Integrand size = 38, antiderivative size = 158 \[ \int x^2 \left (a+b x^3\right )^2 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right ) \, dx=\frac {1}{4} a^2 d x^4+\frac {1}{5} a^2 e x^5+\frac {1}{6} a^2 f x^6+\frac {1}{7} a (2 b d+a g) x^7+\frac {1}{8} a (2 b e+a h) x^8+\frac {2}{9} a b f x^9+\frac {1}{10} b (b d+2 a g) x^{10}+\frac {1}{11} b (b e+2 a h) x^{11}+\frac {1}{12} b^2 f x^{12}+\frac {1}{13} b^2 g x^{13}+\frac {1}{14} b^2 h x^{14}+\frac {c \left (a+b x^3\right )^3}{9 b} \]
1/4*a^2*d*x^4+1/5*a^2*e*x^5+1/6*a^2*f*x^6+1/7*a*(a*g+2*b*d)*x^7+1/8*a*(a*h +2*b*e)*x^8+2/9*a*b*f*x^9+1/10*b*(2*a*g+b*d)*x^10+1/11*b*(2*a*h+b*e)*x^11+ 1/12*b^2*f*x^12+1/13*b^2*g*x^13+1/14*b^2*h*x^14+1/9*c*(b*x^3+a)^3/b
Time = 0.08 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.95 \[ \int x^2 \left (a+b x^3\right )^2 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right ) \, dx=a^2 \left (\frac {c x^3}{3}+\frac {d x^4}{4}+\frac {e x^5}{5}+\frac {f x^6}{6}+\frac {g x^7}{7}+\frac {h x^8}{8}\right )+a b \left (\frac {c x^6}{3}+\frac {2 d x^7}{7}+\frac {e x^8}{4}+\frac {2 f x^9}{9}+\frac {g x^{10}}{5}+\frac {2 h x^{11}}{11}\right )+\frac {b^2 x^9 \left (20020 c+3 x \left (6006 d+5460 e x+55 x^2 \left (91 f+84 g x+78 h x^2\right )\right )\right )}{180180} \]
a^2*((c*x^3)/3 + (d*x^4)/4 + (e*x^5)/5 + (f*x^6)/6 + (g*x^7)/7 + (h*x^8)/8 ) + a*b*((c*x^6)/3 + (2*d*x^7)/7 + (e*x^8)/4 + (2*f*x^9)/9 + (g*x^10)/5 + (2*h*x^11)/11) + (b^2*x^9*(20020*c + 3*x*(6006*d + 5460*e*x + 55*x^2*(91*f + 84*g*x + 78*h*x^2))))/180180
Time = 0.39 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {2017, 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.
| \(\displaystyle \int x^2 \left (a+b x^3\right )^2 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right ) \, dx\) |
| \(\Big \downarrow \) 2017 |
| \(\displaystyle \int \left (b x^3+a\right )^2 \left (x^2 \left (h x^5+g x^4+f x^3+e x^2+d x+c\right )-c x^2\right )dx+\frac {c \left (a+b x^3\right )^3}{9 b}\) |
| \(\Big \downarrow \) 2389 |
| \(\displaystyle \int \left (b^2 h x^{13}+b^2 g x^{12}+b^2 f x^{11}+b (b e+2 a h) x^{10}+b (b d+2 a g) x^9+2 a b f x^8+a (2 b e+a h) x^7+a (2 b d+a g) x^6+a^2 f x^5+a^2 e x^4+a^2 d x^3\right )dx+\frac {c \left (a+b x^3\right )^3}{9 b}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{4} a^2 d x^4+\frac {1}{5} a^2 e x^5+\frac {1}{6} a^2 f x^6+\frac {c \left (a+b x^3\right )^3}{9 b}+\frac {1}{10} b x^{10} (2 a g+b d)+\frac {1}{7} a x^7 (a g+2 b d)+\frac {1}{11} b x^{11} (2 a h+b e)+\frac {1}{8} a x^8 (a h+2 b e)+\frac {2}{9} a b f x^9+\frac {1}{12} b^2 f x^{12}+\frac {1}{13} b^2 g x^{13}+\frac {1}{14} b^2 h x^{14}\) |
(a^2*d*x^4)/4 + (a^2*e*x^5)/5 + (a^2*f*x^6)/6 + (a*(2*b*d + a*g)*x^7)/7 + (a*(2*b*e + a*h)*x^8)/8 + (2*a*b*f*x^9)/9 + (b*(b*d + 2*a*g)*x^10)/10 + (b *(b*e + 2*a*h)*x^11)/11 + (b^2*f*x^12)/12 + (b^2*g*x^13)/13 + (b^2*h*x^14) /14 + (c*(a + b*x^3)^3)/(9*b)
