Integrand size = 20, antiderivative size = 196 \[ \int \frac {A+B x^3}{x^3 \left (a+b x^3\right )^2} \, dx=\frac {-5 A b+2 a B}{6 a^2 b x^2}+\frac {A b-a B}{3 a b x^2 \left (a+b x^3\right )}+\frac {(5 A b-2 a B) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{8/3} \sqrt [3]{b}}-\frac {(5 A b-2 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{8/3} \sqrt [3]{b}}+\frac {(5 A b-2 a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 a^{8/3} \sqrt [3]{b}} \]
1/6*(-5*A*b+2*B*a)/a^2/b/x^2+1/3*(A*b-B*a)/a/b/x^2/(b*x^3+a)-1/9*(5*A*b-2* B*a)*ln(a^(1/3)+b^(1/3)*x)/a^(8/3)/b^(1/3)+1/18*(5*A*b-2*B*a)*ln(a^(2/3)-a ^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/a^(8/3)/b^(1/3)+1/9*(5*A*b-2*B*a)*arctan(1/3 *(a^(1/3)-2*b^(1/3)*x)/a^(1/3)*3^(1/2))/a^(8/3)/b^(1/3)*3^(1/2)
Time = 0.13 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.83 \[ \int \frac {A+B x^3}{x^3 \left (a+b x^3\right )^2} \, dx=\frac {-\frac {9 a^{2/3} A}{x^2}+\frac {6 a^{2/3} (-A b+a B) x}{a+b x^3}+\frac {2 \sqrt {3} (5 A b-2 a B) \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}+\frac {2 (-5 A b+2 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{b}}+\frac {(5 A b-2 a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{\sqrt [3]{b}}}{18 a^{8/3}} \]
((-9*a^(2/3)*A)/x^2 + (6*a^(2/3)*(-(A*b) + a*B)*x)/(a + b*x^3) + (2*Sqrt[3 ]*(5*A*b - 2*a*B)*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]])/b^(1/3) + ( 2*(-5*A*b + 2*a*B)*Log[a^(1/3) + b^(1/3)*x])/b^(1/3) + ((5*A*b - 2*a*B)*Lo g[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/b^(1/3))/(18*a^(8/3))
Time = 0.35 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.91, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {957, 847, 750, 16, 1142, 25, 27, 1082, 217, 1103}
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.
| \(\displaystyle \int \frac {A+B x^3}{x^3 \left (a+b x^3\right )^2} \, dx\) |
| \(\Big \downarrow \) 957 |
| \(\displaystyle \frac {(5 A b-2 a B) \int \frac {1}{x^3 \left (b x^3+a\right )}dx}{3 a b}+\frac {A b-a B}{3 a b x^2 \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 847 |
| \(\displaystyle \frac {(5 A b-2 a B) \left (-\frac {b \int \frac {1}{b x^3+a}dx}{a}-\frac {1}{2 a x^2}\right )}{3 a b}+\frac {A b-a B}{3 a b x^2 \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 750 |
| \(\displaystyle \frac {(5 A b-2 a B) \left (-\frac {b \left (\frac {\int \frac {2 \sqrt [3]{a}-\sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 a^{2/3}}+\frac {\int \frac {1}{\sqrt [3]{b} x+\sqrt [3]{a}}dx}{3 a^{2/3}}\right )}{a}-\frac {1}{2 a x^2}\right )}{3 a b}+\frac {A b-a B}{3 a b x^2 \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {(5 A b-2 a B) \left (-\frac {b \left (\frac {\int \frac {2 \sqrt [3]{a}-\sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{a}-\frac {1}{2 a x^2}\right )}{3 a b}+\frac {A b-a B}{3 a b x^2 \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 1142 |
