\(\int x \sqrt [3]{a+b x^3} \, dx\) [517]

Optimal result

Integrand size = 13, antiderivative size = 94 \[ \int x \sqrt [3]{a+b x^3} \, dx=\frac {1}{3} x^2 \sqrt [3]{a+b x^3}-\frac {a \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} b^{2/3}}-\frac {a \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{6 b^{2/3}} \]

[Out]

1/3*x^2*(b*x^3+a)^(1/3)-1/6*a*ln(b^(1/3)*x-(b*x^3+a)^(1/3))/b^(2/3)-1/9*a*arctan(1/3*(1+2*b^(1/3)*x/(b*x^3+a)^
(1/3))*3^(1/2))/b^(2/3)*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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {285, 337} \[ \int x \sqrt [3]{a+b x^3} \, dx=-\frac {a \arctan \left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{3 \sqrt {3} b^{2/3}}-\frac {a \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{6 b^{2/3}}+\frac {1}{3} x^2 \sqrt [3]{a+b x^3} \]

[In]

Int[x*(a + b*x^3)^(1/3),x]

[Out]

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

Rule 285

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 337

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

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

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.50 \[ \int x \sqrt [3]{a+b x^3} \, dx=\frac {6 b^{2/3} x^2 \sqrt [3]{a+b x^3}-2 \sqrt {3} a \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{a+b x^3}}\right )-2 a \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )+a \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )}{18 b^{2/3}} \]

[In]

Integrate[x*(a + b*x^3)^(1/3),x]

[Out]

(6*b^(2/3)*x^2*(a + b*x^3)^(1/3) - 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^(2/3))

Maple [A] (verified)

Time = 4.10 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.29

method	result	size
pseudoelliptic	\(\frac {6 \left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{2} b^{\frac {2}{3}}+2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{3}} x +2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{3}} x}\right ) a -2 \ln \left (\frac {-b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right ) a +\ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right ) a}{18 b^{\frac {2}{3}}}\)	\(121\)

[In]

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

[Out]

1/18*(6*(b*x^3+a)^(1/3)*x^2*b^(2/3)+2*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-2*
ln((-b^(1/3)*x+(b*x^3+a)^(1/3))/x)*a+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)/b^(2/3
)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 193 vs. \(2 (71) = 142\).

Time = 0.29 (sec) , antiderivative size = 193, normalized size of antiderivative = 2.05 \[ \int x \sqrt [3]{a+b x^3} \, dx=\frac {6 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{2} x^{2} + 2 \, \sqrt {3} a b \sqrt {-\left (-b^{2}\right )^{\frac {1}{3}}} \arctan \left (-\frac {{\left (\sqrt {3} \left (-b^{2}\right )^{\frac {1}{3}} b x - 2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b^{2}\right )^{\frac {2}{3}}\right )} \sqrt {-\left (-b^{2}\right )^{\frac {1}{3}}}}{3 \, b^{2} x}\right ) - 2 \, \left (-b^{2}\right )^{\frac {2}{3}} a \log \left (-\frac {\left (-b^{2}\right )^{\frac {2}{3}} x - {\left (b x^{3} + a\right )}^{\frac {1}{3}} b}{x}\right ) + \left (-b^{2}\right )^{\frac {2}{3}} a \log \left (-\frac {\left (-b^{2}\right )^{\frac {1}{3}} b x^{2} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b^{2}\right )^{\frac {2}{3}} x - {\left (b x^{3} + a\right )}^{\frac {2}{3}} b}{x^{2}}\right )}{18 \, b^{2}} \]

[In]

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

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.65 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.41 \[ \int x \sqrt [3]{a+b x^3} \, dx=\frac {\sqrt [3]{a} x^{2} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {5}{3}\right )} \]

[In]

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

[Out]

a**(1/3)*x**2*gamma(2/3)*hyper((-1/3, 2/3), (5/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(5/3))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.43 \[ \int x \sqrt [3]{a+b x^3} \, dx=\frac {\sqrt {3} a \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{9 \, b^{\frac {2}{3}}} + \frac {a \log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{18 \, b^{\frac {2}{3}}} - \frac {a \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}{9 \, b^{\frac {2}{3}}} - \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} a}{3 \, {\left (b - \frac {b x^{3} + a}{x^{3}}\right )} x} \]

[In]

integrate(x*(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^(2/3) + 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^(2/3) - 1/9*a*log(-b^(1/3) + (b*x^3 + a)^(1/3)/x)/b^(2/3)
- 1/3*(b*x^3 + a)^(1/3)*a/((b - (b*x^3 + a)/x^3)*x)

Giac [F]

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

[In]

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

[Out]

integrate((b*x^3 + a)^(1/3)*x, x)

Mupad [F(-1)]

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

[In]

int(x*(a + b*x^3)^(1/3),x)

[Out]

int(x*(a + b*x^3)^(1/3), x)