Integrand size = 30, antiderivative size = 298 \[ \int \frac {x^4 \left (c+d x^3+e x^6+f x^9\right )}{\left (a+b x^3\right )^2} \, dx=\frac {\left (b^2 d-2 a b e+3 a^2 f\right ) x^2}{2 b^4}+\frac {(b e-2 a f) x^5}{5 b^3}+\frac {f x^8}{8 b^2}-\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) x^2}{3 b^4 \left (a+b x^3\right )}-\frac {\left (2 b^3 c-5 a b^2 d+8 a^2 b e-11 a^3 f\right ) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} \sqrt [3]{a} b^{14/3}}-\frac {\left (2 b^3 c-5 a b^2 d+8 a^2 b e-11 a^3 f\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 \sqrt [3]{a} b^{14/3}}+\frac {\left (2 b^3 c-5 a b^2 d+8 a^2 b e-11 a^3 f\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 \sqrt [3]{a} b^{14/3}} \]
1/2*(3*a^2*f-2*a*b*e+b^2*d)*x^2/b^4+1/5*(-2*a*f+b*e)*x^5/b^3+1/8*f*x^8/b^2 -1/3*(-a^3*f+a^2*b*e-a*b^2*d+b^3*c)*x^2/b^4/(b*x^3+a)-1/9*(-11*a^3*f+8*a^2 *b*e-5*a*b^2*d+2*b^3*c)*ln(a^(1/3)+b^(1/3)*x)/a^(1/3)/b^(14/3)+1/18*(-11*a ^3*f+8*a^2*b*e-5*a*b^2*d+2*b^3*c)*ln(a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2 )/a^(1/3)/b^(14/3)-1/9*(-11*a^3*f+8*a^2*b*e-5*a*b^2*d+2*b^3*c)*arctan(1/3* (a^(1/3)-2*b^(1/3)*x)/a^(1/3)*3^(1/2))/a^(1/3)/b^(14/3)*3^(1/2)
Time = 0.18 (sec) , antiderivative size = 282, normalized size of antiderivative = 0.95 \[ \int \frac {x^4 \left (c+d x^3+e x^6+f x^9\right )}{\left (a+b x^3\right )^2} \, dx=\frac {180 b^{2/3} \left (b^2 d-2 a b e+3 a^2 f\right ) x^2+72 b^{5/3} (b e-2 a f) x^5+45 b^{8/3} f x^8-\frac {120 b^{2/3} \left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) x^2}{a+b x^3}+\frac {40 \sqrt {3} \left (-2 b^3 c+5 a b^2 d-8 a^2 b e+11 a^3 f\right ) \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{a}}+\frac {40 \left (-2 b^3 c+5 a b^2 d-8 a^2 b e+11 a^3 f\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a}}+\frac {20 \left (2 b^3 c-5 a b^2 d+8 a^2 b e-11 a^3 f\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{\sqrt [3]{a}}}{360 b^{14/3}} \]
(180*b^(2/3)*(b^2*d - 2*a*b*e + 3*a^2*f)*x^2 + 72*b^(5/3)*(b*e - 2*a*f)*x^ 5 + 45*b^(8/3)*f*x^8 - (120*b^(2/3)*(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*x^ 2)/(a + b*x^3) + (40*Sqrt[3]*(-2*b^3*c + 5*a*b^2*d - 8*a^2*b*e + 11*a^3*f) *ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]])/a^(1/3) + (40*(-2*b^3*c + 5* a*b^2*d - 8*a^2*b*e + 11*a^3*f)*Log[a^(1/3) + b^(1/3)*x])/a^(1/3) + (20*(2 *b^3*c - 5*a*b^2*d + 8*a^2*b*e - 11*a^3*f)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/a^(1/3))/(360*b^(14/3))
Time = 0.79 (sec) , antiderivative size = 315, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {2367, 25, 2029, 2375, 27, 1812, 2009}
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.
