Integrand size = 38, antiderivative size = 198 \[ \int \frac {\left (a+b x^3\right )^3 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x^2} \, dx=-\frac {a^3 c}{x}+a^3 e x+\frac {1}{2} a^2 (3 b c+a f) x^2+a^2 b d x^3+\frac {1}{4} a^2 (3 b e+a h) x^4+\frac {3}{5} a b (b c+a f) x^5+\frac {1}{2} a b^2 d x^6+\frac {3}{7} a b (b e+a h) x^7+\frac {1}{8} b^2 (b c+3 a f) x^8+\frac {1}{9} b^3 d x^9+\frac {1}{10} b^2 (b e+3 a h) x^{10}+\frac {1}{11} b^3 f x^{11}+\frac {1}{13} b^3 h x^{13}+\frac {g \left (a+b x^3\right )^4}{12 b}+a^3 d \log (x) \]
-a^3*c/x+a^3*e*x+1/2*a^2*(a*f+3*b*c)*x^2+a^2*b*d*x^3+1/4*a^2*(a*h+3*b*e)*x ^4+3/5*a*b*(a*f+b*c)*x^5+1/2*a*b^2*d*x^6+3/7*a*b*(a*h+b*e)*x^7+1/8*b^2*(3* a*f+b*c)*x^8+1/9*b^3*d*x^9+1/10*b^2*(3*a*h+b*e)*x^10+1/11*b^3*f*x^11+1/13* b^3*h*x^13+1/12*g*(b*x^3+a)^4/b+a^3*d*ln(x)
Time = 0.16 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.87 \[ \int \frac {\left (a+b x^3\right )^3 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x^2} \, dx=a^3 \left (-\frac {c}{x}+e x+\frac {1}{12} x^2 \left (6 f+4 g x+3 h x^2\right )\right )+\frac {b^3 x^8 \left (6435 c+5720 d x+6 x^2 \left (858 e+780 f x+715 g x^2+660 h x^3\right )\right )}{51480}+\frac {1}{140} a^2 b x^2 \left (210 c+x \left (140 d+x \left (105 e+84 f x+70 g x^2+60 h x^3\right )\right )\right )+\frac {1}{840} a b^2 x^5 \left (504 c+x \left (420 d+x \left (360 e+315 f x+280 g x^2+252 h x^3\right )\right )\right )+a^3 d \log (x) \]
a^3*(-(c/x) + e*x + (x^2*(6*f + 4*g*x + 3*h*x^2))/12) + (b^3*x^8*(6435*c + 5720*d*x + 6*x^2*(858*e + 780*f*x + 715*g*x^2 + 660*h*x^3)))/51480 + (a^2 *b*x^2*(210*c + x*(140*d + x*(105*e + 84*f*x + 70*g*x^2 + 60*h*x^3))))/140 + (a*b^2*x^5*(504*c + x*(420*d + x*(360*e + 315*f*x + 280*g*x^2 + 252*h*x ^3))))/840 + a^3*d*Log[x]
Time = 0.46 (sec) , antiderivative size = 198, 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 = {2018, 2360, 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.
