Integrand size = 30, antiderivative size = 175 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^{10} \left (a+b x^3\right )^2} \, dx=-\frac {c}{9 a^2 x^9}+\frac {2 b c-a d}{6 a^3 x^6}-\frac {3 b^2 c-2 a b d+a^2 e}{3 a^4 x^3}-\frac {b^3 c-a b^2 d+a^2 b e-a^3 f}{3 a^4 \left (a+b x^3\right )}-\frac {\left (4 b^3 c-3 a b^2 d+2 a^2 b e-a^3 f\right ) \log (x)}{a^5}+\frac {\left (4 b^3 c-3 a b^2 d+2 a^2 b e-a^3 f\right ) \log \left (a+b x^3\right )}{3 a^5} \]
-1/9*c/a^2/x^9+1/6*(-a*d+2*b*c)/a^3/x^6+1/3*(-a^2*e+2*a*b*d-3*b^2*c)/a^4/x ^3+1/3*(a^3*f-a^2*b*e+a*b^2*d-b^3*c)/a^4/(b*x^3+a)-(-a^3*f+2*a^2*b*e-3*a*b ^2*d+4*b^3*c)*ln(x)/a^5+1/3*(-a^3*f+2*a^2*b*e-3*a*b^2*d+4*b^3*c)*ln(b*x^3+ a)/a^5
Time = 0.11 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.91 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^{10} \left (a+b x^3\right )^2} \, dx=\frac {-\frac {2 a^3 c}{x^9}-\frac {3 a^2 (-2 b c+a d)}{x^6}-\frac {6 a \left (3 b^2 c-2 a b d+a^2 e\right )}{x^3}+\frac {6 a \left (-b^3 c+a b^2 d-a^2 b e+a^3 f\right )}{a+b x^3}+18 \left (-4 b^3 c+3 a b^2 d-2 a^2 b e+a^3 f\right ) \log (x)+6 \left (4 b^3 c-3 a b^2 d+2 a^2 b e-a^3 f\right ) \log \left (a+b x^3\right )}{18 a^5} \]
((-2*a^3*c)/x^9 - (3*a^2*(-2*b*c + a*d))/x^6 - (6*a*(3*b^2*c - 2*a*b*d + a ^2*e))/x^3 + (6*a*(-(b^3*c) + a*b^2*d - a^2*b*e + a^3*f))/(a + b*x^3) + 18 *(-4*b^3*c + 3*a*b^2*d - 2*a^2*b*e + a^3*f)*Log[x] + 6*(4*b^3*c - 3*a*b^2* d + 2*a^2*b*e - a^3*f)*Log[a + b*x^3])/(18*a^5)
Time = 0.44 (sec) , antiderivative size = 174, 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.
| \(\displaystyle \int \frac {c+d x^3+e x^6+f x^9}{x^{10} \left (a+b x^3\right )^2} \, dx\) |
| \(\Big \downarrow \) 2361 |
| \(\displaystyle \frac {1}{3} \int \frac {f x^9+e x^6+d x^3+c}{x^{12} \left (b x^3+a\right )^2}dx^3\) |
| \(\Big \downarrow \) 2123 |
| \(\displaystyle \frac {1}{3} \int \left (\frac {c}{a^2 x^{12}}-\frac {b \left (f a^3-2 b e a^2+3 b^2 d a-4 b^3 c\right )}{a^5 \left (b x^3+a\right )}-\frac {b \left (f a^3-b e a^2+b^2 d a-b^3 c\right )}{a^4 \left (b x^3+a\right )^2}+\frac {f a^3-2 b e a^2+3 b^2 d a-4 b^3 c}{a^5 x^3}+\frac {e a^2-2 b d a+3 b^2 c}{a^4 x^6}+\frac {a d-2 b c}{a^3 x^9}\right )dx^3\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{3} \left (\frac {2 b c-a d}{2 a^3 x^6}-\frac {c}{3 a^2 x^9}-\frac {a^2 e-2 a b d+3 b^2 c}{a^4 x^3}-\frac {\log \left (x^3\right ) \left (a^3 (-f)+2 a^2 b e-3 a b^2 d+4 b^3 c\right )}{a^5}+\frac {\log \left (a+b x^3\right ) \left (a^3 (-f)+2 a^2 b e-3 a b^2 d+4 b^3 c\right )}{a^5}-\frac {a^3 (-f)+a^2 b e-a b^2 d+b^3 c}{a^4 \left (a+b x^3\right )}\right )\) |
