Integrand size = 38, antiderivative size = 147 \[ \int \frac {\left (a+b x^3\right )^2 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x^3} \, dx=-\frac {a^2 c}{2 x^2}-\frac {a^2 d}{x}+a (2 b c+a f) x+\frac {1}{2} a (2 b d+a g) x^2+\frac {2}{3} a b e x^3+\frac {1}{4} b (b c+2 a f) x^4+\frac {1}{5} b (b d+2 a g) x^5+\frac {1}{6} b^2 e x^6+\frac {1}{7} b^2 f x^7+\frac {1}{8} b^2 g x^8+\frac {h \left (a+b x^3\right )^3}{9 b}+a^2 e \log (x) \] Output:
-1/2*a^2*c/x^2-a^2*d/x+a*(a*f+2*b*c)*x+1/2*a*(a*g+2*b*d)*x^2+2/3*a*b*e*x^3 +1/4*b*(2*a*f+b*c)*x^4+1/5*b*(2*a*g+b*d)*x^5+1/6*b^2*e*x^6+1/7*b^2*f*x^7+1 /8*b^2*g*x^8+1/9*h*(b*x^3+a)^3/b+a^2*e*ln(x)
Time = 0.06 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.86 \[ \int \frac {\left (a+b x^3\right )^2 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x^3} \, dx=\frac {a^2 \left (-3 c-6 d x+x^3 \left (6 f+3 g x+2 h x^2\right )\right )}{6 x^2}+\frac {1}{30} a b x \left (60 c+x \left (30 d+x \left (20 e+15 f x+12 g x^2+10 h x^3\right )\right )\right )+\frac {b^2 x^4 (630 c+x (504 d+5 x (84 e+x (72 f+7 x (9 g+8 h x)))))}{2520}+a^2 e \log (x) \] Input:
Integrate[((a + b*x^3)^2*(c + d*x + e*x^2 + f*x^3 + g*x^4 + h*x^5))/x^3,x]
Output:
(a^2*(-3*c - 6*d*x + x^3*(6*f + 3*g*x + 2*h*x^2)))/(6*x^2) + (a*b*x*(60*c + x*(30*d + x*(20*e + 15*f*x + 12*g*x^2 + 10*h*x^3))))/30 + (b^2*x^4*(630* c + x*(504*d + 5*x*(84*e + x*(72*f + 7*x*(9*g + 8*h*x))))))/2520 + a^2*e*L og[x]
Time = 0.67 (sec) , antiderivative size = 147, 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.
| \(\displaystyle \int \frac {\left (a+b x^3\right )^2 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x^3} \, dx\) |
| \(\Big \downarrow \) 2018 |
| \(\displaystyle \int \frac {\left (b x^3+a\right )^2 \left (g x^4+f x^3+e x^2+d x+c\right )}{x^3}dx+\frac {h \left (a+b x^3\right )^3}{9 b}\) |
| \(\Big \downarrow \) 2360 |
| \(\displaystyle \int \left (b^2 g x^7+b^2 f x^6+b^2 e x^5+b (b d+2 a g) x^4+b (b c+2 a f) x^3+2 a b e x^2+a (2 b d+a g) x+a (2 b c+a f)+\frac {a^2 e}{x}+\frac {a^2 d}{x^2}+\frac {a^2 c}{x^3}\right )dx+\frac {h \left (a+b x^3\right )^3}{9 b}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {a^2 c}{2 x^2}-\frac {a^2 d}{x}+a^2 e \log (x)+\frac {1}{4} b x^4 (2 a f+b c)+a x (a f+2 b c)+\frac {1}{5} b x^5 (2 a g+b d)+\frac {1}{2} a x^2 (a g+2 b d)+\frac {2}{3} a b e x^3+\frac {h \left (a+b x^3\right )^3}{9 b}+\frac {1}{6} b^2 e x^6+\frac {1}{7} b^2 f x^7+\frac {1}{8} b^2 g x^8\) |
Input:
Int[((a + b*x^3)^2*(c + d*x + e*x^2 + f*x^3 + g*x^4 + h*x^5))/x^3,x]
Output:
-1/2*(a^2*c)/x^2 - (a^2*d)/x + a*(2*b*c + a*f)*x + (a*(2*b*d + a*g)*x^2)/2 + (2*a*b*e*x^3)/3 + (b*(b*c + 2*a*f)*x^4)/4 + (b*(b*d + 2*a*g)*x^5)/5 + ( b^2*e*x^6)/6 + (b^2*f*x^7)/7 + (b^2*g*x^8)/8 + (h*(a + b*x^3)^3)/(9*b) + a ^2*e*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 = 0.10 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.01
