Integrand size = 38, antiderivative size = 200 \[ \int \frac {\left (a+b x^3\right )^3 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x} \, dx=a^3 d x+\frac {1}{2} a^3 e x^2+a^2 b c x^3+\frac {1}{4} a^2 (3 b d+a g) x^4+\frac {1}{5} a^2 (3 b e+a h) x^5+\frac {1}{2} a b^2 c x^6+\frac {3}{7} a b (b d+a g) x^7+\frac {3}{8} a b (b e+a h) x^8+\frac {1}{9} b^3 c x^9+\frac {1}{10} b^2 (b d+3 a g) x^{10}+\frac {1}{11} b^2 (b e+3 a h) x^{11}+\frac {1}{13} b^3 g x^{13}+\frac {1}{14} b^3 h x^{14}+\frac {f \left (a+b x^3\right )^4}{12 b}+a^3 c \log (x) \] Output:
a^3*d*x+1/2*a^3*e*x^2+a^2*b*c*x^3+1/4*a^2*(a*g+3*b*d)*x^4+1/5*a^2*(a*h+3*b *e)*x^5+1/2*a*b^2*c*x^6+3/7*a*b*(a*g+b*d)*x^7+3/8*a*b*(a*h+b*e)*x^8+1/9*b^ 3*c*x^9+1/10*b^2*(3*a*g+b*d)*x^10+1/11*b^2*(3*a*h+b*e)*x^11+1/13*b^3*g*x^1 3+1/14*b^3*h*x^14+1/12*f*(b*x^3+a)^4/b+a^3*c*ln(x)
Time = 0.06 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.07 \[ \int \frac {\left (a+b x^3\right )^3 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x} \, dx=a^3 d x+\frac {1}{2} a^3 e x^2+\frac {1}{3} a^2 (3 b c+a f) x^3+\frac {1}{4} a^2 (3 b d+a g) x^4+\frac {1}{5} a^2 (3 b e+a h) x^5+\frac {1}{2} a b (b c+a f) x^6+\frac {3}{7} a b (b d+a g) x^7+\frac {3}{8} a b (b e+a h) x^8+\frac {1}{9} b^2 (b c+3 a f) x^9+\frac {1}{10} b^2 (b d+3 a g) x^{10}+\frac {1}{11} b^2 (b e+3 a h) x^{11}+\frac {1}{12} b^3 f x^{12}+\frac {1}{13} b^3 g x^{13}+\frac {1}{14} b^3 h x^{14}+a^3 c \log (x) \] Input:
Integrate[((a + b*x^3)^3*(c + d*x + e*x^2 + f*x^3 + g*x^4 + h*x^5))/x,x]
Output:
a^3*d*x + (a^3*e*x^2)/2 + (a^2*(3*b*c + a*f)*x^3)/3 + (a^2*(3*b*d + a*g)*x ^4)/4 + (a^2*(3*b*e + a*h)*x^5)/5 + (a*b*(b*c + a*f)*x^6)/2 + (3*a*b*(b*d + a*g)*x^7)/7 + (3*a*b*(b*e + a*h)*x^8)/8 + (b^2*(b*c + 3*a*f)*x^9)/9 + (b ^2*(b*d + 3*a*g)*x^10)/10 + (b^2*(b*e + 3*a*h)*x^11)/11 + (b^3*f*x^12)/12 + (b^3*g*x^13)/13 + (b^3*h*x^14)/14 + a^3*c*Log[x]
Time = 0.70 (sec) , antiderivative size = 200, 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.