| \(\displaystyle \frac {(5 A b-2 a B) \left (-\frac {b \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx-\frac {\int -\frac {\sqrt [3]{b} \left (\sqrt [3]{a}-2 \sqrt [3]{b} x\right )}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{2 \sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{a}-\frac {1}{2 a x^2}\right )}{3 a b}+\frac {A b-a B}{3 a b x^2 \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {(5 A b-2 a B) \left (-\frac {b \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx+\frac {\int \frac {\sqrt [3]{b} \left (\sqrt [3]{a}-2 \sqrt [3]{b} x\right )}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{2 \sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{a}-\frac {1}{2 a x^2}\right )}{3 a b}+\frac {A b-a B}{3 a b x^2 \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(5 A b-2 a B) \left (-\frac {b \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx+\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{a}-\frac {1}{2 a x^2}\right )}{3 a b}+\frac {A b-a B}{3 a b x^2 \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {(5 A b-2 a B) \left (-\frac {b \left (\frac {\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx+\frac {3 \int \frac {1}{-\left (1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )^2-3}d\left (1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{\sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{a}-\frac {1}{2 a x^2}\right )}{3 a b}+\frac {A b-a B}{3 a b x^2 \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {(5 A b-2 a B) \left (-\frac {b \left (\frac {\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{a}-\frac {1}{2 a x^2}\right )}{3 a b}+\frac {A b-a B}{3 a b x^2 \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {(5 A b-2 a B) \left (-\frac {b \left (\frac {-\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{2 \sqrt [3]{b}}-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{a}-\frac {1}{2 a x^2}\right )}{3 a b}+\frac {A b-a B}{3 a b x^2 \left (a+b x^3\right )}\) |
(A*b - a*B)/(3*a*b*x^2*(a + b*x^3)) + ((5*A*b - 2*a*B)*(-1/2*1/(a*x^2) - ( b*(Log[a^(1/3) + b^(1/3)*x]/(3*a^(2/3)*b^(1/3)) + (-((Sqrt[3]*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]])/b^(1/3)) - Log[a^(2/3) - a^(1/3)*b^(1/3)* x + b^(2/3)*x^2]/(2*b^(1/3)))/(3*a^(2/3))))/a))/(3*a*b)
3.1.83.3.1 Defintions of rubi rules used
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Simp[1/(3*Rt[a, 3]^2) Int[1/ (Rt[a, 3] + Rt[b, 3]*x), x], x] + Simp[1/(3*Rt[a, 3]^2) Int[(2*Rt[a, 3] - Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x] /; FreeQ[{a, b}, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x )^(m + 1)*((a + b*x^n)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))) Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a , b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p , x]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _)), x_Symbol] :> Simp[(-(b*c - a*d))*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a *b*e*n*(p + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b*n* (p + 1)) Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && (( !IntegerQ[p + 1/2] && N eQ[p, -5/4]) || !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] && LeQ[-1 , m, (-n)*(p + 1)]))
