Integrand size = 20, antiderivative size = 246 \[ \int \frac {A+B x^3}{x^6 \left (a+b x^3\right )^3} \, dx=-\frac {4 (11 A b-5 a B)}{45 a^3 b x^5}+\frac {2 (11 A b-5 a B)}{9 a^4 x^2}+\frac {A b-a B}{6 a b x^5 \left (a+b x^3\right )^2}+\frac {11 A b-5 a B}{18 a^2 b x^5 \left (a+b x^3\right )}-\frac {4 b^{2/3} (11 A b-5 a B) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{9 \sqrt {3} a^{14/3}}+\frac {4 b^{2/3} (11 A b-5 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 a^{14/3}}-\frac {2 b^{2/3} (11 A b-5 a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{27 a^{14/3}} \]
-4/45*(11*A*b-5*B*a)/a^3/b/x^5+2/9*(11*A*b-5*B*a)/a^4/x^2+1/6*(A*b-B*a)/a/ b/x^5/(b*x^3+a)^2+1/18*(11*A*b-5*B*a)/a^2/b/x^5/(b*x^3+a)+4/27*b^(2/3)*(11 *A*b-5*B*a)*ln(a^(1/3)+b^(1/3)*x)/a^(14/3)-2/27*b^(2/3)*(11*A*b-5*B*a)*ln( a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/a^(14/3)-4/27*b^(2/3)*(11*A*b-5*B*a )*arctan(1/3*(a^(1/3)-2*b^(1/3)*x)/a^(1/3)*3^(1/2))/a^(14/3)*3^(1/2)
Time = 0.22 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.85 \[ \int \frac {A+B x^3}{x^6 \left (a+b x^3\right )^3} \, dx=\frac {-\frac {54 a^{5/3} A}{x^5}-\frac {135 a^{2/3} (-3 A b+a B)}{x^2}-\frac {45 a^{5/3} b (-A b+a B) x}{\left (a+b x^3\right )^2}-\frac {15 a^{2/3} b (-17 A b+11 a B) x}{a+b x^3}-40 \sqrt {3} b^{2/3} (11 A b-5 a B) \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )+40 b^{2/3} (11 A b-5 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )+20 b^{2/3} (-11 A b+5 a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{270 a^{14/3}} \]
((-54*a^(5/3)*A)/x^5 - (135*a^(2/3)*(-3*A*b + a*B))/x^2 - (45*a^(5/3)*b*(- (A*b) + a*B)*x)/(a + b*x^3)^2 - (15*a^(2/3)*b*(-17*A*b + 11*a*B)*x)/(a + b *x^3) - 40*Sqrt[3]*b^(2/3)*(11*A*b - 5*a*B)*ArcTan[(1 - (2*b^(1/3)*x)/a^(1 /3))/Sqrt[3]] + 40*b^(2/3)*(11*A*b - 5*a*B)*Log[a^(1/3) + b^(1/3)*x] + 20* b^(2/3)*(-11*A*b + 5*a*B)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/ (270*a^(14/3))
Time = 0.42 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.90, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {957, 819, 847, 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.
\(\displaystyle \int \frac {A+B x^3}{x^6 \left (a+b x^3\right )^3} \, dx\) |
\(\Big \downarrow \) 957 |
\(\displaystyle \frac {(11 A b-5 a B) \int \frac {1}{x^6 \left (b x^3+a\right )^2}dx}{6 a b}+\frac {A b-a B}{6 a b x^5 \left (a+b x^3\right )^2}\) |
\(\Big \downarrow \) 819 |
\(\displaystyle \frac {(11 A b-5 a B) \left (\frac {8 \int \frac {1}{x^6 \left (b x^3+a\right )}dx}{3 a}+\frac {1}{3 a x^5 \left (a+b x^3\right )}\right )}{6 a b}+\frac {A b-a B}{6 a b x^5 \left (a+b x^3\right )^2}\) |
\(\Big \downarrow \) 847 |
\(\displaystyle \frac {(11 A b-5 a B) \left (\frac {8 \left (-\frac {b \int \frac {1}{x^3 \left (b x^3+a\right )}dx}{a}-\frac {1}{5 a x^5}\right )}{3 a}+\frac {1}{3 a x^5 \left (a+b x^3\right )}\right )}{6 a b}+\frac {A b-a B}{6 a b x^5 \left (a+b x^3\right )^2}\) |
\(\Big \downarrow \) 847 |
\(\displaystyle \frac {(11 A b-5 a B) \left (\frac {8 \left (-\frac {b \left (-\frac {b \int \frac {1}{b x^3+a}dx}{a}-\frac {1}{2 a x^2}\right )}{a}-\frac {1}{5 a x^5}\right )}{3 a}+\frac {1}{3 a x^5 \left (a+b x^3\right )}\right )}{6 a b}+\frac {A b-a B}{6 a b x^5 \left (a+b x^3\right )^2}\) |