| method | result | size |
| norman | \(\frac {\left (\frac {1}{2} a^{2} g +d a b \right ) x^{4}+\left (\frac {1}{3} a^{2} h +\frac {2}{3} a b e \right ) x^{5}+\left (\frac {1}{2} a b f +\frac {1}{4} b^{2} c \right ) x^{6}+\left (\frac {2}{5} a b g +\frac {1}{5} b^{2} d \right ) x^{7}+\left (\frac {1}{3} a b h +\frac {1}{6} e \,b^{2}\right ) x^{8}+\left (f \,a^{2}+2 a b c \right ) x^{3}-\frac {a^{2} c}{2}-a^{2} d x +\frac {b^{2} g \,x^{10}}{8}+\frac {b^{2} h \,x^{11}}{9}+\frac {f \,x^{9} b^{2}}{7}}{x^{2}}+a^{2} e \ln \left (x \right )\) | \(148\) |
| default | \(\frac {b^{2} h \,x^{9}}{9}+\frac {b^{2} g \,x^{8}}{8}+\frac {b^{2} f \,x^{7}}{7}+\frac {a b h \,x^{6}}{3}+\frac {b^{2} e \,x^{6}}{6}+\frac {2 a b g \,x^{5}}{5}+\frac {b^{2} d \,x^{5}}{5}+\frac {a b f \,x^{4}}{2}+\frac {b^{2} c \,x^{4}}{4}+\frac {a^{2} h \,x^{3}}{3}+\frac {2 a b e \,x^{3}}{3}+\frac {a^{2} g \,x^{2}}{2}+a b d \,x^{2}+f \,a^{2} x +2 a b c x -\frac {a^{2} c}{2 x^{2}}+a^{2} e \ln \left (x \right )-\frac {a^{2} d}{x}\) | \(150\) |
| risch | \(\frac {b^{2} h \,x^{9}}{9}+\frac {b^{2} g \,x^{8}}{8}+\frac {b^{2} f \,x^{7}}{7}+\frac {a b h \,x^{6}}{3}+\frac {b^{2} e \,x^{6}}{6}+\frac {2 a b g \,x^{5}}{5}+\frac {b^{2} d \,x^{5}}{5}+\frac {a b f \,x^{4}}{2}+\frac {b^{2} c \,x^{4}}{4}+\frac {a^{2} h \,x^{3}}{3}+\frac {2 a b e \,x^{3}}{3}+\frac {a^{2} g \,x^{2}}{2}+a b d \,x^{2}+f \,a^{2} x +2 a b c x +\frac {-a^{2} d x -\frac {1}{2} a^{2} c}{x^{2}}+a^{2} e \ln \left (x \right )\) | \(150\) |
| parallelrisch | \(\frac {280 b^{2} h \,x^{11}+315 b^{2} g \,x^{10}+360 f \,x^{9} b^{2}+840 a b h \,x^{8}+420 b^{2} e \,x^{8}+1008 a b g \,x^{7}+504 b^{2} d \,x^{7}+1260 a b f \,x^{6}+630 b^{2} c \,x^{6}+840 a^{2} h \,x^{5}+1680 a b e \,x^{5}+1260 a^{2} g \,x^{4}+2520 a b d \,x^{4}+2520 a^{2} e \ln \left (x \right ) x^{2}+2520 a^{2} f \,x^{3}+5040 a b c \,x^{3}-2520 a^{2} d x -1260 a^{2} c}{2520 x^{2}}\) | \(160\) |
Input:
int((b*x^3+a)^2*(h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/x^3,x,method=_RETURNVERBOS E)
Output:
((1/2*a^2*g+d*a*b)*x^4+(1/3*a^2*h+2/3*a*b*e)*x^5+(1/2*a*b*f+1/4*b^2*c)*x^6 +(2/5*a*b*g+1/5*b^2*d)*x^7+(1/3*a*b*h+1/6*e*b^2)*x^8+(a^2*f+2*a*b*c)*x^3-1 /2*a^2*c-a^2*d*x+1/8*b^2*g*x^10+1/9*b^2*h*x^11+1/7*f*x^9*b^2)/x^2+a^2*e*ln (x)
Time = 0.08 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.04 \[ \int \frac {\left (a+b x^3\right )^2 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x^3} \, dx=\frac {280 \, b^{2} h x^{11} + 315 \, b^{2} g x^{10} + 360 \, b^{2} f x^{9} + 420 \, {\left (b^{2} e + 2 \, a b h\right )} x^{8} + 504 \, {\left (b^{2} d + 2 \, a b g\right )} x^{7} + 630 \, {\left (b^{2} c + 2 \, a b f\right )} x^{6} + 840 \, {\left (2 \, a b e + a^{2} h\right )} x^{5} + 2520 \, a^{2} e x^{2} \log \left (x\right ) + 1260 \, {\left (2 \, a b d + a^{2} g\right )} x^{4} - 2520 \, a^{2} d x + 2520 \, {\left (2 \, a b c + a^{2} f\right )} x^{3} - 1260 \, a^{2} c}{2520 \, x^{2}} \] Input:
