Integrand size = 30, antiderivative size = 128 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^{10} \left (a+b x^3\right )} \, dx=-\frac {c}{9 a x^9}+\frac {b c-a d}{6 a^2 x^6}-\frac {b^2 c-a b d+a^2 e}{3 a^3 x^3}-\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \log (x)}{a^4}+\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \log \left (a+b x^3\right )}{3 a^4} \]
-1/9*c/a/x^9+1/6*(-a*d+b*c)/a^2/x^6+1/3*(-a^2*e+a*b*d-b^2*c)/a^3/x^3-(-a^3 *f+a^2*b*e-a*b^2*d+b^3*c)*ln(x)/a^4+1/3*(-a^3*f+a^2*b*e-a*b^2*d+b^3*c)*ln( b*x^3+a)/a^4
Time = 0.07 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.00 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^{10} \left (a+b x^3\right )} \, dx=-\frac {c}{9 a x^9}+\frac {b c-a d}{6 a^2 x^6}+\frac {-b^2 c+a b d-a^2 e}{3 a^3 x^3}+\frac {\left (-b^3 c+a b^2 d-a^2 b e+a^3 f\right ) \log (x)}{a^4}+\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \log \left (a+b x^3\right )}{3 a^4} \]
-1/9*c/(a*x^9) + (b*c - a*d)/(6*a^2*x^6) + (-(b^2*c) + a*b*d - a^2*e)/(3*a ^3*x^3) + ((-(b^3*c) + a*b^2*d - a^2*b*e + a^3*f)*Log[x])/a^4 + ((b^3*c - a*b^2*d + a^2*b*e - a^3*f)*Log[a + b*x^3])/(3*a^4)
Time = 0.37 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.01, 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.
\(\displaystyle \int \frac {c+d x^3+e x^6+f x^9}{x^{10} \left (a+b x^3\right )} \, dx\) |
\(\Big \downarrow \) 2361 |
\(\displaystyle \frac {1}{3} \int \frac {f x^9+e x^6+d x^3+c}{x^{12} \left (b x^3+a\right )}dx^3\) |
\(\Big \downarrow \) 2123 |
\(\displaystyle \frac {1}{3} \int \left (\frac {c}{a x^{12}}-\frac {b \left (f a^3-b e a^2+b^2 d a-b^3 c\right )}{a^4 \left (b x^3+a\right )}+\frac {f a^3-b e a^2+b^2 d a-b^3 c}{a^4 x^3}+\frac {e a^2-b d a+b^2 c}{a^3 x^6}+\frac {a d-b c}{a^2 x^9}\right )dx^3\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (\frac {b c-a d}{2 a^2 x^6}-\frac {a^2 e-a b d+b^2 c}{a^3 x^3}-\frac {\log \left (x^3\right ) \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{a^4}+\frac {\log \left (a+b x^3\right ) \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{a^4}-\frac {c}{3 a x^9}\right )\) |
(-1/3*c/(a*x^9) + (b*c - a*d)/(2*a^2*x^6) - (b^2*c - a*b*d + a^2*e)/(a^3*x ^3) - ((b^3*c - a*b^2*d + a^2*b*e - a^3*f)*Log[x^3])/a^4 + ((b^3*c - a*b^2 *d + a^2*b*e - a^3*f)*Log[a + b*x^3])/a^4)/3
3.3.30.3.1 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 = 1.51 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.94
method | result | size |
default | \(-\frac {c}{9 a \,x^{9}}-\frac {a d -b c}{6 a^{2} x^{6}}-\frac {a^{2} e -a b d +b^{2} c}{3 a^{3} x^{3}}+\frac {\left (f \,a^{3}-a^{2} b e +a \,b^{2} d -b^{3} c \right ) \ln \left (x \right )}{a^{4}}-\frac {\left (f \,a^{3}-a^{2} b e +a \,b^{2} d -b^{3} c \right ) \ln \left (b \,x^{3}+a \right )}{3 a^{4}}\) | \(120\) |
norman | \(\frac {-\frac {c}{9 a}-\frac {\left (a d -b c \right ) x^{3}}{6 a^{2}}-\frac {\left (a^{2} e -a b d +b^{2} c \right ) x^{6}}{3 a^{3}}}{x^{9}}+\frac {\left (f \,a^{3}-a^{2} b e +a \,b^{2} d -b^{3} c \right ) \ln \left (x \right )}{a^{4}}-\frac {\left (f \,a^{3}-a^{2} b e +a \,b^{2} d -b^{3} c \right ) \ln \left (b \,x^{3}+a \right )}{3 a^{4}}\) | \(122\) |
risch | \(\frac {-\frac {c}{9 a}-\frac {\left (a d -b c \right ) x^{3}}{6 a^{2}}-\frac {\left (a^{2} e -a b d +b^{2} c \right ) x^{6}}{3 a^{3}}}{x^{9}}+\frac {\ln \left (x \right ) f}{a}-\frac {\ln \left (x \right ) b e}{a^{2}}+\frac {\ln \left (x \right ) b^{2} d}{a^{3}}-\frac {\ln \left (x \right ) b^{3} c}{a^{4}}-\frac {\ln \left (b \,x^{3}+a \right ) f}{3 a}+\frac {\ln \left (b \,x^{3}+a \right ) b e}{3 a^{2}}-\frac {\ln \left (b \,x^{3}+a \right ) b^{2} d}{3 a^{3}}+\frac {\ln \left (b \,x^{3}+a \right ) b^{3} c}{3 a^{4}}\) | \(153\) |
