Integrand size = 28, antiderivative size = 150 \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )^2}{x} \, dx=a^2 B x+\frac {1}{2} a (2 A b+a C) x^2+\frac {2}{3} a b B x^3+\frac {1}{4} \left (A \left (b^2+2 a c\right )+2 a b C\right ) x^4+\frac {1}{5} B \left (b^2+2 a c\right ) x^5+\frac {1}{6} \left (2 A b c+\left (b^2+2 a c\right ) C\right ) x^6+\frac {2}{7} b B c x^7+\frac {1}{8} c (A c+2 b C) x^8+\frac {1}{9} B c^2 x^9+\frac {1}{10} c^2 C x^{10}+a^2 A \log (x) \]
a^2*B*x+1/2*a*(2*A*b+C*a)*x^2+2/3*a*b*B*x^3+1/4*(A*(2*a*c+b^2)+2*a*b*C)*x^ 4+1/5*B*(2*a*c+b^2)*x^5+1/6*(2*A*b*c+(2*a*c+b^2)*C)*x^6+2/7*b*B*c*x^7+1/8* c*(A*c+2*C*b)*x^8+1/9*B*c^2*x^9+1/10*c^2*C*x^10+a^2*A*ln(x)
Time = 0.03 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.00 \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )^2}{x} \, dx=a^2 B x+\frac {1}{2} a (2 A b+a C) x^2+\frac {2}{3} a b B x^3+\frac {1}{4} \left (A b^2+2 a A c+2 a b C\right ) x^4+\frac {1}{5} B \left (b^2+2 a c\right ) x^5+\frac {1}{6} \left (2 A b c+b^2 C+2 a c C\right ) x^6+\frac {2}{7} b B c x^7+\frac {1}{8} c (A c+2 b C) x^8+\frac {1}{9} B c^2 x^9+\frac {1}{10} c^2 C x^{10}+a^2 A \log (x) \]
a^2*B*x + (a*(2*A*b + a*C)*x^2)/2 + (2*a*b*B*x^3)/3 + ((A*b^2 + 2*a*A*c + 2*a*b*C)*x^4)/4 + (B*(b^2 + 2*a*c)*x^5)/5 + ((2*A*b*c + b^2*C + 2*a*c*C)*x ^6)/6 + (2*b*B*c*x^7)/7 + (c*(A*c + 2*b*C)*x^8)/8 + (B*c^2*x^9)/9 + (c^2*C *x^10)/10 + a^2*A*Log[x]
Time = 0.35 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {2159, 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.
\(\displaystyle \int \frac {\left (a+b x^2+c x^4\right )^2 \left (A+B x+C x^2\right )}{x} \, dx\) |
\(\Big \downarrow \) 2159 |
\(\displaystyle \int \left (\frac {a^2 A}{x}+a^2 B+x^5 \left (C \left (2 a c+b^2\right )+2 A b c\right )+x^3 \left (A \left (2 a c+b^2\right )+2 a b C\right )+a x (a C+2 A b)+B x^4 \left (2 a c+b^2\right )+2 a b B x^2+c x^7 (A c+2 b C)+2 b B c x^6+B c^2 x^8+c^2 C x^9\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle a^2 A \log (x)+a^2 B x+\frac {1}{6} x^6 \left (C \left (2 a c+b^2\right )+2 A b c\right )+\frac {1}{4} x^4 \left (A \left (2 a c+b^2\right )+2 a b C\right )+\frac {1}{2} a x^2 (a C+2 A b)+\frac {1}{5} B x^5 \left (2 a c+b^2\right )+\frac {2}{3} a b B x^3+\frac {1}{8} c x^8 (A c+2 b C)+\frac {2}{7} b B c x^7+\frac {1}{9} B c^2 x^9+\frac {1}{10} c^2 C x^{10}\) |
a^2*B*x + (a*(2*A*b + a*C)*x^2)/2 + (2*a*b*B*x^3)/3 + ((A*(b^2 + 2*a*c) + 2*a*b*C)*x^4)/4 + (B*(b^2 + 2*a*c)*x^5)/5 + ((2*A*b*c + (b^2 + 2*a*c)*C)*x ^6)/6 + (2*b*B*c*x^7)/7 + (c*(A*c + 2*b*C)*x^8)/8 + (B*c^2*x^9)/9 + (c^2*C *x^10)/10 + a^2*A*Log[x]