| \(\displaystyle \int \frac {\left (a+b x^3\right )^3 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x} \, dx\) |
| \(\Big \downarrow \) 2018 |
| \(\displaystyle \int \frac {\left (b x^3+a\right )^3 \left (h x^5+g x^4+e x^2+d x+c\right )}{x}dx+\frac {f \left (a+b x^3\right )^4}{12 b}\) |
| \(\Big \downarrow \) 2360 |
| \(\displaystyle \int \left (b^3 h x^{13}+b^3 g x^{12}+b^2 (b e+3 a h) x^{10}+b^2 (b d+3 a g) x^9+b^3 c x^8+3 a b (b e+a h) x^7+3 a b (b d+a g) x^6+3 a b^2 c x^5+a^2 (3 b e+a h) x^4+a^2 (3 b d+a g) x^3+3 a^2 b c x^2+a^3 e x+a^3 d+\frac {a^3 c}{x}\right )dx+\frac {f \left (a+b x^3\right )^4}{12 b}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle a^3 c \log (x)+a^3 d x+\frac {1}{2} a^3 e x^2+a^2 b c x^3+\frac {1}{4} a^2 x^4 (a g+3 b d)+\frac {1}{5} a^2 x^5 (a h+3 b e)+\frac {1}{2} a b^2 c x^6+\frac {1}{10} b^2 x^{10} (3 a g+b d)+\frac {1}{11} b^2 x^{11} (3 a h+b e)+\frac {3}{7} a b x^7 (a g+b d)+\frac {3}{8} a b x^8 (a h+b e)+\frac {f \left (a+b x^3\right )^4}{12 b}+\frac {1}{9} b^3 c x^9+\frac {1}{13} b^3 g x^{13}+\frac {1}{14} b^3 h x^{14}\) |
Input:
Int[((a + b*x^3)^3*(c + d*x + e*x^2 + f*x^3 + g*x^4 + h*x^5))/x,x]
Output:
a^3*d*x + (a^3*e*x^2)/2 + a^2*b*c*x^3 + (a^2*(3*b*d + a*g)*x^4)/4 + (a^2*( 3*b*e + a*h)*x^5)/5 + (a*b^2*c*x^6)/2 + (3*a*b*(b*d + a*g)*x^7)/7 + (3*a*b *(b*e + a*h)*x^8)/8 + (b^3*c*x^9)/9 + (b^2*(b*d + 3*a*g)*x^10)/10 + (b^2*( b*e + 3*a*h)*x^11)/11 + (b^3*g*x^13)/13 + (b^3*h*x^14)/14 + (f*(a + b*x^3) ^4)/(12*b) + a^3*c*Log[x]
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 = 0.11 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.08
| method | result | size |
| norman | \(\left (\frac {1}{4} a^{3} g +\frac {3}{4} a^{2} b d \right ) x^{4}+\left (\frac {1}{5} a^{3} h +\frac {3}{5} e \,a^{2} b \right ) x^{5}+\left (\frac {1}{3} f \,a^{3}+c \,a^{2} b \right ) x^{3}+\left (\frac {3}{10} a \,b^{2} g +\frac {1}{10} b^{3} d \right ) x^{10}+\left (\frac {3}{11} a \,b^{2} h +\frac {1}{11} e \,b^{3}\right ) x^{11}+\left (\frac {3}{7} a^{2} b g +\frac {3}{7} d a \,b^{2}\right ) x^{7}+\left (\frac {3}{8} a^{2} b h +\frac {3}{8} e a \,b^{2}\right ) x^{8}+\left (\frac {1}{3} f a \,b^{2}+\frac {1}{9} b^{3} c \right ) x^{9}+\left (\frac {1}{2} f \,a^{2} b +\frac {1}{2} c a \,b^{2}\right ) x^{6}+a^{3} d x +\frac {a^{3} e \,x^{2}}{2}+\frac {b^{3} g \,x^{13}}{13}+\frac {b^{3} h \,x^{14}}{14}+\frac {f \,x^{12} b^{3}}{12}+a^{3} c \ln \left (x \right )\) | \(215\) |
