Integrand size = 38, antiderivative size = 152 \[ \int \frac {\left (a+b x^3\right )^2 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x^4} \, dx=-\frac {a^2 c}{3 x^3}-\frac {a^2 d}{2 x^2}-\frac {a^2 e}{x}+a (2 b d+a g) x+\frac {1}{2} a (2 b e+a h) x^2+\frac {1}{3} b (b c+2 a f) x^3+\frac {1}{4} b (b d+2 a g) x^4+\frac {1}{5} b (b e+2 a h) x^5+\frac {1}{6} b^2 f x^6+\frac {1}{7} b^2 g x^7+\frac {1}{8} b^2 h x^8+a (2 b c+a f) \log (x) \] Output:
-1/3*a^2*c/x^3-1/2*a^2*d/x^2-a^2*e/x+a*(a*g+2*b*d)*x+1/2*a*(a*h+2*b*e)*x^2 +1/3*b*(2*a*f+b*c)*x^3+1/4*b*(2*a*g+b*d)*x^4+1/5*b*(2*a*h+b*e)*x^5+1/6*b^2 *f*x^6+1/7*b^2*g*x^7+1/8*b^2*h*x^8+a*(a*f+2*b*c)*ln(x)
Time = 0.05 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.81 \[ \int \frac {\left (a+b x^3\right )^2 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x^4} \, dx=-\frac {a^2 \left (2 c+3 x \left (d+2 e x-x^3 (2 g+h x)\right )\right )}{6 x^3}+\frac {1}{30} a b x \left (60 d+x \left (30 e+x \left (20 f+15 g x+12 h x^2\right )\right )\right )+\frac {1}{840} b^2 x^3 \left (280 c+x \left (210 d+x \left (168 e+140 f x+120 g x^2+105 h x^3\right )\right )\right )+a (2 b c+a f) \log (x) \] Input:
Integrate[((a + b*x^3)^2*(c + d*x + e*x^2 + f*x^3 + g*x^4 + h*x^5))/x^4,x]
Output:
-1/6*(a^2*(2*c + 3*x*(d + 2*e*x - x^3*(2*g + h*x))))/x^3 + (a*b*x*(60*d + x*(30*e + x*(20*f + 15*g*x + 12*h*x^2))))/30 + (b^2*x^3*(280*c + x*(210*d + x*(168*e + 140*f*x + 120*g*x^2 + 105*h*x^3))))/840 + a*(2*b*c + a*f)*Log [x]
Time = 0.62 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {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.
\(\displaystyle \int \frac {\left (a+b x^3\right )^2 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x^4} \, dx\) |
\(\Big \downarrow \) 2360 |
\(\displaystyle \int \left (\frac {a^2 c}{x^4}+\frac {a^2 d}{x^3}+\frac {a^2 e}{x^2}+b x^2 (2 a f+b c)+\frac {a (a f+2 b c)}{x}+b x^3 (2 a g+b d)+a (a g+2 b d)+b x^4 (2 a h+b e)+a x (a h+2 b e)+b^2 f x^5+b^2 g x^6+b^2 h x^7\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^2 c}{3 x^3}-\frac {a^2 d}{2 x^2}-\frac {a^2 e}{x}+\frac {1}{3} b x^3 (2 a f+b c)+a \log (x) (a f+2 b c)+\frac {1}{4} b x^4 (2 a g+b d)+a x (a g+2 b d)+\frac {1}{5} b x^5 (2 a h+b e)+\frac {1}{2} a x^2 (a h+2 b e)+\frac {1}{6} b^2 f x^6+\frac {1}{7} b^2 g x^7+\frac {1}{8} b^2 h x^8\) |
Input:
Int[((a + b*x^3)^2*(c + d*x + e*x^2 + f*x^3 + g*x^4 + h*x^5))/x^4,x]
Output:
-1/3*(a^2*c)/x^3 - (a^2*d)/(2*x^2) - (a^2*e)/x + a*(2*b*d + a*g)*x + (a*(2 *b*e + a*h)*x^2)/2 + (b*(b*c + 2*a*f)*x^3)/3 + (b*(b*d + 2*a*g)*x^4)/4 + ( b*(b*e + 2*a*h)*x^5)/5 + (b^2*f*x^6)/6 + (b^2*g*x^7)/7 + (b^2*h*x^8)/8 + a *(2*b*c + a*f)*Log[x]
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 = 147, normalized size of antiderivative = 0.97
method | result | size |
