sqrt {c x}}{\sqrt {3 a-2 a x^2}} \, dx\\ &=-\frac {c (c x)^{3/2} \sqrt {3 a-2 a x^2}}{5 a}+\frac {\left (9 c^2 \sqrt {c x}\right ) \int \frac {\sqrt {x}}{\sqrt {3 a-2 a x^2}} \, dx}{10 \sqrt {x}}\\ &=-\frac {c (c x)^{3/2} \sqrt {3 a-2 a x^2}}{5 a}+\frac {\left (9 c^2 \sqrt {c x} \sqrt {1-\frac {2 x^2}{3}}\right ) \int \frac {\sqrt {x}}{\sqrt {1-\frac {2 x^2}{3}}} \, dx}{10 \sqrt {x} \sqrt {3 a-2 a x^2}}\\ &=-\frac {c (c x)^{3/2} \sqrt {3 a-2 a x^2}}{5 a}-\frac {\left (9 \left (\frac {3}{2}\right )^{3/4} c^2 \sqrt {c x} \sqrt {1-\frac {2 x^2}{3}}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {1-2 x^2}}{\sqrt {1-x^2}} \, dx,x,\frac {\sqrt {1-\sqrt {\frac {2}{3}} x}}{\sqrt {2}}\right )}{5 \sqrt {x} \sqrt {3 a-2 a x^2}}\\ &=-\frac {c (c x)^{3/2} \sqrt {3 a-2 a x^2}}{5 a}-\frac {9 \sqrt [4]{3} c^2 \sqrt {c x} \sqrt {3-2 x^2} E\left (\left .\sin ^{-1}\left (\frac {\sqrt {3-\sqrt {6} x}}{\sqrt {6}}\right )\right |2\right )}{5\ 2^{3/4} \sqrt {x} \sqrt {3 a-2 a x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 61, normalized size = 0.57 \[ \frac {c (c x)^{3/2} \left (\sqrt {9-6 x^2} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};\frac {2 x^2}{3}\right )+2 x^2-3\right )}{5 \sqrt {a \left (3-2 x^2\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(c*x)^(3/2)*(-3 + 2*x^2 + Sqrt[9 - 6*x^2]*Hypergeometric2F1[1/2, 3/4, 7/4, (2*x^2)/3]))/(5*Sqrt[a*(3 - 2*x^
2)])

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fricas [F]  time = 0.92, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-2 \, a x^{2} + 3 \, a} \sqrt {c x} c^{2} x^{2}}{2 \, a x^{2} - 3 \, a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c x\right )^{\frac {5}{2}}}{\sqrt {-2 \, a x^{2} + 3 \, a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.04, size = 235, normalized size = 2.20 \[ \frac {\sqrt {c x}\, \sqrt {-\left (2 x^{2}-3\right ) a}\, \left (-16 x^{4}+24 x^{2}+6 \sqrt {2}\, \sqrt {\left (2 x +\sqrt {2}\, \sqrt {3}\right ) \sqrt {2}\, \sqrt {3}}\, \sqrt {\left (-2 x +\sqrt {2}\, \sqrt {3}\right ) \sqrt {2}\, \sqrt {3}}\, \sqrt {3}\, \sqrt {-\sqrt {2}\, \sqrt {3}\, x}\, \EllipticE \left (\frac {\sqrt {3}\, \sqrt {2}\, \sqrt {\left (2 x +\sqrt {2}\, \sqrt {3}\right ) \sqrt {2}\, \sqrt {3}}}{6}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {2}\, \sqrt {\left (2 x +\sqrt {2}\, \sqrt {3}\right ) \sqrt {2}\, \sqrt {3}}\, \sqrt {\left (-2 x +\sqrt {2}\, \sqrt {3}\right ) \sqrt {2}\, \sqrt {3}}\, \sqrt {3}\, \sqrt {-\sqrt {2}\, \sqrt {3}\, x}\, \EllipticF \left (\frac {\sqrt {3}\, \sqrt {2}\, \sqrt {\left (2 x +\sqrt {2}\, \sqrt {3}\right ) \sqrt {2}\, \sqrt {3}}}{6}, \frac {\sqrt {2}}{2}\right )\right ) c^{2}}{40 \left (2 x^{2}-3\right ) a x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/40*c^2/x*(c*x)^(1/2)*(-(2*x^2-3)*a)^(1/2)/a*(6*2^(1/2)*((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)*3^(1/2)*(-2^(1/2)*3^(1/2)*x)^(1/2)*EllipticE(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))-3*2^(1/2)*((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)*3^(1/2)*(-2^(1/2)*3^(1/2)*x)^(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))-16*x^4+24*x^2)/(2*x^2-3)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c x\right )^{\frac {5}{2}}}{\sqrt {-2 \, a x^{2} + 3 \, a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c\,x\right )}^{5/2}}{\sqrt {3\,a-2\,a\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 6.44, size = 51, normalized size = 0.48 \[ \frac {\sqrt {3} c^{\frac {5}{2}} x^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {2 x^{2} e^{2 i \pi }}{3}} \right )}}{6 \sqrt {a} \Gamma \left (\frac {11}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(3)*c**(5/2)*x**(7/2)*gamma(7/4)*hyper((1/2, 7/4), (11/4,), 2*x**2*exp_polar(2*I*pi)/3)/(6*sqrt(a)*gamma(1
1/4))

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