3.4.85.3.1 Defintions of rubi rules used
Int[(Px_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[Coeff[Px, x, n - 1]*((a + b*x^n)^(p + 1)/(b*n*(p + 1))), x] + Int[(Px - Coeff[Px, x, n - 1] *x^(n - 1))*(a + b*x^n)^p, x] /; FreeQ[{a, b}, x] && PolyQ[Px, x] && IGtQ[p , 1] && IGtQ[n, 1] && NeQ[Coeff[Px, x, n - 1], 0] && NeQ[Px, Coeff[Px, x, n - 1]*x^(n - 1)] && !MatchQ[Px, (Qx_.)*((c_) + (d_.)*x^(m_))^(q_) /; FreeQ [{c, d}, x] && PolyQ[Qx, x] && IGtQ[q, 1] && IGtQ[m, 1] && NeQ[Coeff[Qx*(a + b*x^n)^p, x, m - 1], 0] && GtQ[m*q, n*p]]
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 = 2.08 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.96
| method | result | size |
| default | \(\frac {b^{2} h \,x^{14}}{14}+\frac {b^{2} g \,x^{13}}{13}+\frac {b^{2} f \,x^{12}}{12}+\frac {\left (2 a b h +b^{2} e \right ) x^{11}}{11}+\frac {\left (2 a b g +b^{2} d \right ) x^{10}}{10}+\frac {\left (2 a f b +b^{2} c \right ) x^{9}}{9}+\frac {\left (a^{2} h +2 a e b \right ) x^{8}}{8}+\frac {\left (a^{2} g +2 a b d \right ) x^{7}}{7}+\frac {\left (a^{2} f +2 a b c \right ) x^{6}}{6}+\frac {a^{2} e \,x^{5}}{5}+\frac {a^{2} d \,x^{4}}{4}+\frac {a^{2} c \,x^{3}}{3}\) | \(152\) |
| norman | \(\frac {a^{2} c \,x^{3}}{3}+\frac {a^{2} d \,x^{4}}{4}+\frac {a^{2} e \,x^{5}}{5}+\left (\frac {1}{6} a^{2} f +\frac {1}{3} a b c \right ) x^{6}+\left (\frac {1}{7} a^{2} g +\frac {2}{7} a b d \right ) x^{7}+\left (\frac {1}{8} a^{2} h +\frac {1}{4} a e b \right ) x^{8}+\left (\frac {2}{9} a f b +\frac {1}{9} b^{2} c \right ) x^{9}+\left (\frac {1}{5} a b g +\frac {1}{10} b^{2} d \right ) x^{10}+\left (\frac {2}{11} a b h +\frac {1}{11} b^{2} e \right ) x^{11}+\frac {b^{2} f \,x^{12}}{12}+\frac {b^{2} g \,x^{13}}{13}+\frac {b^{2} h \,x^{14}}{14}\) | \(152\) |
| gosper | \(\frac {1}{3} a^{2} c \,x^{3}+\frac {1}{4} a^{2} d \,x^{4}+\frac {1}{5} a^{2} e \,x^{5}+\frac {1}{6} a^{2} f \,x^{6}+\frac {1}{3} a b c \,x^{6}+\frac {1}{7} x^{7} a^{2} g +\frac {2}{7} a d \,x^{7} b +\frac {1}{8} x^{8} a^{2} h +\frac {1}{4} a b e \,x^{8}+\frac {2}{9} a b f \,x^{9}+\frac {1}{9} b^{2} c \,x^{9}+\frac {1}{5} x^{10} a b g +\frac {1}{10} b^{2} d \,x^{10}+\frac {2}{11} x^{11} a b h +\frac {1}{11} b^{2} e \,x^{11}+\frac {1}{12} b^{2} f \,x^{12}+\frac {1}{13} b^{2} g \,x^{13}+\frac {1}{14} b^{2} h \,x^{14}\) | \(158\) |
| risch | \(\frac {1}{3} a^{2} c \,x^{3}+\frac {1}{4} a^{2} d \,x^{4}+\frac {1}{5} a^{2} e \,x^{5}+\frac {1}{6} a^{2} f \,x^{6}+\frac {1}{3} a b c \,x^{6}+\frac {1}{7} x^{7} a^{2} g +\frac {2}{7} a d \,x^{7} b +\frac {1}{8} x^{8} a^{2} h +\frac {1}{4} a b e \,x^{8}+\frac {2}{9} a b f \,x^{9}+\frac {1}{9} b^{2} c \,x^{9}+\frac {1}{5} x^{10} a b g +\frac {1}{10} b^{2} d \,x^{10}+\frac {2}{11} x^{11} a b h +\frac {1}{11} b^{2} e \,x^{11}+\frac {1}{12} b^{2} f \,x^{12}+\frac {1}{13} b^{2} g \,x^{13}+\frac {1}{14} b^{2} h \,x^{14}\) | \(158\) |
