Integrand size = 30, antiderivative size = 279 \[ \int \frac {x^4 \left (c+d x^3+e x^6+f x^9\right )}{a+b x^3} \, dx=\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) x^2}{2 b^4}+\frac {\left (b^2 d-a b e+a^2 f\right ) x^5}{5 b^3}+\frac {(b e-a f) x^8}{8 b^2}+\frac {f x^{11}}{11 b}+\frac {a^{2/3} \left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} b^{14/3}}+\frac {a^{2/3} \left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 b^{14/3}}-\frac {a^{2/3} \left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 b^{14/3}} \]
1/2*(-a^3*f+a^2*b*e-a*b^2*d+b^3*c)*x^2/b^4+1/5*(a^2*f-a*b*e+b^2*d)*x^5/b^3 +1/8*(-a*f+b*e)*x^8/b^2+1/11*f*x^11/b+1/3*a^(2/3)*(-a^3*f+a^2*b*e-a*b^2*d+ b^3*c)*ln(a^(1/3)+b^(1/3)*x)/b^(14/3)-1/6*a^(2/3)*(-a^3*f+a^2*b*e-a*b^2*d+ b^3*c)*ln(a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/b^(14/3)+1/3*a^(2/3)*(-a^ 3*f+a^2*b*e-a*b^2*d+b^3*c)*arctan(1/3*(a^(1/3)-2*b^(1/3)*x)/a^(1/3)*3^(1/2 ))/b^(14/3)*3^(1/2)
Time = 0.11 (sec) , antiderivative size = 266, normalized size of antiderivative = 0.95 \[ \int \frac {x^4 \left (c+d x^3+e x^6+f x^9\right )}{a+b x^3} \, dx=\frac {660 b^{2/3} \left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) x^2+264 b^{5/3} \left (b^2 d-a b e+a^2 f\right ) x^5+165 b^{8/3} (b e-a f) x^8+120 b^{11/3} f x^{11}-440 \sqrt {3} a^{2/3} \left (-b^3 c+a b^2 d-a^2 b e+a^3 f\right ) \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )-440 a^{2/3} \left (-b^3 c+a b^2 d-a^2 b e+a^3 f\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )+220 a^{2/3} \left (-b^3 c+a b^2 d-a^2 b e+a^3 f\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{1320 b^{14/3}} \]
(660*b^(2/3)*(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*x^2 + 264*b^(5/3)*(b^2*d - a*b*e + a^2*f)*x^5 + 165*b^(8/3)*(b*e - a*f)*x^8 + 120*b^(11/3)*f*x^11 - 440*Sqrt[3]*a^(2/3)*(-(b^3*c) + a*b^2*d - a^2*b*e + a^3*f)*ArcTan[(1 - (2 *b^(1/3)*x)/a^(1/3))/Sqrt[3]] - 440*a^(2/3)*(-(b^3*c) + a*b^2*d - a^2*b*e + a^3*f)*Log[a^(1/3) + b^(1/3)*x] + 220*a^(2/3)*(-(b^3*c) + a*b^2*d - a^2* b*e + a^3*f)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(1320*b^(14/3 ))
Time = 0.47 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {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.