| \(\displaystyle \int \frac {\left (a+b x^3\right )^3 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x^2} \, dx\) |
| \(\Big \downarrow \) 2018 |
| \(\displaystyle \int \frac {\left (b x^3+a\right )^3 \left (h x^5+f x^3+e x^2+d x+c\right )}{x^2}dx+\frac {g \left (a+b x^3\right )^4}{12 b}\) |
| \(\Big \downarrow \) 2360 |
| \(\displaystyle \int \left (b^3 h x^{12}+b^3 f x^{10}+b^2 (b e+3 a h) x^9+b^3 d x^8+b^2 (b c+3 a f) x^7+3 a b (b e+a h) x^6+3 a b^2 d x^5+3 a b (b c+a f) x^4+a^2 (3 b e+a h) x^3+3 a^2 b d x^2+a^2 (3 b c+a f) x+a^3 e+\frac {a^3 d}{x}+\frac {a^3 c}{x^2}\right )dx+\frac {g \left (a+b x^3\right )^4}{12 b}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {a^3 c}{x}+a^3 d \log (x)+a^3 e x+\frac {1}{2} a^2 x^2 (a f+3 b c)+a^2 b d x^3+\frac {1}{4} a^2 x^4 (a h+3 b e)+\frac {1}{8} b^2 x^8 (3 a f+b c)+\frac {1}{2} a b^2 d x^6+\frac {1}{10} b^2 x^{10} (3 a h+b e)+\frac {3}{5} a b x^5 (a f+b c)+\frac {3}{7} a b x^7 (a h+b e)+\frac {g \left (a+b x^3\right )^4}{12 b}+\frac {1}{9} b^3 d x^9+\frac {1}{11} b^3 f x^{11}+\frac {1}{13} b^3 h x^{13}\) |
-((a^3*c)/x) + a^3*e*x + (a^2*(3*b*c + a*f)*x^2)/2 + a^2*b*d*x^3 + (a^2*(3 *b*e + a*h)*x^4)/4 + (3*a*b*(b*c + a*f)*x^5)/5 + (a*b^2*d*x^6)/2 + (3*a*b* (b*e + a*h)*x^7)/7 + (b^2*(b*c + 3*a*f)*x^8)/8 + (b^3*d*x^9)/9 + (b^2*(b*e + 3*a*h)*x^10)/10 + (b^3*f*x^11)/11 + (b^3*h*x^13)/13 + (g*(a + b*x^3)^4) /(12*b) + a^3*d*Log[x]
3.4.99.3.1 Defintions of rubi rules used
Int[(Px_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[Coef f[Px, x, n - m - 1]*((a + b*x^n)^(p + 1)/(b*n*(p + 1))), x] + Int[(Px - Coe ff[Px, x, n - m - 1]*x^(n - m - 1))*x^m*(a + b*x^n)^p, x] /; FreeQ[{a, b, m , n}, x] && PolyQ[Px, x] && IGtQ[p, 1] && IGtQ[n - m, 0] && NeQ[Coeff[Px, x , n - m - 1], 0]
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])
Time = 1.50 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.11
| method | result | size |
| norman | \(\frac {\left (\frac {1}{3} a^{3} g +d \,a^{2} b \right ) x^{4}+\left (\frac {1}{4} a^{3} h +\frac {3}{4} a^{2} b e \right ) x^{5}+\left (\frac {1}{2} f \,a^{3}+\frac {3}{2} a^{2} b c \right ) x^{3}+\left (\frac {3}{8} a \,b^{2} f +\frac {1}{8} b^{3} c \right ) x^{9}+\left (\frac {1}{3} a \,b^{2} g +\frac {1}{9} b^{3} d \right ) x^{10}+\left (\frac {3}{10} a \,b^{2} h +\frac {1}{10} b^{3} e \right ) x^{11}+\left (\frac {1}{2} a^{2} b g +\frac {1}{2} a \,b^{2} d \right ) x^{7}+\left (\frac {3}{7} a^{2} b h +\frac {3}{7} a \,b^{2} e \right ) x^{8}+\left (\frac {3}{5} f \,a^{2} b +\frac {3}{5} a \,b^{2} c \right ) x^{6}+a^{3} e \,x^{2}-c \,a^{3}+\frac {b^{3} f \,x^{12}}{11}+\frac {b^{3} g \,x^{13}}{12}+\frac {b^{3} h \,x^{14}}{13}}{x}+a^{3} d \ln \left (x \right )\) | \(219\) |
| default | \(\frac {b^{3} h \,x^{13}}{13}+\frac {b^{3} g \,x^{12}}{12}+\frac {b^{3} f \,x^{11}}{11}+\frac {3 a \,b^{2} h \,x^{10}}{10}+\frac {b^{3} e \,x^{10}}{10}+\frac {a \,b^{2} g \,x^{9}}{3}+\frac {b^{3} d \,x^{9}}{9}+\frac {3 x^{8} a \,b^{2} f}{8}+\frac {b^{3} c \,x^{8}}{8}+\frac {3 a^{2} b h \,x^{7}}{7}+\frac {3 a \,b^{2} e \,x^{7}}{7}+\frac {a^{2} b g \,x^{6}}{2}+\frac {a \,b^{2} d \,x^{6}}{2}+\frac {3 a^{2} b f \,x^{5}}{5}+\frac {3 a \,b^{2} c \,x^{5}}{5}+\frac {a^{3} h \,x^{4}}{4}+\frac {3 a^{2} b e \,x^{4}}{4}+\frac {a^{3} g \,x^{3}}{3}+a^{2} b d \,x^{3}+\frac {f \,a^{3} x^{2}}{2}+\frac {3 a^{2} b c \,x^{2}}{2}+a^{3} e x +a^{3} d \ln \left (x \right )-\frac {a^{3} c}{x}\) | \(224\) |
| risch | \(\frac {b^{3} h \,x^{13}}{13}+\frac {b^{3} g \,x^{12}}{12}+\frac {b^{3} f \,x^{11}}{11}+\frac {3 a \,b^{2} h \,x^{10}}{10}+\frac {b^{3} e \,x^{10}}{10}+\frac {a \,b^{2} g \,x^{9}}{3}+\frac {b^{3} d \,x^{9}}{9}+\frac {3 x^{8} a \,b^{2} f}{8}+\frac {b^{3} c \,x^{8}}{8}+\frac {3 a^{2} b h \,x^{7}}{7}+\frac {3 a \,b^{2} e \,x^{7}}{7}+\frac {a^{2} b g \,x^{6}}{2}+\frac {a \,b^{2} d \,x^{6}}{2}+\frac {3 a^{2} b f \,x^{5}}{5}+\frac {3 a \,b^{2} c \,x^{5}}{5}+\frac {a^{3} h \,x^{4}}{4}+\frac {3 a^{2} b e \,x^{4}}{4}+\frac {a^{3} g \,x^{3}}{3}+a^{2} b d \,x^{3}+\frac {f \,a^{3} x^{2}}{2}+\frac {3 a^{2} b c \,x^{2}}{2}+a^{3} e x +a^{3} d \ln \left (x \right )-\frac {a^{3} c}{x}\) | \(224\) |
| parallelrisch | \(\frac {32760 b^{3} f \,x^{12}+154440 a^{2} b h \,x^{8}+360360 a^{2} b d \,x^{4}+120120 a \,b^{2} g \,x^{10}+180180 f \,a^{3} x^{3}+45045 b^{3} c \,x^{9}+120120 a^{3} g \,x^{4}+90090 a^{3} h \,x^{5}-360360 c \,a^{3}+108108 a \,b^{2} h \,x^{11}+135135 a \,b^{2} f \,x^{9}+216216 x^{6} f \,a^{2} b +180180 a^{2} b g \,x^{7}+270270 a^{2} b e \,x^{5}+180180 a \,b^{2} d \,x^{7}+154440 a \,b^{2} e \,x^{8}+540540 a^{2} x^{3} b c +216216 a \,b^{2} c \,x^{6}+30030 b^{3} g \,x^{13}+27720 b^{3} h \,x^{14}+360360 a^{3} d \ln \left (x \right ) x +360360 a^{3} e \,x^{2}+40040 b^{3} d \,x^{10}+36036 b^{3} e \,x^{11}}{360360 x}\) | \(232\) |
