\(\int \frac {(a+b x^3)^{2/3}}{x^3} \, dx\) [539]

Optimal result

Integrand size = 15, antiderivative size = 88 \[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^3} \, dx=-\frac {\left (a+b x^3\right )^{2/3}}{2 x^2}+\frac {b^{2/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{2} b^{2/3} \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 88, 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 = {283, 245} \[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^3} \, dx=\frac {b^{2/3} \arctan \left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{2} b^{2/3} \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )-\frac {\left (a+b x^3\right )^{2/3}}{2 x^2} \]

[In]

Int[(a + b*x^3)^(2/3)/x^3,x]

[Out]

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

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 283

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

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

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.65 \[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^3} \, dx=-\frac {\left (a+b x^3\right )^{2/3}}{2 x^2}+\frac {b^{2/3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{a+b x^3}}\right )}{\sqrt {3}}-\frac {1}{3} b^{2/3} \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )+\frac {1}{6} b^{2/3} \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right ) \]

[In]

Integrate[(a + b*x^3)^(2/3)/x^3,x]

[Out]

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

Maple [A] (verified)

Time = 3.98 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.48

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

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 194 vs. \(2 (68) = 136\).

Time = 227.95 (sec) , antiderivative size = 194, normalized size of antiderivative = 2.20 \[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^3} \, dx=-\frac {2 \, \sqrt {3} \left (-b^{2}\right )^{\frac {1}{3}} x^{2} \arctan \left (\frac {2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b^{2}\right )^{\frac {1}{3}} b x^{2} + 2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {2}{3}} \left (-b^{2}\right )^{\frac {2}{3}} x - \sqrt {3} a b}{3 \, {\left (2 \, b^{2} x^{3} + a b\right )}}\right ) + \left (-b^{2}\right )^{\frac {1}{3}} x^{2} \log \left (\left (-b^{2}\right )^{\frac {2}{3}} x^{2} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} b x - {\left (b x^{3} + a\right )}^{\frac {2}{3}} \left (-b^{2}\right )^{\frac {1}{3}}\right ) - 2 \, \left (-b^{2}\right )^{\frac {1}{3}} x^{2} \log \left (-b x - {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b^{2}\right )^{\frac {1}{3}}\right ) + 3 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{6 \, x^{2}} \]

[In]

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

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.67 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.48 \[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^3} \, dx=\frac {a^{\frac {2}{3}} \Gamma \left (- \frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, - \frac {2}{3} \\ \frac {1}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 x^{2} \Gamma \left (\frac {1}{3}\right )} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.30 \[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^3} \, dx=-\frac {1}{3} \, \sqrt {3} b^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, b^{\frac {1}{3}}}\right ) + \frac {1}{6} \, b^{\frac {2}{3}} \log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right ) - \frac {1}{3} \, b^{\frac {2}{3}} \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) - \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{2 \, x^{2}} \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^3} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{3}} \,d x } \]

[In]

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

[Out]

integrate((b*x^3 + a)^(2/3)/x^3, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^3} \, dx=\int \frac {{\left (b\,x^3+a\right )}^{2/3}}{x^3} \,d x \]

[In]

int((a + b*x^3)^(2/3)/x^3,x)

[Out]

int((a + b*x^3)^(2/3)/x^3, x)