\(\int \frac {b+x^3}{\sqrt [3]{a+x^3}} \, dx\) [1744]

Optimal result

Integrand size = 15, antiderivative size = 118 \[ \int \frac {b+x^3}{\sqrt [3]{a+x^3}} \, dx=\frac {1}{3} x \left (a+x^3\right )^{2/3}+\frac {1}{9} \left (-\sqrt {3} a+3 \sqrt {3} b\right ) \arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{a+x^3}}\right )+\frac {1}{9} (a-3 b) \log \left (-x+\sqrt [3]{a+x^3}\right )+\frac {1}{18} (-a+3 b) \log \left (x^2+x \sqrt [3]{a+x^3}+\left (a+x^3\right )^{2/3}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.62, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {396, 245} \[ \int \frac {b+x^3}{\sqrt [3]{a+x^3}} \, dx=-\frac {(a-3 b) \arctan \left (\frac {\frac {2 x}{\sqrt [3]{a+x^3}}+1}{\sqrt {3}}\right )}{3 \sqrt {3}}+\frac {1}{6} (a-3 b) \log \left (\sqrt [3]{a+x^3}-x\right )+\frac {1}{3} x \left (a+x^3\right )^{2/3} \]

[In]

Int[(b + x^3)/(a + x^3)^(1/3),x]

[Out]

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

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 396

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

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

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.89 \[ \int \frac {b+x^3}{\sqrt [3]{a+x^3}} \, dx=\frac {1}{18} \left (6 x \left (a+x^3\right )^{2/3}-2 \sqrt {3} (a-3 b) \arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{a+x^3}}\right )+2 (a-3 b) \log \left (-x+\sqrt [3]{a+x^3}\right )-(a-3 b) \log \left (x^2+x \sqrt [3]{a+x^3}+\left (a+x^3\right )^{2/3}\right )\right ) \]

[In]

Integrate[(b + x^3)/(a + x^3)^(1/3),x]

[Out]

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

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.10

method	result	size
pseudoelliptic	\(-\frac {\left (\frac {\sqrt {3}\, \left (a -3 b \right ) \arctan \left (\frac {\sqrt {3}\, \left (x +2 \left (x^{3}+a \right )^{\frac {1}{3}}\right )}{3 x}\right )}{3}+x \left (x^{3}+a \right )^{\frac {2}{3}}-\frac {\left (a -3 b \right ) \left (\ln \left (\frac {x^{2}+x \left (x^{3}+a \right )^{\frac {1}{3}}+\left (x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )-2 \ln \left (\frac {-x +\left (x^{3}+a \right )^{\frac {1}{3}}}{x}\right )\right )}{6}\right ) a}{3 \left (x -\left (x^{3}+a \right )^{\frac {1}{3}}\right ) \left (x^{2}+x \left (x^{3}+a \right )^{\frac {1}{3}}+\left (x^{3}+a \right )^{\frac {2}{3}}\right )}\)	\(130\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.86 \[ \int \frac {b+x^3}{\sqrt [3]{a+x^3}} \, dx=\frac {1}{9} \, \sqrt {3} {\left (a - 3 \, b\right )} \arctan \left (\frac {\sqrt {3} x + 2 \, \sqrt {3} {\left (x^{3} + a\right )}^{\frac {1}{3}}}{3 \, x}\right ) + \frac {1}{9} \, {\left (a - 3 \, b\right )} \log \left (-\frac {x - {\left (x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) - \frac {1}{18} \, {\left (a - 3 \, b\right )} \log \left (\frac {x^{2} + {\left (x^{3} + a\right )}^{\frac {1}{3}} x + {\left (x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right ) + \frac {1}{3} \, {\left (x^{3} + a\right )}^{\frac {2}{3}} x \]

[In]

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

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.02 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.62 \[ \int \frac {b+x^3}{\sqrt [3]{a+x^3}} \, dx=\frac {b x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} \Gamma \left (\frac {4}{3}\right )} + \frac {x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} \Gamma \left (\frac {7}{3}\right )} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.44 \[ \int \frac {b+x^3}{\sqrt [3]{a+x^3}} \, dx=\frac {1}{9} \, \sqrt {3} a \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (\frac {2 \, {\left (x^{3} + a\right )}^{\frac {1}{3}}}{x} + 1\right )}\right ) - \frac {1}{6} \, {\left (2 \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (\frac {2 \, {\left (x^{3} + a\right )}^{\frac {1}{3}}}{x} + 1\right )}\right ) - \log \left (\frac {{\left (x^{3} + a\right )}^{\frac {1}{3}}}{x} + \frac {{\left (x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}} + 1\right ) + 2 \, \log \left (\frac {{\left (x^{3} + a\right )}^{\frac {1}{3}}}{x} - 1\right )\right )} b - \frac {1}{18} \, a \log \left (\frac {{\left (x^{3} + a\right )}^{\frac {1}{3}}}{x} + \frac {{\left (x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}} + 1\right ) + \frac {1}{9} \, a \log \left (\frac {{\left (x^{3} + a\right )}^{\frac {1}{3}}}{x} - 1\right ) + \frac {{\left (x^{3} + a\right )}^{\frac {2}{3}} a}{3 \, x^{2} {\left (\frac {x^{3} + a}{x^{3}} - 1\right )}} \]

[In]

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

[Out]

1/9*sqrt(3)*a*arctan(1/3*sqrt(3)*(2*(x^3 + a)^(1/3)/x + 1)) - 1/6*(2*sqrt(3)*arctan(1/3*sqrt(3)*(2*(x^3 + a)^(
1/3)/x + 1)) - log((x^3 + a)^(1/3)/x + (x^3 + a)^(2/3)/x^2 + 1) + 2*log((x^3 + a)^(1/3)/x - 1))*b - 1/18*a*log
((x^3 + a)^(1/3)/x + (x^3 + a)^(2/3)/x^2 + 1) + 1/9*a*log((x^3 + a)^(1/3)/x - 1) + 1/3*(x^3 + a)^(2/3)*a/(x^2*
((x^3 + a)/x^3 - 1))

Giac [F]

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

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

[In]

int((b + x^3)/(a + x^3)^(1/3),x)

[Out]

int((b + x^3)/(a + x^3)^(1/3), x)