\(\Big \downarrow \) 750 |
\(\displaystyle \frac {(11 A b-5 a B) \left (\frac {8 \left (-\frac {b \left (-\frac {b \left (\frac {\int \frac {2 \sqrt [3]{a}-\sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 a^{2/3}}+\frac {\int \frac {1}{\sqrt [3]{b} x+\sqrt [3]{a}}dx}{3 a^{2/3}}\right )}{a}-\frac {1}{2 a x^2}\right )}{a}-\frac {1}{5 a x^5}\right )}{3 a}+\frac {1}{3 a x^5 \left (a+b x^3\right )}\right )}{6 a b}+\frac {A b-a B}{6 a b x^5 \left (a+b x^3\right )^2}\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {(11 A b-5 a B) \left (\frac {8 \left (-\frac {b \left (-\frac {b \left (\frac {\int \frac {2 \sqrt [3]{a}-\sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{a}-\frac {1}{2 a x^2}\right )}{a}-\frac {1}{5 a x^5}\right )}{3 a}+\frac {1}{3 a x^5 \left (a+b x^3\right )}\right )}{6 a b}+\frac {A b-a B}{6 a b x^5 \left (a+b x^3\right )^2}\) |
\(\Big \downarrow \) 1142 |
\(\displaystyle \frac {(11 A b-5 a B) \left (\frac {8 \left (-\frac {b \left (-\frac {b \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx-\frac {\int -\frac {\sqrt [3]{b} \left (\sqrt [3]{a}-2 \sqrt [3]{b} x\right )}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{2 \sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{a}-\frac {1}{2 a x^2}\right )}{a}-\frac {1}{5 a x^5}\right )}{3 a}+\frac {1}{3 a x^5 \left (a+b x^3\right )}\right )}{6 a b}+\frac {A b-a B}{6 a b x^5 \left (a+b x^3\right )^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {(11 A b-5 a B) \left (\frac {8 \left (-\frac {b \left (-\frac {b \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx+\frac {\int \frac {\sqrt [3]{b} \left (\sqrt [3]{a}-2 \sqrt [3]{b} x\right )}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{2 \sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{a}-\frac {1}{2 a x^2}\right )}{a}-\frac {1}{5 a x^5}\right )}{3 a}+\frac {1}{3 a x^5 \left (a+b x^3\right )}\right )}{6 a b}+\frac {A b-a B}{6 a b x^5 \left (a+b x^3\right )^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(11 A b-5 a B) \left (\frac {8 \left (-\frac {b \left (-\frac {b \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx+\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{a}-\frac {1}{2 a x^2}\right )}{a}-\frac {1}{5 a x^5}\right )}{3 a}+\frac {1}{3 a x^5 \left (a+b x^3\right )}\right )}{6 a b}+\frac {A b-a B}{6 a b x^5 \left (a+b x^3\right )^2}\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle \frac {(11 A b-5 a B) \left (\frac {8 \left (-\frac {b \left (-\frac {b \left (\frac {\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx+\frac {3 \int \frac {1}{-\left (1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )^2-3}d\left (1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{\sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{a}-\frac {1}{2 a x^2}\right )}{a}-\frac {1}{5 a x^5}\right )}{3 a}+\frac {1}{3 a x^5 \left (a+b x^3\right )}\right )}{6 a b}+\frac {A b-a B}{6 a b x^5 \left (a+b x^3\right )^2}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {(11 A b-5 a B) \left (\frac {8 \left (-\frac {b \left (-\frac {b \left (\frac {\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{a}-\frac {1}{2 a x^2}\right )}{a}-\frac {1}{5 a x^5}\right )}{3 a}+\frac {1}{3 a x^5 \left (a+b x^3\right )}\right )}{6 a b}+\frac {A b-a B}{6 a b x^5 \left (a+b x^3\right )^2}\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {(11 A b-5 a B) \left (\frac {8 \left (-\frac {b \left (-\frac {b \left (\frac {-\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{2 \sqrt [3]{b}}-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{a}-\frac {1}{2 a x^2}\right )}{a}-\frac {1}{5 a x^5}\right )}{3 a}+\frac {1}{3 a x^5 \left (a+b x^3\right )}\right )}{6 a b}+\frac {A b-a B}{6 a b x^5 \left (a+b x^3\right )^2}\) |
(A*b - a*B)/(6*a*b*x^5*(a + b*x^3)^2) + ((11*A*b - 5*a*B)*(1/(3*a*x^5*(a + b*x^3)) + (8*(-1/5*1/(a*x^5) - (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))/a))/(3*a)))/(6*a*b)
3.2.6.3.1 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*n*(p + 1))), x] + Simp[(m + n*(p + 1) + 1)/(a*n*(p + 1)) Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a , b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p , 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 = 4.13 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.71
method | result | size |
default | \(-\frac {A}{5 a^{3} x^{5}}-\frac {-3 A b +B a}{2 x^{2} a^{4}}+\frac {b \left (\frac {\left (\frac {17}{18} b^{2} A -\frac {11}{18} a b B \right ) x^{4}+\frac {a \left (10 A b -7 B a \right ) x}{9}}{\left (b \,x^{3}+a \right )^{2}}+\frac {4 \left (11 A b -5 B a \right ) \left (\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right )}{9}\right )}{a^{4}}\) | \(174\) |
risch | \(\frac {\frac {2 b^{2} \left (11 A b -5 B a \right ) x^{9}}{9 a^{4}}+\frac {16 b \left (11 A b -5 B a \right ) x^{6}}{45 a^{3}}+\frac {\left (11 A b -5 B a \right ) x^{3}}{10 a^{2}}-\frac {A}{5 a}}{x^{5} \left (b \,x^{3}+a \right )^{2}}+\frac {4 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{14} \textit {\_Z}^{3}-1331 A^{3} b^{5}+1815 A^{2} B a \,b^{4}-825 A \,B^{2} a^{2} b^{3}+125 B^{3} a^{3} b^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (-4 \textit {\_R}^{3} a^{14}+3993 A^{3} b^{5}-5445 A^{2} B a \,b^{4}+2475 A \,B^{2} a^{2} b^{3}-375 B^{3} a^{3} b^{2}\right ) x +\left (-121 A^{2} a^{5} b^{3}+110 A B \,a^{6} b^{2}-25 B^{2} a^{7} b \right ) \textit {\_R} \right )\right )}{27}\) | \(221\) |
-1/5*A/a^3/x^5-1/2*(-3*A*b+B*a)/x^2/a^4+1/a^4*b*(((17/18*b^2*A-11/18*a*b*B )*x^4+1/9*a*(10*A*b-7*B*a)*x)/(b*x^3+a)^2+4/9*(11*A*b-5*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 = 384, normalized size of antiderivative = 1.56 \[ \int \frac {A+B x^3}{x^6 \left (a+b x^3\right )^3} \, dx=-\frac {60 \, {\left (5 \, B a b^{2} - 11 \, A b^{3}\right )} x^{9} + 96 \, {\left (5 \, B a^{2} b - 11 \, A a b^{2}\right )} x^{6} + 54 \, A a^{3} + 27 \, {\left (5 \, B a^{3} - 11 \, A a^{2} b\right )} x^{3} + 40 \, \sqrt {3} {\left ({\left (5 \, B a b^{2} - 11 \, A b^{3}\right )} x^{11} + 2 \, {\left (5 \, B a^{2} b - 11 \, A a b^{2}\right )} x^{8} + {\left (5 \, B a^{3} - 11 \, A a^{2} b\right )} x^{5}\right )} \left (\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} a x \left (\frac {b^{2}}{a^{2}}\right )^{\frac {2}{3}} - \sqrt {3} b}{3 \, b}\right ) - 20 \, {\left ({\left (5 \, B