Integrand size = 30, antiderivative size = 226 \[ \int \frac {x^{11} \left (c+d x^3+e x^6+f x^9\right )}{\left (a+b x^3\right )^3} \, dx=\frac {\left (b^3 c-3 a b^2 d+6 a^2 b e-10 a^3 f\right ) x^3}{3 b^6}+\frac {\left (b^2 d-3 a b e+6 a^2 f\right ) x^6}{6 b^5}+\frac {(b e-3 a f) x^9}{9 b^4}+\frac {f x^{12}}{12 b^3}+\frac {a^3 \left (b^3 c-a b^2 d+a^2 b e-a^3 f\right )}{6 b^7 \left (a+b x^3\right )^2}-\frac {a^2 \left (3 b^3 c-4 a b^2 d+5 a^2 b e-6 a^3 f\right )}{3 b^7 \left (a+b x^3\right )}-\frac {a \left (3 b^3 c-6 a b^2 d+10 a^2 b e-15 a^3 f\right ) \log \left (a+b x^3\right )}{3 b^7} \] Output:
1/3*(-10*a^3*f+6*a^2*b*e-3*a*b^2*d+b^3*c)*x^3/b^6+1/6*(6*a^2*f-3*a*b*e+b^2 *d)*x^6/b^5+1/9*(-3*a*f+b*e)*x^9/b^4+1/12*f*x^12/b^3+1/6*a^3*(-a^3*f+a^2*b *e-a*b^2*d+b^3*c)/b^7/(b*x^3+a)^2-1/3*a^2*(-6*a^3*f+5*a^2*b*e-4*a*b^2*d+3* b^3*c)/b^7/(b*x^3+a)-1/3*a*(-15*a^3*f+10*a^2*b*e-6*a*b^2*d+3*b^3*c)*ln(b*x ^3+a)/b^7
Time = 0.08 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.92 \[ \int \frac {x^{11} \left (c+d x^3+e x^6+f x^9\right )}{\left (a+b x^3\right )^3} \, dx=\frac {12 b \left (b^3 c-3 a b^2 d+6 a^2 b e-10 a^3 f\right ) x^3+6 b^2 \left (b^2 d-3 a b e+6 a^2 f\right ) x^6+4 b^3 (b e-3 a f) x^9+3 b^4 f x^{12}+\frac {6 a^3 \left (b^3 c-a b^2 d+a^2 b e-a^3 f\right )}{\left (a+b x^3\right )^2}+\frac {12 a^2 \left (-3 b^3 c+4 a b^2 d-5 a^2 b e+6 a^3 f\right )}{a+b x^3}+12 a \left (-3 b^3 c+6 a b^2 d-10 a^2 b e+15 a^3 f\right ) \log \left (a+b x^3\right )}{36 b^7} \] Input:
Integrate[(x^11*(c + d*x^3 + e*x^6 + f*x^9))/(a + b*x^3)^3,x]
Output:
(12*b*(b^3*c - 3*a*b^2*d + 6*a^2*b*e - 10*a^3*f)*x^3 + 6*b^2*(b^2*d - 3*a* b*e + 6*a^2*f)*x^6 + 4*b^3*(b*e - 3*a*f)*x^9 + 3*b^4*f*x^12 + (6*a^3*(b^3* c - a*b^2*d + a^2*b*e - a^3*f))/(a + b*x^3)^2 + (12*a^2*(-3*b^3*c + 4*a*b^ 2*d - 5*a^2*b*e + 6*a^3*f))/(a + b*x^3) + 12*a*(-3*b^3*c + 6*a*b^2*d - 10* a^2*b*e + 15*a^3*f)*Log[a + b*x^3])/(36*b^7)
Time = 0.92 (sec) , antiderivative size = 223, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2361, 2123, 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.