| \(\displaystyle \int \frac {x^4 \left (c+d x^3+e x^6+f x^9\right )}{a+b x^3} \, dx\) |
| \(\Big \downarrow \) 2375 |
| \(\displaystyle \frac {\int \frac {11 x^4 \left ((b e-a f) x^6+b d x^3+b c\right )}{b x^3+a}dx}{11 b}+\frac {f x^{11}}{11 b}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {x^4 \left ((b e-a f) x^6+b d x^3+b c\right )}{b x^3+a}dx}{b}+\frac {f x^{11}}{11 b}\) |
| \(\Big \downarrow \) 1812 |
| \(\displaystyle \frac {\int \left (\frac {(b e-a f) x^7}{b}+\frac {\left (f a^2-b e a+b^2 d\right ) x^4}{b^2}+\frac {\left (-f a^3+b e a^2-b^2 d a+b^3 c\right ) x}{b^3}+\frac {\left (f a^4-b e a^3+b^2 d a^2-b^3 c a\right ) x}{b^3 \left (b x^3+a\right )}\right )dx}{b}+\frac {f x^{11}}{11 b}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {x^5 \left (a^2 f-a b e+b^2 d\right )}{5 b^2}+\frac {x^2 \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{2 b^3}+\frac {a^{2/3} \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right ) \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{\sqrt {3} b^{11/3}}-\frac {a^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right ) \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{6 b^{11/3}}+\frac {a^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{3 b^{11/3}}+\frac {x^8 (b e-a f)}{8 b}}{b}+\frac {f x^{11}}{11 b}\) |
(f*x^11)/(11*b) + (((b^3*c - a*b^2*d + a^2*b*e - a^3*f)*x^2)/(2*b^3) + ((b ^2*d - a*b*e + a^2*f)*x^5)/(5*b^2) + ((b*e - a*f)*x^8)/(8*b) + (a^(2/3)*(b ^3*c - a*b^2*d + a^2*b*e - a^3*f)*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]* a^(1/3))])/(Sqrt[3]*b^(11/3)) + (a^(2/3)*(b^3*c - a*b^2*d + a^2*b*e - a^3* f)*Log[a^(1/3) + b^(1/3)*x])/(3*b^(11/3)) - (a^(2/3)*(b^3*c - a*b^2*d + a^ 2*b*e - a^3*f)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(6*b^(11/3) ))/b
3.3.36.3.1 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[(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 = 1.53 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.56
| method | result | size |
| risch | \(\frac {f \,x^{11}}{11 b}-\frac {x^{8} f a}{8 b^{2}}+\frac {x^{8} e}{8 b}+\frac {x^{5} f \,a^{2}}{5 b^{3}}-\frac {x^{5} a e}{5 b^{2}}+\frac {d \,x^{5}}{5 b}-\frac {x^{2} f \,a^{3}}{2 b^{4}}+\frac {x^{2} a^{2} e}{2 b^{3}}-\frac {x^{2} a d}{2 b^{2}}+\frac {c \,x^{2}}{2 b}+\frac {a \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}+a \right )}{\sum }\frac {\left (f \,a^{3}-a^{2} b e +a \,b^{2} d -b^{3} c \right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}}\right )}{3 b^{5}}\) | \(156\) |
| default | \(-\frac {-\frac {b^{3} f \,x^{11}}{11}+\frac {\left (f a \,b^{2}-b^{3} e \right ) x^{8}}{8}+\frac {\left (-f \,a^{2} b +a \,b^{2} e -b^{3} d \right ) x^{5}}{5}+\frac {x^{2} \left (f \,a^{3}-a^{2} b e +a \,b^{2} d -b^{3} c \right )}{2}}{b^{4}}+\frac {\left (-\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right ) a \left (f \,a^{3}-a^{2} b e +a \,b^{2} d -b^{3} c \right )}{b^{4}}\) | \(210\) |
