3.782 \(\int \frac{1}{(c x)^{8/3} (a+b x^2)^{2/3}} \, dx\)

Optimal. Leaf size=394 \[ -\frac{3\ 3^{3/4} b \sqrt [3]{c x} \sqrt [3]{a+b x^2} \left (c^{2/3}-\frac{\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right ) \sqrt{\frac{\frac{b^{2/3} (c x)^{4/3}}{\left (a+b x^2\right )^{2/3}}+\frac{\sqrt [3]{b} c^{2/3} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}+c^{4/3}}{\left (c^{2/3}-\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )^2}} \text{EllipticF}\left (\cos ^{-1}\left (\frac{c^{2/3}-\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}}{c^{2/3}-\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}}\right ),\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{10 a^2 c^{11/3} \sqrt{-\frac{\sqrt [3]{b} (c x)^{2/3} \left (c^{2/3}-\frac{\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )}{\sqrt [3]{a+b x^2} \left (c^{2/3}-\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )^2}}}-\frac{3 \sqrt [3]{a+b x^2}}{5 a c (c x)^{5/3}} \]

[Out]

(-3*(a + b*x^2)^(1/3))/(5*a*c*(c*x)^(5/3)) - (3*3^(3/4)*b*(c*x)^(1/3)*(a + b*x^2)^(1/3)*(c^(2/3) - (b^(1/3)*(c
*x)^(2/3))/(a + b*x^2)^(1/3))*Sqrt[(c^(4/3) + (b^(2/3)*(c*x)^(4/3))/(a + b*x^2)^(2/3) + (b^(1/3)*c^(2/3)*(c*x)
^(2/3))/(a + b*x^2)^(1/3))/(c^(2/3) - ((1 + Sqrt[3])*b^(1/3)*(c*x)^(2/3))/(a + b*x^2)^(1/3))^2]*EllipticF[ArcC
os[(c^(2/3) - ((1 - Sqrt[3])*b^(1/3)*(c*x)^(2/3))/(a + b*x^2)^(1/3))/(c^(2/3) - ((1 + Sqrt[3])*b^(1/3)*(c*x)^(
2/3))/(a + b*x^2)^(1/3))], (2 + Sqrt[3])/4])/(10*a^2*c^(11/3)*Sqrt[-((b^(1/3)*(c*x)^(2/3)*(c^(2/3) - (b^(1/3)*
(c*x)^(2/3))/(a + b*x^2)^(1/3)))/((a + b*x^2)^(1/3)*(c^(2/3) - ((1 + Sqrt[3])*b^(1/3)*(c*x)^(2/3))/(a + b*x^2)
^(1/3))^2))])

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Rubi [A]  time = 0.641643, antiderivative size = 394, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {325, 329, 241, 225} \[ -\frac{3\ 3^{3/4} b \sqrt [3]{c x} \sqrt [3]{a+b x^2} \left (c^{2/3}-\frac{\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right ) \sqrt{\frac{\frac{b^{2/3} (c x)^{4/3}}{\left (a+b x^2\right )^{2/3}}+\frac{\sqrt [3]{b} c^{2/3} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}+c^{4/3}}{\left (c^{2/3}-\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )^2}} F\left (\cos ^{-1}\left (\frac{c^{2/3}-\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{b x^2+a}}}{c^{2/3}-\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{b x^2+a}}}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{10 a^2 c^{11/3} \sqrt{-\frac{\sqrt [3]{b} (c x)^{2/3} \left (c^{2/3}-\frac{\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )}{\sqrt [3]{a+b x^2} \left (c^{2/3}-\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )^2}}}-\frac{3 \sqrt [3]{a+b x^2}}{5 a c (c x)^{5/3}} \]

Antiderivative was successfully verified.

[In]

Int[1/((c*x)^(8/3)*(a + b*x^2)^(2/3)),x]

[Out]

(-3*(a + b*x^2)^(1/3))/(5*a*c*(c*x)^(5/3)) - (3*3^(3/4)*b*(c*x)^(1/3)*(a + b*x^2)^(1/3)*(c^(2/3) - (b^(1/3)*(c
*x)^(2/3))/(a + b*x^2)^(1/3))*Sqrt[(c^(4/3) + (b^(2/3)*(c*x)^(4/3))/(a + b*x^2)^(2/3) + (b^(1/3)*c^(2/3)*(c*x)
^(2/3))/(a + b*x^2)^(1/3))/(c^(2/3) - ((1 + Sqrt[3])*b^(1/3)*(c*x)^(2/3))/(a + b*x^2)^(1/3))^2]*EllipticF[ArcC
os[(c^(2/3) - ((1 - Sqrt[3])*b^(1/3)*(c*x)^(2/3))/(a + b*x^2)^(1/3))/(c^(2/3) - ((1 + Sqrt[3])*b^(1/3)*(c*x)^(
2/3))/(a + b*x^2)^(1/3))], (2 + Sqrt[3])/4])/(10*a^2*c^(11/3)*Sqrt[-((b^(1/3)*(c*x)^(2/3)*(c^(2/3) - (b^(1/3)*
(c*x)^(2/3))/(a + b*x^2)^(1/3)))/((a + b*x^2)^(1/3)*(c^(2/3) - ((1 + Sqrt[3])*b^(1/3)*(c*x)^(2/3))/(a + b*x^2)
^(1/3))^2))])