\(\displaystyle \int \frac {x^{11} \left (c+d x^3+e x^6+f x^9\right )}{\left (a+b x^3\right )^3} \, dx\) |
\(\Big \downarrow \) 2361 |
\(\displaystyle \frac {1}{3} \int \frac {x^9 \left (f x^9+e x^6+d x^3+c\right )}{\left (b x^3+a\right )^3}dx^3\) |
\(\Big \downarrow \) 2123 |
\(\displaystyle \frac {1}{3} \int \left (\frac {f x^9}{b^3}+\frac {(b e-3 a f) x^6}{b^4}+\frac {\left (6 f a^2-3 b e a+b^2 d\right ) x^3}{b^5}+\frac {-10 f a^3+6 b e a^2-3 b^2 d a+b^3 c}{b^6}+\frac {a \left (15 f a^3-10 b e a^2+6 b^2 d a-3 b^3 c\right )}{b^6 \left (b x^3+a\right )}-\frac {a^2 \left (6 f a^3-5 b e a^2+4 b^2 d a-3 b^3 c\right )}{b^6 \left (b x^3+a\right )^2}+\frac {a^3 \left (f a^3-b e a^2+b^2 d a-b^3 c\right )}{b^6 \left (b x^3+a\right )^3}\right )dx^3\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (\frac {x^6 \left (6 a^2 f-3 a b e+b^2 d\right )}{2 b^5}-\frac {a^2 \left (-6 a^3 f+5 a^2 b e-4 a b^2 d+3 b^3 c\right )}{b^7 \left (a+b x^3\right )}+\frac {a^3 \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{2 b^7 \left (a+b x^3\right )^2}-\frac {a \log \left (a+b x^3\right ) \left (-15 a^3 f+10 a^2 b e-6 a b^2 d+3 b^3 c\right )}{b^7}+\frac {x^3 \left (-10 a^3 f+6 a^2 b e-3 a b^2 d+b^3 c\right )}{b^6}+\frac {x^9 (b e-3 a f)}{3 b^4}+\frac {f x^{12}}{4 b^3}\right )\) |
Input:
Int[(x^11*(c + d*x^3 + e*x^6 + f*x^9))/(a + b*x^3)^3,x]
Output:
(((b^3*c - 3*a*b^2*d + 6*a^2*b*e - 10*a^3*f)*x^3)/b^6 + ((b^2*d - 3*a*b*e + 6*a^2*f)*x^6)/(2*b^5) + ((b*e - 3*a*f)*x^9)/(3*b^4) + (f*x^12)/(4*b^3) + (a^3*(b^3*c - a*b^2*d + a^2*b*e - a^3*f))/(2*b^7*(a + b*x^3)^2) - (a^2*(3 *b^3*c - 4*a*b^2*d + 5*a^2*b*e - 6*a^3*f))/(b^7*(a + b*x^3)) - (a*(3*b^3*c - 6*a*b^2*d + 10*a^2*b*e - 15*a^3*f)*Log[a + b*x^3])/b^7)/3
Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c , d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2])
Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*SubstFor[x^n, Pq, x]*(a + b*x)^p, x ], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && PolyQ[Pq, x^n] && IntegerQ[S implify[(m + 1)/n]]
Time = 0.12 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.96
method | result | size |
