Integrand size = 30, antiderivative size = 214 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^{13} \left (a+b x^3\right )^2} \, dx=-\frac {c}{12 a^2 x^{12}}+\frac {2 b c-a d}{9 a^3 x^9}-\frac {3 b^2 c-2 a b d+a^2 e}{6 a^4 x^6}+\frac {4 b^3 c-3 a b^2 d+2 a^2 b e-a^3 f}{3 a^5 x^3}+\frac {b \left (b^3 c-a b^2 d+a^2 b e-a^3 f\right )}{3 a^5 \left (a+b x^3\right )}+\frac {b \left (5 b^3 c-4 a b^2 d+3 a^2 b e-2 a^3 f\right ) \log (x)}{a^6}-\frac {b \left (5 b^3 c-4 a b^2 d+3 a^2 b e-2 a^3 f\right ) \log \left (a+b x^3\right )}{3 a^6} \]
-1/12*c/a^2/x^12+1/9*(-a*d+2*b*c)/a^3/x^9+1/6*(-a^2*e+2*a*b*d-3*b^2*c)/a^4 /x^6+1/3*(-a^3*f+2*a^2*b*e-3*a*b^2*d+4*b^3*c)/a^5/x^3+1/3*b*(-a^3*f+a^2*b* e-a*b^2*d+b^3*c)/a^5/(b*x^3+a)+b*(-2*a^3*f+3*a^2*b*e-4*a*b^2*d+5*b^3*c)*ln (x)/a^6-1/3*b*(-2*a^3*f+3*a^2*b*e-4*a*b^2*d+5*b^3*c)*ln(b*x^3+a)/a^6
Time = 0.22 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.93 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^{13} \left (a+b x^3\right )^2} \, dx=-\frac {\frac {3 a^4 c}{x^{12}}+\frac {4 a^3 (-2 b c+a d)}{x^9}+\frac {6 a^2 \left (3 b^2 c-2 a b d+a^2 e\right )}{x^6}+\frac {12 a \left (-4 b^3 c+3 a b^2 d-2 a^2 b e+a^3 f\right )}{x^3}+\frac {12 a b \left (-b^3 c+a b^2 d-a^2 b e+a^3 f\right )}{a+b x^3}-36 b \left (5 b^3 c-4 a b^2 d+3 a^2 b e-2 a^3 f\right ) \log (x)+12 b \left (5 b^3 c-4 a b^2 d+3 a^2 b e-2 a^3 f\right ) \log \left (a+b x^3\right )}{36 a^6} \]
-1/36*((3*a^4*c)/x^12 + (4*a^3*(-2*b*c + a*d))/x^9 + (6*a^2*(3*b^2*c - 2*a *b*d + a^2*e))/x^6 + (12*a*(-4*b^3*c + 3*a*b^2*d - 2*a^2*b*e + a^3*f))/x^3 + (12*a*b*(-(b^3*c) + a*b^2*d - a^2*b*e + a^3*f))/(a + b*x^3) - 36*b*(5*b ^3*c - 4*a*b^2*d + 3*a^2*b*e - 2*a^3*f)*Log[x] + 12*b*(5*b^3*c - 4*a*b^2*d + 3*a^2*b*e - 2*a^3*f)*Log[a + b*x^3])/a^6
Time = 0.49 (sec) , antiderivative size = 212, 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.
