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

Optimal. Leaf size=284 \[ -\frac{16 \sqrt{2-\sqrt{3}} b \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt{\frac{a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}{\left (1-\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}\right ),4 \sqrt{3}-7\right )}{9 \sqrt [4]{3} x \sqrt{-\frac{\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}}-\frac{\left (a+b x^2\right )^{4/3}}{3 x^3}-\frac{8 b \sqrt [3]{a+b x^2}}{9 x} \]

[Out]

(-8*b*(a + b*x^2)^(1/3))/(9*x) - (a + b*x^2)^(4/3)/(3*x^3) - (16*Sqrt[2 - Sqrt[3]]*b*(a^(1/3) - (a + b*x^2)^(1
/3))*Sqrt[(a^(2/3) + a^(1/3)*(a + b*x^2)^(1/3) + (a + b*x^2)^(2/3))/((1 - Sqrt[3])*a^(1/3) - (a + b*x^2)^(1/3)
)^2]*EllipticF[ArcSin[((1 + Sqrt[3])*a^(1/3) - (a + b*x^2)^(1/3))/((1 - Sqrt[3])*a^(1/3) - (a + b*x^2)^(1/3))]
, -7 + 4*Sqrt[3]])/(9*3^(1/4)*x*Sqrt[-((a^(1/3)*(a^(1/3) - (a + b*x^2)^(1/3)))/((1 - Sqrt[3])*a^(1/3) - (a + b
*x^2)^(1/3))^2)])

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Rubi [A]  time = 0.150323, antiderivative size = 284, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {277, 236, 219} \[ -\frac{16 \sqrt{2-\sqrt{3}} b \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt{\frac{a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} F\left (\sin ^{-1}\left (\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{b x^2+a}}{\left (1-\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{b x^2+a}}\right )|-7+4 \sqrt{3}\right )}{9 \sqrt [4]{3} x \sqrt{-\frac{\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}}-\frac{\left (a+b x^2\right )^{4/3}}{3 x^3}-\frac{8 b \sqrt [3]{a+b x^2}}{9 x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(4/3)/x^4,x]

[Out]

(-8*b*(a + b*x^2)^(1/3))/(9*x) - (a + b*x^2)^(4/3)/(3*x^3) - (16*Sqrt[2 - Sqrt[3]]*b*(a^(1/3) - (a + b*x^2)^(1
/3))*Sqrt[(a^(2/3) + a^(1/3)*(a + b*x^2)^(1/3) + (a + b*x^2)^(2/3))/((1 - Sqrt[3])*a^(1/3) - (a + b*x^2)^(1/3)
)^2]*EllipticF[ArcSin[((1 + Sqrt[3])*a^(1/3) - (a + b*x^2)^(1/3))/((1 - Sqrt[3])*a^(1/3) - (a + b*x^2)^(1/3))]
, -7 + 4*Sqrt[3]])/(9*3^(1/4)*x*Sqrt[-((a^(1/3)*(a^(1/3) - (a + b*x^2)^(1/3)))/((1 - Sqrt[3])*a^(1/3) - (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 236

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

Rule 219

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(2*Sqr
t[2 - Sqrt[3]]*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 - Sqrt[3])*s + r*x)^2]*EllipticF[ArcSin[((1 + Sqrt[3
])*s + r*x)/((1 - Sqrt[3])*s + r*x)], -7 + 4*Sqrt[3]])/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[-((s*(s + r*x))/((1 - S
qrt[3])*s + r*x)^2)]), x]] /; FreeQ[{a, b}, x] && NegQ[a]

Rubi steps

\begin{align*} \int \frac{\left (a+b x^2\right )^{4/3}}{x^4} \, dx &=-\frac{\left (a+b x^2\right )^{4/3}}{3 x^3}+\frac{1}{9} (8 b) \int \frac{\sqrt [3]{a+b x^2}}{x^2} \, dx\\ &=-\frac{8 b \sqrt [3]{a+b x^2}}{9 x}-\frac{\left (a+b x^2\right )^{4/3}}{3 x^3}+\frac{1}{27} \left (16 b^2\right ) \int \frac{1}{\left (a+b x^2\right )^{2/3}} \, dx\\ &=-\frac{8 b \sqrt [3]{a+b x^2}}{9 x}-\frac{\left (a+b x^2\right )^{4/3}}{3 x^3}+\frac{\left (8 b \sqrt{b x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-a+x^3}} \, dx,x,\sqrt [3]{a+b x^2}\right )}{9 x}\\ &=-\frac{8 b \sqrt [3]{a+b x^2}}{9 x}-\frac{\left (a+b x^2\right )^{4/3}}{3 x^3}-\frac{16 \sqrt{2-\sqrt{3}} b \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt{\frac{a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} F\left (\sin ^{-1}\left (\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}{\left (1-\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}\right )|-7+4 \sqrt{3}\right )}{9 \sqrt [4]{3} x \sqrt{-\frac{\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}}\\ \end{align*}

Mathematica [C]  time = 0.0090037, size = 52, normalized size = 0.18 \[ -\frac{a \sqrt [3]{a+b x^2} \, _2F_1\left (-\frac{3}{2},-\frac{4}{3};-\frac{1}{2};-\frac{b x^2}{a}\right )}{3 x^3 \sqrt [3]{\frac{b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(4/3)/x^4,x]

[Out]

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

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Maple [F]  time = 0.027, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{4}} \left ( b{x}^{2}+a \right ) ^{{\frac{4}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((b*x^2+a)^(4/3)/x^4,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}}}{x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^2 + a)^(4/3)/x^4, x)

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Sympy [A]  time = 1.11716, size = 34, normalized size = 0.12 \begin{align*} - \frac{a^{\frac{4}{3}}{{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{2}, - \frac{4}{3} \\ - \frac{1}{2} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{3 x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**(4/3)*hyper((-3/2, -4/3), (-1/2,), b*x**2*exp_polar(I*pi)/a)/(3*x**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}}}{x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^2 + a)^(4/3)/x^4, x)