norman | \(\frac {\frac {a^{2} \left (15 a^{4} f -10 a^{3} e b +6 a^{2} d \,b^{2}-3 a \,b^{3} c \right )}{2 b^{7}}-\frac {\left (15 f \,a^{3}-10 e \,a^{2} b +6 d a \,b^{2}-3 b^{3} c \right ) x^{9}}{9 b^{4}}+\frac {f \,x^{18}}{12 b}-\frac {\left (3 a f -2 b e \right ) x^{15}}{18 b^{2}}+\frac {\left (15 f \,a^{2}-10 a b e +6 b^{2} d \right ) x^{12}}{36 b^{3}}+\frac {2 a \left (15 a^{4} f -10 a^{3} e b +6 a^{2} d \,b^{2}-3 a \,b^{3} c \right ) x^{3}}{3 b^{6}}}{\left (b \,x^{3}+a \right )^{2}}+\frac {a \left (15 f \,a^{3}-10 e \,a^{2} b +6 d a \,b^{2}-3 b^{3} c \right ) \ln \left (b \,x^{3}+a \right )}{3 b^{7}}\) | \(218\) |
default | \(-\frac {-\frac {f \,x^{12} b^{3}}{12}+\frac {\left (3 f a \,b^{2}-e \,b^{3}\right ) x^{9}}{9}+\frac {\left (-6 f \,a^{2} b +3 e a \,b^{2}-b^{3} d \right ) x^{6}}{6}+\frac {\left (10 f \,a^{3}-6 e \,a^{2} b +3 d a \,b^{2}-b^{3} c \right ) x^{3}}{3}}{b^{6}}+\frac {a \left (\frac {a \left (6 f \,a^{3}-5 e \,a^{2} b +4 d a \,b^{2}-3 b^{3} c \right )}{b \left (b \,x^{3}+a \right )}+\frac {\left (15 f \,a^{3}-10 e \,a^{2} b +6 d a \,b^{2}-3 b^{3} c \right ) \ln \left (b \,x^{3}+a \right )}{b}-\frac {a^{2} \left (f \,a^{3}-e \,a^{2} b +d a \,b^{2}-b^{3} c \right )}{2 b \left (b \,x^{3}+a \right )^{2}}\right )}{3 b^{6}}\) | \(223\) |
risch | \(\frac {f \,x^{12}}{12 b^{3}}-\frac {a f \,x^{9}}{3 b^{4}}+\frac {e \,x^{9}}{9 b^{3}}+\frac {a^{2} f \,x^{6}}{b^{5}}-\frac {a e \,x^{6}}{2 b^{4}}+\frac {d \,x^{6}}{6 b^{3}}-\frac {10 f \,a^{3} x^{3}}{3 b^{6}}+\frac {2 a^{2} e \,x^{3}}{b^{5}}-\frac {a d \,x^{3}}{b^{4}}+\frac {c \,x^{3}}{3 b^{3}}+\frac {\left (2 a^{5} f -\frac {5}{3} a^{4} b e +\frac {4}{3} a^{3} b^{2} d -a^{2} b^{3} c \right ) x^{3}+\frac {a^{3} \left (11 f \,a^{3}-9 e \,a^{2} b +7 d a \,b^{2}-5 b^{3} c \right )}{6 b}}{b^{6} \left (b \,x^{3}+a \right )^{2}}+\frac {5 a^{4} \ln \left (b \,x^{3}+a \right ) f}{b^{7}}-\frac {10 a^{3} \ln \left (b \,x^{3}+a \right ) e}{3 b^{6}}+\frac {2 a^{2} \ln \left (b \,x^{3}+a \right ) d}{b^{5}}-\frac {a \ln \left (b \,x^{3}+a \right ) c}{b^{4}}\) | \(254\) |
parallelrisch | \(\frac {-36 \ln \left (b \,x^{3}+a \right ) a^{3} b^{3} c +360 \ln \left (b \,x^{3}+a \right ) x^{3} a^{5} b f +108 a^{4} b^{2} d +360 x^{3} a^{5} b f -240 x^{3} a^{4} b^{2} e +144 x^{3} a^{3} b^{3} d -72 x^{3} a^{2} b^{4} c -120 \ln \left (b \,x^{3}+a \right ) a^{5} b e +72 \ln \left (b \,x^{3}+a \right ) a^{4} b^{2} d +144 \ln \left (b \,x^{3}+a \right ) x^{3} a^{3} b^{3} d -72 \ln \left (b \,x^{3}+a \right ) x^{3} a^{2} b^{4} c -180 a^{5} b e -54 a^{3} b^{3} c +270 a^{6} f -24 x^{9} a \,b^{5} d -6 x^{15} a \,b^{5} f +15 x^{12} a^{2} b^{4} f -10 x^{12} a \,b^{5} e -60 x^{9} a^{3} b^{3} f +40 x^{9} a^{2} b^{4} e +72 \ln \left (b \,x^{3}+a \right ) x^{6} a^{2} b^{4} d -36 \ln \left (b \,x^{3}+a \right ) x^{6} a \,b^{5} c +3 f \,x^{18} b^{6}+12 x^{9} b^{6} c +180 \ln \left (b \,x^{3}+a \right ) a^{6} f -120 \ln \left (b \,x^{3}+a \right ) x^{6} a^{3} b^{3} e -240 \ln \left (b \,x^{3}+a \right ) x^{3} a^{4} b^{2} e +4 x^{15} b^{6} e +6 x^{12} b^{6} d +180 \ln \left (b \,x^{3}+a \right ) x^{6} a^{4} b^{2} f}{36 b^{7} \left (b \,x^{3}+a \right )^{2}}\) | \(414\) |
Input:
int(x^11*(f*x^9+e*x^6+d*x^3+c)/(b*x^3+a)^3,x,method=_RETURNVERBOSE)
Output:
(1/2*a^2*(15*a^4*f-10*a^3*b*e+6*a^2*b^2*d-3*a*b^3*c)/b^7-1/9/b^4*(15*a^3*f -10*a^2*b*e+6*a*b^2*d-3*b^3*c)*x^9+1/12*f*x^18/b-1/18*(3*a*f-2*b*e)/b^2*x^ 15+1/36*(15*a^2*f-10*a*b*e+6*b^2*d)/b^3*x^12+2/3*a*(15*a^4*f-10*a^3*b*e+6* a^2*b^2*d-3*a*b^3*c)/b^6*x^3)/(b*x^3+a)^2+1/3*a*(15*a^3*f-10*a^2*b*e+6*a*b ^2*d-3*b^3*c)/b^7*ln(b*x^3+a)
Time = 0.08 (sec) , antiderivative size = 353, normalized size of antiderivative = 1.56 \[ \int \frac {x^{11} \left (c+d x^3+e x^6+f x^9\right )}{\left (a+b x^3\right )^3} \, dx=\frac {3 \, b^{6} f x^{18} + 2 \, {\left (2 \, b^{6} e - 3 \, a b^{5} f\right )} x^{15} + {\left (6 \, b^{6} d - 10 \, a b^{5} e + 15 \, a^{2} b^{4} f\right )} x^{12} + 4 \, {\left (3 \, b^{6} c - 6 \, a b^{5} d + 10 \, a^{2} b^{4} e - 15 \, a^{3} b^{3} f\right )} x^{9} - 30 \, a^{3} b^{3} c + 42 \, a^{4} b^{2} d - 54 \, a^{5} b e + 66 \, a^{6} f + 6 \, {\left (4 \, a b^{5} c - 11 \, a^{2} b^{4} d + 21 \, a^{3} b^{3} e - 34 \, a^{4} b^{2} f\right )} x^{6} - 12 \, {\left (2 \, a^{2} b^{4} c - a^{3} b^{3} d - a^{4} b^{2} e + 4 \, a^{5} b f\right )} x^{3} - 12 \, {\left (3 \, a^{3} b^{3} c - 6 \, a^{4} b^{2} d + 10 \, a^{5} b e - 15 \, a^{6} f + {\left (3 \, a b^{5} c - 6 \, a^{2} b^{4} d + 10 \, a^{3} b^{3} e - 15 \, a^{4} b^{2} f\right )} x^{6} + 2 \, {\left (3 \, a^{2} b^{4} c - 6 \, a^{3} b^{3} d + 10 \, a^{4} b^{2} e - 15 \, a^{5} b f\right )} x^{3}\right )} \log \left (b x^{3} + a\right )}{36 \, {\left (b^{9} x^{6} + 2 \, a b^{8} x^{3} + a^{2} b^{7}\right )}} \] Input:
integrate(x^11*(f*x^9+e*x^6+d*x^3+c)/(b*x^3+a)^3,x, algorithm="fricas")