| \(\displaystyle \int \frac {x^4 \left (c+d x^3+e x^6+f x^9\right )}{\left (a+b x^3\right )^2} \, dx\) |
| \(\Big \downarrow \) 2367 |
| \(\displaystyle -\frac {\int -\frac {3 a b^4 f x^{10}+3 a b^3 (b e-a f) x^7+3 a b^2 \left (f a^2-b e a+b^2 d\right ) x^4+2 a b \left (-f a^3+b e a^2-b^2 d a+b^3 c\right ) x}{b x^3+a}dx}{3 a b^5}-\frac {x^2 \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{3 b^4 \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {3 a b^4 f x^{10}+3 a b^3 (b e-a f) x^7+3 a b^2 \left (f a^2-b e a+b^2 d\right ) x^4+2 a b \left (-f a^3+b e a^2-b^2 d a+b^3 c\right ) x}{b x^3+a}dx}{3 a b^5}-\frac {x^2 \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{3 b^4 \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 2029 |
| \(\displaystyle \frac {\int \frac {x \left (3 a b^4 f x^9+3 a b^3 (b e-a f) x^6+3 a b^2 \left (f a^2-b e a+b^2 d\right ) x^3+2 a b \left (-f a^3+b e a^2-b^2 d a+b^3 c\right )\right )}{b x^3+a}dx}{3 a b^5}-\frac {x^2 \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{3 b^4 \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 2375 |
| \(\displaystyle \frac {\frac {\int \frac {8 x \left (3 a b^4 (b e-2 a f) x^6+3 a b^3 \left (f a^2-b e a+b^2 d\right ) x^3+2 a b^2 \left (-f a^3+b e a^2-b^2 d a+b^3 c\right )\right )}{b x^3+a}dx}{8 b}+\frac {3}{8} a b^3 f x^8}{3 a b^5}-\frac {x^2 \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{3 b^4 \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {x \left (3 a b^4 (b e-2 a f) x^6+3 a b^3 \left (f a^2-b e a+b^2 d\right ) x^3+2 a b^2 \left (-f a^3+b e a^2-b^2 d a+b^3 c\right )\right )}{b x^3+a}dx}{b}+\frac {3}{8} a b^3 f x^8}{3 a b^5}-\frac {x^2 \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{3 b^4 \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 1812 |
| \(\displaystyle \frac {\frac {\int \left (3 a b^3 (b e-2 a f) x^4+3 a b^2 \left (3 f a^2-2 b e a+b^2 d\right ) x+\frac {\left (2 a c b^5-5 a^2 d b^4+8 a^3 e b^3-11 a^4 f b^2\right ) x}{b x^3+a}\right )dx}{b}+\frac {3}{8} a b^3 f x^8}{3 a b^5}-\frac {x^2 \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{3 b^4 \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {\frac {3}{2} a b^2 x^2 \left (3 a^2 f-2 a b e+b^2 d\right )-\frac {a^{2/3} b^{4/3} \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right ) \left (-11 a^3 f+8 a^2 b e-5 a b^2 d+2 b^3 c\right )}{\sqrt {3}}+\frac {1}{6} a^{2/3} b^{4/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right ) \left (-11 a^3 f+8 a^2 b e-5 a b^2 d+2 b^3 c\right )-\frac {1}{3} a^{2/3} b^{4/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \left (-11 a^3 f+8 a^2 b e-5 a b^2 d+2 b^3 c\right )+\frac {3}{5} a b^3 x^5 (b e-2 a f)}{b}+\frac {3}{8} a b^3 f x^8}{3 a b^5}-\frac {x^2 \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{3 b^4 \left (a+b x^3\right )}\) |
