Integrand size = 38, antiderivative size = 198 \[ \int \frac {\left (a+b x^3\right )^3 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x^3} \, dx=-\frac {a^3 c}{2 x^2}-\frac {a^3 d}{x}+a^2 (3 b c+a f) x+\frac {1}{2} a^2 (3 b d+a g) x^2+a^2 b e x^3+\frac {3}{4} a b (b c+a f) x^4+\frac {3}{5} a b (b d+a g) x^5+\frac {1}{2} a b^2 e x^6+\frac {1}{7} b^2 (b c+3 a f) x^7+\frac {1}{8} b^2 (b d+3 a g) x^8+\frac {1}{9} b^3 e x^9+\frac {1}{10} b^3 f x^{10}+\frac {1}{11} b^3 g x^{11}+\frac {h \left (a+b x^3\right )^4}{12 b}+a^3 e \log (x) \]
-1/2*a^3*c/x^2-a^3*d/x+a^2*(a*f+3*b*c)*x+1/2*a^2*(a*g+3*b*d)*x^2+a^2*b*e*x ^3+3/4*a*b*(a*f+b*c)*x^4+3/5*a*b*(a*g+b*d)*x^5+1/2*a*b^2*e*x^6+1/7*b^2*(3* a*f+b*c)*x^7+1/8*b^2*(3*a*g+b*d)*x^8+1/9*b^3*e*x^9+1/10*b^3*f*x^10+1/11*b^ 3*g*x^11+1/12*h*(b*x^3+a)^4/b+a^3*e*ln(x)
Time = 0.14 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.88 \[ \int \frac {\left (a+b x^3\right )^3 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x^3} \, dx=\frac {a^3 \left (-3 c-6 d x+x^3 \left (6 f+3 g x+2 h x^2\right )\right )}{6 x^2}+\frac {b^3 x^7 \left (3960 c+7 x \left (495 d+440 e x+6 x^2 \left (66 f+60 g x+55 h x^2\right )\right )\right )}{27720}+\frac {1}{20} a^2 b x \left (60 c+x \left (30 d+x \left (20 e+15 f x+12 g x^2+10 h x^3\right )\right )\right )+\frac {1}{840} a b^2 x^4 (630 c+x (504 d+5 x (84 e+x (72 f+7 x (9 g+8 h x)))))+a^3 e \log (x) \]
(a^3*(-3*c - 6*d*x + x^3*(6*f + 3*g*x + 2*h*x^2)))/(6*x^2) + (b^3*x^7*(396 0*c + 7*x*(495*d + 440*e*x + 6*x^2*(66*f + 60*g*x + 55*h*x^2))))/27720 + ( a^2*b*x*(60*c + x*(30*d + x*(20*e + 15*f*x + 12*g*x^2 + 10*h*x^3))))/20 + (a*b^2*x^4*(630*c + x*(504*d + 5*x*(84*e + x*(72*f + 7*x*(9*g + 8*h*x))))) )/840 + a^3*e*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.
| \(\displaystyle \int \frac {\left (a+b x^3\right )^3 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x^3} \, dx\) |
| \(\Big \downarrow \) 2018 |
| \(\displaystyle \int \frac {\left (b x^3+a\right )^3 \left (g x^4+f x^3+e x^2+d x+c\right )}{x^3}dx+\frac {h \left (a+b x^3\right )^4}{12 b}\) |
| \(\Big \downarrow \) 2360 |
| \(\displaystyle \int \left (b^3 g x^{10}+b^3 f x^9+b^3 e x^8+b^2 (b d+3 a g) x^7+b^2 (b c+3 a f) x^6+3 a b^2 e x^5+3 a b (b d+a g) x^4+3 a b (b c+a f) x^3+3 a^2 b e x^2+a^2 (3 b d+a g) x+a^2 (3 b c+a f)+\frac {a^3 e}{x}+\frac {a^3 d}{x^2}+\frac {a^3 c}{x^3}\right )dx+\frac {h \left (a+b x^3\right )^4}{12 b}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {a^3 c}{2 x^2}-\frac {a^3 d}{x}+a^3 e \log (x)+a^2 x (a f+3 b c)+\frac {1}{2} a^2 x^2 (a g+3 b d)+a^2 b e x^3+\frac {1}{7} b^2 x^7 (3 a f+b c)+\frac {1}{8} b^2 x^8 (3 a g+b d)+\frac {1}{2} a b^2 e x^6+\frac {3}{4} a b x^4 (a f+b c)+\frac {3}{5} a b x^5 (a g+b d)+\frac {h \left (a+b x^3\right )^4}{12 b}+\frac {1}{9} b^3 e x^9+\frac {1}{10} b^3 f x^{10}+\frac {1}{11} b^3 g x^{11}\) |
