3.765 \(\int \frac{(a+b x^2)^{4/3}}{(c x)^{8/3}} \, dx\)

Optimal. Leaf size=414 \[ \frac{8 b \sqrt [3]{c x} \sqrt [3]{a+b x^2} \left (c^{2/3}-\frac{\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right ) \sqrt{\frac{\frac{b^{2/3} (c x)^{4/3}}{\left (a+b x^2\right )^{2/3}}+\frac{\sqrt [3]{b} c^{2/3} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}+c^{4/3}}{\left (c^{2/3}-\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )^2}} \text{EllipticF}\left (\cos ^{-1}\left (\frac{c^{2/3}-\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}}{c^{2/3}-\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}}\right ),\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{5 \sqrt [4]{3} c^{11/3} \sqrt{-\frac{\sqrt [3]{b} (c x)^{2/3} \left (c^{2/3}-\frac{\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )}{\sqrt [3]{a+b x^2} \left (c^{2/3}-\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )^2}}}+\frac{8 b \sqrt [3]{c x} \sqrt [3]{a+b x^2}}{5 c^3}-\frac{3 \left (a+b x^2\right )^{4/3}}{5 c (c x)^{5/3}} \]

[Out]

(8*b*(c*x)^(1/3)*(a + b*x^2)^(1/3))/(5*c^3) - (3*(a + b*x^2)^(4/3))/(5*c*(c*x)^(5/3)) + (8*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[ArcCos[(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])/(5*3^(1/4)*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 + Sqr
t[3])*b^(1/3)*(c*x)^(2/3))/(a + b*x^2)^(1/3))^2))])

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Rubi [A]  time = 0.697927, antiderivative size = 414, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {277, 279, 329, 241, 225} \[ \frac{8 b \sqrt [3]{c x} \sqrt [3]{a+b x^2} \left (c^{2/3}-\frac{\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right ) \sqrt{\frac{\frac{b^{2/3} (c x)^{4/3}}{\left (a+b x^2\right )^{2/3}}+\frac{\sqrt [3]{b} c^{2/3} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}+c^{4/3}}{\left (c^{2/3}-\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )^2}} F\left (\cos ^{-1}\left (\frac{c^{2/3}-\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{b x^2+a}}}{c^{2/3}-\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{b x^2+a}}}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{5 \sqrt [4]{3} c^{11/3} \sqrt{-\frac{\sqrt [3]{b} (c x)^{2/3} \left (c^{2/3}-\frac{\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )}{\sqrt [3]{a+b x^2} \left (c^{2/3}-\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )^2}}}+\frac{8 b \sqrt [3]{c x} \sqrt [3]{a+b x^2}}{5 c^3}-\frac{3 \left (a+b x^2\right )^{4/3}}{5 c (c x)^{5/3}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(4/3)/(c*x)^(8/3),x]

[Out]

(8*b*(c*x)^(1/3)*(a + b*x^2)^(1/3))/(5*c^3) - (3*(a + b*x^2)^(4/3))/(5*c*(c*x)^(5/3)) + (8*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[ArcCos[(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])/(5*3^(1/4)*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 + Sqr
t[3])*b^(1/3)*(c*x)^(2/3))/(a + b*x^2)^(1/3))^2))])

Rule 277

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

Rule 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
n*p + 1)), x] + Dist[(a*n*p)/(m + n*p + 1), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 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 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{\left (a+b x^2\right )^{4/3}}{(c x)^{8/3}} \, dx &=-\frac{3 \left (a+b x^2\right )^{4/3}}{5 c (c x)^{5/3}}+\frac{(8 b) \int \frac{\sqrt [3]{a+b x^2}}{(c x)^{2/3}} \, dx}{5 c^2}\\ &=\frac{8 b \sqrt [3]{c x} \sqrt [3]{a+b x^2}}{5 c^3}-\frac{3 \left (a+b x^2\right )^{4/3}}{5 c (c x)^{5/3}}+\frac{(16 a b) \int \frac{1}{(c x)^{2/3} \left (a+b x^2\right )^{2/3}} \, dx}{15 c^2}\\ &=\frac{8 b \sqrt [3]{c x} \sqrt [3]{a+b x^2}}{5 c^3}-\frac{3 \left (a+b x^2\right )^{4/3}}{5 c (c x)^{5/3}}+\frac{(16 a 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 c^3}\\ &=\frac{8 b \sqrt [3]{c x} \sqrt [3]{a+b x^2}}{5 c^3}-\frac{3 \left (a+b x^2\right )^{4/3}}{5 c (c x)^{5/3}}+\frac{(16 a 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 c^3 \sqrt{\frac{a}{a+b x^2}} \sqrt{a+b x^2}}\\ &=\frac{8 b \sqrt [3]{c x} \sqrt [3]{a+b x^2}}{5 c^3}-\frac{3 \left (a+b x^2\right )^{4/3}}{5 c (c x)^{5/3}}+\frac{8 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 )}{5 \sqrt [4]{3} 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 = 0.0125823, size = 57, normalized size = 0.14 \[ -\frac{3 a x \sqrt [3]{a+b x^2} \, _2F_1\left (-\frac{4}{3},-\frac{5}{6};\frac{1}{6};-\frac{b x^2}{a}\right )}{5 (c x)^{8/3} \sqrt [3]{\frac{b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(4/3)/(c*x)^(8/3),x]

[Out]

(-3*a*x*(a + b*x^2)^(1/3)*Hypergeometric2F1[-4/3, -5/6, 1/6, -((b*x^2)/a)])/(5*(c*x)^(8/3)*(1 + (b*x^2)/a)^(1/
3))

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Maple [F]  time = 0.02, size = 0, normalized size = 0. \begin{align*} \int{ \left ( b{x}^{2}+a \right ) ^{{\frac{4}{3}}} \left ( cx \right ) ^{-{\frac{8}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(4/3)/(c*x)^(8/3),x)

[Out]

int((b*x^2+a)^(4/3)/(c*x)^(8/3),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{2} + a\right )}^{\frac{4}{3}}}{\left (c x\right )^{\frac{8}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(4/3)/(c*x)^(8/3),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(4/3)/(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{4}{3}} \left (c x\right )^{\frac{1}{3}}}{c^{3} x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(4/3)/(c*x)^(8/3),x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^(4/3)*(c*x)^(1/3)/(c^3*x^3), x)

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Sympy [C]  time = 71.727, size = 32, normalized size = 0.08 \begin{align*} \frac{b^{\frac{4}{3}} x{{}_{2}F_{1}\left (\begin{matrix} - \frac{4}{3}, - \frac{1}{2} \\ \frac{1}{2} \end{matrix}\middle |{\frac{a e^{i \pi }}{b x^{2}}} \right )}}{c^{\frac{8}{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(4/3)/(c*x)**(8/3),x)

[Out]

b**(4/3)*x*hyper((-4/3, -1/2), (1/2,), a*exp_polar(I*pi)/(b*x**2))/c**(8/3)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{2} + a\right )}^{\frac{4}{3}}}{\left (c x\right )^{\frac{8}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(4/3)/(c*x)^(8/3),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^(4/3)/(c*x)^(8/3), x)