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b )], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /; Fre eQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[(2*c*d - b*e)/(2*c) Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
Time = 4.04 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.70
| method | result | size |
| default | \(-\frac {A}{2 a^{2} x^{2}}-\frac {\frac {\left (\frac {A b}{3}-\frac {B a}{3}\right ) x}{b \,x^{3}+a}+\frac {\left (5 A b -2 B a \right ) \left (\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right )}{3}}{a^{2}}\) | \(138\) |
| risch | \(\frac {-\frac {\left (5 A b -2 B a \right ) x^{3}}{6 a^{2}}-\frac {A}{2 a}}{x^{2} \left (b \,x^{3}+a \right )}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{8} b \,\textit {\_Z}^{3}+125 A^{3} b^{3}-150 A^{2} B a \,b^{2}+60 A \,B^{2} a^{2} b -8 B^{3} a^{3}\right )}{\sum }\textit {\_R} \ln \left (\left (-4 \textit {\_R}^{3} a^{8} b -375 A^{3} b^{3}+450 A^{2} B a \,b^{2}-180 A \,B^{2} a^{2} b +24 B^{3} a^{3}\right ) x +\left (-25 A^{2} b^{2} a^{3}+20 A B b \,a^{4}-4 B^{2} a^{5}\right ) \textit {\_R} \right )\right )}{9}\) | \(172\) |
-1/2*A/a^2/x^2-1/a^2*((1/3*A*b-1/3*B*a)*x/(b*x^3+a)+1/3*(5*A*b-2*B*a)*(1/3 /b/(a/b)^(2/3)*ln(x+(a/b)^(1/3))-1/6/b/(a/b)^(2/3)*ln(x^2-(a/b)^(1/3)*x+(a /b)^(2/3))+1/3/b/(a/b)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*x-1 ))))
Time = 0.27 (sec) , antiderivative size = 618, normalized size of antiderivative = 3.15 \[ \int \frac {A+B x^3}{x^3 \left (a+b x^3\right )^2} \, dx=\left [-\frac {9 \, A a^{3} b - 3 \, {\left (2 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{3} + 3 \, \sqrt {\frac {1}{3}} {\left ({\left (2 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{5} + {\left (2 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{2}\right )} \sqrt {\frac {\left (-a^{2} b\right )^{\frac {1}{3}}}{b}} \log \left (\frac {2 \, a b x^{3} + 3 \, \left (-a^{2} b\right )^{\frac {1}{3}} a x - a^{2} - 3 \, \sqrt {\frac {1}{3}} {\left (2 \, a b x^{2} + \left (-a^{2} b\right )^{\frac {2}{3}} x + \left (-a^{2} b\right )^{\frac {1}{3}} a\right )} \sqrt {\frac {\left (-a^{2} b\right )^{\frac {1}{3}}}{b}}}{b x^{3} + a}\right ) + {\left ({\left (2 \, B a b - 5 \, A b^{2}\right )} x^{5} + {\left (2 \, B a^{2} - 5 \, A a b\right )} x^{2}\right )} \left (-a^{2} b\right )^{\frac {2}{3}} \log \left (a b x^{2} - \left (-a^{2} b\right )^{\frac {2}{3}} x - \left (-a^{2} b\right )^{\frac {1}{3}} a\right ) - 2 \, {\left ({\left (2 \, B a b - 5 \, A b^{2}\right )} x^{5} + {\left (2 \, B a^{2} - 5 \, A a b\right )} x^{2}\right )} \left (-a^{2} b\right )^{\frac {2}{3}} \log \left (a b x + \left (-a^{2} b\right )^{\frac {2}{3}}\right )}{18 \, {\left (a^{4} b^{2} x^{5} + a^{5} b