-1/3*((b^3*c - a*b^2*d + a^2*b*e - a^3*f)*x^2)/(b^4*(a + b*x^3)) + ((3*a*b ^3*f*x^8)/8 + ((3*a*b^2*(b^2*d - 2*a*b*e + 3*a^2*f)*x^2)/2 + (3*a*b^3*(b*e - 2*a*f)*x^5)/5 - (a^(2/3)*b^(4/3)*(2*b^3*c - 5*a*b^2*d + 8*a^2*b*e - 11* a^3*f)*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/Sqrt[3] - (a^(2/ 3)*b^(4/3)*(2*b^3*c - 5*a*b^2*d + 8*a^2*b*e - 11*a^3*f)*Log[a^(1/3) + b^(1 /3)*x])/3 + (a^(2/3)*b^(4/3)*(2*b^3*c - 5*a*b^2*d + 8*a^2*b*e - 11*a^3*f)* Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/6)/b)/(3*a*b^5)
3.3.63.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((f_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.)*( (d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^n)^q*(a + b*x^n + c*x^(2*n))^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && EqQ[n2, 2*n] && IGtQ[n, 0] && IGtQ[p, 0]
Int[(Fx_.)*((d_.)*(x_)^(q_.) + (a_.)*(x_)^(r_.) + (b_.)*(x_)^(s_.) + (c_.)* (x_)^(t_.))^(p_.), x_Symbol] :> Int[x^(p*r)*(a + b*x^(s - r) + c*x^(t - r) + d*x^(q - r))^p*Fx, x] /; FreeQ[{a, b, c, d, r, s, t, q}, x] && IntegerQ[p ] && PosQ[s - r] && PosQ[t - r] && PosQ[q - r] && !(EqQ[p, 1] && EqQ[u, 1] )
Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> With[{q = m + Expon[Pq, x]}, Module[{Q = PolynomialQuotient[b^(Floor[(q - 1)/n] + 1) *x^m*Pq, a + b*x^n, x], R = PolynomialRemainder[b^(Floor[(q - 1)/n] + 1)*x^ m*Pq, a + b*x^n, x]}, Simp[(-x)*R*((a + b*x^n)^(p + 1)/(a*n*(p + 1)*b^(Floo r[(q - 1)/n] + 1))), x] + Simp[1/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1)) I nt[(a + b*x^n)^(p + 1)*ExpandToSum[a*n*(p + 1)*Q + n*(p + 1)*R + D[x*R, x], x], x], x]] /; GeQ[q, n]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[n, 0 ] && LtQ[p, -1] && IGtQ[m, 0]
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Wi th[{q = Expon[Pq, x]}, With[{Pqq = Coeff[Pq, x, q]}, Simp[Pqq*(c*x)^(m + q - n + 1)*((a + b*x^n)^(p + 1)/(b*c^(q - n + 1)*(m + q + n*p + 1))), x] + Si mp[1/(b*(m + q + n*p + 1)) Int[(c*x)^m*ExpandToSum[b*(m + q + n*p + 1)*(P q - Pqq*x^q) - a*Pqq*(m + q - n + 1)*x^(q - n), x]*(a + b*x^n)^p, x], x]] / ; NeQ[m + q + n*p + 1, 0] && q - n >= 0 && (IntegerQ[2*p] || IntegerQ[p + ( q + 1)/(2*n)])] /; FreeQ[{a, b, c, m, p}, x] && PolyQ[Pq, x] && IGtQ[n, 0]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.54 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.53
| method | result | size |
| risch | \(\frac {f \,x^{8}}{8 b^{2}}-\frac {2 x^{5} a f}{5 b^{3}}+\frac {x^{5} e}{5 b^{2}}+\frac {3 a^{2} f \,x^{2}}{2 b^{4}}-\frac {a e \,x^{2}}{b^{3}}+\frac {d \,x^{2}}{2 b^{2}}+\frac {\left (\frac {1}{3} f \,a^{3}-\frac {1}{3} a^{2} b e +\frac {1}{3} a \,b^{2} d -\frac {1}{3} b^{3} c \right ) x^{2}}{b^{4} \left (b \,x^{3}+a \right )}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}+a \right )}{\sum }\frac {\left (-11 f \,a^{3}+8 a^{2} b e -5 a \,b^{2} d +2 b^{3} c \right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}}}{9 b^{5}}\) | \(157\) |