integrate((b*x^3+a)^2*(h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/x^3,x, algorithm="fr icas")
Output:
1/2520*(280*b^2*h*x^11 + 315*b^2*g*x^10 + 360*b^2*f*x^9 + 420*(b^2*e + 2*a *b*h)*x^8 + 504*(b^2*d + 2*a*b*g)*x^7 + 630*(b^2*c + 2*a*b*f)*x^6 + 840*(2 *a*b*e + a^2*h)*x^5 + 2520*a^2*e*x^2*log(x) + 1260*(2*a*b*d + a^2*g)*x^4 - 2520*a^2*d*x + 2520*(2*a*b*c + a^2*f)*x^3 - 1260*a^2*c)/x^2
Time = 0.22 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.07 \[ \int \frac {\left (a+b x^3\right )^2 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x^3} \, dx=a^{2} e \log {\left (x \right )} + \frac {b^{2} f x^{7}}{7} + \frac {b^{2} g x^{8}}{8} + \frac {b^{2} h x^{9}}{9} + x^{6} \left (\frac {a b h}{3} + \frac {b^{2} e}{6}\right ) + x^{5} \cdot \left (\frac {2 a b g}{5} + \frac {b^{2} d}{5}\right ) + x^{4} \left (\frac {a b f}{2} + \frac {b^{2} c}{4}\right ) + x^{3} \left (\frac {a^{2} h}{3} + \frac {2 a b e}{3}\right ) + x^{2} \left (\frac {a^{2} g}{2} + a b d\right ) + x \left (a^{2} f + 2 a b c\right ) + \frac {- a^{2} c - 2 a^{2} d x}{2 x^{2}} \] Input:
integrate((b*x**3+a)**2*(h*x**5+g*x**4+f*x**3+e*x**2+d*x+c)/x**3,x)
Output:
a**2*e*log(x) + b**2*f*x**7/7 + b**2*g*x**8/8 + b**2*h*x**9/9 + x**6*(a*b* h/3 + b**2*e/6) + x**5*(2*a*b*g/5 + b**2*d/5) + x**4*(a*b*f/2 + b**2*c/4) + x**3*(a**2*h/3 + 2*a*b*e/3) + x**2*(a**2*g/2 + a*b*d) + x*(a**2*f + 2*a* b*c) + (-a**2*c - 2*a**2*d*x)/(2*x**2)
Time = 0.04 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.99 \[ \int \frac {\left (a+b x^3\right )^2 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x^3} \, dx=\frac {1}{9} \, b^{2} h x^{9} + \frac {1}{8} \, b^{2} g x^{8} + \frac {1}{7} \, b^{2} f x^{7} + \frac {1}{6} \, {\left (b^{2} e + 2 \, a b h\right )} x^{6} + \frac {1}{5} \, {\left (b^{2} d + 2 \, a b g\right )} x^{5} + \frac {1}{4} \, {\left (b^{2} c + 2 \, a b f\right )} x^{4} + \frac {1}{3} \, {\left (2 \, a b e + a^{2} h\right )} x^{3} + a^{2} e \log \left (x\right ) + \frac {1}{2} \, {\left (2 \, a b d + a^{2} g\right )} x^{2} + {\left (2 \, a b c + a^{2} f\right )} x - \frac {2 \, a^{2} d x + a^{2} c}{2 \, x^{2}} \] Input:
integrate((b*x^3+a)^2*(h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/x^3,x, algorithm="ma xima")
Output:
1/9*b^2*h*x^9 + 1/8*b^2*g*x^8 + 1/7*b^2*f*x^7 + 1/6*(b^2*e + 2*a*b*h)*x^6 + 1/5*(b^2*d + 2*a*b*g)*x^5 + 1/4*(b^2*c + 2*a*b*f)*x^4 + 1/3*(2*a*b*e + a ^2*h)*x^3 + a^2*e*log(x) + 1/2*(2*a*b*d + a^2*g)*x^2 + (2*a*b*c + a^2*f)*x - 1/2*(2*a^2*d*x + a^2*c)/x^2