1/11*f*x^11/b-1/8/b^2*x^8*f*a+1/8/b*x^8*e+1/5/b^3*x^5*f*a^2-1/5/b^2*x^5*a* e+1/5*d*x^5/b-1/2/b^4*x^2*f*a^3+1/2/b^3*x^2*a^2*e-1/2/b^2*x^2*a*d+1/2*c*x^ 2/b+1/3/b^5*a*sum((a^3*f-a^2*b*e+a*b^2*d-b^3*c)/_R*ln(x-_R),_R=RootOf(_Z^3 *b+a))
Time = 0.30 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.01 \[ \int \frac {x^4 \left (c+d x^3+e x^6+f x^9\right )}{a+b x^3} \, dx=\frac {120 \, b^{3} f x^{11} + 165 \, {\left (b^{3} e - a b^{2} f\right )} x^{8} + 264 \, {\left (b^{3} d - a b^{2} e + a^{2} b f\right )} x^{5} + 660 \, {\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} x^{2} - 440 \, \sqrt {3} {\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \left (-\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} b x \left (-\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} + \sqrt {3} a}{3 \, a}\right ) + 220 \, {\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \left (-\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \log \left (a x^{2} - b x \left (-\frac {a^{2}}{b^{2}}\right )^{\frac {2}{3}} - a \left (-\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}}\right ) - 440 \, {\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \left (-\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \log \left (a x + b \left (-\frac {a^{2}}{b^{2}}\right )^{\frac {2}{3}}\right )}{1320 \, b^{4}} \]
1/1320*(120*b^3*f*x^11 + 165*(b^3*e - a*b^2*f)*x^8 + 264*(b^3*d - a*b^2*e + a^2*b*f)*x^5 + 660*(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*x^2 - 440*sqrt(3) *(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*(-a^2/b^2)^(1/3)*arctan(1/3*(2*sqrt(3 )*b*x*(-a^2/b^2)^(1/3) + sqrt(3)*a)/a) + 220*(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*(-a^2/b^2)^(1/3)*log(a*x^2 - b*x*(-a^2/b^2)^(2/3) - a*(-a^2/b^2)^(1 /3)) - 440*(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*(-a^2/b^2)^(1/3)*log(a*x + b*(-a^2/b^2)^(2/3)))/b^4
Time = 0.77 (sec) , antiderivative size = 469, normalized size of antiderivative = 1.68 \[ \int \frac {x^4 \left (c+d x^3+e x^6+f x^9\right )}{a+b x^3} \, dx=x^{8} \left (- \frac {a f}{8 b^{2}} + \frac {e}{8 b}\right ) + x^{5} \left (\frac {a^{2} f}{5 b^{3}} - \frac {a e}{5 b^{2}} + \frac {d}{5 b}\right ) + x^{2} \left (- \frac {a^{3} f}{2 b^{4}} + \frac {a^{2} e}{2 b^{3}} - \frac {a d}{2 b^{2}} + \frac {c}{2 b}\right ) + \operatorname {RootSum} {\left (27 t^{3} b^{14} + a^{11} f^{3} - 3 a^{10} b e f^{2} + 3 a^{9} b^{2} d f^{2} + 3 a^{9} b^{2} e^{2} f - 3 a^{8} b^{3} c f^{2} - 6 a^{8} b^{3} d e f - a^{8} b^{3} e^{3} + 6 a^{7} b^{4} c e f + 3 a^{7} b^{4} d^{2} f + 3 a^{7} b^{4} d e^{2} - 6 a^{6} b^{5} c d f - 3 a^{6} b^{5} c e^{2} - 3 a^{6} b^{5} d^{2} e + 3 a^{5} b^{6} c^{2} f + 6 a^{5} b^{6} c d e + a^{5} b^{6} d^{3} - 3 a^{4} b^{7} c^{2} e - 3 a^{4} b^{7} c d^{2} + 3 a^{3} b^{8} c^{2} d - a^{2} b^{9} c^{3}, \left ( t \mapsto t \log {\left (\frac {9 t^{2} b^{9}}{a^{7} f^{2} - 2 a^{6} b e f + 2 a^{5} b^{2} d f + a^{5} b^{2} e^{2} - 2 a^{4} b^{3} c f - 2 a^{4} b^{3} d e + 2 a^{3} b^{4} c e + a^{3} b^{4} d^{2} - 2 a^{2} b^{5} c d + a b^{6} c^{2}} + x \right )} \right )\right )} + \frac {f x^{11}}{11 b} \]