3.1.14.3.1 Defintions of rubi rules used
Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p _.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x ], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
Time = 0.04 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.92
method | result | size |
norman | \(\left (\frac {1}{8} A \,c^{2}+\frac {1}{4} C b c \right ) x^{8}+\left (A a b +\frac {1}{2} C \,a^{2}\right ) x^{2}+\left (\frac {2}{5} B a c +\frac {1}{5} B \,b^{2}\right ) x^{5}+\left (\frac {1}{2} A a c +\frac {1}{4} A \,b^{2}+\frac {1}{2} a b C \right ) x^{4}+\left (\frac {1}{3} A b c +\frac {1}{3} a c C +\frac {1}{6} b^{2} C \right ) x^{6}+B \,a^{2} x +\frac {B \,c^{2} x^{9}}{9}+\frac {c^{2} C \,x^{10}}{10}+\frac {2 B a b \,x^{3}}{3}+\frac {2 b B c \,x^{7}}{7}+a^{2} A \ln \left (x \right )\) | \(138\) |
default | \(\frac {c^{2} C \,x^{10}}{10}+\frac {B \,c^{2} x^{9}}{9}+\frac {A \,c^{2} x^{8}}{8}+\frac {C b c \,x^{8}}{4}+\frac {2 b B c \,x^{7}}{7}+\frac {A b c \,x^{6}}{3}+\frac {C a c \,x^{6}}{3}+\frac {C \,b^{2} x^{6}}{6}+\frac {2 B a c \,x^{5}}{5}+\frac {B \,b^{2} x^{5}}{5}+\frac {A a c \,x^{4}}{2}+\frac {A \,b^{2} x^{4}}{4}+\frac {C a b \,x^{4}}{2}+\frac {2 B a b \,x^{3}}{3}+A a b \,x^{2}+\frac {C \,a^{2} x^{2}}{2}+B \,a^{2} x +a^{2} A \ln \left (x \right )\) | \(149\) |
risch | \(\frac {c^{2} C \,x^{10}}{10}+\frac {B \,c^{2} x^{9}}{9}+\frac {A \,c^{2} x^{8}}{8}+\frac {C b c \,x^{8}}{4}+\frac {2 b B c \,x^{7}}{7}+\frac {A b c \,x^{6}}{3}+\frac {C a c \,x^{6}}{3}+\frac {C \,b^{2} x^{6}}{6}+\frac {2 B a c \,x^{5}}{5}+\frac {B \,b^{2} x^{5}}{5}+\frac {A a c \,x^{4}}{2}+\frac {A \,b^{2} x^{4}}{4}+\frac {C a b \,x^{4}}{2}+\frac {2 B a b \,x^{3}}{3}+A a b \,x^{2}+\frac {C \,a^{2} x^{2}}{2}+B \,a^{2} x +a^{2} A \ln \left (x \right )\) | \(149\) |
parallelrisch | \(\frac {c^{2} C \,x^{10}}{10}+\frac {B \,c^{2} x^{9}}{9}+\frac {A \,c^{2} x^{8}}{8}+\frac {C b c \,x^{8}}{4}+\frac {2 b B c \,x^{7}}{7}+\frac {A b c \,x^{6}}{3}+\frac {C a c \,x^{6}}{3}+\frac {C \,b^{2} x^{6}}{6}+\frac {2 B a c \,x^{5}}{5}+\frac {B \,b^{2} x^{5}}{5}+\frac {A a c \,x^{4}}{2}+\frac {A \,b^{2} x^{4}}{4}+\frac {C a b \,x^{4}}{2}+\frac {2 B a b \,x^{3}}{3}+A a b \,x^{2}+\frac {C \,a^{2} x^{2}}{2}+B \,a^{2} x +a^{2} A \ln \left (x \right )\) | \(149\) |
(1/8*A*c^2+1/4*C*b*c)*x^8+(A*a*b+1/2*C*a^2)*x^2+(2/5*B*a*c+1/5*B*b^2)*x^5+ (1/2*A*a*c+1/4*A*b^2+1/2*a*b*C)*x^4+(1/3*A*b*c+1/3*a*c*C+1/6*b^2*C)*x^6+B* a^2*x+1/9*B*c^2*x^9+1/10*c^2*C*x^10+2/3*B*a*b*x^3+2/7*b*B*c*x^7+a^2*A*ln(x )