| \(\displaystyle \int \frac {c+d x^3+e x^6+f x^9}{x^{13} \left (a+b x^3\right )^2} \, dx\) |
| \(\Big \downarrow \) 2361 |
| \(\displaystyle \frac {1}{3} \int \frac {f x^9+e x^6+d x^3+c}{x^{15} \left (b x^3+a\right )^2}dx^3\) |
| \(\Big \downarrow \) 2123 |
| \(\displaystyle \frac {1}{3} \int \left (\frac {\left (2 f a^3-3 b e a^2+4 b^2 d a-5 b^3 c\right ) b^2}{a^6 \left (b x^3+a\right )}+\frac {\left (f a^3-b e a^2+b^2 d a-b^3 c\right ) b^2}{a^5 \left (b x^3+a\right )^2}-\frac {\left (2 f a^3-3 b e a^2+4 b^2 d a-5 b^3 c\right ) b}{a^6 x^3}+\frac {f a^3-2 b e a^2+3 b^2 d a-4 b^3 c}{a^5 x^6}+\frac {e a^2-2 b d a+3 b^2 c}{a^4 x^9}+\frac {a d-2 b c}{a^3 x^{12}}+\frac {c}{a^2 x^{15}}\right )dx^3\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{3} \left (\frac {2 b c-a d}{3 a^3 x^9}-\frac {c}{4 a^2 x^{12}}-\frac {a^2 e-2 a b d+3 b^2 c}{2 a^4 x^6}+\frac {b \log \left (x^3\right ) \left (-2 a^3 f+3 a^2 b e-4 a b^2 d+5 b^3 c\right )}{a^6}-\frac {b \log \left (a+b x^3\right ) \left (-2 a^3 f+3 a^2 b e-4 a b^2 d+5 b^3 c\right )}{a^6}+\frac {b \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{a^5 \left (a+b x^3\right )}+\frac {a^3 (-f)+2 a^2 b e-3 a b^2 d+4 b^3 c}{a^5 x^3}\right )\) |
(-1/4*c/(a^2*x^12) + (2*b*c - a*d)/(3*a^3*x^9) - (3*b^2*c - 2*a*b*d + a^2* e)/(2*a^4*x^6) + (4*b^3*c - 3*a*b^2*d + 2*a^2*b*e - a^3*f)/(a^5*x^3) + (b* (b^3*c - a*b^2*d + a^2*b*e - a^3*f))/(a^5*(a + b*x^3)) + (b*(5*b^3*c - 4*a *b^2*d + 3*a^2*b*e - 2*a^3*f)*Log[x^3])/a^6 - (b*(5*b^3*c - 4*a*b^2*d + 3* a^2*b*e - 2*a^3*f)*Log[a + b*x^3])/a^6)/3
3.3.59.3.1 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 = 1.52 (sec) , antiderivative size = 209, normalized size of antiderivative = 0.98
| method | result | size |
| default | \(-\frac {c}{12 a^{2} x^{12}}-\frac {a d -2 b c}{9 a^{3} x^{9}}-\frac {a^{2} e -2 a b d +3 b^{2} c}{6 a^{4} x^{6}}-\frac {f \,a^{3}-2 a^{2} b e +3 a \,b^{2} d -4 b^{3} c}{3 a^{5} x^{3}}-\frac {b \left (2 f \,a^{3}-3 a^{2} b e +4 a \,b^{2} d -5 b^{3} c \right ) \ln \left (x \right )}{a^{6}}+\frac {b^{2} \left (\frac {\left (2 f \,a^{3}-3 a^{2} b e +4 a \,b^{2} d -5 b^{3} c \right ) \ln \left (b \,x^{3}+a \right )}{b}-\frac {a \left (f \,a^{3}-a^{2} b e +a \,b^{2} d -b^{3} c \right )}{b \left (b \,x^{3}+a \right )}\right )}{3 a^{6}}\) | \(209\) |
| norman | \(\frac {-\frac {c}{12 a}-\frac {\left (2 f \,a^{3}-3 a^{2} b e +4 a \,b^{2} d -5 b^{3} c \right ) x^{9}}{6 a^{4}}-\frac {\left (4 a d -5 b c \right ) x^{3}}{36 a^{2}}-\frac {\left (3 a^{2} e -4 a b d +5 b^{2} c \right ) x^{6}}{18 a^{3}}+\frac {b \left (2 a^{3} b f -3 a^{2} e \,b^{2}+4 a \,b^{3} d -5 b^{4} c \right ) x^{15}}{3 a^{6}}}{x^{12} \left (b \,x^{3}+a \right )}-\frac {b \left (2 f \,a^{3}-3 a^{2} b e +4 a \,b^{2} d -5 b^{3} c \right ) \ln \left (x \right )}{a^{6}}+\frac {b \left (2 f \,a^{3}-3 a^{2} b e +4 a \,b^{2} d -5 b^{3} c \right ) \ln \left (b \,x^{3}+a \right )}{3 a^{6}}\) | \(215\) |