a b^{2} - 11 \, A b^{3}\right )} x^{11} + 2 \, {\left (5 \, B a^{2} b - 11 \, A a b^{2}\right )} x^{8} + {\left (5 \, B a^{3} - 11 \, A a^{2} b\right )} x^{5}\right )} \left (\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \log \left (b^{2} x^{2} - a b x \left (\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} + a^{2} \left (\frac {b^{2}}{a^{2}}\right )^{\frac {2}{3}}\right ) + 40 \, {\left ({\left (5 \, B a b^{2} - 11 \, A b^{3}\right )} x^{11} + 2 \, {\left (5 \, B a^{2} b - 11 \, A a b^{2}\right )} x^{8} + {\left (5 \, B a^{3} - 11 \, A a^{2} b\right )} x^{5}\right )} \left (\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \log \left (b x + a \left (\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}}\right )}{270 \, {\left (a^{4} b^{2} x^{11} + 2 \, a^{5} b x^{8} + a^{6} x^{5}\right )}} \]
-1/270*(60*(5*B*a*b^2 - 11*A*b^3)*x^9 + 96*(5*B*a^2*b - 11*A*a*b^2)*x^6 + 54*A*a^3 + 27*(5*B*a^3 - 11*A*a^2*b)*x^3 + 40*sqrt(3)*((5*B*a*b^2 - 11*A*b ^3)*x^11 + 2*(5*B*a^2*b - 11*A*a*b^2)*x^8 + (5*B*a^3 - 11*A*a^2*b)*x^5)*(b ^2/a^2)^(1/3)*arctan(1/3*(2*sqrt(3)*a*x*(b^2/a^2)^(2/3) - sqrt(3)*b)/b) - 20*((5*B*a*b^2 - 11*A*b^3)*x^11 + 2*(5*B*a^2*b - 11*A*a*b^2)*x^8 + (5*B*a^ 3 - 11*A*a^2*b)*x^5)*(b^2/a^2)^(1/3)*log(b^2*x^2 - a*b*x*(b^2/a^2)^(1/3) + a^2*(b^2/a^2)^(2/3)) + 40*((5*B*a*b^2 - 11*A*b^3)*x^11 + 2*(5*B*a^2*b - 1 1*A*a*b^2)*x^8 + (5*B*a^3 - 11*A*a^2*b)*x^5)*(b^2/a^2)^(1/3)*log(b*x + a*( b^2/a^2)^(1/3)))/(a^4*b^2*x^11 + 2*a^5*b*x^8 + a^6*x^5)
Time = 0.59 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.70 \[ \int \frac {A+B x^3}{x^6 \left (a+b x^3\right )^3} \, dx=\operatorname {RootSum} {\left (19683 t^{3} a^{14} - 85184 A^{3} b^{5} + 116160 A^{2} B a b^{4} - 52800 A B^{2} a^{2} b^{3} + 8000 B^{3} a^{3} b^{2}, \left ( t \mapsto t \log {\left (- \frac {27 t a^{5}}{- 44 A b^{2} + 20 B a b} + x \right )} \right )\right )} + \frac {- 18 A a^{3} + x^{9} \cdot \left (220 A b^{3} - 100 B a b^{2}\right ) + x^{6} \cdot \left (352 A a b^{2} - 160 B a^{2} b\right ) + x^{3} \cdot \left (99 A a^{2} b - 45 B a^{3}\right )}{90 a^{6} x^{5} + 180 a^{5} b x^{8} + 90 a^{4} b^{2} x^{11}} \]
RootSum(19683*_t**3*a**14 - 85184*A**3*b**5 + 116160*A**2*B*a*b**4 - 52800 *A*B**2*a**2*b**3 + 8000*B**3*a**3*b**2, Lambda(_t, _t*log(-27*_t*a**5/(-4 4*A*b**2 + 20*B*a*b) + x))) + (-18*A*a**3 + x**9*(220*A*b**3 - 100*B*a*b** 2) + x**6*(352*A*a*b**2 - 160*B*a**2*b) + x**3*(99*A*a**2*b - 45*B*a**3))/ (90*a**6*x**5 + 180*a**5*b*x**8 + 90*a**4*b**2*x**11)
Time = 0.29 (sec) , antiderivative size = 221, normalized size of antiderivative = 0.90 \[ \int \frac {A+B x^3}{x^6 \left (a+b x^3\right )^3} \, dx=-\frac {20 \, {\left (5 \, B a b^{2} - 11 \, A b^{3}\right )} x^{9} + 32 \, {\left (5 \, B a^{2} b - 11 \, A a b^{2}\right )} x^{6} + 18 \, A a^{3} + 9 \, {\left (5 \, B a^{3} - 11 \, A a^{2} b\right )} x^{3}}{90 \, {\left (a^{4} b^{2} x^{11} + 2 \, a^{5} b x^{8} + a^{6} x^{5}\right )}} - \frac {4 \, \sqrt {3} {\left (5 \, B a - 11 \, A b\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{27 \, a^{4} \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {2 \, {\left (5 \, B a - 11 \, A b\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{27 \, a^{4} \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {4 \, {\left (5 \, B a - 11 \, A b\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{27 \, a^{4} \left (\frac {a}{b}\right )^{\frac {2}{3}}} \]