default | \(\frac {b^{2} h \,x^{8}}{8}+\frac {b^{2} g \,x^{7}}{7}+\frac {b^{2} f \,x^{6}}{6}+\frac {2 a b h \,x^{5}}{5}+\frac {b^{2} e \,x^{5}}{5}+\frac {a b g \,x^{4}}{2}+\frac {b^{2} d \,x^{4}}{4}+\frac {2 a b f \,x^{3}}{3}+\frac {b^{2} c \,x^{3}}{3}+\frac {a^{2} h \,x^{2}}{2}+a \,x^{2} e b +a^{2} g x +2 a b d x -\frac {a^{2} c}{3 x^{3}}-\frac {a^{2} d}{2 x^{2}}+a \left (a f +2 c b \right ) \ln \left (x \right )-\frac {a^{2} e}{x}\) | \(147\) |
norman | \(\frac {\left (\frac {1}{2} a^{2} h +a b e \right ) x^{5}+\left (\frac {2}{3} a b f +\frac {1}{3} b^{2} c \right ) x^{6}+\left (\frac {1}{2} a b g +\frac {1}{4} b^{2} d \right ) x^{7}+\left (\frac {2}{5} a b h +\frac {1}{5} e \,b^{2}\right ) x^{8}+\left (a^{2} g +2 d a b \right ) x^{4}-\frac {a^{2} c}{3}-\frac {a^{2} d x}{2}-a^{2} e \,x^{2}+\frac {b^{2} g \,x^{10}}{7}+\frac {b^{2} h \,x^{11}}{8}+\frac {f \,x^{9} b^{2}}{6}}{x^{3}}+\left (f \,a^{2}+2 a b c \right ) \ln \left (x \right )\) | \(148\) |
risch | \(\frac {b^{2} h \,x^{8}}{8}+\frac {b^{2} g \,x^{7}}{7}+\frac {b^{2} f \,x^{6}}{6}+\frac {2 a b h \,x^{5}}{5}+\frac {b^{2} e \,x^{5}}{5}+\frac {a b g \,x^{4}}{2}+\frac {b^{2} d \,x^{4}}{4}+\frac {2 a b f \,x^{3}}{3}+\frac {b^{2} c \,x^{3}}{3}+\frac {a^{2} h \,x^{2}}{2}+a \,x^{2} e b +a^{2} g x +2 a b d x +\frac {-a^{2} e \,x^{2}-\frac {1}{2} a^{2} d x -\frac {1}{3} a^{2} c}{x^{3}}+\ln \left (x \right ) a^{2} f +2 \ln \left (x \right ) a b c\) | \(149\) |
parallelrisch | \(\frac {105 b^{2} h \,x^{11}+120 b^{2} g \,x^{10}+140 f \,x^{9} b^{2}+336 a b h \,x^{8}+168 b^{2} e \,x^{8}+420 a b g \,x^{7}+210 b^{2} d \,x^{7}+560 a b f \,x^{6}+280 b^{2} c \,x^{6}+420 a^{2} h \,x^{5}+840 a b e \,x^{5}+840 \ln \left (x \right ) x^{3} a^{2} f +1680 \ln \left (x \right ) x^{3} a b c +840 a^{2} g \,x^{4}+1680 a b d \,x^{4}-840 a^{2} e \,x^{2}-420 a^{2} d x -280 a^{2} c}{840 x^{3}}\) | \(162\) |
Input:
int((b*x^3+a)^2*(h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/x^4,x,method=_RETURNVERBOS E)
Output:
1/8*b^2*h*x^8+1/7*b^2*g*x^7+1/6*b^2*f*x^6+2/5*a*b*h*x^5+1/5*b^2*e*x^5+1/2* a*b*g*x^4+1/4*b^2*d*x^4+2/3*a*b*f*x^3+1/3*b^2*c*x^3+1/2*a^2*h*x^2+a*x^2*e* b+a^2*g*x+2*a*b*d*x-1/3*a^2*c/x^3-1/2*a^2*d/x^2+a*(a*f+2*b*c)*ln(x)-a^2*e/ x
Time = 0.08 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.01 \[ \int \frac {\left (a+b x^3\right )^2 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x^4} \, dx=\frac {105 \, b^{2} h x^{11} + 120 \, b^{2} g x^{10} + 140 \, b^{2} f x^{9} + 168 \, {\left (b^{2} e + 2 \, a b h\right )} x^{8} + 210 \, {\left (b^{2} d + 2 \, a b g\right )} x^{7} + 280 \, {\left (b^{2} c + 2 \, a b f\right )} x^{6} + 420 \, {\left (2 \, a b e + a^{2} h\right )} x^{5} - 840 \, a^{2} e x^{2} + 840 \, {\left (2 \, a b d + a^{2} g\right )} x^{4} + 840 \, {\left (2 \, a b c + a^{2} f\right )} x^{3} \log \left (x\right ) - 420 \, a^{2} d x - 280 \, a^{2} c}{840 \, x^{3}} \] Input:
