\(\int \frac {x^3}{\sqrt [3]{a+b x^3}} \, dx\) [555]

Optimal result

Integrand size = 15, antiderivative size = 94 \[ \int \frac {x^3}{\sqrt [3]{a+b x^3}} \, dx=\frac {x \left (a+b x^3\right )^{2/3}}{3 b}-\frac {a \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} b^{4/3}}+\frac {a \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{6 b^{4/3}} \]

[Out]

1/3*x*(b*x^3+a)^(2/3)/b+1/6*a*ln(-b^(1/3)*x+(b*x^3+a)^(1/3))/b^(4/3)-1/9*a*arctan(1/3*(1+2*b^(1/3)*x/(b*x^3+a)
^(1/3))*3^(1/2))/b^(4/3)*3^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {327, 245} \[ \int \frac {x^3}{\sqrt [3]{a+b x^3}} \, dx=-\frac {a \arctan \left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{3 \sqrt {3} b^{4/3}}+\frac {a \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{6 b^{4/3}}+\frac {x \left (a+b x^3\right )^{2/3}}{3 b} \]

[In]

Int[x^3/(a + b*x^3)^(1/3),x]

[Out]

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

Rule 245

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 327

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

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.49 \[ \int \frac {x^3}{\sqrt [3]{a+b x^3}} \, dx=\frac {6 \sqrt [3]{b} x \left (a+b x^3\right )^{2/3}-2 \sqrt {3} a \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{a+b x^3}}\right )+2 a \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )-a \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )}{18 b^{4/3}} \]

[In]

Integrate[x^3/(a + b*x^3)^(1/3),x]

[Out]

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

Maple [A] (verified)

Time = 4.00 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.26

method	result	size
pseudoelliptic	\(\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{3}} x +2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{3}} x}\right ) a +3 \left (b \,x^{3}+a \right )^{\frac {2}{3}} x \,b^{\frac {1}{3}}+\ln \left (\frac {-b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right ) a -\frac {\ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right ) a}{2}}{9 b^{\frac {4}{3}}}\)	\(118\)

[In]

int(x^3/(b*x^3+a)^(1/3),x,method=_RETURNVERBOSE)

[Out]

1/9/b^(4/3)*(3^(1/2)*arctan(1/3*3^(1/2)*(b^(1/3)*x+2*(b*x^3+a)^(1/3))/b^(1/3)/x)*a+3*(b*x^3+a)^(2/3)*x*b^(1/3)
+ln((-b^(1/3)*x+(b*x^3+a)^(1/3))/x)*a-1/2*ln((b^(2/3)*x^2+b^(1/3)*(b*x^3+a)^(1/3)*x+(b*x^3+a)^(2/3))/x^2)*a)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 309, normalized size of antiderivative = 3.29 \[ \int \frac {x^3}{\sqrt [3]{a+b x^3}} \, dx=\left [\frac {3 \, \sqrt {\frac {1}{3}} a b \sqrt {-\frac {1}{b^{\frac {2}{3}}}} \log \left (3 \, b x^{3} - 3 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {2}{3}} x^{2} - 3 \, \sqrt {\frac {1}{3}} {\left (b^{\frac {4}{3}} x^{3} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} b x^{2} - 2 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b^{\frac {2}{3}} x\right )} \sqrt {-\frac {1}{b^{\frac {2}{3}}}} + 2 \, a\right ) + 6 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b x + 2 \, a b^{\frac {2}{3}} \log \left (-\frac {b^{\frac {1}{3}} x - {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) - a b^{\frac {2}{3}} \log \left (\frac {b^{\frac {2}{3}} x^{2} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{18 \, b^{2}}, \frac {6 \, \sqrt {\frac {1}{3}} a b^{\frac {2}{3}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (b^{\frac {1}{3}} x + 2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}\right )}}{b^{\frac {1}{3}} x}\right ) + 6 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b x + 2 \, a b^{\frac {2}{3}} \log \left (-\frac {b^{\frac {1}{3}} x - {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) - a b^{\frac {2}{3}} \log \left (\frac {b^{\frac {2}{3}} x^{2} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{18 \, b^{2}}\right ] \]

[In]

integrate(x^3/(b*x^3+a)^(1/3),x, algorithm="fricas")

[Out]

[1/18*(3*sqrt(1/3)*a*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) + 6*(b*x^3 + a)^(2/3)*b*
x + 2*a*b^(2/3)*log(-(b^(1/3)*x - (b*x^3 + a)^(1/3))/x) - a*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))/b^2, 1/18*(6*sqrt(1/3)*a*b^(2/3)*arctan(sqrt(1/3)*(b^(1/3)*x + 2*(b*x^3 + a)^(
1/3))/(b^(1/3)*x)) + 6*(b*x^3 + a)^(2/3)*b*x + 2*a*b^(2/3)*log(-(b^(1/3)*x - (b*x^3 + a)^(1/3))/x) - a*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))/b^2]

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.68 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.39 \[ \int \frac {x^3}{\sqrt [3]{a+b x^3}} \, dx=\frac {x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} \Gamma \left (\frac {7}{3}\right )} \]

[In]

integrate(x**3/(b*x**3+a)**(1/3),x)

[Out]

x**4*gamma(4/3)*hyper((1/3, 4/3), (7/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(1/3)*gamma(7/3))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.46 \[ \int \frac {x^3}{\sqrt [3]{a+b x^3}} \, dx=\frac {\sqrt {3} a \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{9 \, b^{\frac {4}{3}}} - \frac {a \log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{18 \, b^{\frac {4}{3}}} + \frac {a \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}{9 \, b^{\frac {4}{3}}} - \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}} a}{3 \, {\left (b^{2} - \frac {{\left (b x^{3} + a\right )} b}{x^{3}}\right )} x^{2}} \]

[In]

integrate(x^3/(b*x^3+a)^(1/3),x, algorithm="maxima")

[Out]

1/9*sqrt(3)*a*arctan(1/3*sqrt(3)*(b^(1/3) + 2*(b*x^3 + a)^(1/3)/x)/b^(1/3))/b^(4/3) - 1/18*a*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) + 1/9*a*log(-b^(1/3) + (b*x^3 + a)^(1/3)/x)/b^(4/3)
- 1/3*(b*x^3 + a)^(2/3)*a/((b^2 - (b*x^3 + a)*b/x^3)*x^2)

Giac [F]

\[ \int \frac {x^3}{\sqrt [3]{a+b x^3}} \, dx=\int { \frac {x^{3}}{{\left (b x^{3} + a\right )}^{\frac {1}{3}}} \,d x } \]

[In]

integrate(x^3/(b*x^3+a)^(1/3),x, algorithm="giac")

[Out]

integrate(x^3/(b*x^3 + a)^(1/3), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3}{\sqrt [3]{a+b x^3}} \, dx=\int \frac {x^3}{{\left (b\,x^3+a\right )}^{1/3}} \,d x \]

[In]

int(x^3/(a + b*x^3)^(1/3),x)

[Out]

int(x^3/(a + b*x^3)^(1/3), x)