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

Optimal. Leaf size=117 \[ \frac{a^2 \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{18 b^{4/3}}-\frac{a^2 \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{9 \sqrt{3} b^{4/3}}+\frac{1}{6} x^4 \left (a+b x^3\right )^{2/3}+\frac{a x \left (a+b x^3\right )^{2/3}}{9 b} \]

[Out]

(a*x*(a + b*x^3)^(2/3))/(9*b) + (x^4*(a + b*x^3)^(2/3))/6 - (a^2*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/
Sqrt[3]])/(9*Sqrt[3]*b^(4/3)) + (a^2*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)])/(18*b^(4/3))

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Rubi [A]  time = 0.0356335, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {279, 321, 239} \[ \frac{a^2 \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{18 b^{4/3}}-\frac{a^2 \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{9 \sqrt{3} b^{4/3}}+\frac{1}{6} x^4 \left (a+b x^3\right )^{2/3}+\frac{a x \left (a+b x^3\right )^{2/3}}{9 b} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*x^3)^(2/3),x]

[Out]

(a*x*(a + b*x^3)^(2/3))/(9*b) + (x^4*(a + b*x^3)^(2/3))/6 - (a^2*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/
Sqrt[3]])/(9*Sqrt[3]*b^(4/3)) + (a^2*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)])/(18*b^(4/3))

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 321

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

Rule 239

Int[((a_) + (b_.)*(x_)^3)^(-1/3), x_Symbol] :> Simp[ArcTan[(1 + (2*Rt[b, 3]*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(Sq
rt[3]*Rt[b, 3]), x] - Simp[Log[(a + b*x^3)^(1/3) - Rt[b, 3]*x]/(2*Rt[b, 3]), x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int x^3 \left (a+b x^3\right )^{2/3} \, dx &=\frac{1}{6} x^4 \left (a+b x^3\right )^{2/3}+\frac{1}{3} a \int \frac{x^3}{\sqrt [3]{a+b x^3}} \, dx\\ &=\frac{a x \left (a+b x^3\right )^{2/3}}{9 b}+\frac{1}{6} x^4 \left (a+b x^3\right )^{2/3}-\frac{a^2 \int \frac{1}{\sqrt [3]{a+b x^3}} \, dx}{9 b}\\ &=\frac{a x \left (a+b x^3\right )^{2/3}}{9 b}+\frac{1}{6} x^4 \left (a+b x^3\right )^{2/3}-\frac{a^2 \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt{3}}\right )}{9 \sqrt{3} b^{4/3}}+\frac{a^2 \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{18 b^{4/3}}\\ \end{align*}

Mathematica [C]  time = 0.0514558, size = 62, normalized size = 0.53 \[ \frac{x \left (a+b x^3\right )^{2/3} \left (-\frac{a \, _2F_1\left (-\frac{2}{3},\frac{1}{3};\frac{4}{3};-\frac{b x^3}{a}\right )}{\left (\frac{b x^3}{a}+1\right )^{2/3}}+a+b x^3\right )}{6 b} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*x^3)^(2/3),x]

[Out]