| default | \(\frac {\frac {b^{2} f \,x^{8}}{8}+\frac {\left (-2 a f b +b^{2} e \right ) x^{5}}{5}+\frac {x^{2} \left (3 a^{2} f -2 a e b +b^{2} d \right )}{2}}{b^{4}}-\frac {\frac {\left (-\frac {1}{3} f \,a^{3}+\frac {1}{3} a^{2} b e -\frac {1}{3} a \,b^{2} d +\frac {1}{3} b^{3} c \right ) x^{2}}{b \,x^{3}+a}+\left (\frac {11}{3} f \,a^{3}-\frac {8}{3} a^{2} b e +\frac {5}{3} a \,b^{2} d -\frac {2}{3} b^{3} c \right ) \left (-\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{b^{4}}\) | \(218\) |
1/8*f*x^8/b^2-2/5/b^3*x^5*a*f+1/5/b^2*x^5*e+3/2/b^4*a^2*f*x^2-1/b^3*a*e*x^ 2+1/2*d*x^2/b^2+(1/3*f*a^3-1/3*a^2*b*e+1/3*a*b^2*d-1/3*b^3*c)*x^2/b^4/(b*x ^3+a)+1/9/b^5*sum((-11*a^3*f+8*a^2*b*e-5*a*b^2*d+2*b^3*c)/_R*ln(x-_R),_R=R ootOf(_Z^3*b+a))
Time = 0.37 (sec) , antiderivative size = 920, normalized size of antiderivative = 3.09 \[ \int \frac {x^4 \left (c+d x^3+e x^6+f x^9\right )}{\left (a+b x^3\right )^2} \, dx=\left [\frac {45 \, a b^{5} f x^{11} + 9 \, {\left (8 \, a b^{5} e - 11 \, a^{2} b^{4} f\right )} x^{8} + 36 \, {\left (5 \, a b^{5} d - 8 \, a^{2} b^{4} e + 11 \, a^{3} b^{3} f\right )} x^{5} - 60 \, {\left (2 \, a b^{5} c - 5 \, a^{2} b^{4} d + 8 \, a^{3} b^{3} e - 11 \, a^{4} b^{2} f\right )} x^{2} - 60 \, \sqrt {\frac {1}{3}} {\left (2 \, a^{2} b^{4} c - 5 \, a^{3} b^{3} d + 8 \, a^{4} b^{2} e - 11 \, a^{5} b f + {\left (2 \, a b^{5} c - 5 \, a^{2} b^{4} d + 8 \, a^{3} b^{3} e - 11 \, a^{4} b^{2} f\right )} x^{3}\right )} \sqrt {-\frac {\left (a b^{2}\right )^{\frac {1}{3}}}{a}} \log \left (\frac {2 \, b^{2} x^{3} - a b - 3 \, \sqrt {\frac {1}{3}} {\left (a b x + 2 \, \left (a b^{2}\right )^{\frac {2}{3}} x^{2} - \left (a b^{2}\right )^{\frac {1}{3}} a\right )} \sqrt {-\frac {\left (a b^{2}\right )^{\frac {1}{3}}}{a}} - 3 \, \left (a b^{2}\right )^{\frac {2}{3}} x}{b x^{3} + a}\right ) + 20 \, {\left (2 \, a b^{3} c - 5 \, a^{2} b^{2} d + 8 \, a^{3} b e - 11 \, a^{4} f + {\left (2 \, b^{4} c - 5 \, a b^{3} d + 8 \, a^{2} b^{2} e - 11 \, a^{3} b f\right )} x^{3}\right )} \left (a b^{2}\right )^{\frac {2}{3}} \log \left (b^{2} x^{2} - \left (a b^{2}\right )^{\frac {1}{3}} b x + \left (a b^{2}\right )^{\frac {2}{3}}\right ) - 40 \, {\left (2 \, a b^{3} c - 5 \, a^{2} b^{2} d + 8 \, a^{3} b e - 11 \, a^{4} f + {\left (2 \, b^{4} c - 5 \, a b^{3} d + 8 \, a^{2} b^{2} e - 11 \, a^{3} b f\right )} x^{3}\right )} \left (a b^{2}\right )^{\frac {2}{3}} \log \left (b x + \left (a b^{2}\right )^{\frac {1}{3}}\right )}{360 \, {\left (a b^{7} x^{3} + a^{2} b^{6}\right )}}, \frac {45 \, a b^{5} f x^{11} + 9 \, {\left (8 \, a b^{5} e - 11 \, a^{2} b^{4} f\right )} x^{8} + 36 \, {\left (5 \, a b^{5} d - 8 \, a^{2} b^{4} e + 11 \, a^{3} b^{3} f\right )} x^{5} - 60 \, {\left (2 \, a b^{5} c - 5 \, a^{2} b^{4} d + 8 \, a^{3} b^{3} e - 11 \, a^{4} b^{2} f\right )} x^{2} - 120 \, \sqrt {\frac {1}{3}} {\left (2 \, a^{2} b^{4} c - 5 \, a^{3} b^{3} d + 8 \, a^{4} b^{2} e - 11 \, a^{5} b f + {\left (2 \, a b^{5} c - 5 \, a^{2} b^{4} d + 8 \, a^{3} b^{3} e - 11 \, a^{4} b^{2} f\right )} x^{3}\right )} \sqrt {\frac {\left (a b^{2}\right )^{\frac {1}{3}}}{a}} \arctan \left (-\frac {\sqrt {\frac {1}{3}} {\left (2 \, b x - \left (a b^{2}\right )^{\frac {1}{3}}\right )} \sqrt {\frac {\left (a b^{2}\right )^{\frac {1}{3}}}{a}}}{b}\right ) + 20 \, {\left (2 \, a b^{3} c - 5 \, a^{2} b^{2} d + 8 \, a^{3} b e - 11 \, a^{4} f + {\left (2 \, b^{4} c - 5 \, a b^{3} d + 8 \, a^{2} b^{2} e - 11 \, a^{3} b f\right )} x^{3}\right )} \left (a b^{2}\right )^{\frac {2}{3}} \log \left (b^{2} x^{2} - \left (a b^{2}\right )^{\frac {1}{3}} b x + \left (a b^{2}\right )^{\frac {2}{3}}\right ) - 40 \, {\left (2 \, a b^{3} c - 5 \, a^{2} b^{2} d + 8 \, a^{3} b e - 11 \, a^{4} f + {\left (2 \, b^{4} c - 5 \, a b^{3} d + 8 \, a^{2} b^{2} e - 11 \, a^{3} b f\right )} x^{3}\right )} \left (a b^{2}\right )^{\frac {2}{3}} \log \left (b x + \left (a b^{2}\right )^{\frac {1}{3}}\right )}{360 \, {\left (a b^{7} x^{3} + a^{2} b^{6}\right )}}\right ] \]
[1/360*(45*a*b^5*f*x^11 + 9*(8*a*b^5*e - 11*a^2*b^4*f)*x^8 + 36*(5*a*b^5*d - 8*a^2*b^4*e + 11*a^3*b^3*f)*x^5 - 60*(2*a*b^5*c - 5*a^2*b^4*d + 8*a^3*b ^3*e - 11*a^4*b^2*f)*x^2 - 60*sqrt(1/3)*(2*a^2*b^4*c - 5*a^3*b^3*d + 8*a^4 *b^2*e - 11*a^5*b*f + (2*a*b^5*c - 5*a^2*b^4*d + 8*a^3*b^3*e - 11*a^4*b^2* f)*x^3)*sqrt(-(a*b^2)^(1/3)/a)*log((2*b^2*x^3 - a*b - 3*sqrt(1/3)*(a*b*x + 2*(a*b^2)^(2/3)*x^2 - (a*b^2)^(1/3)*a)*sqrt(-(a*b^2)^(1/3)/a) - 3*(a*b^2) ^(2/3)*x)/(b*x^3 + a)) + 20*(2*a*b^3*c - 5*a^2*b^2*d + 8*a^3*b*e - 11*a^4* f + (2*b^4*c - 5*a*b^3*d + 8*a^2*b^2*e - 11*a^3*b*f)*x^3)*(a*b^2)^(2/3)*lo g(b^2*x^2 - (a*b^2)^(1/3)*b*x + (a*b^2)^(2/3)) - 40*(2*a*b^3*c - 5*a^2*b^2 *d + 8*a^3*b*e - 11*a^4*f + (2*b^4*c - 5*a*b^3*d + 8*a^2*b^2*e - 11*a^3*b* f)*x^3)*(a*b^2)^(2/3)*log(b*x + (a*b^2)^(1/3)))/(a*b^7*x^3 + a^2*b^6), 1/3 60*(45*a*b^5*f*x^11 + 9*(8*a*b^5*e - 11*a^2*b^4*f)*x^8 + 36*(5*a*b^5*d - 8 *a^2*b^4*e + 11*a^3*b^3*f)*x^5 - 60*(2*a*b^5*c - 5*a^2*b^4*d + 8*a^3*b^3*e - 11*a^4*b^2*f)*x^2 - 120*sqrt(1/3)*(2*a^2*b^4*c - 5*a^3*b^3*d + 8*a^4*b^ 2*e - 11*a^5*b*f + (2*a*b^5*c - 5*a^2*b^4*d + 8*a^3*b^3*e - 11*a^4*b^2*f)* x^3)*sqrt((a*b^2)^(1/3)/a)*arctan(-sqrt(1/3)*(2*b*x - (a*b^2)^(1/3))*sqrt( (a*b^2)^(1/3)/a)/b) + 20*(2*a*b^3*c - 5*a^2*b^2*d + 8*a^3*b*e - 11*a^4*f + (2*b^4*c - 5*a*b^3*d + 8*a^2*b^2*e - 11*a^3*b*f)*x^3)*(a*b^2)^(2/3)*log(b ^2*x^2 - (a*b^2)^(1/3)*b*x + (a*b^2)^(2/3)) - 40*(2*a*b^3*c - 5*a^2*b^2*d + 8*a^3*b*e - 11*a^4*f + (2*b^4*c - 5*a*b^3*d + 8*a^2*b^2*e - 11*a^3*b*...