x^{2}\right )}}, -\frac {9 \, A a^{3} b - 3 \, {\left (2 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{3} - 6 \, \sqrt {\frac {1}{3}} {\left ({\left (2 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{5} + {\left (2 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{2}\right )} \sqrt {-\frac {\left (-a^{2} b\right )^{\frac {1}{3}}}{b}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, \left (-a^{2} b\right )^{\frac {2}{3}} x + \left (-a^{2} b\right )^{\frac {1}{3}} a\right )} \sqrt {-\frac {\left (-a^{2} b\right )^{\frac {1}{3}}}{b}}}{a^{2}}\right ) + {\left ({\left (2 \, B a b - 5 \, A b^{2}\right )} x^{5} + {\left (2 \, B a^{2} - 5 \, A a b\right )} x^{2}\right )} \left (-a^{2} b\right )^{\frac {2}{3}} \log \left (a b x^{2} - \left (-a^{2} b\right )^{\frac {2}{3}} x - \left (-a^{2} b\right )^{\frac {1}{3}} a\right ) - 2 \, {\left ({\left (2 \, B a b - 5 \, A b^{2}\right )} x^{5} + {\left (2 \, B a^{2} - 5 \, A a b\right )} x^{2}\right )} \left (-a^{2} b\right )^{\frac {2}{3}} \log \left (a b x + \left (-a^{2} b\right )^{\frac {2}{3}}\right )}{18 \, {\left (a^{4} b^{2} x^{5} + a^{5} b x^{2}\right )}}\right ] \]
[-1/18*(9*A*a^3*b - 3*(2*B*a^3*b - 5*A*a^2*b^2)*x^3 + 3*sqrt(1/3)*((2*B*a^ 2*b^2 - 5*A*a*b^3)*x^5 + (2*B*a^3*b - 5*A*a^2*b^2)*x^2)*sqrt((-a^2*b)^(1/3 )/b)*log((2*a*b*x^3 + 3*(-a^2*b)^(1/3)*a*x - a^2 - 3*sqrt(1/3)*(2*a*b*x^2 + (-a^2*b)^(2/3)*x + (-a^2*b)^(1/3)*a)*sqrt((-a^2*b)^(1/3)/b))/(b*x^3 + a) ) + ((2*B*a*b - 5*A*b^2)*x^5 + (2*B*a^2 - 5*A*a*b)*x^2)*(-a^2*b)^(2/3)*log (a*b*x^2 - (-a^2*b)^(2/3)*x - (-a^2*b)^(1/3)*a) - 2*((2*B*a*b - 5*A*b^2)*x ^5 + (2*B*a^2 - 5*A*a*b)*x^2)*(-a^2*b)^(2/3)*log(a*b*x + (-a^2*b)^(2/3)))/ (a^4*b^2*x^5 + a^5*b*x^2), -1/18*(9*A*a^3*b - 3*(2*B*a^3*b - 5*A*a^2*b^2)* x^3 - 6*sqrt(1/3)*((2*B*a^2*b^2 - 5*A*a*b^3)*x^5 + (2*B*a^3*b - 5*A*a^2*b^ 2)*x^2)*sqrt(-(-a^2*b)^(1/3)/b)*arctan(sqrt(1/3)*(2*(-a^2*b)^(2/3)*x + (-a ^2*b)^(1/3)*a)*sqrt(-(-a^2*b)^(1/3)/b)/a^2) + ((2*B*a*b - 5*A*b^2)*x^5 + ( 2*B*a^2 - 5*A*a*b)*x^2)*(-a^2*b)^(2/3)*log(a*b*x^2 - (-a^2*b)^(2/3)*x - (- a^2*b)^(1/3)*a) - 2*((2*B*a*b - 5*A*b^2)*x^5 + (2*B*a^2 - 5*A*a*b)*x^2)*(- a^2*b)^(2/3)*log(a*b*x + (-a^2*b)^(2/3)))/(a^4*b^2*x^5 + a^5*b*x^2)]
Time = 0.40 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.56 \[ \int \frac {A+B x^3}{x^3 \left (a+b x^3\right )^2} \, dx=\frac {- 3 A a + x^{3} \left (- 5 A b + 2 B a\right )}{6 a^{3} x^{2} + 6 a^{2} b x^{5}} + \operatorname {RootSum} {\left (729 t^{3} a^{8} b + 125 A^{3} b^{3} - 150 A^{2} B a b^{2} + 60 A B^{2} a^{2} b - 8 B^{3} a^{3}, \left ( t \mapsto t \log {\left (\frac {9 t a^{3}}{- 5 A b + 2 B a} + x \right )} \right )\right )} \]
(-3*A*a + x**3*(-5*A*b + 2*B*a))/(6*a**3*x**2 + 6*a**2*b*x**5) + RootSum(7 29*_t**3*a**8*b + 125*A**3*b**3 - 150*A**2*B*a*b**2 + 60*A*B**2*a**2*b - 8 *B**3*a**3, Lambda(_t, _t*log(9*_t*a**3/(-5*A*b + 2*B*a) + x)))