integrate((b*x^3+a)^2*(h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/x^4,x, algorithm="fr icas")
Output:
1/840*(105*b^2*h*x^11 + 120*b^2*g*x^10 + 140*b^2*f*x^9 + 168*(b^2*e + 2*a* b*h)*x^8 + 210*(b^2*d + 2*a*b*g)*x^7 + 280*(b^2*c + 2*a*b*f)*x^6 + 420*(2* a*b*e + a^2*h)*x^5 - 840*a^2*e*x^2 + 840*(2*a*b*d + a^2*g)*x^4 + 840*(2*a* b*c + a^2*f)*x^3*log(x) - 420*a^2*d*x - 280*a^2*c)/x^3
Time = 0.44 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.04 \[ \int \frac {\left (a+b x^3\right )^2 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x^4} \, dx=a \left (a f + 2 b c\right ) \log {\left (x \right )} + \frac {b^{2} f x^{6}}{6} + \frac {b^{2} g x^{7}}{7} + \frac {b^{2} h x^{8}}{8} + x^{5} \cdot \left (\frac {2 a b h}{5} + \frac {b^{2} e}{5}\right ) + x^{4} \left (\frac {a b g}{2} + \frac {b^{2} d}{4}\right ) + x^{3} \cdot \left (\frac {2 a b f}{3} + \frac {b^{2} c}{3}\right ) + x^{2} \left (\frac {a^{2} h}{2} + a b e\right ) + x \left (a^{2} g + 2 a b d\right ) + \frac {- 2 a^{2} c - 3 a^{2} d x - 6 a^{2} e x^{2}}{6 x^{3}} \] Input:
integrate((b*x**3+a)**2*(h*x**5+g*x**4+f*x**3+e*x**2+d*x+c)/x**4,x)
Output:
a*(a*f + 2*b*c)*log(x) + b**2*f*x**6/6 + b**2*g*x**7/7 + b**2*h*x**8/8 + x **5*(2*a*b*h/5 + b**2*e/5) + x**4*(a*b*g/2 + b**2*d/4) + x**3*(2*a*b*f/3 + b**2*c/3) + x**2*(a**2*h/2 + a*b*e) + x*(a**2*g + 2*a*b*d) + (-2*a**2*c - 3*a**2*d*x - 6*a**2*e*x**2)/(6*x**3)
Time = 0.03 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.97 \[ \int \frac {\left (a+b x^3\right )^2 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x^4} \, dx=\frac {1}{8} \, b^{2} h x^{8} + \frac {1}{7} \, b^{2} g x^{7} + \frac {1}{6} \, b^{2} f x^{6} + \frac {1}{5} \, {\left (b^{2} e + 2 \, a b h\right )} x^{5} + \frac {1}{4} \, {\left (b^{2} d + 2 \, a b g\right )} x^{4} + \frac {1}{3} \, {\left (b^{2} c + 2 \, a b f\right )} x^{3} + \frac {1}{2} \, {\left (2 \, a b e + a^{2} h\right )} x^{2} + {\left (2 \, a b d + a^{2} g\right )} x + {\left (2 \, a b c + a^{2} f\right )} \log \left (x\right ) - \frac {6 \, a^{2} e x^{2} + 3 \, a^{2} d x + 2 \, a^{2} c}{6 \, x^{3}} \] Input:
integrate((b*x^3+a)^2*(h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/x^4,x, algorithm="ma xima")
Output:
1/8*b^2*h*x^8 + 1/7*b^2*g*x^7 + 1/6*b^2*f*x^6 + 1/5*(b^2*e + 2*a*b*h)*x^5 + 1/4*(b^2*d + 2*a*b*g)*x^4 + 1/3*(b^2*c + 2*a*b*f)*x^3 + 1/2*(2*a*b*e + a ^2*h)*x^2 + (2*a*b*d + a^2*g)*x + (2*a*b*c + a^2*f)*log(x) - 1/6*(6*a^2*e* x^2 + 3*a^2*d*x + 2*a^2*c)/x^3
