Integrand size = 30, antiderivative size = 109 \[ \int \frac {x^2 \left (c+d x^3+e x^6+f x^9\right )}{\left (a+b x^3\right )^3} \, dx=\frac {f x^3}{3 b^3}-\frac {b^3 c-a b^2 d+a^2 b e-a^3 f}{6 b^4 \left (a+b x^3\right )^2}-\frac {b^2 d-2 a b e+3 a^2 f}{3 b^4 \left (a+b x^3\right )}+\frac {(b e-3 a f) \log \left (a+b x^3\right )}{3 b^4} \]
1/3*f*x^3/b^3+1/6*(a^3*f-a^2*b*e+a*b^2*d-b^3*c)/b^4/(b*x^3+a)^2+1/3*(-3*a^ 2*f+2*a*b*e-b^2*d)/b^4/(b*x^3+a)+1/3*(-3*a*f+b*e)*ln(b*x^3+a)/b^4
Time = 0.06 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.96 \[ \int \frac {x^2 \left (c+d x^3+e x^6+f x^9\right )}{\left (a+b x^3\right )^3} \, dx=\frac {-5 a^3 f+a^2 b \left (3 e-4 f x^3\right )+a b^2 \left (-d+4 e x^3+4 f x^6\right )-b^3 \left (c+2 d x^3-2 f x^9\right )+2 (b e-3 a f) \left (a+b x^3\right )^2 \log \left (a+b x^3\right )}{6 b^4 \left (a+b x^3\right )^2} \]
(-5*a^3*f + a^2*b*(3*e - 4*f*x^3) + a*b^2*(-d + 4*e*x^3 + 4*f*x^6) - b^3*( c + 2*d*x^3 - 2*f*x^9) + 2*(b*e - 3*a*f)*(a + b*x^3)^2*Log[a + b*x^3])/(6* b^4*(a + b*x^3)^2)
Time = 0.36 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.96, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2359, 2389, 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.
\(\displaystyle \int \frac {x^2 \left (c+d x^3+e x^6+f x^9\right )}{\left (a+b x^3\right )^3} \, dx\) |
\(\Big \downarrow \) 2359 |
\(\displaystyle \frac {1}{3} \int \frac {f x^9+e x^6+d x^3+c}{\left (b x^3+a\right )^3}dx^3\) |
\(\Big \downarrow \) 2389 |
\(\displaystyle \frac {1}{3} \int \left (\frac {f}{b^3}+\frac {b e-3 a f}{b^3 \left (b x^3+a\right )}+\frac {3 f a^2-2 b e a+b^2 d}{b^3 \left (b x^3+a\right )^2}+\frac {-f a^3+b e a^2-b^2 d a+b^3 c}{b^3 \left (b x^3+a\right )^3}\right )dx^3\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (-\frac {3 a^2 f-2 a b e+b^2 d}{b^4 \left (a+b x^3\right )}-\frac {a^3 (-f)+a^2 b e-a b^2 d+b^3 c}{2 b^4 \left (a+b x^3\right )^2}+\frac {(b e-3 a f) \log \left (a+b x^3\right )}{b^4}+\frac {f x^3}{b^3}\right )\) |
((f*x^3)/b^3 - (b^3*c - a*b^2*d + a^2*b*e - a^3*f)/(2*b^4*(a + b*x^3)^2) - (b^2*d - 2*a*b*e + 3*a^2*f)/(b^4*(a + b*x^3)) + ((b*e - 3*a*f)*Log[a + b* x^3])/b^4)/3
3.3.80.3.1 Defintions of rubi rules used
Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Simp[1/(m + 1) Subst[Int[SubstFor[x^(m + 1), Pq, x]*(a + b*x^Simplify[n/(m + 1)])^ p, x], x, x^(m + 1)], x] /; FreeQ[{a, b, m, n, p}, x] && NeQ[m, -1] && IGtQ [Simplify[n/(m + 1)], 0] && PolyQ[Pq, x^(m + 1)]
Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand [Pq*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p , 0] || EqQ[n, 1])
Time = 1.52 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.91
method | result | size |
norman | \(\frac {-\frac {9 f \,a^{3}-3 a^{2} b e +a \,b^{2} d +b^{3} c}{6 b^{4}}+\frac {f \,x^{9}}{3 b}-\frac {\left (6 a^{2} f -2 a e b +b^{2} d \right ) x^{3}}{3 b^{3}}}{\left (b \,x^{3}+a \right )^{2}}-\frac {\left (3 a f -b e \right ) \ln \left (b \,x^{3}+a \right )}{3 b^{4}}\) | \(99\) |
risch | \(\frac {f \,x^{3}}{3 b^{3}}+\frac {\left (-a^{2} f +\frac {2}{3} a e b -\frac {1}{3} b^{2} d \right ) x^{3}-\frac {5 f \,a^{3}-3 a^{2} b e +a \,b^{2} d +b^{3} c}{6 b}}{b^{3} \left (b \,x^{3}+a \right )^{2}}-\frac {\ln \left (b \,x^{3}+a \right ) a f}{b^{4}}+\frac {\ln \left (b \,x^{3}+a \right ) e}{3 b^{3}}\) | \(106\) |
