Integrand size = 22, antiderivative size = 94 \[ \int \frac {A+B x^3}{x^2 \left (a+b x^3\right )^{2/3}} \, dx=-\frac {A \sqrt [3]{a+b x^3}}{a x}-\frac {B \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} b^{2/3}}-\frac {B \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{2 b^{2/3}} \] Output:
-A*(b*x^3+a)^(1/3)/a/x-1/3*B*arctan(1/3*(1+2*b^(1/3)*x/(b*x^3+a)^(1/3))*3^ (1/2))*3^(1/2)/b^(2/3)-1/2*B*ln(b^(1/3)*x-(b*x^3+a)^(1/3))/b^(2/3)
Time = 0.55 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.61 \[ \int \frac {A+B x^3}{x^2 \left (a+b x^3\right )^{2/3}} \, dx=-\frac {A \sqrt [3]{a+b x^3}}{a x}-\frac {B \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{a+b x^3}}\right )}{\sqrt {3} b^{2/3}}-\frac {B \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{3 b^{2/3}}+\frac {B \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )}{6 b^{2/3}} \] Input:
Integrate[(A + B*x^3)/(x^2*(a + b*x^3)^(2/3)),x]
Output:
-((A*(a + b*x^3)^(1/3))/(a*x)) - (B*ArcTan[(Sqrt[3]*b^(1/3)*x)/(b^(1/3)*x + 2*(a + b*x^3)^(1/3))])/(Sqrt[3]*b^(2/3)) - (B*Log[-(b^(1/3)*x) + (a + b* x^3)^(1/3)])/(3*b^(2/3)) + (B*Log[b^(2/3)*x^2 + b^(1/3)*x*(a + b*x^3)^(1/3 ) + (a + b*x^3)^(2/3)])/(6*b^(2/3))
Time = 0.32 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.01, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {953, 853}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {A+B x^3}{x^2 \left (a+b x^3\right )^{2/3}} \, dx\) |
\(\Big \downarrow \) 953 |
\(\displaystyle B \int \frac {x}{\left (b x^3+a\right )^{2/3}}dx-\frac {A \sqrt [3]{a+b x^3}}{a x}\) |
\(\Big \downarrow \) 853 |
\(\displaystyle B \left (-\frac {\arctan \left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} b^{2/3}}-\frac {\log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{2 b^{2/3}}\right )-\frac {A \sqrt [3]{a+b x^3}}{a x}\) |
Input:
Int[(A + B*x^3)/(x^2*(a + b*x^3)^(2/3)),x]
Output:
-((A*(a + b*x^3)^(1/3))/(a*x)) + B*(-(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*b^(2/3)))
Int[(x_)/((a_) + (b_.)*(x_)^3)^(2/3), x_Symbol] :> With[{q = Rt[b, 3]}, Sim p[-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]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _)), x_Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Simp[d/e^n Int[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n*(p + 1) + 1, 0] && (IntegerQ[n] || GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && G tQ[m + n, -1]))
Time = 1.21 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.35
method | result | size |
pseudoelliptic | \(-\frac {-\frac {B \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{b^{\frac {1}{3}}}+x \right )}{3 x}\right ) a x}{3}+\frac {B \ln \left (\frac {-b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right ) a x}{3}-\frac {B \ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right ) a x}{6}+A \left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{\frac {2}{3}}}{b^{\frac {2}{3}} x a}\) | \(127\) |
Input:
int((B*x^3+A)/x^2/(b*x^3+a)^(2/3),x,method=_RETURNVERBOSE)
Output:
-(-1/3*B*3^(1/2)*arctan(1/3*3^(1/2)*(2*(b*x^3+a)^(1/3)/b^(1/3)+x)/x)*a*x+1 /3*B*ln((-b^(1/3)*x+(b*x^3+a)^(1/3))/x)*a*x-1/6*B*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*x+A*(b*x^3+a)^(1/3)*b^(2/3))/b^( 2/3)/x/a
Timed out. \[ \int \frac {A+B x^3}{x^2 \left (a+b x^3\right )^{2/3}} \, dx=\text {Timed out} \] Input:
integrate((B*x^3+A)/x^2/(b*x^3+a)^(2/3),x, algorithm="fricas")
Output:
Timed out
Result contains complex when optimal does not.
Time = 1.38 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.78 \[ \int \frac {A+B x^3}{x^2 \left (a+b x^3\right )^{2/3}} \, dx=\frac {A \sqrt [3]{b} \sqrt [3]{\frac {a}{b x^{3}} + 1} \Gamma \left (- \frac {1}{3}\right )}{3 a \Gamma \left (\frac {2}{3}\right )} + \frac {B x^{2} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {2}{3}} \Gamma \left (\frac {5}{3}\right )} \] Input:
integrate((B*x**3+A)/x**2/(b*x**3+a)**(2/3),x)
Output:
A*b**(1/3)*(a/(b*x**3) + 1)**(1/3)*gamma(-1/3)/(3*a*gamma(2/3)) + B*x**2*g amma(2/3)*hyper((2/3, 2/3), (5/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(2/3)* gamma(5/3))
Time = 0.12 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.29 \[ \int \frac {A+B x^3}{x^2 \left (a+b x^3\right )^{2/3}} \, dx=\frac {1}{6} \, B {\left (\frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{b^{\frac {2}{3}}} + \frac {\log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{b^{\frac {2}{3}}} - \frac {2 \, \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}{b^{\frac {2}{3}}}\right )} - \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} A}{a x} \] Input:
integrate((B*x^3+A)/x^2/(b*x^3+a)^(2/3),x, algorithm="maxima")
Output:
1/6*B*(2*sqrt(3)*arctan(1/3*sqrt(3)*(b^(1/3) + 2*(b*x^3 + a)^(1/3)/x)/b^(1 /3))/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)/b^(2/3) - 2*log(-b^(1/3) + (b*x^3 + a)^(1/3)/x)/b^(2/3)) - (b*x^3 + a)^(1/3)*A/(a*x)
\[ \int \frac {A+B x^3}{x^2 \left (a+b x^3\right )^{2/3}} \, dx=\int { \frac {B x^{3} + A}{{\left (b x^{3} + a\right )}^{\frac {2}{3}} x^{2}} \,d x } \] Input:
integrate((B*x^3+A)/x^2/(b*x^3+a)^(2/3),x, algorithm="giac")
Output:
integrate((B*x^3 + A)/((b*x^3 + a)^(2/3)*x^2), x)
Timed out. \[ \int \frac {A+B x^3}{x^2 \left (a+b x^3\right )^{2/3}} \, dx=\int \frac {B\,x^3+A}{x^2\,{\left (b\,x^3+a\right )}^{2/3}} \,d x \] Input:
int((A + B*x^3)/(x^2*(a + b*x^3)^(2/3)),x)
Output:
int((A + B*x^3)/(x^2*(a + b*x^3)^(2/3)), x)
\[ \int \frac {A+B x^3}{x^2 \left (a+b x^3\right )^{2/3}} \, dx=\left (\int \frac {x}{\left (b \,x^{3}+a \right )^{\frac {2}{3}}}d x \right ) b +\left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {2}{3}} x^{2}}d x \right ) a \] Input:
int((B*x^3+A)/x^2/(b*x^3+a)^(2/3),x)
Output:
int(x/(a + b*x**3)**(2/3),x)*b + int(1/((a + b*x**3)**(2/3)*x**2),x)*a