Integrand size = 30, antiderivative size = 134 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^4 \left (a+b x^3\right )^3} \, dx=-\frac {c}{3 a^3 x^3}-\frac {b^3 c-a b^2 d+a^2 b e-a^3 f}{6 a^2 b^2 \left (a+b x^3\right )^2}-\frac {2 b^3 c-a b^2 d+a^3 f}{3 a^3 b^2 \left (a+b x^3\right )}-\frac {(3 b c-a d) \log (x)}{a^4}+\frac {(3 b c-a d) \log \left (a+b x^3\right )}{3 a^4} \]
-1/3*c/a^3/x^3+1/6*(a^3*f-a^2*b*e+a*b^2*d-b^3*c)/a^2/b^2/(b*x^3+a)^2+1/3*( -a^3*f+a*b^2*d-2*b^3*c)/a^3/b^2/(b*x^3+a)-(-a*d+3*b*c)*ln(x)/a^4+1/3*(-a*d +3*b*c)*ln(b*x^3+a)/a^4
Time = 0.09 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.90 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^4 \left (a+b x^3\right )^3} \, dx=\frac {-\frac {2 a c}{x^3}+\frac {a^2 \left (-b^3 c+a b^2 d-a^2 b e+a^3 f\right )}{b^2 \left (a+b x^3\right )^2}-\frac {2 a \left (2 b^3 c-a b^2 d+a^3 f\right )}{b^2 \left (a+b x^3\right )}+6 (-3 b c+a d) \log (x)+2 (3 b c-a d) \log \left (a+b x^3\right )}{6 a^4} \]
((-2*a*c)/x^3 + (a^2*(-(b^3*c) + a*b^2*d - a^2*b*e + a^3*f))/(b^2*(a + b*x ^3)^2) - (2*a*(2*b^3*c - a*b^2*d + a^3*f))/(b^2*(a + b*x^3)) + 6*(-3*b*c + a*d)*Log[x] + 2*(3*b*c - a*d)*Log[a + b*x^3])/(6*a^4)
Time = 0.39 (sec) , antiderivative size = 133, 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.
\(\displaystyle \int \frac {c+d x^3+e x^6+f x^9}{x^4 \left (a+b x^3\right )^3} \, dx\) |
\(\Big \downarrow \) 2361 |
\(\displaystyle \frac {1}{3} \int \frac {f x^9+e x^6+d x^3+c}{x^6 \left (b x^3+a\right )^3}dx^3\) |
\(\Big \downarrow \) 2123 |
\(\displaystyle \frac {1}{3} \int \left (\frac {c}{a^3 x^6}-\frac {b (a d-3 b c)}{a^4 \left (b x^3+a\right )}+\frac {f a^3-b^2 d a+2 b^3 c}{a^3 b \left (b x^3+a\right )^2}+\frac {a d-3 b c}{a^4 x^3}+\frac {-f a^3+b e a^2-b^2 d a+b^3 c}{a^2 b \left (b x^3+a\right )^3}\right )dx^3\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (-\frac {\log \left (x^3\right ) (3 b c-a d)}{a^4}+\frac {(3 b c-a d) \log \left (a+b x^3\right )}{a^4}-\frac {a^3 f-a b^2 d+2 b^3 c}{a^3 b^2 \left (a+b x^3\right )}-\frac {c}{a^3 x^3}-\frac {a^3 (-f)+a^2 b e-a b^2 d+b^3 c}{2 a^2 b^2 \left (a+b x^3\right )^2}\right )\) |
(-(c/(a^3*x^3)) - (b^3*c - a*b^2*d + a^2*b*e - a^3*f)/(2*a^2*b^2*(a + b*x^ 3)^2) - (2*b^3*c - a*b^2*d + a^3*f)/(a^3*b^2*(a + b*x^3)) - ((3*b*c - a*d) *Log[x^3])/a^4 + ((3*b*c - a*d)*Log[a + b*x^3])/a^4)/3
3.3.82.3.1 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 = 1.53 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.86
method | result | size |
norman | \(\frac {-\frac {c}{3 a}+\frac {\left (a^{2} e -2 a b d +6 b^{2} c \right ) x^{6}}{3 a^{3}}+\frac {\left (f \,a^{3}+a^{2} b e -3 a \,b^{2} d +9 b^{3} c \right ) x^{9}}{6 a^{4}}}{x^{3} \left (b \,x^{3}+a \right )^{2}}+\frac {\left (a d -3 b c \right ) \ln \left (x \right )}{a^{4}}-\frac {\left (a d -3 b c \right ) \ln \left (b \,x^{3}+a \right )}{3 a^{4}}\) | \(115\) |