x**8*(-a*f/(8*b**2) + e/(8*b)) + x**5*(a**2*f/(5*b**3) - a*e/(5*b**2) + d/ (5*b)) + x**2*(-a**3*f/(2*b**4) + a**2*e/(2*b**3) - a*d/(2*b**2) + c/(2*b) ) + RootSum(27*_t**3*b**14 + a**11*f**3 - 3*a**10*b*e*f**2 + 3*a**9*b**2*d *f**2 + 3*a**9*b**2*e**2*f - 3*a**8*b**3*c*f**2 - 6*a**8*b**3*d*e*f - a**8 *b**3*e**3 + 6*a**7*b**4*c*e*f + 3*a**7*b**4*d**2*f + 3*a**7*b**4*d*e**2 - 6*a**6*b**5*c*d*f - 3*a**6*b**5*c*e**2 - 3*a**6*b**5*d**2*e + 3*a**5*b**6 *c**2*f + 6*a**5*b**6*c*d*e + a**5*b**6*d**3 - 3*a**4*b**7*c**2*e - 3*a**4 *b**7*c*d**2 + 3*a**3*b**8*c**2*d - a**2*b**9*c**3, Lambda(_t, _t*log(9*_t **2*b**9/(a**7*f**2 - 2*a**6*b*e*f + 2*a**5*b**2*d*f + a**5*b**2*e**2 - 2* a**4*b**3*c*f - 2*a**4*b**3*d*e + 2*a**3*b**4*c*e + a**3*b**4*d**2 - 2*a** 2*b**5*c*d + a*b**6*c**2) + x))) + f*x**11/(11*b)
Time = 0.29 (sec) , antiderivative size = 269, normalized size of antiderivative = 0.96 \[ \int \frac {x^4 \left (c+d x^3+e x^6+f x^9\right )}{a+b x^3} \, dx=-\frac {\sqrt {3} {\left (a b^{3} c - a^{2} b^{2} d + a^{3} b e - a^{4} f\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, b^{5} \left (\frac {a}{b}\right )^{\frac {1}{3}}} + \frac {40 \, b^{3} f x^{11} + 55 \, {\left (b^{3} e - a b^{2} f\right )} x^{8} + 88 \, {\left (b^{3} d - a b^{2} e + a^{2} b f\right )} x^{5} + 220 \, {\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} x^{2}}{440 \, b^{4}} - \frac {{\left (a b^{3} c - a^{2} b^{2} d + a^{3} b e - a^{4} f\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, b^{5} \left (\frac {a}{b}\right )^{\frac {1}{3}}} + \frac {{\left (a b^{3} c - a^{2} b^{2} d + a^{3} b e - a^{4} f\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \, b^{5} \left (\frac {a}{b}\right )^{\frac {1}{3}}} \]
-1/3*sqrt(3)*(a*b^3*c - a^2*b^2*d + a^3*b*e - a^4*f)*arctan(1/3*sqrt(3)*(2 *x - (a/b)^(1/3))/(a/b)^(1/3))/(b^5*(a/b)^(1/3)) + 1/440*(40*b^3*f*x^11 + 55*(b^3*e - a*b^2*f)*x^8 + 88*(b^3*d - a*b^2*e + a^2*b*f)*x^5 + 220*(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*x^2)/b^4 - 1/6*(a*b^3*c - a^2*b^2*d + a^3*b* e - a^4*f)*log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(b^5*(a/b)^(1/3)) + 1/3* (a*b^3*c - a^2*b^2*d + a^3*b*e - a^4*f)*log(x + (a/b)^(1/3))/(b^5*(a/b)^(1 /3))
Time = 0.29 (sec) , antiderivative size = 380, normalized size of antiderivative = 1.36 \[ \int \frac {x^4 \left (c+d x^3+e x^6+f x^9\right )}{a+b x^3} \, dx=\frac {\sqrt {3} {\left (\left (-a b^{2}\right )^{\frac {2}{3}} b^{3} c - \left (-a b^{2}\right )^{\frac {2}{3}} a b^{2} d + \left (-a b^{2}\right )^{\frac {2}{3}} a^{2} b e - \left (-a b^{2}\right )^{\frac {2}{3}} a^{3} f\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, b^{6}} - \frac {{\left (\left (-a b^{2}\right )^{\frac {2}{3}} b^{3} c - \left (-a b^{2}\right )^{\frac {2}{3}} a b^{2} d + \left (-a b^{2}\right )^{\frac {2}{3}} a^{2} b e - \left (-a b^{2}\right )^{\frac {2}{3}} a^{3} f\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, b^{6}} + \frac {{\left (a b^{10} c \left (-\frac {a}{b}\right )^{\frac {1}{3}} - a^{2} b^{9} d \left (-\frac {a}{b}\right )^{\frac {1}{3}} + a^{3} b^{8} e \left (-\frac {a}{b}\right )^{\frac {1}{3}} - a^{4} b^{7} f \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{3 \, a b^{11}} + \frac {40 \, b^{10} f x^{11} + 55 \, b^{10} e x^{8} - 55 \, a b^{9} f x^{8} + 88 \, b^{10} d x^{5} - 88 \, a b^{9} e x^{5} + 88 \, a^{2} b^{8} f x^{5} + 220 \, b^{10} c x^{2} - 220 \, a b^{9} d x^{2} + 220 \, a^{2} b^{8} e x^{2} - 220 \, a^{3} b^{7} f x^{2}}{440 \, b^{11}} \]
1/3*sqrt(3)*((-a*b^2)^(2/3)*b^3*c - (-a*b^2)^(2/3)*a*b^2*d + (-a*b^2)^(2/3 )*a^2*b*e - (-a*b^2)^(2/3)*a^3*f)*arctan(1/3*sqrt(3)*(2*x + (-a/b)^(1/3))/ (-a/b)^(1/3))/b^6 - 1/6*((-a*b^2)^(2/3)*b^3*c - (-a*b^2)^(2/3)*a*b^2*d + ( -a*b^2)^(2/3)*a^2*b*e - (-a*b^2)^(2/3)*a^3*f)*log(x^2 + x*(-a/b)^(1/3) + ( -a/b)^(2/3))/b^6 + 1/3*(a*b^10*c*(-a/b)^(1/3) - a^2*b^9*d*(-a/b)^(1/3) + a ^3*b^8*e*(-a/b)^(1/3) - a^4*b^7*f*(-a/b)^(1/3))*(-a/b)^(1/3)*log(abs(x - ( -a/b)^(1/3)))/(a*b^11) + 1/440*(40*b^10*f*x^11 + 55*b^10*e*x^8 - 55*a*b^9* f*x^8 + 88*b^10*d*x^5 - 88*a*b^9*e*x^5 + 88*a^2*b^8*f*x^5 + 220*b^10*c*x^2 - 220*a*b^9*d*x^2 + 220*a^2*b^8*e*x^2 - 220*a^3*b^7*f*x^2)/b^11
Time = 10.58 (sec) , antiderivative size = 267, normalized size of antiderivative = 0.96 \[ \int \frac {x^4 \left (c+d x^3+e x^6+f x^9\right )}{a+b x^3} \, dx=x^8\,\left (\frac {e}{8\,b}-\frac {a\,f}{8\,b^2}\right )+x^5\,\left (\frac {d}{5\,b}-\frac {a\,\left (\frac {e}{b}-\frac {a\,f}{b^2}\right )}{5\,b}\right )+x^2\,\left (\frac {c}{2\,b}-\frac {a\,\left (\frac {d}{b}-\frac {a\,\left (\frac {e}{b}-\frac {a\,f}{b^2}\right )}{b}\right )}{2\,b}\right )+\frac {f\,x^{11}}{11\,b}+\frac {a^{2/3}\,\ln \left (b^{1/3}\,x+a^{1/3}\right )\,\left (-f\,a^3+e\,a^2\,b-d\,a\,b^2+c\,b^3\right )}{3\,b^{14/3}}-\frac {a^{2/3}\,\ln \left (2\,b^{1/3}\,x-a^{1/3}+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-f\,a^3+e\,a^2\,b-d\,a\,b^2+c\,b^3\right )}{3\,b^{14/3}}+\frac {a^{2/3}\,\ln \left (a^{1/3}-2\,b^{1/3}\,x+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-f\,a^3+e\,a^2\,b-d\,a\,b^2+c\,b^3\right )}{3\,b^{14/3}} \]
x^8*(e/(8*b) - (a*f)/(8*b^2)) + x^5*(d/(5*b) - (a*(e/b - (a*f)/b^2))/(5*b) ) + x^2*(c/(2*b) - (a*(d/b - (a*(e/b - (a*f)/b^2))/b))/(2*b)) + (f*x^11)/( 11*b) + (a^(2/3)*log(b^(1/3)*x + a^(1/3))*(b^3*c - a^3*f - a*b^2*d + a^2*b *e))/(3*b^(14/3)) - (a^(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)*(b^3*c - a^3*f - a*b^2*d + a^2*b*e))/(3*b^(14/3) ) + (a^(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)*(b^3*c - a^3*f - a*b^2*d + a^2*b*e))/(3*b^(14/3))