Time = 0.12 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.99 \[ \int \frac {\left (a+b x^3\right )^2 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x^4} \, dx=\frac {1}{8} \, b^{2} h x^{8} + \frac {1}{7} \, b^{2} g x^{7} + \frac {1}{6} \, b^{2} f x^{6} + \frac {1}{5} \, b^{2} e x^{5} + \frac {2}{5} \, a b h x^{5} + \frac {1}{4} \, b^{2} d x^{4} + \frac {1}{2} \, a b g x^{4} + \frac {1}{3} \, b^{2} c x^{3} + \frac {2}{3} \, a b f x^{3} + a b e x^{2} + \frac {1}{2} \, a^{2} h x^{2} + 2 \, a b d x + a^{2} g x + {\left (2 \, a b c + a^{2} f\right )} \log \left ({\left | x \right |}\right ) - \frac {6 \, a^{2} e x^{2} + 3 \, a^{2} d x + 2 \, a^{2} c}{6 \, x^{3}} \] Input:
integrate((b*x^3+a)^2*(h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/x^4,x, algorithm="gi ac")
Output:
1/8*b^2*h*x^8 + 1/7*b^2*g*x^7 + 1/6*b^2*f*x^6 + 1/5*b^2*e*x^5 + 2/5*a*b*h* x^5 + 1/4*b^2*d*x^4 + 1/2*a*b*g*x^4 + 1/3*b^2*c*x^3 + 2/3*a*b*f*x^3 + a*b* e*x^2 + 1/2*a^2*h*x^2 + 2*a*b*d*x + a^2*g*x + (2*a*b*c + a^2*f)*log(abs(x) ) - 1/6*(6*a^2*e*x^2 + 3*a^2*d*x + 2*a^2*c)/x^3
Time = 5.97 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.95 \[ \int \frac {\left (a+b x^3\right )^2 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x^4} \, dx=x\,\left (g\,a^2+2\,b\,d\,a\right )-\frac {e\,a^2\,x^2+\frac {d\,a^2\,x}{2}+\frac {c\,a^2}{3}}{x^3}+x^3\,\left (\frac {c\,b^2}{3}+\frac {2\,a\,f\,b}{3}\right )+x^4\,\left (\frac {d\,b^2}{4}+\frac {a\,g\,b}{2}\right )+x^2\,\left (\frac {h\,a^2}{2}+b\,e\,a\right )+x^5\,\left (\frac {e\,b^2}{5}+\frac {2\,a\,h\,b}{5}\right )+\ln \left (x\right )\,\left (f\,a^2+2\,b\,c\,a\right )+\frac {b^2\,f\,x^6}{6}+\frac {b^2\,g\,x^7}{7}+\frac {b^2\,h\,x^8}{8} \] Input:
int(((a + b*x^3)^2*(c + d*x + e*x^2 + f*x^3 + g*x^4 + h*x^5))/x^4,x)
Output:
x*(a^2*g + 2*a*b*d) - ((a^2*c)/3 + a^2*e*x^2 + (a^2*d*x)/2)/x^3 + x^3*((b^ 2*c)/3 + (2*a*b*f)/3) + x^4*((b^2*d)/4 + (a*b*g)/2) + x^2*((a^2*h)/2 + a*b *e) + x^5*((b^2*e)/5 + (2*a*b*h)/5) + log(x)*(a^2*f + 2*a*b*c) + (b^2*f*x^ 6)/6 + (b^2*g*x^7)/7 + (b^2*h*x^8)/8
Time = 0.16 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.06 \[ \int \frac {\left (a+b x^3\right )^2 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x^4} \, dx=\frac {840 \,\mathrm {log}\left (x \right ) a^{2} f \,x^{3}+1680 \,\mathrm {log}\left (x \right ) a b c \,x^{3}-280 a^{2} c -420 a^{2} d x -840 a^{2} e \,x^{2}+840 a^{2} g \,x^{4}+420 a^{2} h \,x^{5}+1680 a b d \,x^{4}+840 a b e \,x^{5}+560 a b f \,x^{6}+420 a b g \,x^{7}+336 a b h \,x^{8}+280 b^{2} c \,x^{6}+210 b^{2} d \,x^{7}+168 b^{2} e \,x^{8}+140 b^{2} f \,x^{9}+120 b^{2} g \,x^{10}+105 b^{2} h \,x^{11}}{840 x^{3}} \] Input:
int((b*x^3+a)^2*(h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/x^4,x)
Output:
(840*log(x)*a**2*f*x**3 + 1680*log(x)*a*b*c*x**3 - 280*a**2*c - 420*a**2*d *x - 840*a**2*e*x**2 + 840*a**2*g*x**4 + 420*a**2*h*x**5 + 1680*a*b*d*x**4 + 840*a*b*e*x**5 + 560*a*b*f*x**6 + 420*a*b*g*x**7 + 336*a*b*h*x**8 + 280 *b**2*c*x**6 + 210*b**2*d*x**7 + 168*b**2*e*x**8 + 140*b**2*f*x**9 + 120*b **2*g*x**10 + 105*b**2*h*x**11)/(840*x**3)