| default | \(\frac {b^{3} h \,x^{14}}{14}+\frac {b^{3} g \,x^{13}}{13}+\frac {f \,x^{12} b^{3}}{12}+\frac {3 a \,b^{2} h \,x^{11}}{11}+\frac {b^{3} e \,x^{11}}{11}+\frac {3 a \,b^{2} g \,x^{10}}{10}+\frac {b^{3} d \,x^{10}}{10}+\frac {a \,b^{2} f \,x^{9}}{3}+\frac {b^{3} c \,x^{9}}{9}+\frac {3 a^{2} b h \,x^{8}}{8}+\frac {3 a \,b^{2} e \,x^{8}}{8}+\frac {3 a^{2} b g \,x^{7}}{7}+\frac {3 a \,b^{2} d \,x^{7}}{7}+\frac {a^{2} b f \,x^{6}}{2}+\frac {a \,b^{2} c \,x^{6}}{2}+\frac {a^{3} h \,x^{5}}{5}+\frac {3 a^{2} b e \,x^{5}}{5}+\frac {a^{3} g \,x^{4}}{4}+\frac {3 a^{2} b d \,x^{4}}{4}+\frac {a^{3} f \,x^{3}}{3}+a^{2} b c \,x^{3}+\frac {a^{3} e \,x^{2}}{2}+a^{3} d x +a^{3} c \ln \left (x \right )\) | \(224\) |
| risch | \(\frac {b^{3} h \,x^{14}}{14}+\frac {b^{3} g \,x^{13}}{13}+\frac {f \,x^{12} b^{3}}{12}+\frac {3 a \,b^{2} h \,x^{11}}{11}+\frac {b^{3} e \,x^{11}}{11}+\frac {3 a \,b^{2} g \,x^{10}}{10}+\frac {b^{3} d \,x^{10}}{10}+\frac {a \,b^{2} f \,x^{9}}{3}+\frac {b^{3} c \,x^{9}}{9}+\frac {3 a^{2} b h \,x^{8}}{8}+\frac {3 a \,b^{2} e \,x^{8}}{8}+\frac {3 a^{2} b g \,x^{7}}{7}+\frac {3 a \,b^{2} d \,x^{7}}{7}+\frac {a^{2} b f \,x^{6}}{2}+\frac {a \,b^{2} c \,x^{6}}{2}+\frac {a^{3} h \,x^{5}}{5}+\frac {3 a^{2} b e \,x^{5}}{5}+\frac {a^{3} g \,x^{4}}{4}+\frac {3 a^{2} b d \,x^{4}}{4}+\frac {a^{3} f \,x^{3}}{3}+a^{2} b c \,x^{3}+\frac {a^{3} e \,x^{2}}{2}+a^{3} d x +a^{3} c \ln \left (x \right )\) | \(224\) |
| parallelrisch | \(\frac {b^{3} h \,x^{14}}{14}+\frac {b^{3} g \,x^{13}}{13}+\frac {f \,x^{12} b^{3}}{12}+\frac {3 a \,b^{2} h \,x^{11}}{11}+\frac {b^{3} e \,x^{11}}{11}+\frac {3 a \,b^{2} g \,x^{10}}{10}+\frac {b^{3} d \,x^{10}}{10}+\frac {a \,b^{2} f \,x^{9}}{3}+\frac {b^{3} c \,x^{9}}{9}+\frac {3 a^{2} b h \,x^{8}}{8}+\frac {3 a \,b^{2} e \,x^{8}}{8}+\frac {3 a^{2} b g \,x^{7}}{7}+\frac {3 a \,b^{2} d \,x^{7}}{7}+\frac {a^{2} b f \,x^{6}}{2}+\frac {a \,b^{2} c \,x^{6}}{2}+\frac {a^{3} h \,x^{5}}{5}+\frac {3 a^{2} b e \,x^{5}}{5}+\frac {a^{3} g \,x^{4}}{4}+\frac {3 a^{2} b d \,x^{4}}{4}+\frac {a^{3} f \,x^{3}}{3}+a^{2} b c \,x^{3}+\frac {a^{3} e \,x^{2}}{2}+a^{3} d x +a^{3} c \ln \left (x \right )\) | \(224\) |