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 241

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a/(a + b*x^n))^(p + 1/n)*(a + b*x^n)^(p + 1/n), Subst[In
t[1/(1 - b*x^n)^(p + 1/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p,
0] && NeQ[p, -2^(-1)] && LtQ[Denominator[p + 1/n], Denominator[p]]

Rule 225

Int[1/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(x*(s
+ r*x^2)*Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/(s + (1 + Sqrt[3])*r*x^2)^2]*EllipticF[ArcCos[(s + (1 - Sqrt[3])*r*x^2
)/(s + (1 + Sqrt[3])*r*x^2)], (2 + Sqrt[3])/4])/(2*3^(1/4)*s*Sqrt[a + b*x^6]*Sqrt[(r*x^2*(s + r*x^2))/(s + (1
+ Sqrt[3])*r*x^2)^2]), x]] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{1}{(c x)^{8/3} \left (a+b x^2\right )^{2/3}} \, dx &=-\frac{3 \sqrt [3]{a+b x^2}}{5 a c (c x)^{5/3}}-\frac{(3 b) \int \frac{1}{(c x)^{2/3} \left (a+b x^2\right )^{2/3}} \, dx}{5 a c^2}\\ &=-\frac{3 \sqrt [3]{a+b x^2}}{5 a c (c x)^{5/3}}-\frac{(9 b) \operatorname{Subst}\left (\int \frac{1}{\left (a+\frac{b x^6}{c^2}\right )^{2/3}} \, dx,x,\sqrt [3]{c x}\right )}{5 a c^3}\\ &=-\frac{3 \sqrt [3]{a+b x^2}}{5 a c (c x)^{5/3}}-\frac{(9 b) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{b x^6}{c^2}}} \, dx,x,\frac{\sqrt [3]{c x}}{\sqrt [6]{a+b x^2}}\right )}{5 a c^3 \sqrt{\frac{a}{a+b x^2}} \sqrt{a+b x^2}}\\ &=-\frac{3 \sqrt [3]{a+b x^2}}{5 a c (c x)^{5/3}}-\frac{3\ 3^{3/4} b \sqrt [3]{c x} \sqrt [3]{a+b x^2} \left (c^{2/3}-\frac{\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right ) \sqrt{\frac{c^{4/3}+\frac{b^{2/3} (c x)^{4/3}}{\left (a+b x^2\right )^{2/3}}+\frac{\sqrt [3]{b} c^{2/3} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}}{\left (c^{2/3}-\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )^2}} F\left (\cos ^{-1}\left (\frac{c^{2/3}-\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}}{c^{2/3}-\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{10 a^2 c^{11/3} \sqrt{-\frac{\sqrt [3]{b} (c x)^{2/3} \left (c^{2/3}-\frac{\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )}{\sqrt [3]{a+b x^2} \left (c^{2/3}-\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )^2}}}\\ \end{align*}

Mathematica [C]  time = 0.012114, size = 56, normalized size = 0.14 \[ -\frac{3 x \left (\frac{b x^2}{a}+1\right )^{2/3} \, _2F_1\left (-\frac{5}{6},\frac{2}{3};\frac{1}{6};-\frac{b x^2}{a}\right )}{5 (c x)^{8/3} \left (a+b x^2\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((c*x)^(8/3)*(a + b*x^2)^(2/3)),x]

[Out]

(-3*x*(1 + (b*x^2)/a)^(2/3)*Hypergeometric2F1[-5/6, 2/3, 1/6, -((b*x^2)/a)])/(5*(c*x)^(8/3)*(a + b*x^2)^(2/3))

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Maple [F]  time = 0.019, size = 0, normalized size = 0. \begin{align*} \int{ \left ( cx \right ) ^{-{\frac{8}{3}}} \left ( b{x}^{2}+a \right ) ^{-{\frac{2}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*x)^(8/3)/(b*x^2+a)^(2/3),x)

[Out]

int(1/(c*x)^(8/3)/(b*x^2+a)^(2/3),x)

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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{2}{3}} \left (c x\right )^{\frac{8}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(8/3)/(b*x^2+a)^(2/3),x, algorithm="maxima")

[Out]

integrate(1/((b*x^2 + a)^(2/3)*(c*x)^(8/3)), 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{1}{3}} \left (c x\right )^{\frac{1}{3}}}{b c^{3} x^{5} + a c^{3} x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(8/3)/(b*x^2+a)^(2/3),x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^(1/3)*(c*x)^(1/3)/(b*c^3*x^5 + a*c^3*x^3), x)

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Sympy [C]  time = 116.922, size = 48, normalized size = 0.12 \begin{align*} \frac{\Gamma \left (- \frac{5}{6}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{5}{6}, \frac{2}{3} \\ \frac{1}{6} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac{2}{3}} c^{\frac{8}{3}} x^{\frac{5}{3}} \Gamma \left (\frac{1}{6}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)**(8/3)/(b*x**2+a)**(2/3),x)

[Out]

gamma(-5/6)*hyper((-5/6, 2/3), (1/6,), b*x**2*exp_polar(I*pi)/a)/(2*a**(2/3)*c**(8/3)*x**(5/3)*gamma(1/6))

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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{2}{3}} \left (c x\right )^{\frac{8}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(8/3)/(b*x^2+a)^(2/3),x, algorithm="giac")

[Out]

integrate(1/((b*x^2 + a)^(2/3)*(c*x)^(8/3)), x)