default | \(-\frac {c}{3 a^{3} x^{3}}+\frac {\left (a d -3 b c \right ) \ln \left (x \right )}{a^{4}}+\frac {\left (-a d +3 b c \right ) \ln \left (b \,x^{3}+a \right )+\frac {a^{2} \left (f \,a^{3}-a^{2} b e +a \,b^{2} d -b^{3} c \right )}{2 b^{2} \left (b \,x^{3}+a \right )^{2}}-\frac {a \left (f \,a^{3}-a \,b^{2} d +2 b^{3} c \right )}{b^{2} \left (b \,x^{3}+a \right )}}{3 a^{4}}\) | \(125\) |
risch | \(\frac {-\frac {\left (f \,a^{3}-a \,b^{2} d +3 b^{3} c \right ) x^{6}}{3 a^{3} b}-\frac {\left (f \,a^{3}+a^{2} b e -3 a \,b^{2} d +9 b^{3} c \right ) x^{3}}{6 a^{2} b^{2}}-\frac {c}{3 a}}{x^{3} \left (b \,x^{3}+a \right )^{2}}+\frac {d \ln \left (x \right )}{a^{3}}-\frac {3 b c \ln \left (x \right )}{a^{4}}-\frac {d \ln \left (b \,x^{3}+a \right )}{3 a^{3}}+\frac {b c \ln \left (b \,x^{3}+a \right )}{a^{4}}\) | \(132\) |
parallelrisch | \(\frac {6 \ln \left (x \right ) x^{9} a \,b^{2} d -18 \ln \left (x \right ) x^{9} b^{3} c -2 \ln \left (b \,x^{3}+a \right ) x^{9} a \,b^{2} d +6 \ln \left (b \,x^{3}+a \right ) x^{9} b^{3} c +x^{9} a^{3} f +x^{9} a^{2} b e -3 x^{9} a \,b^{2} d +9 b^{3} c \,x^{9}+12 \ln \left (x \right ) x^{6} a^{2} b d -36 \ln \left (x \right ) x^{6} a \,b^{2} c -4 \ln \left (b \,x^{3}+a \right ) x^{6} a^{2} b d +12 \ln \left (b \,x^{3}+a \right ) x^{6} a \,b^{2} c +2 a^{3} e \,x^{6}-4 a^{2} b d \,x^{6}+12 a \,b^{2} c \,x^{6}+6 \ln \left (x \right ) x^{3} a^{3} d -18 \ln \left (x \right ) x^{3} a^{2} b c -2 \ln \left (b \,x^{3}+a \right ) x^{3} a^{3} d +6 \ln \left (b \,x^{3}+a \right ) x^{3} a^{2} b c -2 c \,a^{3}}{6 a^{4} x^{3} \left (b \,x^{3}+a \right )^{2}}\) | \(266\) |
(-1/3*c/a+1/3*(a^2*e-2*a*b*d+6*b^2*c)/a^3*x^6+1/6*(a^3*f+a^2*b*e-3*a*b^2*d +9*b^3*c)/a^4*x^9)/x^3/(b*x^3+a)^2+(a*d-3*b*c)/a^4*ln(x)-1/3*(a*d-3*b*c)/a ^4*ln(b*x^3+a)
Time = 0.26 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.87 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^4 \left (a+b x^3\right )^3} \, dx=-\frac {2 \, {\left (3 \, a b^{4} c - a^{2} b^{3} d + a^{4} b f\right )} x^{6} + 2 \, a^{3} b^{2} c + {\left (9 \, a^{2} b^{3} c - 3 \, a^{3} b^{2} d + a^{4} b e + a^{5} f\right )} x^{3} - 2 \, {\left ({\left (3 \, b^{5} c - a b^{4} d\right )} x^{9} + 2 \, {\left (3 \, a b^{4} c - a^{2} b^{3} d\right )} x^{6} + {\left (3 \, a^{2} b^{3} c - a^{3} b^{2} d\right )} x^{3}\right )} \log \left (b x^{3} + a\right ) + 6 \, {\left ({\left (3 \, b^{5} c - a b^{4} d\right )} x^{9} + 2 \, {\left (3 \, a b^{4} c - a^{2} b^{3} d\right )} x^{6} + {\left (3 \, a^{2} b^{3} c - a^{3} b^{2} d\right )} x^{3}\right )} \log \left (x\right )}{6 \, {\left (a^{4} b^{4} x^{9} + 2 \, a^{5} b^{3} x^{6} + a^{6} b^{2} x^{3}\right )}} \]