Input:
int((b*x^3+a)^3*(h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/x,x,method=_RETURNVERBOSE)
Output:
(1/4*a^3*g+3/4*a^2*b*d)*x^4+(1/5*a^3*h+3/5*e*a^2*b)*x^5+(1/3*f*a^3+c*a^2*b )*x^3+(3/10*a*b^2*g+1/10*b^3*d)*x^10+(3/11*a*b^2*h+1/11*e*b^3)*x^11+(3/7*a ^2*b*g+3/7*d*a*b^2)*x^7+(3/8*a^2*b*h+3/8*e*a*b^2)*x^8+(1/3*f*a*b^2+1/9*b^3 *c)*x^9+(1/2*f*a^2*b+1/2*c*a*b^2)*x^6+a^3*d*x+1/2*a^3*e*x^2+1/13*b^3*g*x^1 3+1/14*b^3*h*x^14+1/12*f*x^12*b^3+a^3*c*ln(x)
Time = 0.07 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.06 \[ \int \frac {\left (a+b x^3\right )^3 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x} \, dx=\frac {1}{14} \, b^{3} h x^{14} + \frac {1}{13} \, b^{3} g x^{13} + \frac {1}{12} \, b^{3} f x^{12} + \frac {1}{11} \, {\left (b^{3} e + 3 \, a b^{2} h\right )} x^{11} + \frac {1}{10} \, {\left (b^{3} d + 3 \, a b^{2} g\right )} x^{10} + \frac {1}{9} \, {\left (b^{3} c + 3 \, a b^{2} f\right )} x^{9} + \frac {3}{8} \, {\left (a b^{2} e + a^{2} b h\right )} x^{8} + \frac {3}{7} \, {\left (a b^{2} d + a^{2} b g\right )} x^{7} + \frac {1}{2} \, {\left (a b^{2} c + a^{2} b f\right )} x^{6} + \frac {1}{2} \, a^{3} e x^{2} + \frac {1}{5} \, {\left (3 \, a^{2} b e + a^{3} h\right )} x^{5} + a^{3} d x + \frac {1}{4} \, {\left (3 \, a^{2} b d + a^{3} g\right )} x^{4} + a^{3} c \log \left (x\right ) + \frac {1}{3} \, {\left (3 \, a^{2} b c + a^{3} f\right )} x^{3} \] Input:
integrate((b*x^3+a)^3*(h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/x,x, algorithm="fric as")
Output:
1/14*b^3*h*x^14 + 1/13*b^3*g*x^13 + 1/12*b^3*f*x^12 + 1/11*(b^3*e + 3*a*b^ 2*h)*x^11 + 1/10*(b^3*d + 3*a*b^2*g)*x^10 + 1/9*(b^3*c + 3*a*b^2*f)*x^9 + 3/8*(a*b^2*e + a^2*b*h)*x^8 + 3/7*(a*b^2*d + a^2*b*g)*x^7 + 1/2*(a*b^2*c + a^2*b*f)*x^6 + 1/2*a^3*e*x^2 + 1/5*(3*a^2*b*e + a^3*h)*x^5 + a^3*d*x + 1/ 4*(3*a^2*b*d + a^3*g)*x^4 + a^3*c*log(x) + 1/3*(3*a^2*b*c + a^3*f)*x^3
