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

Optimal. Leaf size=285 \[ -\frac{16\ 3^{3/4} \sqrt{2-\sqrt{3}} a^2 \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt{\frac{a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}{\left (1-\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}\right ),4 \sqrt{3}-7\right )}{55 b x \sqrt{-\frac{\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}}+\frac{24}{55} a x \sqrt [3]{a+b x^2}+\frac{3}{11} x \left (a+b x^2\right )^{4/3} \]

[Out]

(24*a*x*(a + b*x^2)^(1/3))/55 + (3*x*(a + b*x^2)^(4/3))/11 - (16*3^(3/4)*Sqrt[2 - Sqrt[3]]*a^2*(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]])/(55*b*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 = 0.166135, antiderivative size = 285, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {195, 236, 219} \[ -\frac{16\ 3^{3/4} \sqrt{2-\sqrt{3}} a^2 \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt{\frac{a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} F\left (\sin ^{-1}\left (\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{b x^2+a}}{\left (1-\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{b x^2+a}}\right )|-7+4 \sqrt{3}\right )}{55 b x \sqrt{-\frac{\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}}+\frac{24}{55} a x \sqrt [3]{a+b x^2}+\frac{3}{11} x \left (a+b x^2\right )^{4/3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(4/3),x]

[Out]

(24*a*x*(a + b*x^2)^(1/3))/55 + (3*x*(a + b*x^2)^(4/3))/11 - (16*3^(3/4)*Sqrt[2 - Sqrt[3]]*a^2*(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]])/(55*b*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 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 \left (a+b x^2\right )^{4/3} \, dx &=\frac{3}{11} x \left (a+b x^2\right )^{4/3}+\frac{1}{11} (8 a) \int \sqrt [3]{a+b x^2} \, dx\\ &=\frac{24}{55} a x \sqrt [3]{a+b x^2}+\frac{3}{11} x \left (a+b x^2\right )^{4/3}+\frac{1}{55} \left (16 a^2\right ) \int \frac{1}{\left (a+b x^2\right )^{2/3}} \, dx\\ &=\frac{24}{55} a x \sqrt [3]{a+b x^2}+\frac{3}{11} x \left (a+b x^2\right )^{4/3}+\frac{\left (24 a^2 \sqrt{b x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-a+x^3}} \, dx,x,\sqrt [3]{a+b x^2}\right )}{55 b x}\\ &=\frac{24}{55} a x \sqrt [3]{a+b x^2}+\frac{3}{11} x \left (a+b x^2\right )^{4/3}-\frac{16\ 3^{3/4} \sqrt{2-\sqrt{3}} a^2 \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 )}{55 b 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 = 0.0061998, size = 47, normalized size = 0.16 \[ \frac{a x \sqrt [3]{a+b x^2} \, _2F_1\left (-\frac{4}{3},\frac{1}{2};\frac{3}{2};-\frac{b x^2}{a}\right )}{\sqrt [3]{\frac{b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(4/3),x]

[Out]

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

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Maple [F]  time = 0.033, size = 0, normalized size = 0. \begin{align*} \int \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)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^2 + a)^(4/3), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b x^{2} + a\right )}^{\frac{4}{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^2 + a)^(4/3), x)

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Sympy [A]  time = 0.921562, size = 26, normalized size = 0.09 \begin{align*} a^{\frac{4}{3}} x{{}_{2}F_{1}\left (\begin{matrix} - \frac{4}{3}, \frac{1}{2} \\ \frac{3}{2} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^2 + a)^(4/3), x)