((1/3*a^3*g+d*a^2*b)*x^4+(1/4*a^3*h+3/4*a^2*b*e)*x^5+(1/2*f*a^3+3/2*a^2*b* c)*x^3+(3/8*a*b^2*f+1/8*b^3*c)*x^9+(1/3*a*b^2*g+1/9*b^3*d)*x^10+(3/10*a*b^ 2*h+1/10*b^3*e)*x^11+(1/2*a^2*b*g+1/2*a*b^2*d)*x^7+(3/7*a^2*b*h+3/7*a*b^2* e)*x^8+(3/5*f*a^2*b+3/5*a*b^2*c)*x^6+a^3*e*x^2-c*a^3+1/11*b^3*f*x^12+1/12* b^3*g*x^13+1/13*b^3*h*x^14)/x+a^3*d*ln(x)
Time = 0.38 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.11 \[ \int \frac {\left (a+b x^3\right )^3 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x^2} \, dx=\frac {27720 \, b^{3} h x^{14} + 30030 \, b^{3} g x^{13} + 32760 \, b^{3} f x^{12} + 36036 \, {\left (b^{3} e + 3 \, a b^{2} h\right )} x^{11} + 40040 \, {\left (b^{3} d + 3 \, a b^{2} g\right )} x^{10} + 45045 \, {\left (b^{3} c + 3 \, a b^{2} f\right )} x^{9} + 154440 \, {\left (a b^{2} e + a^{2} b h\right )} x^{8} + 180180 \, {\left (a b^{2} d + a^{2} b g\right )} x^{7} + 216216 \, {\left (a b^{2} c + a^{2} b f\right )} x^{6} + 360360 \, a^{3} e x^{2} + 90090 \, {\left (3 \, a^{2} b e + a^{3} h\right )} x^{5} + 360360 \, a^{3} d x \log \left (x\right ) + 120120 \, {\left (3 \, a^{2} b d + a^{3} g\right )} x^{4} - 360360 \, a^{3} c + 180180 \, {\left (3 \, a^{2} b c + a^{3} f\right )} x^{3}}{360360 \, x} \]
1/360360*(27720*b^3*h*x^14 + 30030*b^3*g*x^13 + 32760*b^3*f*x^12 + 36036*( b^3*e + 3*a*b^2*h)*x^11 + 40040*(b^3*d + 3*a*b^2*g)*x^10 + 45045*(b^3*c + 3*a*b^2*f)*x^9 + 154440*(a*b^2*e + a^2*b*h)*x^8 + 180180*(a*b^2*d + a^2*b* g)*x^7 + 216216*(a*b^2*c + a^2*b*f)*x^6 + 360360*a^3*e*x^2 + 90090*(3*a^2* b*e + a^3*h)*x^5 + 360360*a^3*d*x*log(x) + 120120*(3*a^2*b*d + a^3*g)*x^4 - 360360*a^3*c + 180180*(3*a^2*b*c + a^3*f)*x^3)/x
Time = 0.19 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.19 \[ \int \frac {\left (a+b x^3\right )^3 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x^2} \, dx=- \frac {a^{3} c}{x} + a^{3} d \log {\left (x \right )} + a^{3} e x + \frac {b^{3} f x^{11}}{11} + \frac {b^{3} g x^{12}}{12} + \frac {b^{3} h x^{13}}{13} + x^{10} \cdot \left (\frac {3 a b^{2} h}{10} + \frac {b^{3} e}{10}\right ) + x^{9} \left (\frac {a b^{2} g}{3} + \frac {b^{3} d}{9}\right ) + x^{8} \cdot \left (\frac {3 a b^{2} f}{8} + \frac {b^{3} c}{8}\right ) + x^{7} \cdot \left (\frac {3 a^{2} b h}{7} + \frac {3 a b^{2} e}{7}\right ) + x^{6} \left (\frac {a^{2} b g}{2} + \frac {a b^{2} d}{2}\right ) + x^{5} \cdot \left (\frac {3 a^{2} b f}{5} + \frac {3 a b^{2} c}{5}\right ) + x^{4} \left (\frac {a^{3} h}{4} + \frac {3 a^{2} b e}{4}\right ) + x^{3} \left (\frac {a^{3} g}{3} + a^{2} b d\right ) + x^{2} \left (\frac {a^{3} f}{2} + \frac {3 a^{2} b c}{2}\right ) \]
-a**3*c/x + a**3*d*log(x) + a**3*e*x + b**3*f*x**11/11 + b**3*g*x**12/12 + b**3*h*x**13/13 + x**10*(3*a*b**2*h/10 + b**3*e/10) + x**9*(a*b**2*g/3 + b**3*d/9) + x**8*(3*a*b**2*f/8 + b**3*c/8) + x**7*(3*a**2*b*h/7 + 3*a*b**2 *e/7) + x**6*(a**2*b*g/2 + a*b**2*d/2) + x**5*(3*a**2*b*f/5 + 3*a*b**2*c/5 ) + x**4*(a**3*h/4 + 3*a**2*b*e/4) + x**3*(a**3*g/3 + a**2*b*d) + x**2*(a* *3*f/2 + 3*a**2*b*c/2)
Time = 0.21 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.07 \[ \int \frac {\left (a+b x^3\right )^3 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x^2} \, dx=\frac {1}{13} \, b^{3} h x^{13} + \frac {1}{12} \, b^{3} g x^{12} + \frac {1}{11} \, b^{3} f x^{11} + \frac {1}{10} \, {\left (b^{3} e + 3 \, a b^{2} h\right )} x^{10} + \frac {1}{9} \, {\left (b^{3} d + 3 \, a b^{2} g\right )} x^{9} + \frac {1}{8} \, {\left (b^{3} c + 3 \, a b^{2} f\right )} x^{8} + \frac {3}{7} \, {\left (a b^{2} e + a^{2} b h\right )} x^{7} + \frac {1}{2} \, {\left (a b^{2} d + a^{2} b g\right )} x^{6} + \frac {3}{5} \, {\left (a b^{2} c + a^{2} b f\right )} x^{5} + a^{3} e x + \frac {1}{4} \, {\left (3 \, a^{2} b e + a^{3} h\right )} x^{4} + a^{3} d \log \left (x\right ) + \frac {1}{3} \, {\left (3 \, a^{2} b d + a^{3} g\right )} x^{3} - \frac {a^{3} c}{x} + \frac {1}{2} \, {\left (3 \, a^{2} b c + a^{3} f\right )} x^{2} \]
1/13*b^3*h*x^13 + 1/12*b^3*g*x^12 + 1/11*b^3*f*x^11 + 1/10*(b^3*e + 3*a*b^ 2*h)*x^10 + 1/9*(b^3*d + 3*a*b^2*g)*x^9 + 1/8*(b^3*c + 3*a*b^2*f)*x^8 + 3/ 7*(a*b^2*e + a^2*b*h)*x^7 + 1/2*(a*b^2*d + a^2*b*g)*x^6 + 3/5*(a*b^2*c + a ^2*b*f)*x^5 + a^3*e*x + 1/4*(3*a^2*b*e + a^3*h)*x^4 + a^3*d*log(x) + 1/3*( 3*a^2*b*d + a^3*g)*x^3 - a^3*c/x + 1/2*(3*a^2*b*c + a^3*f)*x^2