Time = 0.12 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.02 \[ \int \frac {\left (a+b x^3\right )^2 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x^3} \, dx=\frac {1}{9} \, b^{2} h x^{9} + \frac {1}{8} \, b^{2} g x^{8} + \frac {1}{7} \, b^{2} f x^{7} + \frac {1}{6} \, b^{2} e x^{6} + \frac {1}{3} \, a b h x^{6} + \frac {1}{5} \, b^{2} d x^{5} + \frac {2}{5} \, a b g x^{5} + \frac {1}{4} \, b^{2} c x^{4} + \frac {1}{2} \, a b f x^{4} + \frac {2}{3} \, a b e x^{3} + \frac {1}{3} \, a^{2} h x^{3} + a b d x^{2} + \frac {1}{2} \, a^{2} g x^{2} + 2 \, a b c x + a^{2} f x + a^{2} e \log \left ({\left | x \right |}\right ) - \frac {2 \, a^{2} d x + a^{2} c}{2 \, x^{2}} \] Input:
integrate((b*x^3+a)^2*(h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/x^3,x, algorithm="gi ac")
Output:
1/9*b^2*h*x^9 + 1/8*b^2*g*x^8 + 1/7*b^2*f*x^7 + 1/6*b^2*e*x^6 + 1/3*a*b*h* x^6 + 1/5*b^2*d*x^5 + 2/5*a*b*g*x^5 + 1/4*b^2*c*x^4 + 1/2*a*b*f*x^4 + 2/3* a*b*e*x^3 + 1/3*a^2*h*x^3 + a*b*d*x^2 + 1/2*a^2*g*x^2 + 2*a*b*c*x + a^2*f* x + a^2*e*log(abs(x)) - 1/2*(2*a^2*d*x + a^2*c)/x^2
Time = 5.80 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.99 \[ \int \frac {\left (a+b x^3\right )^2 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x^3} \, dx=x\,\left (f\,a^2+2\,b\,c\,a\right )-\frac {\frac {a^2\,c}{2}+a^2\,d\,x}{x^2}+x^4\,\left (\frac {c\,b^2}{4}+\frac {a\,f\,b}{2}\right )+x^2\,\left (\frac {g\,a^2}{2}+b\,d\,a\right )+x^5\,\left (\frac {d\,b^2}{5}+\frac {2\,a\,g\,b}{5}\right )+x^3\,\left (\frac {h\,a^2}{3}+\frac {2\,b\,e\,a}{3}\right )+x^6\,\left (\frac {e\,b^2}{6}+\frac {a\,h\,b}{3}\right )+\frac {b^2\,f\,x^7}{7}+\frac {b^2\,g\,x^8}{8}+\frac {b^2\,h\,x^9}{9}+a^2\,e\,\ln \left (x\right ) \] Input:
int(((a + b*x^3)^2*(c + d*x + e*x^2 + f*x^3 + g*x^4 + h*x^5))/x^3,x)
Output:
x*(a^2*f + 2*a*b*c) - ((a^2*c)/2 + a^2*d*x)/x^2 + x^4*((b^2*c)/4 + (a*b*f) /2) + x^2*((a^2*g)/2 + a*b*d) + x^5*((b^2*d)/5 + (2*a*b*g)/5) + x^3*((a^2* h)/3 + (2*a*b*e)/3) + x^6*((b^2*e)/6 + (a*b*h)/3) + (b^2*f*x^7)/7 + (b^2*g *x^8)/8 + (b^2*h*x^9)/9 + a^2*e*log(x)
Time = 0.21 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.08 \[ \int \frac {\left (a+b x^3\right )^2 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x^3} \, dx=\frac {2520 \,\mathrm {log}\left (x \right ) a^{2} e \,x^{2}-1260 a^{2} c -2520 a^{2} d x +2520 a^{2} f \,x^{3}+1260 a^{2} g \,x^{4}+840 a^{2} h \,x^{5}+5040 a b c \,x^{3}+2520 a b d \,x^{4}+1680 a b e \,x^{5}+1260 a b f \,x^{6}+1008 a b g \,x^{7}+840 a b h \,x^{8}+630 b^{2} c \,x^{6}+504 b^{2} d \,x^{7}+420 b^{2} e \,x^{8}+360 b^{2} f \,x^{9}+315 b^{2} g \,x^{10}+280 b^{2} h \,x^{11}}{2520 x^{2}} \] Input:
int((b*x^3+a)^2*(h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/x^3,x)
Output:
(2520*log(x)*a**2*e*x**2 - 1260*a**2*c - 2520*a**2*d*x + 2520*a**2*f*x**3 + 1260*a**2*g*x**4 + 840*a**2*h*x**5 + 5040*a*b*c*x**3 + 2520*a*b*d*x**4 + 1680*a*b*e*x**5 + 1260*a*b*f*x**6 + 1008*a*b*g*x**7 + 840*a*b*h*x**8 + 63 0*b**2*c*x**6 + 504*b**2*d*x**7 + 420*b**2*e*x**8 + 360*b**2*f*x**9 + 315* b**2*g*x**10 + 280*b**2*h*x**11)/(2520*x**2)