-1/6*(2*(3*a*b^4*c - a^2*b^3*d + a^4*b*f)*x^6 + 2*a^3*b^2*c + (9*a^2*b^3*c - 3*a^3*b^2*d + a^4*b*e + a^5*f)*x^3 - 2*((3*b^5*c - a*b^4*d)*x^9 + 2*(3* a*b^4*c - a^2*b^3*d)*x^6 + (3*a^2*b^3*c - a^3*b^2*d)*x^3)*log(b*x^3 + a) + 6*((3*b^5*c - a*b^4*d)*x^9 + 2*(3*a*b^4*c - a^2*b^3*d)*x^6 + (3*a^2*b^3*c - a^3*b^2*d)*x^3)*log(x))/(a^4*b^4*x^9 + 2*a^5*b^3*x^6 + a^6*b^2*x^3)
Timed out. \[ \int \frac {c+d x^3+e x^6+f x^9}{x^4 \left (a+b x^3\right )^3} \, dx=\text {Timed out} \]
Time = 0.21 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.07 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^4 \left (a+b x^3\right )^3} \, dx=-\frac {2 \, {\left (3 \, b^{4} c - a b^{3} d + a^{3} b f\right )} x^{6} + 2 \, a^{2} b^{2} c + {\left (9 \, a b^{3} c - 3 \, a^{2} b^{2} d + a^{3} b e + a^{4} f\right )} x^{3}}{6 \, {\left (a^{3} b^{4} x^{9} + 2 \, a^{4} b^{3} x^{6} + a^{5} b^{2} x^{3}\right )}} + \frac {{\left (3 \, b c - a d\right )} \log \left (b x^{3} + a\right )}{3 \, a^{4}} - \frac {{\left (3 \, b c - a d\right )} \log \left (x^{3}\right )}{3 \, a^{4}} \]
-1/6*(2*(3*b^4*c - a*b^3*d + a^3*b*f)*x^6 + 2*a^2*b^2*c + (9*a*b^3*c - 3*a ^2*b^2*d + a^3*b*e + a^4*f)*x^3)/(a^3*b^4*x^9 + 2*a^4*b^3*x^6 + a^5*b^2*x^ 3) + 1/3*(3*b*c - a*d)*log(b*x^3 + a)/a^4 - 1/3*(3*b*c - a*d)*log(x^3)/a^4
Time = 0.27 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.28 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^4 \left (a+b x^3\right )^3} \, dx=-\frac {{\left (3 \, b c - a d\right )} \log \left ({\left | x \right |}\right )}{a^{4}} + \frac {{\left (3 \, b^{2} c - a b d\right )} \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, a^{4} b} + \frac {3 \, b c x^{3} - a d x^{3} - a c}{3 \, a^{4} x^{3}} - \frac {9 \, b^{5} c x^{6} - 3 \, a b^{4} d x^{6} + 22 \, a b^{4} c x^{3} - 8 \, a^{2} b^{3} d x^{3} + 2 \, a^{4} b f x^{3} + 14 \, a^{2} b^{3} c - 6 \, a^{3} b^{2} d + a^{4} b e + a^{5} f}{6 \, {\left (b x^{3} + a\right )}^{2} a^{4} b^{2}} \]
-(3*b*c - a*d)*log(abs(x))/a^4 + 1/3*(3*b^2*c - a*b*d)*log(abs(b*x^3 + a)) /(a^4*b) + 1/3*(3*b*c*x^3 - a*d*x^3 - a*c)/(a^4*x^3) - 1/6*(9*b^5*c*x^6 - 3*a*b^4*d*x^6 + 22*a*b^4*c*x^3 - 8*a^2*b^3*d*x^3 + 2*a^4*b*f*x^3 + 14*a^2* b^3*c - 6*a^3*b^2*d + a^4*b*e + a^5*f)/((b*x^3 + a)^2*a^4*b^2)
Time = 9.15 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.01 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^4 \left (a+b x^3\right )^3} \, dx=\frac {\ln \left (x\right )\,\left (a\,d-3\,b\,c\right )}{a^4}-\frac {\ln \left (b\,x^3+a\right )\,\left (a\,d-3\,b\,c\right )}{3\,a^4}-\frac {\frac {c}{3\,a}+\frac {x^6\,\left (f\,a^3-d\,a\,b^2+3\,c\,b^3\right )}{3\,a^3\,b}+\frac {x^3\,\left (f\,a^3+e\,a^2\,b-3\,d\,a\,b^2+9\,c\,b^3\right )}{6\,a^2\,b^2}}{a^2\,x^3+2\,a\,b\,x^6+b^2\,x^9} \]