Output:
1/36*(3*b^6*f*x^18 + 2*(2*b^6*e - 3*a*b^5*f)*x^15 + (6*b^6*d - 10*a*b^5*e + 15*a^2*b^4*f)*x^12 + 4*(3*b^6*c - 6*a*b^5*d + 10*a^2*b^4*e - 15*a^3*b^3* f)*x^9 - 30*a^3*b^3*c + 42*a^4*b^2*d - 54*a^5*b*e + 66*a^6*f + 6*(4*a*b^5* c - 11*a^2*b^4*d + 21*a^3*b^3*e - 34*a^4*b^2*f)*x^6 - 12*(2*a^2*b^4*c - a^ 3*b^3*d - a^4*b^2*e + 4*a^5*b*f)*x^3 - 12*(3*a^3*b^3*c - 6*a^4*b^2*d + 10* a^5*b*e - 15*a^6*f + (3*a*b^5*c - 6*a^2*b^4*d + 10*a^3*b^3*e - 15*a^4*b^2* f)*x^6 + 2*(3*a^2*b^4*c - 6*a^3*b^3*d + 10*a^4*b^2*e - 15*a^5*b*f)*x^3)*lo g(b*x^3 + a))/(b^9*x^6 + 2*a*b^8*x^3 + a^2*b^7)
Timed out. \[ \int \frac {x^{11} \left (c+d x^3+e x^6+f x^9\right )}{\left (a+b x^3\right )^3} \, dx=\text {Timed out} \] Input:
integrate(x**11*(f*x**9+e*x**6+d*x**3+c)/(b*x**3+a)**3,x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.03 \[ \int \frac {x^{11} \left (c+d x^3+e x^6+f x^9\right )}{\left (a+b x^3\right )^3} \, dx=-\frac {5 \, a^{3} b^{3} c - 7 \, a^{4} b^{2} d + 9 \, a^{5} b e - 11 \, a^{6} f + 2 \, {\left (3 \, a^{2} b^{4} c - 4 \, a^{3} b^{3} d + 5 \, a^{4} b^{2} e - 6 \, a^{5} b f\right )} x^{3}}{6 \, {\left (b^{9} x^{6} + 2 \, a b^{8} x^{3} + a^{2} b^{7}\right )}} + \frac {3 \, b^{3} f x^{12} + 4 \, {\left (b^{3} e - 3 \, a b^{2} f\right )} x^{9} + 6 \, {\left (b^{3} d - 3 \, a b^{2} e + 6 \, a^{2} b f\right )} x^{6} + 12 \, {\left (b^{3} c - 3 \, a b^{2} d + 6 \, a^{2} b e - 10 \, a^{3} f\right )} x^{3}}{36 \, b^{6}} - \frac {{\left (3 \, a b^{3} c - 6 \, a^{2} b^{2} d + 10 \, a^{3} b e - 15 \, a^{4} f\right )} \log \left (b x^{3} + a\right )}{3 \, b^{7}} \] Input:
integrate(x^11*(f*x^9+e*x^6+d*x^3+c)/(b*x^3+a)^3,x, algorithm="maxima")
Output:
-1/6*(5*a^3*b^3*c - 7*a^4*b^2*d + 9*a^5*b*e - 11*a^6*f + 2*(3*a^2*b^4*c - 4*a^3*b^3*d + 5*a^4*b^2*e - 6*a^5*b*f)*x^3)/(b^9*x^6 + 2*a*b^8*x^3 + a^2*b ^7) + 1/36*(3*b^3*f*x^12 + 4*(b^3*e - 3*a*b^2*f)*x^9 + 6*(b^3*d - 3*a*b^2* e + 6*a^2*b*f)*x^6 + 12*(b^3*c - 3*a*b^2*d + 6*a^2*b*e - 10*a^3*f)*x^3)/b^ 6 - 1/3*(3*a*b^3*c - 6*a^2*b^2*d + 10*a^3*b*e - 15*a^4*f)*log(b*x^3 + a)/b ^7