(-1/3*c/(a^2*x^9) + (2*b*c - a*d)/(2*a^3*x^6) - (3*b^2*c - 2*a*b*d + a^2*e )/(a^4*x^3) - (b^3*c - a*b^2*d + a^2*b*e - a^3*f)/(a^4*(a + b*x^3)) - ((4* b^3*c - 3*a*b^2*d + 2*a^2*b*e - a^3*f)*Log[x^3])/a^5 + ((4*b^3*c - 3*a*b^2 *d + 2*a^2*b*e - a^3*f)*Log[a + b*x^3])/a^5)/3
3.3.58.3.1 Defintions of rubi rules used
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 = 1.50 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.97
| method | result | size |
| default | \(-\frac {c}{9 a^{2} x^{9}}-\frac {a d -2 b c}{6 a^{3} x^{6}}-\frac {a^{2} e -2 a b d +3 b^{2} c}{3 a^{4} x^{3}}+\frac {\left (f \,a^{3}-2 a^{2} b e +3 a \,b^{2} d -4 b^{3} c \right ) \ln \left (x \right )}{a^{5}}-\frac {b \left (\frac {\left (f \,a^{3}-2 a^{2} b e +3 a \,b^{2} d -4 b^{3} c \right ) \ln \left (b \,x^{3}+a \right )}{b}-\frac {a \left (f \,a^{3}-a^{2} b e +a \,b^{2} d -b^{3} c \right )}{b \left (b \,x^{3}+a \right )}\right )}{3 a^{5}}\) | \(169\) |
| norman | \(\frac {-\frac {c}{9 a}-\frac {\left (3 a d -4 b c \right ) x^{3}}{18 a^{2}}-\frac {\left (2 a^{2} e -3 a b d +4 b^{2} c \right ) x^{6}}{6 a^{3}}+\frac {b \left (-f \,a^{3}+2 a^{2} b e -3 a \,b^{2} d +4 b^{3} c \right ) x^{12}}{3 a^{5}}}{x^{9} \left (b \,x^{3}+a \right )}+\frac {\left (f \,a^{3}-2 a^{2} b e +3 a \,b^{2} d -4 b^{3} c \right ) \ln \left (x \right )}{a^{5}}-\frac {\left (f \,a^{3}-2 a^{2} b e +3 a \,b^{2} d -4 b^{3} c \right ) \ln \left (b \,x^{3}+a \right )}{3 a^{5}}\) | \(172\) |
| risch | \(\frac {\frac {\left (f \,a^{3}-2 a^{2} b e +3 a \,b^{2} d -4 b^{3} c \right ) x^{9}}{3 a^{4}}-\frac {\left (2 a^{2} e -3 a b d +4 b^{2} c \right ) x^{6}}{6 a^{3}}-\frac {\left (3 a d -4 b c \right ) x^{3}}{18 a^{2}}-\frac {c}{9 a}}{x^{9} \left (b \,x^{3}+a \right )}+\frac {\ln \left (x \right ) f}{a^{2}}-\frac {2 \ln \left (x \right ) b e}{a^{3}}+\frac {3 \ln \left (x \right ) b^{2} d}{a^{4}}-\frac {4 \ln \left (x \right ) b^{3} c}{a^{5}}-\frac {\ln \left (b \,x^{3}+a \right ) f}{3 a^{2}}+\frac {2 \ln \left (b \,x^{3}+a \right ) b e}{3 a^{3}}-\frac {\ln \left (b \,x^{3}+a \right ) b^{2} d}{a^{4}}+\frac {4 \ln \left (b \,x^{3}+a \right ) b^{3} c}{3 a^{5}}\) | \(200\) |
| parallelrisch | \(\frac {-6 x^{6} a^{4} b e +9 x^{6} a^{3} b^{2} d -12 x^{6} a^{2} b^{3} c -3 x^{3} a^{4} b d -72 \ln \left (x \right ) x^{12} b^{5} c +24 \ln \left (b \,x^{3}+a \right ) x^{12} b^{5} c -2 a^{4} b c +4 a^{3} b^{2} c \,x^{3}-36 \ln \left (x \right ) x^{12} a^{2} b^{3} e +54 \ln \left (x \right ) x^{12} a \,b^{4} d -6 \ln \left (b \,x^{3}+a \right ) x^{12} a^{3} b^{2} f +6 x^{9} a^{4} b f -12 x^{9} a^{3} b^{2} e +18 x^{9} a^{2} b^{3} d -24 x^{9} a \,b^{4} c -36 \ln \left (x \right ) x^{9} a^{3} b^{2} e +54 \ln \left (x \right ) x^{9} a^{2} b^{3} d -72 \ln \left (x \right ) x^{9} a \,b^{4} c -6 \ln \left (b \,x^{3}+a \right ) x^{9} a^{4} b f +12 \ln \left (b \,x^{3}+a \right ) x^{9} a^{3} b^{2} e -18 \ln \left (b \,x^{3}+a \right ) x^{9} a^{2} b^{3} d +24 \ln \left (b \,x^{3}+a \right ) x^{9} a \,b^{4} c +18 \ln \left (x \right ) x^{12} a^{3} b^{2} f +12 \ln \left (b \,x^{3}+a \right ) x^{12} a^{2} b^{3} e +18 \ln \left (x \right ) x^{9} a^{4} b f -18 \ln \left (b \,x^{3}+a \right ) x^{12} a \,b^{4} d}{18 a^{5} b \,x^{9} \left (b \,x^{3}+a \right )}\) | \(383\) |
-1/9*c/a^2/x^9-1/6*(a*d-2*b*c)/a^3/x^6-1/3*(a^2*e-2*a*b*d+3*b^2*c)/a^4/x^3 +(a^3*f-2*a^2*b*e+3*a*b^2*d-4*b^3*c)/a^5*ln(x)-1/3/a^5*b*((a^3*f-2*a^2*b*e +3*a*b^2*d-4*b^3*c)/b*ln(b*x^3+a)-a*(a^3*f-a^2*b*e+a*b^2*d-b^3*c)/b/(b*x^3 +a))
Time = 0.31 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.49 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^{10} \left (a+b x^3\right )^2} \, dx=-\frac {6 \, {\left (4 \, a b^{3} c - 3 \, a^{2} b^{2} d + 2 \, a^{3} b e - a^{4} f\right )} x^{9} + 3 \, {\left (4 \, a^{2} b^{2} c - 3 \, a^{3} b d + 2 \, a^{4} e\right )} x^{6} + 2 \, a^{4} c - {\left (4 \, a^{3} b c - 3 \, a^{4} d\right )} x^{3} - 6 \, {\left ({\left (4 \, b^{4} c - 3 \, a b^{3} d + 2 \, a^{2} b^{2} e - a^{3} b f\right )} x^{12} + {\left (4 \, a b^{3} c - 3 \, a^{2} b^{2} d + 2 \, a^{3} b e - a^{4} f\right )} x^{9}\right )} \log \left (b x^{3} + a\right ) + 18 \, {\left ({\left (4 \, b^{4} c - 3 \, a b^{3} d + 2 \, a^{2} b^{2} e - a^{3} b f\right )} x^{12} + {\left (4 \, a b^{3} c - 3 \, a^{2} b^{2} d + 2 \, a^{3} b e - a^{4} f\right )} x^{9}\right )} \log \left (x\right )}{18 \, {\left (a^{5} b x^{12} + a^{6} x^{9}\right )}} \]
-1/18*(6*(4*a*b^3*c - 3*a^2*b^2*d + 2*a^3*b*e - a^4*f)*x^9 + 3*(4*a^2*b^2* c - 3*a^3*b*d + 2*a^4*e)*x^6 + 2*a^4*c - (4*a^3*b*c - 3*a^4*d)*x^3 - 6*((4 *b^4*c - 3*a*b^3*d + 2*a^2*b^2*e - a^3*b*f)*x^12 + (4*a*b^3*c - 3*a^2*b^2* d + 2*a^3*b*e - a^4*f)*x^9)*log(b*x^3 + a) + 18*((4*b^4*c - 3*a*b^3*d + 2* a^2*b^2*e - a^3*b*f)*x^12 + (4*a*b^3*c - 3*a^2*b^2*d + 2*a^3*b*e - a^4*f)* x^9)*log(x))/(a^5*b*x^12 + a^6*x^9)
Timed out. \[ \int \frac {c+d x^3+e x^6+f x^9}{x^{10} \left (a+b x^3\right )^2} \, dx=\text {Timed out} \]