| risch | \(\frac {-\frac {c}{12 a}-\frac {\left (4 a d -5 b c \right ) x^{3}}{36 a^{2}}-\frac {\left (3 a^{2} e -4 a b d +5 b^{2} c \right ) x^{6}}{18 a^{3}}-\frac {\left (2 f \,a^{3}-3 a^{2} b e +4 a \,b^{2} d -5 b^{3} c \right ) x^{9}}{6 a^{4}}-\frac {b \left (2 f \,a^{3}-3 a^{2} b e +4 a \,b^{2} d -5 b^{3} c \right ) x^{12}}{3 a^{5}}}{x^{12} \left (b \,x^{3}+a \right )}-\frac {2 b \ln \left (x \right ) f}{a^{3}}+\frac {3 b^{2} \ln \left (x \right ) e}{a^{4}}-\frac {4 b^{3} \ln \left (x \right ) d}{a^{5}}+\frac {5 b^{4} \ln \left (x \right ) c}{a^{6}}+\frac {2 b \ln \left (-b \,x^{3}-a \right ) f}{3 a^{3}}-\frac {b^{2} \ln \left (-b \,x^{3}-a \right ) e}{a^{4}}+\frac {4 b^{3} \ln \left (-b \,x^{3}-a \right ) d}{3 a^{5}}-\frac {5 b^{4} \ln \left (-b \,x^{3}-a \right ) c}{3 a^{6}}\) | \(256\) |
| parallelrisch | \(-\frac {3 a^{5} b c -180 \ln \left (x \right ) x^{12} a \,b^{5} c -24 \ln \left (b \,x^{3}+a \right ) x^{12} a^{4} b^{2} f +36 \ln \left (b \,x^{3}+a \right ) x^{12} a^{3} b^{3} e -48 \ln \left (b \,x^{3}+a \right ) x^{12} a^{2} b^{4} d +60 \ln \left (b \,x^{3}+a \right ) x^{12} a \,b^{5} c +72 \ln \left (x \right ) x^{15} a^{3} b^{3} f -108 \ln \left (x \right ) x^{15} a^{2} b^{4} e +144 \ln \left (x \right ) x^{15} a \,b^{5} d -24 \ln \left (b \,x^{3}+a \right ) x^{15} a^{3} b^{3} f +36 \ln \left (b \,x^{3}+a \right ) x^{15} a^{2} b^{4} e -48 \ln \left (b \,x^{3}+a \right ) x^{15} a \,b^{5} d +72 \ln \left (x \right ) x^{12} a^{4} b^{2} f -108 \ln \left (x \right ) x^{12} a^{3} b^{3} e +144 \ln \left (x \right ) x^{12} a^{2} b^{4} d +24 x^{12} a^{4} b^{2} f -36 x^{12} a^{3} b^{3} e +48 x^{12} a^{2} b^{4} d -60 x^{12} a \,b^{5} c +12 x^{9} a^{5} b f -18 x^{9} a^{4} b^{2} e +24 x^{9} a^{3} b^{3} d -30 x^{9} a^{2} b^{4} c +6 x^{6} a^{5} b e -8 x^{6} a^{4} b^{2} d +10 x^{6} a^{3} b^{3} c +4 x^{3} a^{5} b d -5 x^{3} a^{4} b^{2} c -180 \ln \left (x \right ) x^{15} b^{6} c +60 \ln \left (b \,x^{3}+a \right ) x^{15} b^{6} c}{36 a^{6} b \,x^{12} \left (b \,x^{3}+a \right )}\) | \(435\) |
-1/12*c/a^2/x^12-1/9*(a*d-2*b*c)/a^3/x^9-1/6*(a^2*e-2*a*b*d+3*b^2*c)/a^4/x ^6-1/3*(a^3*f-2*a^2*b*e+3*a*b^2*d-4*b^3*c)/a^5/x^3-b*(2*a^3*f-3*a^2*b*e+4* a*b^2*d-5*b^3*c)/a^6*ln(x)+1/3*b^2/a^6*((2*a^3*f-3*a^2*b*e+4*a*b^2*d-5*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 = 0.32 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.45 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^{13} \left (a+b x^3\right )^2} \, dx=\frac {12 \, {\left (5 \, a b^{4} c - 