Timed out. \[ \int \frac {x^4 \left (c+d x^3+e x^6+f x^9\right )}{\left (a+b x^3\right )^2} \, dx=\text {Timed out} \]
Time = 0.28 (sec) , antiderivative size = 277, normalized size of antiderivative = 0.93 \[ \int \frac {x^4 \left (c+d x^3+e x^6+f x^9\right )}{\left (a+b x^3\right )^2} \, dx=-\frac {{\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} x^{2}}{3 \, {\left (b^{5} x^{3} + a b^{4}\right )}} + \frac {\sqrt {3} {\left (2 \, b^{3} c - 5 \, a b^{2} d + 8 \, a^{2} b e - 11 \, a^{3} f\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, b^{5} \left (\frac {a}{b}\right )^{\frac {1}{3}}} + \frac {5 \, b^{2} f x^{8} + 8 \, {\left (b^{2} e - 2 \, a b f\right )} x^{5} + 20 \, {\left (b^{2} d - 2 \, a b e + 3 \, a^{2} f\right )} x^{2}}{40 \, b^{4}} + \frac {{\left (2 \, b^{3} c - 5 \, a b^{2} d + 8 \, a^{2} b e - 11 \, a^{3} f\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \, b^{5} \left (\frac {a}{b}\right )^{\frac {1}{3}}} - \frac {{\left (2 \, b^{3} c - 5 \, a b^{2} d + 8 \, a^{2} b e - 11 \, a^{3} f\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 \, b^{5} \left (\frac {a}{b}\right )^{\frac {1}{3}}} \]
-1/3*(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*x^2/(b^5*x^3 + a*b^4) + 1/9*sqrt( 3)*(2*b^3*c - 5*a*b^2*d + 8*a^2*b*e - 11*a^3*f)*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^(1/3))/(b^5*(a/b)^(1/3)) + 1/40*(5*b^2*f*x^8 + 8*(b^2*e - 2*a*b*f)*x^5 + 20*(b^2*d - 2*a*b*e + 3*a^2*f)*x^2)/b^4 + 1/18*(2*b^3*c - 5*a*b^2*d + 8*a^2*b*e - 11*a^3*f)*log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3)) /(b^5*(a/b)^(1/3)) - 1/9*(2*b^3*c - 5*a*b^2*d + 8*a^2*b*e - 11*a^3*f)*log( x + (a/b)^(1/3))/(b^5*(a/b)^(1/3))
Time = 0.27 (sec) , antiderivative size = 338, normalized size of antiderivative = 1.13 \[ \int \frac {x^4 \left (c+d x^3+e x^6+f x^9\right )}{\left (a+b x^3\right )^2} \, dx=\frac {\sqrt {3} {\left (2 \, b^{3} c - 5 \, a b^{2} d + 8 \, a^{2} b e - 11 \, a^{3} f\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, \left (-a b^{2}\right )^{\frac {1}{3}} b^{4}} - \frac {{\left (2 \, b^{3} c - 5 \, a b^{2} d + 8 \, a^{2} b e - 11 \, a^{3} f\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \, \left (-a b^{2}\right )^{\frac {1}{3}} b^{4}} - \frac {{\left (2 \, b^{3} c \left (-\frac {a}{b}\right )^{\frac {1}{3}} - 5 \, a b^{2} d \left (-\frac {a}{b}\right )^{\frac {1}{3}} + 8 \, a^{2} b e \left (-\frac {a}{b}\right )^{\frac {1}{3}} - 11 \, a^{3} f \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{9 \, a b^{4}} - \frac {b^{3} c x^{2} - a b^{2} d x^{2} + a^{2} b e x^{2} - a^{3} f x^{2}}{3 \, {\left (b x^{3} + a\right )} b^{4}} + \frac {5 \, b^{14} f x^{8} + 8 \, b^{14} e x^{5} - 16 \, a b^{13} f x^{5} + 20 \, b^{14} d x^{2} - 40 \, a b^{13} e x^{2} + 60 \, a^{2} b^{12} f x^{2}}{40 \, b^{16}} \]