Time = 0.21 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.03 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^{10} \left (a+b x^3\right )^2} \, dx=-\frac {6 \, {\left (4 \, b^{3} c - 3 \, a b^{2} d + 2 \, a^{2} b e - a^{3} f\right )} x^{9} + 3 \, {\left (4 \, a b^{2} c - 3 \, a^{2} b d + 2 \, a^{3} e\right )} x^{6} + 2 \, a^{3} c - {\left (4 \, a^{2} b c - 3 \, a^{3} d\right )} x^{3}}{18 \, {\left (a^{4} b x^{12} + a^{5} x^{9}\right )}} + \frac {{\left (4 \, b^{3} c - 3 \, a b^{2} d + 2 \, a^{2} b e - a^{3} f\right )} \log \left (b x^{3} + a\right )}{3 \, a^{5}} - \frac {{\left (4 \, b^{3} c - 3 \, a b^{2} d + 2 \, a^{2} b e - a^{3} f\right )} \log \left (x^{3}\right )}{3 \, a^{5}} \]
-1/18*(6*(4*b^3*c - 3*a*b^2*d + 2*a^2*b*e - a^3*f)*x^9 + 3*(4*a*b^2*c - 3* a^2*b*d + 2*a^3*e)*x^6 + 2*a^3*c - (4*a^2*b*c - 3*a^3*d)*x^3)/(a^4*b*x^12 + a^5*x^9) + 1/3*(4*b^3*c - 3*a*b^2*d + 2*a^2*b*e - a^3*f)*log(b*x^3 + a)/ a^5 - 1/3*(4*b^3*c - 3*a*b^2*d + 2*a^2*b*e - a^3*f)*log(x^3)/a^5
Time = 0.28 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.54 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^{10} \left (a+b x^3\right )^2} \, dx=-\frac {{\left (4 \, b^{3} c - 3 \, a b^{2} d + 2 \, a^{2} b e - a^{3} f\right )} \log \left ({\left | x \right |}\right )}{a^{5}} + \frac {{\left (4 \, b^{4} c - 3 \, a b^{3} d + 2 \, a^{2} b^{2} e - a^{3} b f\right )} \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, a^{5} b} - \frac {4 \, b^{4} c x^{3} - 3 \, a b^{3} d x^{3} + 2 \, a^{2} b^{2} e x^{3} - a^{3} b f x^{3} + 5 \, a b^{3} c - 4 \, a^{2} b^{2} d + 3 \, a^{3} b e - 2 \, a^{4} f}{3 \, {\left (b x^{3} + a\right )} a^{5}} + \frac {44 \, b^{3} c x^{9} - 33 \, a b^{2} d x^{9} + 22 \, a^{2} b e x^{9} - 11 \, a^{3} f x^{9} - 18 \, a b^{2} c x^{6} + 12 \, a^{2} b d x^{6} - 6 \, a^{3} e x^{6} + 6 \, a^{2} b c x^{3} - 3 \, a^{3} d x^{3} - 2 \, a^{3} c}{18 \, a^{5} x^{9}} \]
-(4*b^3*c - 3*a*b^2*d + 2*a^2*b*e - a^3*f)*log(abs(x))/a^5 + 1/3*(4*b^4*c - 3*a*b^3*d + 2*a^2*b^2*e - a^3*b*f)*log(abs(b*x^3 + a))/(a^5*b) - 1/3*(4* b^4*c*x^3 - 3*a*b^3*d*x^3 + 2*a^2*b^2*e*x^3 - a^3*b*f*x^3 + 5*a*b^3*c - 4* a^2*b^2*d + 3*a^3*b*e - 2*a^4*f)/((b*x^3 + a)*a^5) + 1/18*(44*b^3*c*x^9 - 33*a*b^2*d*x^9 + 22*a^2*b*e*x^9 - 11*a^3*f*x^9 - 18*a*b^2*c*x^6 + 12*a^2*b *d*x^6 - 6*a^3*e*x^6 + 6*a^2*b*c*x^3 - 3*a^3*d*x^3 - 2*a^3*c)/(a^5*x^9)
Time = 9.65 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.00 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^{10} \left (a+b x^3\right )^2} \, dx=\frac {\ln \left (b\,x^3+a\right )\,\left (-f\,a^3+2\,e\,a^2\,b-3\,d\,a\,b^2+4\,c\,b^3\right )}{3\,a^5}-\frac {\frac {c}{9\,a}+\frac {x^9\,\left (-f\,a^3+2\,e\,a^2\,b-3\,d\,a\,b^2+4\,c\,b^3\right )}{3\,a^4}+\frac {x^3\,\left (3\,a\,d-4\,b\,c\right )}{18\,a^2}+\frac {x^6\,\left (2\,e\,a^2-3\,d\,a\,b+4\,c\,b^2\right )}{6\,a^3}}{b\,x^{12}+a\,x^9}-\frac {\ln \left (x\right )\,\left (-f\,a^3+2\,e\,a^2\,b-3\,d\,a\,b^2+4\,c\,b^3\right )}{a^5} \]