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

Optimal. Leaf size=280 \[ -\frac{16 \sqrt{2-\sqrt{3}} a \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt{\frac{a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}{\left (1-\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}\right ),4 \sqrt{3}-7\right )}{5 \sqrt [4]{3} x \sqrt{-\frac{\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}}-\frac{\left (a+b x^2\right )^{4/3}}{x}+\frac{8}{5} b x \sqrt [3]{a+b x^2} \]

[Out]

(8*b*x*(a + b*x^2)^(1/3))/5 - (a + b*x^2)^(4/3)/x - (16*Sqrt[2 - Sqrt[3]]*a*(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]*Elli
pticF[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]])/(5*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 = 0.15408, antiderivative size = 280, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {277, 195, 236, 219} \[ -\frac{16 \sqrt{2-\sqrt{3}} a \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt{\frac{a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} F\left (\sin ^{-1}\left (\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{b x^2+a}}{\left (1-\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{b x^2+a}}\right )|-7+4 \sqrt{3}\right )}{5 \sqrt [4]{3} x \sqrt{-\frac{\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}}-\frac{\left (a+b x^2\right )^{4/3}}{x}+\frac{8}{5} b x \sqrt [3]{a+b x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(4/3)/x^2,x]

[Out]

(8*b*x*(a + b*x^2)^(1/3))/5 - (a + b*x^2)^(4/3)/x - (16*Sqrt[2 - Sqrt[3]]*a*(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]*Elli
pticF[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]])/(5*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 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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^2} \, dx &=-\frac{\left (a+b x^2\right )^{4/3}}{x}+\frac{1}{3} (8 b) \int \sqrt [3]{a+b x^2} \, dx\\ &=\frac{8}{5} b x \sqrt [3]{a+b x^2}-\frac{\left (a+b x^2\right )^{4/3}}{x}+\frac{1}{15} (16 a b) \int \frac{1}{\left (a+b x^2\right )^{2/3}} \, dx\\ &=\frac{8}{5} b x \sqrt [3]{a+b x^2}-\frac{\left (a+b x^2\right )^{4/3}}{x}+\frac{\left (8 a \sqrt{b x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-a+x^3}} \, dx,x,\sqrt [3]{a+b x^2}\right )}{5 x}\\ &=\frac{8}{5} b x \sqrt [3]{a+b x^2}-\frac{\left (a+b x^2\right )^{4/3}}{x}-\frac{16 \sqrt{2-\sqrt{3}} a \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 )}{5 \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 = 0.0085901, size = 50, normalized size = 0.18 \[ -\frac{a \sqrt [3]{a+b x^2} \, _2F_1\left (-\frac{4}{3},-\frac{1}{2};\frac{1}{2};-\frac{b x^2}{a}\right )}{x \sqrt [3]{\frac{b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(4/3)/x^2,x]

[Out]

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

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Maple [F]  time = 0.026, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{2}} \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^2,x)

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^2 + a)^(4/3)/x^2, x)

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Sympy [A]  time = 1.0987, size = 29, normalized size = 0.1 \begin{align*} - \frac{a^{\frac{4}{3}}{{}_{2}F_{1}\left (\begin{matrix} - \frac{4}{3}, - \frac{1}{2} \\ \frac{1}{2} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^2 + a)^(4/3)/x^2, x)