Time = 0.13 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.29 \[ \int \frac {x^{11} \left (c+d x^3+e x^6+f x^9\right )}{\left (a+b x^3\right )^3} \, dx=-\frac {{\left (3 \, a b^{3} c - 6 \, a^{2} b^{2} d + 10 \, a^{3} b e - 15 \, a^{4} f\right )} \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, b^{7}} + \frac {9 \, a b^{5} c x^{6} - 18 \, a^{2} b^{4} d x^{6} + 30 \, a^{3} b^{3} e x^{6} - 45 \, a^{4} b^{2} f x^{6} + 12 \, a^{2} b^{4} c x^{3} - 28 \, a^{3} b^{3} d x^{3} + 50 \, a^{4} b^{2} e x^{3} - 78 \, a^{5} b f x^{3} + 4 \, a^{3} b^{3} c - 11 \, a^{4} b^{2} d + 21 \, a^{5} b e - 34 \, a^{6} f}{6 \, {\left (b x^{3} + a\right )}^{2} b^{7}} + \frac {3 \, b^{9} f x^{12} + 4 \, b^{9} e x^{9} - 12 \, a b^{8} f x^{9} + 6 \, b^{9} d x^{6} - 18 \, a b^{8} e x^{6} + 36 \, a^{2} b^{7} f x^{6} + 12 \, b^{9} c x^{3} - 36 \, a b^{8} d x^{3} + 72 \, a^{2} b^{7} e x^{3} - 120 \, a^{3} b^{6} f x^{3}}{36 \, b^{12}} \] Input:
integrate(x^11*(f*x^9+e*x^6+d*x^3+c)/(b*x^3+a)^3,x, algorithm="giac")
Output:
-1/3*(3*a*b^3*c - 6*a^2*b^2*d + 10*a^3*b*e - 15*a^4*f)*log(abs(b*x^3 + a)) /b^7 + 1/6*(9*a*b^5*c*x^6 - 18*a^2*b^4*d*x^6 + 30*a^3*b^3*e*x^6 - 45*a^4*b ^2*f*x^6 + 12*a^2*b^4*c*x^3 - 28*a^3*b^3*d*x^3 + 50*a^4*b^2*e*x^3 - 78*a^5 *b*f*x^3 + 4*a^3*b^3*c - 11*a^4*b^2*d + 21*a^5*b*e - 34*a^6*f)/((b*x^3 + a )^2*b^7) + 1/36*(3*b^9*f*x^12 + 4*b^9*e*x^9 - 12*a*b^8*f*x^9 + 6*b^9*d*x^6 - 18*a*b^8*e*x^6 + 36*a^2*b^7*f*x^6 + 12*b^9*c*x^3 - 36*a*b^8*d*x^3 + 72* a^2*b^7*e*x^3 - 120*a^3*b^6*f*x^3)/b^12
Time = 5.86 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.30 \[ \int \frac {x^{11} \left (c+d x^3+e x^6+f x^9\right )}{\left (a+b x^3\right )^3} \, dx=x^9\,\left (\frac {e}{9\,b^3}-\frac {a\,f}{3\,b^4}\right )+x^3\,\left (\frac {c}{3\,b^3}-\frac {a^3\,f}{3\,b^6}-\frac {a^2\,\left (\frac {e}{b^3}-\frac {3\,a\,f}{b^4}\right )}{b^2}+\frac {a\,\left (\frac {3\,a^2\,f}{b^5}-\frac {d}{b^3}+\frac {3\,a\,\left (\frac {e}{b^3}-\frac {3\,a\,f}{b^4}\right )}{b}\right )}{b}\right )-x^6\,\left (\frac {a^2\,f}{2\,b^5}-\frac {d}{6\,b^3}+\frac {a\,\left (\frac {e}{b^3}-\frac {3\,a\,f}{b^4}\right )}{2\,b}\right )+\frac {\frac {11\,f\,a^6-9\,e\,a^5\,b+7\,d\,a^4\,b^2-5\,c\,a^3\,b^3}{6\,b}+x^3\,\left (2\,f\,a^5-\frac {5\,e\,a^4\,b}{3}+\frac {4\,d\,a^3\,b^2}{3}-c\,a^2\,b^3\right )}{a^2\,b^6+2\,a\,b^7\,x^3+b^8\,x^6}+\frac {f\,x^{12}}{12\,b^3}+\frac {\ln \left (b\,x^3+a\right )\,\left (15\,f\,a^4-10\,e\,a^3\,b+6\,d\,a^2\,b^2-3\,c\,a\,b^3\right )}{3\,b^7} \] Input:
int((x^11*(c + d*x^3 + e*x^6 + f*x^9))/(a + b*x^3)^3,x)
Output:
x^9*(e/(9*b^3) - (a*f)/(3*b^4)) + x^3*(c/(3*b^3) - (a^3*f)/(3*b^6) - (a^2* (e/b^3 - (3*a*f)/b^4))/b^2 + (a*((3*a^2*f)/b^5 - d/b^3 + (3*a*(e/b^3 - (3* a*f)/b^4))/b))/b) - x^6*((a^2*f)/(2*b^5) - d/(6*b^3) + (a*(e/b^3 - (3*a*f) /b^4))/(2*b)) + ((11*a^6*f - 5*a^3*b^3*c + 7*a^4*b^2*d - 9*a^5*b*e)/(6*b) + x^3*(2*a^5*f - a^2*b^3*c + (4*a^3*b^2*d)/3 - (5*a^4*b*e)/3))/(a^2*b^6 + b^8*x^6 + 2*a*b^7*x^3) + (f*x^12)/(12*b^3) + (log(a + b*x^3)*(15*a^4*f + 6 *a^2*b^2*d - 3*a*b^3*c - 10*a^3*b*e))/(3*b^7)
Time = 0.21 (sec) , antiderivative size = 823, normalized size of antiderivative = 3.64 \[ \int \frac {x^{11} \left (c+d x^3+e x^6+f x^9\right )}{\left (a+b x^3\right )^3} \, dx =\text {Too large to display} \] Input:
int(x^11*(f*x^9+e*x^6+d*x^3+c)/(b*x^3+a)^3,x)
Output:
(180*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**6*f - 120*log( a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**5*b*e + 360*log(a**(2/3 ) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**5*b*f*x**3 + 72*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**4*b**2*d - 240*log(a**(2/3) - b** (1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**4*b**2*e*x**3 + 180*log(a**(2/3) - b* *(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**4*b**2*f*x**6 - 36*log(a**(2/3) - b* *(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**3*b**3*c + 144*log(a**(2/3) - b**(1/ 3)*a**(1/3)*x + b**(2/3)*x**2)*a**3*b**3*d*x**3 - 120*log(a**(2/3) - b**(1 /3)*a**(1/3)*x + b**(2/3)*x**2)*a**3*b**3*e*x**6 - 72*log(a**(2/3) - b**(1 /3)*a**(1/3)*x + b**(2/3)*x**2)*a**2*b**4*c*x**3 + 72*log(a**(2/3) - b**(1 /3)*a**(1/3)*x + b**(2/3)*x**2)*a**2*b**4*d*x**6 - 36*log(a**(2/3) - b**(1 /3)*a**(1/3)*x + b**(2/3)*x**2)*a*b**5*c*x**6 + 180*log(a**(1/3) + b**(1/3 )*x)*a**6*f - 120*log(a**(1/3) + b**(1/3)*x)*a**5*b*e + 360*log(a**(1/3) + b**(1/3)*x)*a**5*b*f*x**3 + 72*log(a**(1/3) + b**(1/3)*x)*a**4*b**2*d - 2 40*log(a**(1/3) + b**(1/3)*x)*a**4*b**2*e*x**3 + 180*log(a**(1/3) + b**(1/ 3)*x)*a**4*b**2*f*x**6 - 36*log(a**(1/3) + b**(1/3)*x)*a**3*b**3*c + 144*l og(a**(1/3) + b**(1/3)*x)*a**3*b**3*d*x**3 - 120*log(a**(1/3) + b**(1/3)*x )*a**3*b**3*e*x**6 - 72*log(a**(1/3) + b**(1/3)*x)*a**2*b**4*c*x**3 + 72*l og(a**(1/3) + b**(1/3)*x)*a**2*b**4*d*x**6 - 36*log(a**(1/3) + b**(1/3)*x) *a*b**5*c*x**6 + 90*a**6*f - 60*a**5*b*e + 36*a**4*b**2*d - 180*a**4*b*...