1/9*sqrt(3)*(2*b^3*c - 5*a*b^2*d + 8*a^2*b*e - 11*a^3*f)*arctan(1/3*sqrt(3 )*(2*x + (-a/b)^(1/3))/(-a/b)^(1/3))/((-a*b^2)^(1/3)*b^4) - 1/18*(2*b^3*c - 5*a*b^2*d + 8*a^2*b*e - 11*a^3*f)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3 ))/((-a*b^2)^(1/3)*b^4) - 1/9*(2*b^3*c*(-a/b)^(1/3) - 5*a*b^2*d*(-a/b)^(1/ 3) + 8*a^2*b*e*(-a/b)^(1/3) - 11*a^3*f*(-a/b)^(1/3))*(-a/b)^(1/3)*log(abs( x - (-a/b)^(1/3)))/(a*b^4) - 1/3*(b^3*c*x^2 - a*b^2*d*x^2 + a^2*b*e*x^2 - a^3*f*x^2)/((b*x^3 + a)*b^4) + 1/40*(5*b^14*f*x^8 + 8*b^14*e*x^5 - 16*a*b^ 13*f*x^5 + 20*b^14*d*x^2 - 40*a*b^13*e*x^2 + 60*a^2*b^12*f*x^2)/b^16
Time = 10.32 (sec) , antiderivative size = 287, normalized size of antiderivative = 0.96 \[ \int \frac {x^4 \left (c+d x^3+e x^6+f x^9\right )}{\left (a+b x^3\right )^2} \, dx=x^5\,\left (\frac {e}{5\,b^2}-\frac {2\,a\,f}{5\,b^3}\right )-x^2\,\left (\frac {a^2\,f}{2\,b^4}-\frac {d}{2\,b^2}+\frac {a\,\left (\frac {e}{b^2}-\frac {2\,a\,f}{b^3}\right )}{b}\right )+\frac {f\,x^8}{8\,b^2}-\frac {x^2\,\left (-\frac {f\,a^3}{3}+\frac {e\,a^2\,b}{3}-\frac {d\,a\,b^2}{3}+\frac {c\,b^3}{3}\right )}{b^5\,x^3+a\,b^4}-\frac {\ln \left (b^{1/3}\,x+a^{1/3}\right )\,\left (-11\,f\,a^3+8\,e\,a^2\,b-5\,d\,a\,b^2+2\,c\,b^3\right )}{9\,a^{1/3}\,b^{14/3}}+\frac {\ln \left (2\,b^{1/3}\,x-a^{1/3}+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-11\,f\,a^3+8\,e\,a^2\,b-5\,d\,a\,b^2+2\,c\,b^3\right )}{9\,a^{1/3}\,b^{14/3}}-\frac {\ln \left (a^{1/3}-2\,b^{1/3}\,x+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-11\,f\,a^3+8\,e\,a^2\,b-5\,d\,a\,b^2+2\,c\,b^3\right )}{9\,a^{1/3}\,b^{14/3}} \]
x^5*(e/(5*b^2) - (2*a*f)/(5*b^3)) - x^2*((a^2*f)/(2*b^4) - d/(2*b^2) + (a* (e/b^2 - (2*a*f)/b^3))/b) + (f*x^8)/(8*b^2) - (x^2*((b^3*c)/3 - (a^3*f)/3 - (a*b^2*d)/3 + (a^2*b*e)/3))/(a*b^4 + b^5*x^3) - (log(b^(1/3)*x + a^(1/3) )*(2*b^3*c - 11*a^3*f - 5*a*b^2*d + 8*a^2*b*e))/(9*a^(1/3)*b^(14/3)) + (lo g(3^(1/2)*a^(1/3)*1i + 2*b^(1/3)*x - a^(1/3))*((3^(1/2)*1i)/2 + 1/2)*(2*b^ 3*c - 11*a^3*f - 5*a*b^2*d + 8*a^2*b*e))/(9*a^(1/3)*b^(14/3)) - (log(3^(1/ 2)*a^(1/3)*1i - 2*b^(1/3)*x + a^(1/3))*((3^(1/2)*1i)/2 - 1/2)*(2*b^3*c - 1 1*a^3*f - 5*a*b^2*d + 8*a^2*b*e))/(9*a^(1/3)*b^(14/3))