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

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Maple [F]  time = 0.026, size = 0, normalized size = 0. \begin{align*} \int{x}^{3} \left ( b{x}^{3}+a \right ) ^{{\frac{2}{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.98637, size = 919, normalized size = 7.85 \begin{align*} \left [\frac{3 \, \sqrt{\frac{1}{3}} a^{2} b \sqrt{-\frac{1}{b^{\frac{2}{3}}}} \log \left (3 \, b x^{3} - 3 \,{\left (b x^{3} + a\right )}^{\frac{1}{3}} b^{\frac{2}{3}} x^{2} - 3 \, \sqrt{\frac{1}{3}}{\left (b^{\frac{4}{3}} x^{3} +{\left (b x^{3} + a\right )}^{\frac{1}{3}} b x^{2} - 2 \,{\left (b x^{3} + a\right )}^{\frac{2}{3}} b^{\frac{2}{3}} x\right )} \sqrt{-\frac{1}{b^{\frac{2}{3}}}} + 2 \, a\right ) + 2 \, a^{2} b^{\frac{2}{3}} \log \left (-\frac{b^{\frac{1}{3}} x -{\left (b x^{3} + a\right )}^{\frac{1}{3}}}{x}\right ) - a^{2} b^{\frac{2}{3}} \log \left (\frac{b^{\frac{2}{3}} x^{2} +{\left (b x^{3} + a\right )}^{\frac{1}{3}} b^{\frac{1}{3}} x +{\left (b x^{3} + a\right )}^{\frac{2}{3}}}{x^{2}}\right ) + 3 \,{\left (3 \, b^{2} x^{4} + 2 \, a b x\right )}{\left (b x^{3} + a\right )}^{\frac{2}{3}}}{54 \, b^{2}}, \frac{6 \, \sqrt{\frac{1}{3}} a^{2} b^{\frac{2}{3}} \arctan \left (\frac{\sqrt{\frac{1}{3}}{\left (b^{\frac{1}{3}} x + 2 \,{\left (b x^{3} + a\right )}^{\frac{1}{3}}\right )}}{b^{\frac{1}{3}} x}\right ) + 2 \, a^{2} b^{\frac{2}{3}} \log \left (-\frac{b^{\frac{1}{3}} x -{\left (b x^{3} + a\right )}^{\frac{1}{3}}}{x}\right ) - a^{2} b^{\frac{2}{3}} \log \left (\frac{b^{\frac{2}{3}} x^{2} +{\left (b x^{3} + a\right )}^{\frac{1}{3}} b^{\frac{1}{3}} x +{\left (b x^{3} + a\right )}^{\frac{2}{3}}}{x^{2}}\right ) + 3 \,{\left (3 \, b^{2} x^{4} + 2 \, a b x\right )}{\left (b x^{3} + a\right )}^{\frac{2}{3}}}{54 \, b^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/54*(3*sqrt(1/3)*a^2*b*sqrt(-1/b^(2/3))*log(3*b*x^3 - 3*(b*x^3 + a)^(1/3)*b^(2/3)*x^2 - 3*sqrt(1/3)*(b^(4/3)
*x^3 + (b*x^3 + a)^(1/3)*b*x^2 - 2*(b*x^3 + a)^(2/3)*b^(2/3)*x)*sqrt(-1/b^(2/3)) + 2*a) + 2*a^2*b^(2/3)*log(-(
b^(1/3)*x - (b*x^3 + a)^(1/3))/x) - a^2*b^(2/3)*log((b^(2/3)*x^2 + (b*x^3 + a)^(1/3)*b^(1/3)*x + (b*x^3 + a)^(
2/3))/x^2) + 3*(3*b^2*x^4 + 2*a*b*x)*(b*x^3 + a)^(2/3))/b^2, 1/54*(6*sqrt(1/3)*a^2*b^(2/3)*arctan(sqrt(1/3)*(b
^(1/3)*x + 2*(b*x^3 + a)^(1/3))/(b^(1/3)*x)) + 2*a^2*b^(2/3)*log(-(b^(1/3)*x - (b*x^3 + a)^(1/3))/x) - a^2*b^(
2/3)*log((b^(2/3)*x^2 + (b*x^3 + a)^(1/3)*b^(1/3)*x + (b*x^3 + a)^(2/3))/x^2) + 3*(3*b^2*x^4 + 2*a*b*x)*(b*x^3
+ a)^(2/3))/b^2]

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Sympy [C]  time = 2.12514, size = 39, normalized size = 0.33 \begin{align*} \frac{a^{\frac{2}{3}} x^{4} \Gamma \left (\frac{4}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{2}{3}, \frac{4}{3} \\ \frac{7}{3} \end{matrix}\middle |{\frac{b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac{7}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**(2/3)*x**4*gamma(4/3)*hyper((-2/3, 4/3), (7/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(7/3))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^3 + a)^(2/3)*x^3, x)