Time = 0.27 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.88 \[ \int \frac {A+B x^3}{x^3 \left (a+b x^3\right )^2} \, dx=\frac {{\left (2 \, B a - 5 \, A b\right )} x^{3} - 3 \, A a}{6 \, {\left (a^{2} b x^{5} + a^{3} x^{2}\right )}} + \frac {\sqrt {3} {\left (2 \, B a - 5 \, A b\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, a^{2} b \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {{\left (2 \, B a - 5 \, A b\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \, a^{2} b \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {{\left (2 \, B a - 5 \, A b\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 \, a^{2} b \left (\frac {a}{b}\right )^{\frac {2}{3}}} \]
1/6*((2*B*a - 5*A*b)*x^3 - 3*A*a)/(a^2*b*x^5 + a^3*x^2) + 1/9*sqrt(3)*(2*B *a - 5*A*b)*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^(1/3))/(a^2*b*(a/ b)^(2/3)) - 1/18*(2*B*a - 5*A*b)*log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(a ^2*b*(a/b)^(2/3)) + 1/9*(2*B*a - 5*A*b)*log(x + (a/b)^(1/3))/(a^2*b*(a/b)^ (2/3))
Time = 0.28 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.96 \[ \int \frac {A+B x^3}{x^3 \left (a+b x^3\right )^2} \, dx=-\frac {{\left (2 \, B a - 5 \, A b\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{9 \, a^{3}} + \frac {\sqrt {3} {\left (2 \, \left (-a b^{2}\right )^{\frac {1}{3}} B a - 5 \, \left (-a b^{2}\right )^{\frac {1}{3}} A b\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, a^{3} b} + \frac {B a x - A b x}{3 \, {\left (b x^{3} + a\right )} a^{2}} + \frac {{\left (2 \, \left (-a b^{2}\right )^{\frac {1}{3}} B a - 5 \, \left (-a b^{2}\right )^{\frac {1}{3}} A b\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \, a^{3} b} - \frac {A}{2 \, a^{2} x^{2}} \]
-1/9*(2*B*a - 5*A*b)*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/a^3 + 1/9*sqr t(3)*(2*(-a*b^2)^(1/3)*B*a - 5*(-a*b^2)^(1/3)*A*b)*arctan(1/3*sqrt(3)*(2*x + (-a/b)^(1/3))/(-a/b)^(1/3))/(a^3*b) + 1/3*(B*a*x - A*b*x)/((b*x^3 + a)* a^2) + 1/18*(2*(-a*b^2)^(1/3)*B*a - 5*(-a*b^2)^(1/3)*A*b)*log(x^2 + x*(-a/ b)^(1/3) + (-a/b)^(2/3))/(a^3*b) - 1/2*A/(a^2*x^2)
Time = 0.26 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.81 \[ \int \frac {A+B x^3}{x^3 \left (a+b x^3\right )^2} \, dx=-\frac {\frac {A}{2\,a}+\frac {x^3\,\left (5\,A\,b-2\,B\,a\right )}{6\,a^2}}{b\,x^5+a\,x^2}-\frac {\ln \left (b^{1/3}\,x+a^{1/3}\right )\,\left (5\,A\,b-2\,B\,a\right )}{9\,a^{8/3}\,b^{1/3}}+\frac {\ln \left (a^{1/3}-2\,b^{1/3}\,x+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (5\,A\,b-2\,B\,a\right )}{9\,a^{8/3}\,b^{1/3}}-\frac {\ln \left (2\,b^{1/3}\,x-a^{1/3}+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (5\,A\,b-2\,B\,a\right )}{9\,a^{8/3}\,b^{1/3}} \]
(log(3^(1/2)*a^(1/3)*1i - 2*b^(1/3)*x + a^(1/3))*((3^(1/2)*1i)/2 + 1/2)*(5 *A*b - 2*B*a))/(9*a^(8/3)*b^(1/3)) - (log(b^(1/3)*x + a^(1/3))*(5*A*b - 2* B*a))/(9*a^(8/3)*b^(1/3)) - (A/(2*a) + (x^3*(5*A*b - 2*B*a))/(6*a^2))/(a*x ^2 + b*x^5) - (log(3^(1/2)*a^(1/3)*1i + 2*b^(1/3)*x - a^(1/3))*((3^(1/2)*1 i)/2 - 1/2)*(5*A*b - 2*B*a))/(9*a^(8/3)*b^(1/3))