Time = 0.21 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.20 \[ \int \frac {\left (a+b x^3\right )^3 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x} \, dx=a^{3} c \log {\left (x \right )} + a^{3} d x + \frac {a^{3} e x^{2}}{2} + \frac {b^{3} f x^{12}}{12} + \frac {b^{3} g x^{13}}{13} + \frac {b^{3} h x^{14}}{14} + x^{11} \cdot \left (\frac {3 a b^{2} h}{11} + \frac {b^{3} e}{11}\right ) + x^{10} \cdot \left (\frac {3 a b^{2} g}{10} + \frac {b^{3} d}{10}\right ) + x^{9} \left (\frac {a b^{2} f}{3} + \frac {b^{3} c}{9}\right ) + x^{8} \cdot \left (\frac {3 a^{2} b h}{8} + \frac {3 a b^{2} e}{8}\right ) + x^{7} \cdot \left (\frac {3 a^{2} b g}{7} + \frac {3 a b^{2} d}{7}\right ) + x^{6} \left (\frac {a^{2} b f}{2} + \frac {a b^{2} c}{2}\right ) + x^{5} \left (\frac {a^{3} h}{5} + \frac {3 a^{2} b e}{5}\right ) + x^{4} \left (\frac {a^{3} g}{4} + \frac {3 a^{2} b d}{4}\right ) + x^{3} \left (\frac {a^{3} f}{3} + a^{2} b c\right ) \] Input:
integrate((b*x**3+a)**3*(h*x**5+g*x**4+f*x**3+e*x**2+d*x+c)/x,x)
Output:
a**3*c*log(x) + a**3*d*x + a**3*e*x**2/2 + b**3*f*x**12/12 + b**3*g*x**13/ 13 + b**3*h*x**14/14 + x**11*(3*a*b**2*h/11 + b**3*e/11) + x**10*(3*a*b**2 *g/10 + b**3*d/10) + x**9*(a*b**2*f/3 + b**3*c/9) + x**8*(3*a**2*b*h/8 + 3 *a*b**2*e/8) + x**7*(3*a**2*b*g/7 + 3*a*b**2*d/7) + x**6*(a**2*b*f/2 + a*b **2*c/2) + x**5*(a**3*h/5 + 3*a**2*b*e/5) + x**4*(a**3*g/4 + 3*a**2*b*d/4) + x**3*(a**3*f/3 + a**2*b*c)
Time = 0.04 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.06 \[ \int \frac {\left (a+b x^3\right )^3 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x} \, dx=\frac {1}{14} \, b^{3} h x^{14} + \frac {1}{13} \, b^{3} g x^{13} + \frac {1}{12} \, b^{3} f x^{12} + \frac {1}{11} \, {\left (b^{3} e + 3 \, a b^{2} h\right )} x^{11} + \frac {1}{10} \, {\left (b^{3} d + 3 \, a b^{2} g\right )} x^{10} + \frac {1}{9} \, {\left (b^{3} c + 3 \, a b^{2} f\right )} x^{9} + \frac {3}{8} \, {\left (a b^{2} e + a^{2} b h\right )} x^{8} + \frac {3}{7} \, {\left (a b^{2} d + a^{2} b g\right )} x^{7} + \frac {1}{2} \, {\left (a b^{2} c + a^{2} b f\right )} x^{6} + \frac {1}{2} \, a^{3} e x^{2} + \frac {1}{5} \, {\left (3 \, a^{2} b e + a^{3} h\right )} x^{5} + a^{3} d x + \frac {1}{4} \, {\left (3 \, a^{2} b d + a^{3} g\right )} x^{4} + a^{3} c \log \left (x\right ) + \frac {1}{3} \, {\left (3 \, a^{2} b c + a^{3} f\right )} x^{3} \] Input:
integrate((b*x^3+a)^3*(h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/x,x, algorithm="maxi ma")
Output:
1/14*b^3*h*x^14 + 1/13*b^3*g*x^13 + 1/12*b^3*f*x^12 + 1/11*(b^3*e + 3*a*b^ 2*h)*x^11 + 1/10*(b^3*d + 3*a*b^2*g)*x^10 + 1/9*(b^3*c + 3*a*b^2*f)*x^9 + 3/8*(a*b^2*e + a^2*b*h)*x^8 + 3/7*(a*b^2*d + a^2*b*g)*x^7 + 1/2*(a*b^2*c + a^2*b*f)*x^6 + 1/2*a^3*e*x^2 + 1/5*(3*a^2*b*e + a^3*h)*x^5 + a^3*d*x + 1/ 4*(3*a^2*b*d + a^3*g)*x^4 + a^3*c*log(x) + 1/3*(3*a^2*b*c + a^3*f)*x^3
Time = 0.12 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.12 \[ \int \frac {\left (a+b x^3\right )^3 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x} \, dx=\frac {1}{14} \, b^{3} h x^{14} + \frac {1}{13} \, b^{3} g x^{13} + \frac {1}{12} \, b^{3} f x^{12} + \frac {1}{11} \, b^{3} e x^{11} + \frac {3}{11} \, a b^{2} h x^{11} + \frac {1}{10} \, b^{3} d x^{10} + \frac {3}{10} \, a b^{2} g x^{10} + \frac {1}{9} \, b^{3} c x^{9} + \frac {1}{3} \, a b^{2} f x^{9} + \frac {3}{8} \, a b^{2} e x^{8} + \frac {3}{8} \, a^{2} b h x^{8} + \frac {3}{7} \, a b^{2} d x^{7} + \frac {3}{7} \, a^{2} b g x^{7} + \frac {1}{2} \, a b^{2} c x^{6} + \frac {1}{2} \, a^{2} b f x^{6} + \frac {3}{5} \, a^{2} b e x^{5} + \frac {1}{5} \, a^{3} h x^{5} + \frac {3}{4} \, a^{2} b d x^{4} + \frac {1}{4} \, a^{3} g x^{4} + a^{2} b c x^{3} + \frac {1}{3} \, a^{3} f x^{3} + \frac {1}{2} \, a^{3} e x^{2} + a^{3} d x + a^{3} c \log \left ({\left | x \right |}\right ) \] Input:
integrate((b*x^3+a)^3*(h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/x,x, algorithm="giac ")
Output:
1/14*b^3*h*x^14 + 1/13*b^3*g*x^13 + 1/12*b^3*f*x^12 + 1/11*b^3*e*x^11 + 3/ 11*a*b^2*h*x^11 + 1/10*b^3*d*x^10 + 3/10*a*b^2*g*x^10 + 1/9*b^3*c*x^9 + 1/ 3*a*b^2*f*x^9 + 3/8*a*b^2*e*x^8 + 3/8*a^2*b*h*x^8 + 3/7*a*b^2*d*x^7 + 3/7* a^2*b*g*x^7 + 1/2*a*b^2*c*x^6 + 1/2*a^2*b*f*x^6 + 3/5*a^2*b*e*x^5 + 1/5*a^ 3*h*x^5 + 3/4*a^2*b*d*x^4 + 1/4*a^3*g*x^4 + a^2*b*c*x^3 + 1/3*a^3*f*x^3 + 1/2*a^3*e*x^2 + a^3*d*x + a^3*c*log(abs(x))
Time = 0.17 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^3\right )^3 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x} \, dx=x^3\,\left (\frac {f\,a^3}{3}+b\,c\,a^2\right )+x^9\,\left (\frac {c\,b^3}{9}+\frac {a\,f\,b^2}{3}\right )+x^4\,\left (\frac {g\,a^3}{4}+\frac {3\,b\,d\,a^2}{4}\right )+x^{10}\,\left (\frac {d\,b^3}{10}+\frac {3\,a\,g\,b^2}{10}\right )+x^5\,\left (\frac {h\,a^3}{5}+\frac {3\,b\,e\,a^2}{5}\right )+x^{11}\,\left (\frac {e\,b^3}{11}+\frac {3\,a\,h\,b^2}{11}\right )+\frac {a^3\,e\,x^2}{2}+\frac {b^3\,f\,x^{12}}{12}+\frac {b^3\,g\,x^{13}}{13}+\frac {b^3\,h\,x^{14}}{14}+a^3\,c\,\ln \left (x\right )+a^3\,d\,x+\frac {a\,b\,x^6\,\left (b\,c+a\,f\right )}{2}+\frac {3\,a\,b\,x^7\,\left (b\,d+a\,g\right )}{7}+\frac {3\,a\,b\,x^8\,\left (b\,e+a\,h\right )}{8} \] Input:
int(((a + b*x^3)^3*(c + d*x + e*x^2 + f*x^3 + g*x^4 + h*x^5))/x,x)
Output:
x^3*((a^3*f)/3 + a^2*b*c) + x^9*((b^3*c)/9 + (a*b^2*f)/3) + x^4*((a^3*g)/4 + (3*a^2*b*d)/4) + x^10*((b^3*d)/10 + (3*a*b^2*g)/10) + x^5*((a^3*h)/5 + (3*a^2*b*e)/5) + x^11*((b^3*e)/11 + (3*a*b^2*h)/11) + (a^3*e*x^2)/2 + (b^3 *f*x^12)/12 + (b^3*g*x^13)/13 + (b^3*h*x^14)/14 + a^3*c*log(x) + a^3*d*x + (a*b*x^6*(b*c + a*f))/2 + (3*a*b*x^7*(b*d + a*g))/7 + (3*a*b*x^8*(b*e + a *h))/8
Time = 0.16 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.12 \[ \int \frac {\left (a+b x^3\right )^3 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x} \, dx=\mathrm {log}\left (x \right ) a^{3} c +a^{3} d x +\frac {a^{3} e \,x^{2}}{2}+\frac {a^{3} f \,x^{3}}{3}+\frac {a^{3} g \,x^{4}}{4}+\frac {a^{3} h \,x^{5}}{5}+a^{2} b c \,x^{3}+\frac {3 a^{2} b d \,x^{4}}{4}+\frac {3 a^{2} b e \,x^{5}}{5}+\frac {a^{2} b f \,x^{6}}{2}+\frac {3 a^{2} b g \,x^{7}}{7}+\frac {3 a^{2} b h \,x^{8}}{8}+\frac {a \,b^{2} c \,x^{6}}{2}+\frac {3 a \,b^{2} d \,x^{7}}{7}+\frac {3 a \,b^{2} e \,x^{8}}{8}+\frac {a \,b^{2} f \,x^{9}}{3}+\frac {3 a \,b^{2} g \,x^{10}}{10}+\frac {3 a \,b^{2} h \,x^{11}}{11}+\frac {b^{3} c \,x^{9}}{9}+\frac {b^{3} d \,x^{10}}{10}+\frac {b^{3} e \,x^{11}}{11}+\frac {b^{3} f \,x^{12}}{12}+\frac {b^{3} g \,x^{13}}{13}+\frac {b^{3} h \,x^{14}}{14} \] Input:
int((b*x^3+a)^3*(h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/x,x)
Output:
(360360*log(x)*a**3*c + 360360*a**3*d*x + 180180*a**3*e*x**2 + 120120*a**3 *f*x**3 + 90090*a**3*g*x**4 + 72072*a**3*h*x**5 + 360360*a**2*b*c*x**3 + 2 70270*a**2*b*d*x**4 + 216216*a**2*b*e*x**5 + 180180*a**2*b*f*x**6 + 154440 *a**2*b*g*x**7 + 135135*a**2*b*h*x**8 + 180180*a*b**2*c*x**6 + 154440*a*b* *2*d*x**7 + 135135*a*b**2*e*x**8 + 120120*a*b**2*f*x**9 + 108108*a*b**2*g* x**10 + 98280*a*b**2*h*x**11 + 40040*b**3*c*x**9 + 36036*b**3*d*x**10 + 32 760*b**3*e*x**11 + 30030*b**3*f*x**12 + 27720*b**3*g*x**13 + 25740*b**3*h* x**14)/360360