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

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

[Out]

1/9*a*x*(b*x^3+a)^(2/3)/b+1/6*x^4*(b*x^3+a)^(2/3)+1/18*a^2*ln(-b^(1/3)*x+(b*x^3+a)^(1/3))/b^(4/3)-1/27*a^2*arc
tan(1/3*(1+2*b^(1/3)*x/(b*x^3+a)^(1/3))*3^(1/2))/b^(4/3)*3^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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 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]

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]

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*}

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Mathematica [C]  time = 0.06, 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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fricas [A]  time = 0.67, size = 345, normalized size = 2.95 \[ \left [\frac {3 \, \sqrt {\frac {1}{3}} a^{2} b \sqrt {-\frac {1}{b^{\frac {2}{3}}}} \log \left (3 \, b x^{3} - 3 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {2}{3}} x^{2} - 3 \, \sqrt {\frac {1}{3}} {\left (b^{\frac {4}{3}} x^{3} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} b x^{2} - 2 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b^{\frac {2}{3}} x\right )} \sqrt {-\frac {1}{b^{\frac {2}{3}}}} + 2 \, a\right ) + 2 \, a^{2} b^{\frac {2}{3}} \log \left (-\frac {b^{\frac {1}{3}} x - {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) - a^{2} b^{\frac {2}{3}} \log \left (\frac {b^{\frac {2}{3}} x^{2} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right ) + 3 \, {\left (3 \, b^{2} x^{4} + 2 \, a b x\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{54 \, b^{2}}, \frac {6 \, \sqrt {\frac {1}{3}} a^{2} b^{\frac {2}{3}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (b^{\frac {1}{3}} x + 2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}\right )}}{b^{\frac {1}{3}} x}\right ) + 2 \, a^{2} b^{\frac {2}{3}} \log \left (-\frac {b^{\frac {1}{3}} x - {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) - a^{2} b^{\frac {2}{3}} \log \left (\frac {b^{\frac {2}{3}} x^{2} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right ) + 3 \, {\left (3 \, b^{2} x^{4} + 2 \, a b x\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{54 \, b^{2}}\right ] \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x^{3} + a\right )}^{\frac {2}{3}} x^{3}\,{d x} \]

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)

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maple [F]  time = 0.11, size = 0, normalized size = 0.00 \[ \int \left (b \,x^{3}+a \right )^{\frac {2}{3}} x^{3}\, dx \]

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 [B]  time = 2.90, size = 181, normalized size = 1.55 \[ \frac {\sqrt {3} a^{2} \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{27 \, b^{\frac {4}{3}}} - \frac {a^{2} \log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{54 \, b^{\frac {4}{3}}} + \frac {a^{2} \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}{27 \, b^{\frac {4}{3}}} + \frac {\frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}} a^{2} b}{x^{2}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} a^{2}}{x^{5}}}{18 \, {\left (b^{3} - \frac {2 \, {\left (b x^{3} + a\right )} b^{2}}{x^{3}} + \frac {{\left (b x^{3} + a\right )}^{2} b}{x^{6}}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/27*sqrt(3)*a^2*arctan(1/3*sqrt(3)*(b^(1/3) + 2*(b*x^3 + a)^(1/3)/x)/b^(1/3))/b^(4/3) - 1/54*a^2*log(b^(2/3)
+ (b*x^3 + a)^(1/3)*b^(1/3)/x + (b*x^3 + a)^(2/3)/x^2)/b^(4/3) + 1/27*a^2*log(-b^(1/3) + (b*x^3 + a)^(1/3)/x)/
b^(4/3) + 1/18*((b*x^3 + a)^(2/3)*a^2*b/x^2 + 2*(b*x^3 + a)^(5/3)*a^2/x^5)/(b^3 - 2*(b*x^3 + a)*b^2/x^3 + (b*x
^3 + a)^2*b/x^6)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^3\,{\left (b\,x^3+a\right )}^{2/3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b*x^3)^(2/3),x)

[Out]

int(x^3*(a + b*x^3)^(2/3), x)

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sympy [C]  time = 2.50, size = 39, normalized size = 0.33 \[ \frac {a^{\frac {2}{3}} x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} \]

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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