-1/2*(a^3*c)/x^2 - (a^3*d)/x + a^2*(3*b*c + a*f)*x + (a^2*(3*b*d + a*g)*x^ 2)/2 + a^2*b*e*x^3 + (3*a*b*(b*c + a*f)*x^4)/4 + (3*a*b*(b*d + a*g)*x^5)/5 + (a*b^2*e*x^6)/2 + (b^2*(b*c + 3*a*f)*x^7)/7 + (b^2*(b*d + 3*a*g)*x^8)/8 + (b^3*e*x^9)/9 + (b^3*f*x^10)/10 + (b^3*g*x^11)/11 + (h*(a + b*x^3)^4)/( 12*b) + a^3*e*Log[x]
3.4.100.3.1 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 = 1.48 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.10
| method | result | size |
| norman | \(\frac {\left (\frac {1}{2} a^{3} g +\frac {3}{2} d \,a^{2} b \right ) x^{4}+\left (\frac {1}{3} a^{3} h +a^{2} b e \right ) x^{5}+\left (\frac {3}{7} a \,b^{2} f +\frac {1}{7} b^{3} c \right ) x^{9}+\left (\frac {3}{8} a \,b^{2} g +\frac {1}{8} b^{3} d \right ) x^{10}+\left (\frac {1}{3} a \,b^{2} h +\frac {1}{9} b^{3} e \right ) x^{11}+\left (\frac {3}{5} a^{2} b g +\frac {3}{5} a \,b^{2} d \right ) x^{7}+\left (\frac {1}{2} a^{2} b h +\frac {1}{2} a \,b^{2} e \right ) x^{8}+\left (\frac {3}{4} f \,a^{2} b +\frac {3}{4} a \,b^{2} c \right ) x^{6}+\left (f \,a^{3}+3 a^{2} b c \right ) x^{3}-\frac {c \,a^{3}}{2}-a^{3} d x +\frac {b^{3} f \,x^{12}}{10}+\frac {b^{3} g \,x^{13}}{11}+\frac {b^{3} h \,x^{14}}{12}}{x^{2}}+a^{3} e \ln \left (x \right )\) | \(217\) |
| default | \(\frac {b^{3} h \,x^{12}}{12}+\frac {b^{3} g \,x^{11}}{11}+\frac {b^{3} f \,x^{10}}{10}+\frac {a \,b^{2} h \,x^{9}}{3}+\frac {b^{3} e \,x^{9}}{9}+\frac {3 a \,b^{2} g \,x^{8}}{8}+\frac {b^{3} d \,x^{8}}{8}+\frac {3 x^{7} a \,b^{2} f}{7}+\frac {b^{3} c \,x^{7}}{7}+\frac {a^{2} b h \,x^{6}}{2}+\frac {a \,b^{2} e \,x^{6}}{2}+\frac {3 a^{2} b g \,x^{5}}{5}+\frac {3 a \,b^{2} d \,x^{5}}{5}+\frac {3 a^{2} b f \,x^{4}}{4}+\frac {3 a \,b^{2} c \,x^{4}}{4}+\frac {a^{3} h \,x^{3}}{3}+a^{2} b e \,x^{3}+\frac {a^{3} g \,x^{2}}{2}+\frac {3 a^{2} b d \,x^{2}}{2}+f \,a^{3} x +3 a^{2} b c x +a^{3} e \ln \left (x \right )-\frac {a^{3} d}{x}-\frac {a^{3} c}{2 x^{2}}\) | \(222\) |
| risch | \(\frac {b^{3} h \,x^{12}}{12}+\frac {b^{3} g \,x^{11}}{11}+\frac {b^{3} f \,x^{10}}{10}+\frac {a \,b^{2} h \,x^{9}}{3}+\frac {b^{3} e \,x^{9}}{9}+\frac {3 a \,b^{2} g \,x^{8}}{8}+\frac {b^{3} d \,x^{8}}{8}+\frac {3 x^{7} a \,b^{2} f}{7}+\frac {b^{3} c \,x^{7}}{7}+\frac {a^{2} b h \,x^{6}}{2}+\frac {a \,b^{2} e \,x^{6}}{2}+\frac {3 a^{2} b g \,x^{5}}{5}+\frac {3 a \,b^{2} d \,x^{5}}{5}+\frac {3 a^{2} b f \,x^{4}}{4}+\frac {3 a \,b^{2} c \,x^{4}}{4}+\frac {a^{3} h \,x^{3}}{3}+a^{2} b e \,x^{3}+\frac {a^{3} g \,x^{2}}{2}+\frac {3 a^{2} b d \,x^{2}}{2}+f \,a^{3} x +3 a^{2} b c x +\frac {-a^{3} d x -\frac {1}{2} c \,a^{3}}{x^{2}}+a^{3} e \ln \left (x \right )\) | \(222\) |