4 \, a^{2} b^{3} d + 3 \, a^{3} b^{2} e - 2 \, a^{4} b f\right )} x^{12} + 6 \, {\left (5 \, a^{2} b^{3} c - 4 \, a^{3} b^{2} d + 3 \, a^{4} b e - 2 \, a^{5} f\right )} x^{9} - 2 \, {\left (5 \, a^{3} b^{2} c - 4 \, a^{4} b d + 3 \, a^{5} e\right )} x^{6} - 3 \, a^{5} c + {\left (5 \, a^{4} b c - 4 \, a^{5} d\right )} x^{3} - 12 \, {\left ({\left (5 \, b^{5} c - 4 \, a b^{4} d + 3 \, a^{2} b^{3} e - 2 \, a^{3} b^{2} f\right )} x^{15} + {\left (5 \, a b^{4} c - 4 \, a^{2} b^{3} d + 3 \, a^{3} b^{2} e - 2 \, a^{4} b f\right )} x^{12}\right )} \log \left (b x^{3} + a\right ) + 36 \, {\left ({\left (5 \, b^{5} c - 4 \, a b^{4} d + 3 \, a^{2} b^{3} e - 2 \, a^{3} b^{2} f\right )} x^{15} + {\left (5 \, a b^{4} c - 4 \, a^{2} b^{3} d + 3 \, a^{3} b^{2} e - 2 \, a^{4} b f\right )} x^{12}\right )} \log \left (x\right )}{36 \, {\left (a^{6} b x^{15} + a^{7} x^{12}\right )}} \]
1/36*(12*(5*a*b^4*c - 4*a^2*b^3*d + 3*a^3*b^2*e - 2*a^4*b*f)*x^12 + 6*(5*a ^2*b^3*c - 4*a^3*b^2*d + 3*a^4*b*e - 2*a^5*f)*x^9 - 2*(5*a^3*b^2*c - 4*a^4 *b*d + 3*a^5*e)*x^6 - 3*a^5*c + (5*a^4*b*c - 4*a^5*d)*x^3 - 12*((5*b^5*c - 4*a*b^4*d + 3*a^2*b^3*e - 2*a^3*b^2*f)*x^15 + (5*a*b^4*c - 4*a^2*b^3*d + 3*a^3*b^2*e - 2*a^4*b*f)*x^12)*log(b*x^3 + a) + 36*((5*b^5*c - 4*a*b^4*d + 3*a^2*b^3*e - 2*a^3*b^2*f)*x^15 + (5*a*b^4*c - 4*a^2*b^3*d + 3*a^3*b^2*e - 2*a^4*b*f)*x^12)*log(x))/(a^6*b*x^15 + a^7*x^12)
Timed out. \[ \int \frac {c+d x^3+e x^6+f x^9}{x^{13} \left (a+b x^3\right )^2} \, dx=\text {Timed out} \]
Time = 0.20 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.06 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^{13} \left (a+b x^3\right )^2} \, dx=\frac {12 \, {\left (5 \, b^{4} c - 4 \, a b^{3} d + 3 \, a^{2} b^{2} e - 2 \, a^{3} b f\right )} x^{12} + 6 \, {\left (5 \, a b^{3} c - 4 \, a^{2} b^{2} d + 3 \, a^{3} b e - 2 \, a^{4} f\right )} x^{9} - 2 \, {\left (5 \, a^{2} b^{2} c - 4 \, a^{3} b d + 3 \, a^{4} e\right )} x^{6} - 3 \, a^{4} c + {\left (5 \, a^{3} b c - 4 \, a^{4} d\right )} x^{3}}{36 \, {\left (a^{5} b x^{15} + a^{6} x^{12}\right )}} - \frac {{\left (5 \, b^{4} c - 4 \, a b^{3} d + 3 \, a^{2} b^{2} e - 2 \, a^{3} b f\right )} \log \left (b x^{3} + a\right )}{3 \, a^{6}} + \frac {{\left (5 \, b^{4} c - 4 \, a b^{3} d + 3 \, a^{2} b^{2} e - 2 \, a^{3} b f\right )} \log \left (x^{3}\right )}{3 \, a^{6}} \]
1/36*(12*(5*b^4*c - 4*a*b^3*d + 3*a^2*b^2*e - 2*a^3*b*f)*x^12 + 6*(5*a*b^3 *c - 4*a^2*b^2*d + 3*a^3*b*e - 2*a^4*f)*x^9 - 2*(5*a^2*b^2*c - 4*a^3*b*d + 3*a^4*e)*x^6 - 3*a^4*c + (5*a^3*b*c - 4*a^4*d)*x^3)/(a^5*b*x^15 + a^6*x^1 2) - 1/3*(5*b^4*c - 4*a*b^3*d + 3*a^2*b^2*e - 2*a^3*b*f)*log(b*x^3 + a)/a^ 6 + 1/3*(5*b^4*c - 4*a*b^3*d + 3*a^2*b^2*e - 2*a^3*b*f)*log(x^3)/a^6