parallelrisch | \(\frac {18 \ln \left (x \right ) x^{9} a^{3} f -18 \ln \left (x \right ) x^{9} a^{2} b e +18 \ln \left (x \right ) x^{9} a \,b^{2} d -18 \ln \left (x \right ) x^{9} b^{3} c -6 \ln \left (b \,x^{3}+a \right ) x^{9} a^{3} f +6 \ln \left (b \,x^{3}+a \right ) x^{9} a^{2} b e -6 \ln \left (b \,x^{3}+a \right ) x^{9} a \,b^{2} d +6 \ln \left (b \,x^{3}+a \right ) x^{9} b^{3} c -6 a^{3} e \,x^{6}+6 a^{2} b d \,x^{6}-6 a \,b^{2} c \,x^{6}-3 a^{3} d \,x^{3}+3 a^{2} x^{3} b c -2 c \,a^{3}}{18 a^{4} x^{9}}\) | \(180\) |
-1/9*c/a/x^9-1/6*(a*d-b*c)/a^2/x^6-1/3*(a^2*e-a*b*d+b^2*c)/a^3/x^3+(a^3*f- a^2*b*e+a*b^2*d-b^3*c)/a^4*ln(x)-1/3*(a^3*f-a^2*b*e+a*b^2*d-b^3*c)/a^4*ln( b*x^3+a)
Time = 0.32 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.99 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^{10} \left (a+b x^3\right )} \, dx=\frac {6 \, {\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} x^{9} \log \left (b x^{3} + a\right ) - 18 \, {\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} x^{9} \log \left (x\right ) - 6 \, {\left (a b^{2} c - a^{2} b d + a^{3} e\right )} x^{6} - 2 \, a^{3} c + 3 \, {\left (a^{2} b c - a^{3} d\right )} x^{3}}{18 \, a^{4} x^{9}} \]
1/18*(6*(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*x^9*log(b*x^3 + a) - 18*(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*x^9*log(x) - 6*(a*b^2*c - a^2*b*d + a^3*e)*x ^6 - 2*a^3*c + 3*(a^2*b*c - a^3*d)*x^3)/(a^4*x^9)
Timed out. \[ \int \frac {c+d x^3+e x^6+f x^9}{x^{10} \left (a+b x^3\right )} \, dx=\text {Timed out} \]
Time = 0.21 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.98 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^{10} \left (a+b x^3\right )} \, dx=\frac {{\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \log \left (b x^{3} + a\right )}{3 \, a^{4}} - \frac {{\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \log \left (x^{3}\right )}{3 \, a^{4}} - \frac {6 \, {\left (b^{2} c - a b d + a^{2} e\right )} x^{6} - 3 \, {\left (a b c - a^{2} d\right )} x^{3} + 2 \, a^{2} c}{18 \, a^{3} x^{9}} \]
1/3*(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*log(b*x^3 + a)/a^4 - 1/3*(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*log(x^3)/a^4 - 1/18*(6*(b^2*c - a*b*d + a^2*e)* x^6 - 3*(a*b*c - a^2*d)*x^3 + 2*a^2*c)/(a^3*x^9)
Time = 0.27 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.41 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^{10} \left (a+b x^3\right )} \, dx=-\frac {{\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \log \left ({\left | x \right |}\right )}{a^{4}} + \frac {{\left (b^{4} c - a b^{3} d + a^{2} b^{2} e - a^{3} b f\right )} \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, a^{4} b} + \frac {11 \, b^{3} c x^{9} - 11 \, a b^{2} d x^{9} + 11 \, a^{2} b e x^{9} - 11 \, a^{3} f x^{9} - 6 \, a b^{2} c x^{6} + 6 \, a^{2} b d x^{6} - 6 \, a^{3} e x^{6} + 3 \, a^{2} b c x^{3} - 3 \, a^{3} d x^{3} - 2 \, a^{3} c}{18 \, a^{4} x^{9}} \]
-(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*log(abs(x))/a^4 + 1/3*(b^4*c - a*b^3* d + a^2*b^2*e - a^3*b*f)*log(abs(b*x^3 + a))/(a^4*b) + 1/18*(11*b^3*c*x^9 - 11*a*b^2*d*x^9 + 11*a^2*b*e*x^9 - 11*a^3*f*x^9 - 6*a*b^2*c*x^6 + 6*a^2*b *d*x^6 - 6*a^3*e*x^6 + 3*a^2*b*c*x^3 - 3*a^3*d*x^3 - 2*a^3*c)/(a^4*x^9)
Time = 9.23 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.96 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^{10} \left (a+b x^3\right )} \, dx=\frac {\ln \left (b\,x^3+a\right )\,\left (-f\,a^3+e\,a^2\,b-d\,a\,b^2+c\,b^3\right )}{3\,a^4}-\frac {\frac {c}{9\,a}+\frac {x^3\,\left (a\,d-b\,c\right )}{6\,a^2}+\frac {x^6\,\left (e\,a^2-d\,a\,b+c\,b^2\right )}{3\,a^3}}{x^9}-\frac {\ln \left (x\right )\,\left (-f\,a^3+e\,a^2\,b-d\,a\,b^2+c\,b^3\right )}{a^4} \]