| parallelrisch | \(\frac {1}{3} a^{2} c \,x^{3}+\frac {1}{4} a^{2} d \,x^{4}+\frac {1}{5} a^{2} e \,x^{5}+\frac {1}{6} a^{2} f \,x^{6}+\frac {1}{3} a b c \,x^{6}+\frac {1}{7} x^{7} a^{2} g +\frac {2}{7} a d \,x^{7} b +\frac {1}{8} x^{8} a^{2} h +\frac {1}{4} a b e \,x^{8}+\frac {2}{9} a b f \,x^{9}+\frac {1}{9} b^{2} c \,x^{9}+\frac {1}{5} x^{10} a b g +\frac {1}{10} b^{2} d \,x^{10}+\frac {2}{11} x^{11} a b h +\frac {1}{11} b^{2} e \,x^{11}+\frac {1}{12} b^{2} f \,x^{12}+\frac {1}{13} b^{2} g \,x^{13}+\frac {1}{14} b^{2} h \,x^{14}\) | \(158\) |
1/14*b^2*h*x^14+1/13*b^2*g*x^13+1/12*b^2*f*x^12+1/11*(2*a*b*h+b^2*e)*x^11+ 1/10*(2*a*b*g+b^2*d)*x^10+1/9*(2*a*b*f+b^2*c)*x^9+1/8*(a^2*h+2*a*b*e)*x^8+ 1/7*(a^2*g+2*a*b*d)*x^7+1/6*(a^2*f+2*a*b*c)*x^6+1/5*a^2*e*x^5+1/4*a^2*d*x^ 4+1/3*a^2*c*x^3
Time = 0.38 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.96 \[ \int x^2 \left (a+b x^3\right )^2 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right ) \, dx=\frac {1}{14} \, b^{2} h x^{14} + \frac {1}{13} \, b^{2} g x^{13} + \frac {1}{12} \, b^{2} f x^{12} + \frac {1}{11} \, {\left (b^{2} e + 2 \, a b h\right )} x^{11} + \frac {1}{10} \, {\left (b^{2} d + 2 \, a b g\right )} x^{10} + \frac {1}{9} \, {\left (b^{2} c + 2 \, a b f\right )} x^{9} + \frac {1}{8} \, {\left (2 \, a b e + a^{2} h\right )} x^{8} + \frac {1}{5} \, a^{2} e x^{5} + \frac {1}{7} \, {\left (2 \, a b d + a^{2} g\right )} x^{7} + \frac {1}{4} \, a^{2} d x^{4} + \frac {1}{6} \, {\left (2 \, a b c + a^{2} f\right )} x^{6} + \frac {1}{3} \, a^{2} c x^{3} \]
1/14*b^2*h*x^14 + 1/13*b^2*g*x^13 + 1/12*b^2*f*x^12 + 1/11*(b^2*e + 2*a*b* h)*x^11 + 1/10*(b^2*d + 2*a*b*g)*x^10 + 1/9*(b^2*c + 2*a*b*f)*x^9 + 1/8*(2 *a*b*e + a^2*h)*x^8 + 1/5*a^2*e*x^5 + 1/7*(2*a*b*d + a^2*g)*x^7 + 1/4*a^2* d*x^4 + 1/6*(2*a*b*c + a^2*f)*x^6 + 1/3*a^2*c*x^3
Time = 0.04 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.06 \[ \int x^2 \left (a+b x^3\right )^2 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right ) \, dx=\frac {a^{2} c x^{3}}{3} + \frac {a^{2} d x^{4}}{4} + \frac {a^{2} e x^{5}}{5} + \frac {b^{2} f x^{12}}{12} + \frac {b^{2} g x^{13}}{13} + \frac {b^{2} h x^{14}}{14} + x^{11} \cdot \left (\frac {2 a b h}{11} + \frac {b^{2} e}{11}\right ) + x^{10} \left (\frac {a b g}{5} + \frac {b^{2} d}{10}\right ) + x^{9} \cdot \left (\frac {2 a b f}{9} + \frac {b^{2} c}{9}\right ) + x^{8} \left (\frac {a^{2} h}{8} + \frac {a b e}{4}\right ) + x^{7} \left (\frac {a^{2} g}{7} + \frac {2 a b d}{7}\right ) + x^{6} \left (\frac {a^{2} f}{6} + \frac {a b c}{3}\right ) \]
a**2*c*x**3/3 + a**2*d*x**4/4 + a**2*e*x**5/5 + b**2*f*x**12/12 + b**2*g*x **13/13 + b**2*h*x**14/14 + x**11*(2*a*b*h/11 + b**2*e/11) + x**10*(a*b*g/ 5 + b**2*d/10) + x**9*(2*a*b*f/9 + b**2*c/9) + x**8*(a**2*h/8 + a*b*e/4) + x**7*(a**2*g/7 + 2*a*b*d/7) + x**6*(a**2*f/6 + a*b*c/3)