Time = 0.28 (sec) , antiderivative size = 324, normalized size of antiderivative = 1.51 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^{13} \left (a+b x^3\right )^2} \, dx=\frac {{\left (5 \, b^{4} c - 4 \, a b^{3} d + 3 \, a^{2} b^{2} e - 2 \, a^{3} b f\right )} \log \left ({\left | x \right |}\right )}{a^{6}} - \frac {{\left (5 \, b^{5} c - 4 \, a b^{4} d + 3 \, a^{2} b^{3} e - 2 \, a^{3} b^{2} f\right )} \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, a^{6} b} + \frac {5 \, b^{5} c x^{3} - 4 \, a b^{4} d x^{3} + 3 \, a^{2} b^{3} e x^{3} - 2 \, a^{3} b^{2} f x^{3} + 6 \, a b^{4} c - 5 \, a^{2} b^{3} d + 4 \, a^{3} b^{2} e - 3 \, a^{4} b f}{3 \, {\left (b x^{3} + a\right )} a^{6}} - \frac {125 \, b^{4} c x^{12} - 100 \, a b^{3} d x^{12} + 75 \, a^{2} b^{2} e x^{12} - 50 \, a^{3} b f x^{12} - 48 \, a b^{3} c x^{9} + 36 \, a^{2} b^{2} d x^{9} - 24 \, a^{3} b e x^{9} + 12 \, a^{4} f x^{9} + 18 \, a^{2} b^{2} c x^{6} - 12 \, a^{3} b d x^{6} + 6 \, a^{4} e x^{6} - 8 \, a^{3} b c x^{3} + 4 \, a^{4} d x^{3} + 3 \, a^{4} c}{36 \, a^{6} x^{12}} \]
(5*b^4*c - 4*a*b^3*d + 3*a^2*b^2*e - 2*a^3*b*f)*log(abs(x))/a^6 - 1/3*(5*b ^5*c - 4*a*b^4*d + 3*a^2*b^3*e - 2*a^3*b^2*f)*log(abs(b*x^3 + a))/(a^6*b) + 1/3*(5*b^5*c*x^3 - 4*a*b^4*d*x^3 + 3*a^2*b^3*e*x^3 - 2*a^3*b^2*f*x^3 + 6 *a*b^4*c - 5*a^2*b^3*d + 4*a^3*b^2*e - 3*a^4*b*f)/((b*x^3 + a)*a^6) - 1/36 *(125*b^4*c*x^12 - 100*a*b^3*d*x^12 + 75*a^2*b^2*e*x^12 - 50*a^3*b*f*x^12 - 48*a*b^3*c*x^9 + 36*a^2*b^2*d*x^9 - 24*a^3*b*e*x^9 + 12*a^4*f*x^9 + 18*a ^2*b^2*c*x^6 - 12*a^3*b*d*x^6 + 6*a^4*e*x^6 - 8*a^3*b*c*x^3 + 4*a^4*d*x^3 + 3*a^4*c)/(a^6*x^12)
Time = 10.33 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.01 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^{13} \left (a+b x^3\right )^2} \, dx=\frac {\ln \left (x\right )\,\left (-2\,f\,a^3\,b+3\,e\,a^2\,b^2-4\,d\,a\,b^3+5\,c\,b^4\right )}{a^6}-\frac {\ln \left (b\,x^3+a\right )\,\left (-2\,f\,a^3\,b+3\,e\,a^2\,b^2-4\,d\,a\,b^3+5\,c\,b^4\right )}{3\,a^6}-\frac {\frac {c}{12\,a}-\frac {x^9\,\left (-2\,f\,a^3+3\,e\,a^2\,b-4\,d\,a\,b^2+5\,c\,b^3\right )}{6\,a^4}+\frac {x^3\,\left (4\,a\,d-5\,b\,c\right )}{36\,a^2}+\frac {x^6\,\left (3\,e\,a^2-4\,d\,a\,b+5\,c\,b^2\right )}{18\,a^3}-\frac {b\,x^{12}\,\left (-2\,f\,a^3+3\,e\,a^2\,b-4\,d\,a\,b^2+5\,c\,b^3\right )}{3\,a^5}}{b\,x^{15}+a\,x^{12}} \]
(log(x)*(5*b^4*c + 3*a^2*b^2*e - 4*a*b^3*d - 2*a^3*b*f))/a^6 - (log(a + b* x^3)*(5*b^4*c + 3*a^2*b^2*e - 4*a*b^3*d - 2*a^3*b*f))/(3*a^6) - (c/(12*a) - (x^9*(5*b^3*c - 2*a^3*f - 4*a*b^2*d + 3*a^2*b*e))/(6*a^4) + (x^3*(4*a*d - 5*b*c))/(36*a^2) + (x^6*(5*b^2*c + 3*a^2*e - 4*a*b*d))/(18*a^3) - (b*x^1 2*(5*b^3*c - 2*a^3*f - 4*a*b^2*d + 3*a^2*b*e))/(3*a^5))/(a*x^12 + b*x^15)