| parallelrisch | \(\frac {2772 b^{3} f \,x^{12}+13860 a^{2} b h \,x^{8}-27720 a^{3} d x +41580 a^{2} b d \,x^{4}+10395 a \,b^{2} g \,x^{10}+27720 f \,a^{3} x^{3}+3960 b^{3} c \,x^{9}+13860 a^{3} g \,x^{4}+9240 a^{3} h \,x^{5}-13860 c \,a^{3}+9240 a \,b^{2} h \,x^{11}+11880 a \,b^{2} f \,x^{9}+20790 x^{6} f \,a^{2} b +16632 a^{2} b g \,x^{7}+27720 a^{2} b e \,x^{5}+16632 a \,b^{2} d \,x^{7}+13860 a \,b^{2} e \,x^{8}+83160 a^{2} x^{3} b c +20790 a \,b^{2} c \,x^{6}+2520 b^{3} g \,x^{13}+2310 b^{3} h \,x^{14}+27720 e \,a^{3} \ln \left (x \right ) x^{2}+3465 b^{3} d \,x^{10}+3080 b^{3} e \,x^{11}}{27720 x^{2}}\) | \(232\) |
((1/2*a^3*g+3/2*d*a^2*b)*x^4+(1/3*a^3*h+a^2*b*e)*x^5+(3/7*a*b^2*f+1/7*b^3* c)*x^9+(3/8*a*b^2*g+1/8*b^3*d)*x^10+(1/3*a*b^2*h+1/9*b^3*e)*x^11+(3/5*a^2* b*g+3/5*a*b^2*d)*x^7+(1/2*a^2*b*h+1/2*a*b^2*e)*x^8+(3/4*f*a^2*b+3/4*a*b^2* c)*x^6+(a^3*f+3*a^2*b*c)*x^3-1/2*c*a^3-a^3*d*x+1/10*b^3*f*x^12+1/11*b^3*g* x^13+1/12*b^3*h*x^14)/x^2+a^3*e*ln(x)
Time = 0.39 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.11 \[ \int \frac {\left (a+b x^3\right )^3 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x^3} \, dx=\frac {2310 \, b^{3} h x^{14} + 2520 \, b^{3} g x^{13} + 2772 \, b^{3} f x^{12} + 3080 \, {\left (b^{3} e + 3 \, a b^{2} h\right )} x^{11} + 3465 \, {\left (b^{3} d + 3 \, a b^{2} g\right )} x^{10} + 3960 \, {\left (b^{3} c + 3 \, a b^{2} f\right )} x^{9} + 13860 \, {\left (a b^{2} e + a^{2} b h\right )} x^{8} + 16632 \, {\left (a b^{2} d + a^{2} b g\right )} x^{7} + 20790 \, {\left (a b^{2} c + a^{2} b f\right )} x^{6} + 27720 \, a^{3} e x^{2} \log \left (x\right ) + 9240 \, {\left (3 \, a^{2} b e + a^{3} h\right )} x^{5} - 27720 \, a^{3} d x + 13860 \, {\left (3 \, a^{2} b d + a^{3} g\right )} x^{4} - 13860 \, a^{3} c + 27720 \, {\left (3 \, a^{2} b c + a^{3} f\right )} x^{3}}{27720 \, x^{2}} \]
1/27720*(2310*b^3*h*x^14 + 2520*b^3*g*x^13 + 2772*b^3*f*x^12 + 3080*(b^3*e + 3*a*b^2*h)*x^11 + 3465*(b^3*d + 3*a*b^2*g)*x^10 + 3960*(b^3*c + 3*a*b^2 *f)*x^9 + 13860*(a*b^2*e + a^2*b*h)*x^8 + 16632*(a*b^2*d + a^2*b*g)*x^7 + 20790*(a*b^2*c + a^2*b*f)*x^6 + 27720*a^3*e*x^2*log(x) + 9240*(3*a^2*b*e + a^3*h)*x^5 - 27720*a^3*d*x + 13860*(3*a^2*b*d + a^3*g)*x^4 - 13860*a^3*c + 27720*(3*a^2*b*c + a^3*f)*x^3)/x^2