default | \(\frac {f \,x^{3}}{3 b^{3}}-\frac {\frac {\left (3 a f -b e \right ) \ln \left (b \,x^{3}+a \right )}{b}-\frac {f \,a^{3}-a^{2} b e +a \,b^{2} d -b^{3} c}{2 b \left (b \,x^{3}+a \right )^{2}}-\frac {-3 a^{2} f +2 a e b -b^{2} d}{b \left (b \,x^{3}+a \right )}}{3 b^{3}}\) | \(109\) |
parallelrisch | \(-\frac {-2 b^{3} f \,x^{9}+6 \ln \left (b \,x^{3}+a \right ) x^{6} a \,b^{2} f -2 \ln \left (b \,x^{3}+a \right ) x^{6} b^{3} e +12 \ln \left (b \,x^{3}+a \right ) x^{3} a^{2} b f -4 \ln \left (b \,x^{3}+a \right ) x^{3} a \,b^{2} e +12 a^{2} b f \,x^{3}-4 a \,b^{2} e \,x^{3}+2 b^{3} d \,x^{3}+6 \ln \left (b \,x^{3}+a \right ) a^{3} f -2 \ln \left (b \,x^{3}+a \right ) a^{2} b e +9 f \,a^{3}-3 a^{2} b e +a \,b^{2} d +b^{3} c}{6 b^{4} \left (b \,x^{3}+a \right )^{2}}\) | \(178\) |
(-1/6*(9*a^3*f-3*a^2*b*e+a*b^2*d+b^3*c)/b^4+1/3*f*x^9/b-1/3*(6*a^2*f-2*a*b *e+b^2*d)/b^3*x^3)/(b*x^3+a)^2-1/3*(3*a*f-b*e)/b^4*ln(b*x^3+a)
Time = 0.25 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.45 \[ \int \frac {x^2 \left (c+d x^3+e x^6+f x^9\right )}{\left (a+b x^3\right )^3} \, dx=\frac {2 \, b^{3} f x^{9} + 4 \, a b^{2} f x^{6} - b^{3} c - a b^{2} d + 3 \, a^{2} b e - 5 \, a^{3} f - 2 \, {\left (b^{3} d - 2 \, a b^{2} e + 2 \, a^{2} b f\right )} x^{3} + 2 \, {\left ({\left (b^{3} e - 3 \, a b^{2} f\right )} x^{6} + a^{2} b e - 3 \, a^{3} f + 2 \, {\left (a b^{2} e - 3 \, a^{2} b f\right )} x^{3}\right )} \log \left (b x^{3} + a\right )}{6 \, {\left (b^{6} x^{6} + 2 \, a b^{5} x^{3} + a^{2} b^{4}\right )}} \]
1/6*(2*b^3*f*x^9 + 4*a*b^2*f*x^6 - b^3*c - a*b^2*d + 3*a^2*b*e - 5*a^3*f - 2*(b^3*d - 2*a*b^2*e + 2*a^2*b*f)*x^3 + 2*((b^3*e - 3*a*b^2*f)*x^6 + a^2* b*e - 3*a^3*f + 2*(a*b^2*e - 3*a^2*b*f)*x^3)*log(b*x^3 + a))/(b^6*x^6 + 2* a*b^5*x^3 + a^2*b^4)
Timed out. \[ \int \frac {x^2 \left (c+d x^3+e x^6+f x^9\right )}{\left (a+b x^3\right )^3} \, dx=\text {Timed out} \]
Time = 0.20 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00 \[ \int \frac {x^2 \left (c+d x^3+e x^6+f x^9\right )}{\left (a+b x^3\right )^3} \, dx=\frac {f x^{3}}{3 \, b^{3}} - \frac {b^{3} c + a b^{2} d - 3 \, a^{2} b e + 5 \, a^{3} f + 2 \, {\left (b^{3} d - 2 \, a b^{2} e + 3 \, a^{2} b f\right )} x^{3}}{6 \, {\left (b^{6} x^{6} + 2 \, a b^{5} x^{3} + a^{2} b^{4}\right )}} + \frac {{\left (b e - 3 \, a f\right )} \log \left (b x^{3} + a\right )}{3 \, b^{4}} \]
1/3*f*x^3/b^3 - 1/6*(b^3*c + a*b^2*d - 3*a^2*b*e + 5*a^3*f + 2*(b^3*d - 2* a*b^2*e + 3*a^2*b*f)*x^3)/(b^6*x^6 + 2*a*b^5*x^3 + a^2*b^4) + 1/3*(b*e - 3 *a*f)*log(b*x^3 + a)/b^4
Time = 0.27 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.88 \[ \int \frac {x^2 \left (c+d x^3+e x^6+f x^9\right )}{\left (a+b x^3\right )^3} \, dx=\frac {f x^{3}}{3 \, b^{3}} + \frac {{\left (b e - 3 \, a f\right )} \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, b^{4}} - \frac {b^{3} c + a b^{2} d - 3 \, a^{2} b e + 5 \, a^{3} f + 2 \, {\left (b^{3} d - 2 \, a b^{2} e + 3 \, a^{2} b f\right )} x^{3}}{6 \, {\left (b x^{3} + a\right )}^{2} b^{4}} \]
1/3*f*x^3/b^3 + 1/3*(b*e - 3*a*f)*log(abs(b*x^3 + a))/b^4 - 1/6*(b^3*c + a *b^2*d - 3*a^2*b*e + 5*a^3*f + 2*(b^3*d - 2*a*b^2*e + 3*a^2*b*f)*x^3)/((b* x^3 + a)^2*b^4)
Time = 9.16 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.03 \[ \int \frac {x^2 \left (c+d x^3+e x^6+f x^9\right )}{\left (a+b x^3\right )^3} \, dx=\frac {f\,x^3}{3\,b^3}-\frac {x^3\,\left (f\,a^2-\frac {2\,e\,a\,b}{3}+\frac {d\,b^2}{3}\right )+\frac {5\,f\,a^3-3\,e\,a^2\,b+d\,a\,b^2+c\,b^3}{6\,b}}{a^2\,b^3+2\,a\,b^4\,x^3+b^5\,x^6}-\frac {\ln \left (b\,x^3+a\right )\,\left (3\,a\,f-b\,e\right )}{3\,b^4} \]