Time = 0.26 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.13 \[ \int \frac {\left (a+b x^3\right )^3 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x^2} \, dx=\frac {1}{13} \, b^{3} h x^{13} + \frac {1}{12} \, b^{3} g x^{12} + \frac {1}{11} \, b^{3} f x^{11} + \frac {1}{10} \, b^{3} e x^{10} + \frac {3}{10} \, a b^{2} h x^{10} + \frac {1}{9} \, b^{3} d x^{9} + \frac {1}{3} \, a b^{2} g x^{9} + \frac {1}{8} \, b^{3} c x^{8} + \frac {3}{8} \, a b^{2} f x^{8} + \frac {3}{7} \, a b^{2} e x^{7} + \frac {3}{7} \, a^{2} b h x^{7} + \frac {1}{2} \, a b^{2} d x^{6} + \frac {1}{2} \, a^{2} b g x^{6} + \frac {3}{5} \, a b^{2} c x^{5} + \frac {3}{5} \, a^{2} b f x^{5} + \frac {3}{4} \, a^{2} b e x^{4} + \frac {1}{4} \, a^{3} h x^{4} + a^{2} b d x^{3} + \frac {1}{3} \, a^{3} g x^{3} + \frac {3}{2} \, a^{2} b c x^{2} + \frac {1}{2} \, a^{3} f x^{2} + a^{3} e x + a^{3} d \log \left ({\left | x \right |}\right ) - \frac {a^{3} c}{x} \]
1/13*b^3*h*x^13 + 1/12*b^3*g*x^12 + 1/11*b^3*f*x^11 + 1/10*b^3*e*x^10 + 3/ 10*a*b^2*h*x^10 + 1/9*b^3*d*x^9 + 1/3*a*b^2*g*x^9 + 1/8*b^3*c*x^8 + 3/8*a* b^2*f*x^8 + 3/7*a*b^2*e*x^7 + 3/7*a^2*b*h*x^7 + 1/2*a*b^2*d*x^6 + 1/2*a^2* b*g*x^6 + 3/5*a*b^2*c*x^5 + 3/5*a^2*b*f*x^5 + 3/4*a^2*b*e*x^4 + 1/4*a^3*h* x^4 + a^2*b*d*x^3 + 1/3*a^3*g*x^3 + 3/2*a^2*b*c*x^2 + 1/2*a^3*f*x^2 + a^3* e*x + a^3*d*log(abs(x)) - a^3*c/x
Time = 9.55 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.01 \[ \int \frac {\left (a+b x^3\right )^3 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x^2} \, dx=x^2\,\left (\frac {f\,a^3}{2}+\frac {3\,b\,c\,a^2}{2}\right )+x^8\,\left (\frac {c\,b^3}{8}+\frac {3\,a\,f\,b^2}{8}\right )+x^3\,\left (\frac {g\,a^3}{3}+b\,d\,a^2\right )+x^9\,\left (\frac {d\,b^3}{9}+\frac {a\,g\,b^2}{3}\right )+x^4\,\left (\frac {h\,a^3}{4}+\frac {3\,b\,e\,a^2}{4}\right )+x^{10}\,\left (\frac {e\,b^3}{10}+\frac {3\,a\,h\,b^2}{10}\right )-\frac {a^3\,c}{x}+\frac {b^3\,f\,x^{11}}{11}+\frac {b^3\,g\,x^{12}}{12}+\frac {b^3\,h\,x^{13}}{13}+a^3\,d\,\ln \left (x\right )+a^3\,e\,x+\frac {3\,a\,b\,x^5\,\left (b\,c+a\,f\right )}{5}+\frac {a\,b\,x^6\,\left (b\,d+a\,g\right )}{2}+\frac {3\,a\,b\,x^7\,\left (b\,e+a\,h\right )}{7} \]
x^2*((a^3*f)/2 + (3*a^2*b*c)/2) + x^8*((b^3*c)/8 + (3*a*b^2*f)/8) + x^3*(( a^3*g)/3 + a^2*b*d) + x^9*((b^3*d)/9 + (a*b^2*g)/3) + x^4*((a^3*h)/4 + (3* a^2*b*e)/4) + x^10*((b^3*e)/10 + (3*a*b^2*h)/10) - (a^3*c)/x + (b^3*f*x^11 )/11 + (b^3*g*x^12)/12 + (b^3*h*x^13)/13 + a^3*d*log(x) + a^3*e*x + (3*a*b *x^5*(b*c + a*f))/5 + (a*b*x^6*(b*d + a*g))/2 + (3*a*b*x^7*(b*e + a*h))/7