Time = 0.24 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.20 \[ \int \frac {\left (a+b x^3\right )^3 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x^3} \, dx=a^{3} e \log {\left (x \right )} + \frac {b^{3} f x^{10}}{10} + \frac {b^{3} g x^{11}}{11} + \frac {b^{3} h x^{12}}{12} + x^{9} \left (\frac {a b^{2} h}{3} + \frac {b^{3} e}{9}\right ) + x^{8} \cdot \left (\frac {3 a b^{2} g}{8} + \frac {b^{3} d}{8}\right ) + x^{7} \cdot \left (\frac {3 a b^{2} f}{7} + \frac {b^{3} c}{7}\right ) + x^{6} \left (\frac {a^{2} b h}{2} + \frac {a b^{2} e}{2}\right ) + x^{5} \cdot \left (\frac {3 a^{2} b g}{5} + \frac {3 a b^{2} d}{5}\right ) + x^{4} \cdot \left (\frac {3 a^{2} b f}{4} + \frac {3 a b^{2} c}{4}\right ) + x^{3} \left (\frac {a^{3} h}{3} + a^{2} b e\right ) + x^{2} \left (\frac {a^{3} g}{2} + \frac {3 a^{2} b d}{2}\right ) + x \left (a^{3} f + 3 a^{2} b c\right ) + \frac {- a^{3} c - 2 a^{3} d x}{2 x^{2}} \]
a**3*e*log(x) + b**3*f*x**10/10 + b**3*g*x**11/11 + b**3*h*x**12/12 + x**9 *(a*b**2*h/3 + b**3*e/9) + x**8*(3*a*b**2*g/8 + b**3*d/8) + x**7*(3*a*b**2 *f/7 + b**3*c/7) + x**6*(a**2*b*h/2 + a*b**2*e/2) + x**5*(3*a**2*b*g/5 + 3 *a*b**2*d/5) + x**4*(3*a**2*b*f/4 + 3*a*b**2*c/4) + x**3*(a**3*h/3 + a**2* b*e) + x**2*(a**3*g/2 + 3*a**2*b*d/2) + x*(a**3*f + 3*a**2*b*c) + (-a**3*c - 2*a**3*d*x)/(2*x**2)
Time = 0.19 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.07 \[ \int \frac {\left (a+b x^3\right )^3 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x^3} \, dx=\frac {1}{12} \, b^{3} h x^{12} + \frac {1}{11} \, b^{3} g x^{11} + \frac {1}{10} \, b^{3} f x^{10} + \frac {1}{9} \, {\left (b^{3} e + 3 \, a b^{2} h\right )} x^{9} + \frac {1}{8} \, {\left (b^{3} d + 3 \, a b^{2} g\right )} x^{8} + \frac {1}{7} \, {\left (b^{3} c + 3 \, a b^{2} f\right )} x^{7} + \frac {1}{2} \, {\left (a b^{2} e + a^{2} b h\right )} x^{6} + \frac {3}{5} \, {\left (a b^{2} d + a^{2} b g\right )} x^{5} + \frac {3}{4} \, {\left (a b^{2} c + a^{2} b f\right )} x^{4} + a^{3} e \log \left (x\right ) + \frac {1}{3} \, {\left (3 \, a^{2} b e + a^{3} h\right )} x^{3} + \frac {1}{2} \, {\left (3 \, a^{2} b d + a^{3} g\right )} x^{2} + {\left (3 \, a^{2} b c + a^{3} f\right )} x - \frac {2 \, a^{3} d x + a^{3} c}{2 \, x^{2}} \]
1/12*b^3*h*x^12 + 1/11*b^3*g*x^11 + 1/10*b^3*f*x^10 + 1/9*(b^3*e + 3*a*b^2 *h)*x^9 + 1/8*(b^3*d + 3*a*b^2*g)*x^8 + 1/7*(b^3*c + 3*a*b^2*f)*x^7 + 1/2* (a*b^2*e + a^2*b*h)*x^6 + 3/5*(a*b^2*d + a^2*b*g)*x^5 + 3/4*(a*b^2*c + a^2 *b*f)*x^4 + a^3*e*log(x) + 1/3*(3*a^2*b*e + a^3*h)*x^3 + 1/2*(3*a^2*b*d + a^3*g)*x^2 + (3*a^2*b*c + a^3*f)*x - 1/2*(2*a^3*d*x + a^3*c)/x^2