Time = 0.20 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.96 \[ \int x^2 \left (a+b x^3\right )^2 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right ) \, dx=\frac {1}{14} \, b^{2} h x^{14} + \frac {1}{13} \, b^{2} g x^{13} + \frac {1}{12} \, b^{2} f x^{12} + \frac {1}{11} \, {\left (b^{2} e + 2 \, a b h\right )} x^{11} + \frac {1}{10} \, {\left (b^{2} d + 2 \, a b g\right )} x^{10} + \frac {1}{9} \, {\left (b^{2} c + 2 \, a b f\right )} x^{9} + \frac {1}{8} \, {\left (2 \, a b e + a^{2} h\right )} x^{8} + \frac {1}{5} \, a^{2} e x^{5} + \frac {1}{7} \, {\left (2 \, a b d + a^{2} g\right )} x^{7} + \frac {1}{4} \, a^{2} d x^{4} + \frac {1}{6} \, {\left (2 \, a b c + a^{2} f\right )} x^{6} + \frac {1}{3} \, a^{2} c x^{3} \]
1/14*b^2*h*x^14 + 1/13*b^2*g*x^13 + 1/12*b^2*f*x^12 + 1/11*(b^2*e + 2*a*b* h)*x^11 + 1/10*(b^2*d + 2*a*b*g)*x^10 + 1/9*(b^2*c + 2*a*b*f)*x^9 + 1/8*(2 *a*b*e + a^2*h)*x^8 + 1/5*a^2*e*x^5 + 1/7*(2*a*b*d + a^2*g)*x^7 + 1/4*a^2* d*x^4 + 1/6*(2*a*b*c + a^2*f)*x^6 + 1/3*a^2*c*x^3
Time = 0.27 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.99 \[ \int x^2 \left (a+b x^3\right )^2 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right ) \, dx=\frac {1}{14} \, b^{2} h x^{14} + \frac {1}{13} \, b^{2} g x^{13} + \frac {1}{12} \, b^{2} f x^{12} + \frac {1}{11} \, b^{2} e x^{11} + \frac {2}{11} \, a b h x^{11} + \frac {1}{10} \, b^{2} d x^{10} + \frac {1}{5} \, a b g x^{10} + \frac {1}{9} \, b^{2} c x^{9} + \frac {2}{9} \, a b f x^{9} + \frac {1}{4} \, a b e x^{8} + \frac {1}{8} \, a^{2} h x^{8} + \frac {2}{7} \, a b d x^{7} + \frac {1}{7} \, a^{2} g x^{7} + \frac {1}{3} \, a b c x^{6} + \frac {1}{6} \, a^{2} f x^{6} + \frac {1}{5} \, a^{2} e x^{5} + \frac {1}{4} \, a^{2} d x^{4} + \frac {1}{3} \, a^{2} c x^{3} \]
1/14*b^2*h*x^14 + 1/13*b^2*g*x^13 + 1/12*b^2*f*x^12 + 1/11*b^2*e*x^11 + 2/ 11*a*b*h*x^11 + 1/10*b^2*d*x^10 + 1/5*a*b*g*x^10 + 1/9*b^2*c*x^9 + 2/9*a*b *f*x^9 + 1/4*a*b*e*x^8 + 1/8*a^2*h*x^8 + 2/7*a*b*d*x^7 + 1/7*a^2*g*x^7 + 1 /3*a*b*c*x^6 + 1/6*a^2*f*x^6 + 1/5*a^2*e*x^5 + 1/4*a^2*d*x^4 + 1/3*a^2*c*x ^3
Time = 0.11 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.96 \[ \int x^2 \left (a+b x^3\right )^2 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right ) \, dx=x^6\,\left (\frac {f\,a^2}{6}+\frac {b\,c\,a}{3}\right )+x^9\,\left (\frac {c\,b^2}{9}+\frac {2\,a\,f\,b}{9}\right )+x^7\,\left (\frac {g\,a^2}{7}+\frac {2\,b\,d\,a}{7}\right )+x^{10}\,\left (\frac {d\,b^2}{10}+\frac {a\,g\,b}{5}\right )+x^8\,\left (\frac {h\,a^2}{8}+\frac {b\,e\,a}{4}\right )+x^{11}\,\left (\frac {e\,b^2}{11}+\frac {2\,a\,h\,b}{11}\right )+\frac {a^2\,c\,x^3}{3}+\frac {a^2\,d\,x^4}{4}+\frac {a^2\,e\,x^5}{5}+\frac {b^2\,f\,x^{12}}{12}+\frac {b^2\,g\,x^{13}}{13}+\frac {b^2\,h\,x^{14}}{14} \]