-1/90*(20*(5*B*a*b^2 - 11*A*b^3)*x^9 + 32*(5*B*a^2*b - 11*A*a*b^2)*x^6 + 1 8*A*a^3 + 9*(5*B*a^3 - 11*A*a^2*b)*x^3)/(a^4*b^2*x^11 + 2*a^5*b*x^8 + a^6* x^5) - 4/27*sqrt(3)*(5*B*a - 11*A*b)*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3) )/(a/b)^(1/3))/(a^4*(a/b)^(2/3)) + 2/27*(5*B*a - 11*A*b)*log(x^2 - x*(a/b) ^(1/3) + (a/b)^(2/3))/(a^4*(a/b)^(2/3)) - 4/27*(5*B*a - 11*A*b)*log(x + (a /b)^(1/3))/(a^4*(a/b)^(2/3))
Time = 0.28 (sec) , antiderivative size = 229, normalized size of antiderivative = 0.93 \[ \int \frac {A+B x^3}{x^6 \left (a+b x^3\right )^3} \, dx=-\frac {4 \, \sqrt {3} {\left (5 \, \left (-a b^{2}\right )^{\frac {1}{3}} B a - 11 \, \left (-a b^{2}\right )^{\frac {1}{3}} A b\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{27 \, a^{5}} + \frac {4 \, {\left (5 \, B a b - 11 \, A b^{2}\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{27 \, a^{5}} - \frac {2 \, {\left (5 \, \left (-a b^{2}\right )^{\frac {1}{3}} B a - 11 \, \left (-a b^{2}\right )^{\frac {1}{3}} A b\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{27 \, a^{5}} - \frac {11 \, B a b^{2} x^{4} - 17 \, A b^{3} x^{4} + 14 \, B a^{2} b x - 20 \, A a b^{2} x}{18 \, {\left (b x^{3} + a\right )}^{2} a^{4}} - \frac {5 \, B a x^{3} - 15 \, A b x^{3} + 2 \, A a}{10 \, a^{4} x^{5}} \]
-4/27*sqrt(3)*(5*(-a*b^2)^(1/3)*B*a - 11*(-a*b^2)^(1/3)*A*b)*arctan(1/3*sq rt(3)*(2*x + (-a/b)^(1/3))/(-a/b)^(1/3))/a^5 + 4/27*(5*B*a*b - 11*A*b^2)*( -a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/a^5 - 2/27*(5*(-a*b^2)^(1/3)*B*a - 11*(-a*b^2)^(1/3)*A*b)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/a^5 - 1/18 *(11*B*a*b^2*x^4 - 17*A*b^3*x^4 + 14*B*a^2*b*x - 20*A*a*b^2*x)/((b*x^3 + a )^2*a^4) - 1/10*(5*B*a*x^3 - 15*A*b*x^3 + 2*A*a)/(a^4*x^5)
Time = 6.83 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.84 \[ \int \frac {A+B x^3}{x^6 \left (a+b x^3\right )^3} \, dx=\frac {\frac {x^3\,\left (11\,A\,b-5\,B\,a\right )}{10\,a^2}-\frac {A}{5\,a}+\frac {2\,b^2\,x^9\,\left (11\,A\,b-5\,B\,a\right )}{9\,a^4}+\frac {16\,b\,x^6\,\left (11\,A\,b-5\,B\,a\right )}{45\,a^3}}{a^2\,x^5+2\,a\,b\,x^8+b^2\,x^{11}}+\frac {4\,b^{2/3}\,\ln \left (b^{1/3}\,x+a^{1/3}\right )\,\left (11\,A\,b-5\,B\,a\right )}{27\,a^{14/3}}-\frac {4\,b^{2/3}\,\ln \left (a^{1/3}-2\,b^{1/3}\,x+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (11\,A\,b-5\,B\,a\right )}{27\,a^{14/3}}+\frac {4\,b^{2/3}\,\ln \left (2\,b^{1/3}\,x-a^{1/3}+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (11\,A\,b-5\,B\,a\right )}{27\,a^{14/3}} \]
((x^3*(11*A*b - 5*B*a))/(10*a^2) - A/(5*a) + (2*b^2*x^9*(11*A*b - 5*B*a))/ (9*a^4) + (16*b*x^6*(11*A*b - 5*B*a))/(45*a^3))/(a^2*x^5 + b^2*x^11 + 2*a* b*x^8) + (4*b^(2/3)*log(b^(1/3)*x + a^(1/3))*(11*A*b - 5*B*a))/(27*a^(14/3 )) - (4*b^(2/3)*log(3^(1/2)*a^(1/3)*1i - 2*b^(1/3)*x + a^(1/3))*((3^(1/2)* 1i)/2 + 1/2)*(11*A*b - 5*B*a))/(27*a^(14/3)) + (4*b^(2/3)*log(3^(1/2)*a^(1 /3)*1i + 2*b^(1/3)*x - a^(1/3))*((3^(1/2)*1i)/2 - 1/2)*(11*A*b - 5*B*a))/( 27*a^(14/3))