Time = 0.26 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.12 \[ \int \frac {\left (a+b x^3\right )^3 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x^3} \, dx=\frac {1}{12} \, b^{3} h x^{12} + \frac {1}{11} \, b^{3} g x^{11} + \frac {1}{10} \, b^{3} f x^{10} + \frac {1}{9} \, b^{3} e x^{9} + \frac {1}{3} \, a b^{2} h x^{9} + \frac {1}{8} \, b^{3} d x^{8} + \frac {3}{8} \, a b^{2} g x^{8} + \frac {1}{7} \, b^{3} c x^{7} + \frac {3}{7} \, a b^{2} f x^{7} + \frac {1}{2} \, a b^{2} e x^{6} + \frac {1}{2} \, a^{2} b h x^{6} + \frac {3}{5} \, a b^{2} d x^{5} + \frac {3}{5} \, a^{2} b g x^{5} + \frac {3}{4} \, a b^{2} c x^{4} + \frac {3}{4} \, a^{2} b f x^{4} + a^{2} b e x^{3} + \frac {1}{3} \, a^{3} h x^{3} + \frac {3}{2} \, a^{2} b d x^{2} + \frac {1}{2} \, a^{3} g x^{2} + 3 \, a^{2} b c x + a^{3} f x + a^{3} e \log \left ({\left | x \right |}\right ) - \frac {2 \, a^{3} d x + a^{3} c}{2 \, x^{2}} \]
1/12*b^3*h*x^12 + 1/11*b^3*g*x^11 + 1/10*b^3*f*x^10 + 1/9*b^3*e*x^9 + 1/3* a*b^2*h*x^9 + 1/8*b^3*d*x^8 + 3/8*a*b^2*g*x^8 + 1/7*b^3*c*x^7 + 3/7*a*b^2* f*x^7 + 1/2*a*b^2*e*x^6 + 1/2*a^2*b*h*x^6 + 3/5*a*b^2*d*x^5 + 3/5*a^2*b*g* x^5 + 3/4*a*b^2*c*x^4 + 3/4*a^2*b*f*x^4 + a^2*b*e*x^3 + 1/3*a^3*h*x^3 + 3/ 2*a^2*b*d*x^2 + 1/2*a^3*g*x^2 + 3*a^2*b*c*x + a^3*f*x + a^3*e*log(abs(x)) - 1/2*(2*a^3*d*x + a^3*c)/x^2
Time = 0.15 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.01 \[ \int \frac {\left (a+b x^3\right )^3 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x^3} \, dx=x^7\,\left (\frac {c\,b^3}{7}+\frac {3\,a\,f\,b^2}{7}\right )+x^2\,\left (\frac {g\,a^3}{2}+\frac {3\,b\,d\,a^2}{2}\right )+x^8\,\left (\frac {d\,b^3}{8}+\frac {3\,a\,g\,b^2}{8}\right )+x^3\,\left (\frac {h\,a^3}{3}+b\,e\,a^2\right )+x^9\,\left (\frac {e\,b^3}{9}+\frac {a\,h\,b^2}{3}\right )-\frac {\frac {a^3\,c}{2}+a^3\,d\,x}{x^2}+x\,\left (f\,a^3+3\,b\,c\,a^2\right )+\frac {b^3\,f\,x^{10}}{10}+\frac {b^3\,g\,x^{11}}{11}+\frac {b^3\,h\,x^{12}}{12}+a^3\,e\,\ln \left (x\right )+\frac {3\,a\,b\,x^4\,\left (b\,c+a\,f\right )}{4}+\frac {3\,a\,b\,x^5\,\left (b\,d+a\,g\right )}{5}+\frac {a\,b\,x^6\,\left (b\,e+a\,h\right )}{2} \]
x^7*((b^3*c)/7 + (3*a*b^2*f)/7) + x^2*((a^3*g)/2 + (3*a^2*b*d)/2) + x^8*(( b^3*d)/8 + (3*a*b^2*g)/8) + x^3*((a^3*h)/3 + a^2*b*e) + x^9*((b^3*e)/9 + ( a*b^2*h)/3) - ((a^3*c)/2 + a^3*d*x)/x^2 + x*(a^3*f + 3*a^2*b*c) + (b^3*f*x ^10)/10 + (b^3*g*x^11)/11 + (b^3*h*x^12)/12 + a^3*e*log(x) + (3*a*b*x^4*(b *c + a*f))/4 + (3*a*b*x^5*(b*d + a*g))/5 + (a*b*x^6*(b*e + a*h))/2