Time = 0.25 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.92 \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )^2}{x} \, dx=\frac {1}{10} \, C c^{2} x^{10} + \frac {1}{9} \, B c^{2} x^{9} + \frac {2}{7} \, B b c x^{7} + \frac {1}{8} \, {\left (2 \, C b c + A c^{2}\right )} x^{8} + \frac {1}{6} \, {\left (C b^{2} + 2 \, {\left (C a + A b\right )} c\right )} x^{6} + \frac {2}{3} \, B a b x^{3} + \frac {1}{5} \, {\left (B b^{2} + 2 \, B a c\right )} x^{5} + \frac {1}{4} \, {\left (2 \, C a b + A b^{2} + 2 \, A a c\right )} x^{4} + B a^{2} x + A a^{2} \log \left (x\right ) + \frac {1}{2} \, {\left (C a^{2} + 2 \, A a b\right )} x^{2} \]
1/10*C*c^2*x^10 + 1/9*B*c^2*x^9 + 2/7*B*b*c*x^7 + 1/8*(2*C*b*c + A*c^2)*x^ 8 + 1/6*(C*b^2 + 2*(C*a + A*b)*c)*x^6 + 2/3*B*a*b*x^3 + 1/5*(B*b^2 + 2*B*a *c)*x^5 + 1/4*(2*C*a*b + A*b^2 + 2*A*a*c)*x^4 + B*a^2*x + A*a^2*log(x) + 1 /2*(C*a^2 + 2*A*a*b)*x^2
Time = 0.14 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.04 \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )^2}{x} \, dx=A a^{2} \log {\left (x \right )} + B a^{2} x + \frac {2 B a b x^{3}}{3} + \frac {2 B b c x^{7}}{7} + \frac {B c^{2} x^{9}}{9} + \frac {C c^{2} x^{10}}{10} + x^{8} \left (\frac {A c^{2}}{8} + \frac {C b c}{4}\right ) + x^{6} \left (\frac {A b c}{3} + \frac {C a c}{3} + \frac {C b^{2}}{6}\right ) + x^{5} \cdot \left (\frac {2 B a c}{5} + \frac {B b^{2}}{5}\right ) + x^{4} \left (\frac {A a c}{2} + \frac {A b^{2}}{4} + \frac {C a b}{2}\right ) + x^{2} \left (A a b + \frac {C a^{2}}{2}\right ) \]
A*a**2*log(x) + B*a**2*x + 2*B*a*b*x**3/3 + 2*B*b*c*x**7/7 + B*c**2*x**9/9 + C*c**2*x**10/10 + x**8*(A*c**2/8 + C*b*c/4) + x**6*(A*b*c/3 + C*a*c/3 + C*b**2/6) + x**5*(2*B*a*c/5 + B*b**2/5) + x**4*(A*a*c/2 + A*b**2/4 + C*a* b/2) + x**2*(A*a*b + C*a**2/2)
Time = 0.20 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.92 \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )^2}{x} \, dx=\frac {1}{10} \, C c^{2} x^{10} + \frac {1}{9} \, B c^{2} x^{9} + \frac {2}{7} \, B b c x^{7} + \frac {1}{8} \, {\left (2 \, C b c + A c^{2}\right )} x^{8} + \frac {1}{6} \, {\left (C b^{2} + 2 \, {\left (C a + A b\right )} c\right )} x^{6} + \frac {2}{3} \, B a b x^{3} + \frac {1}{5} \, {\left (B b^{2} + 2 \, B a c\right )} x^{5} + \frac {1}{4} \, {\left (2 \, C a b + A b^{2} + 2 \, A a c\right )} x^{4} + B a^{2} x + A a^{2} \log \left (x\right ) + \frac {1}{2} \, {\left (C a^{2} + 2 \, A a b\right )} x^{2} \]
1/10*C*c^2*x^10 + 1/9*B*c^2*x^9 + 2/7*B*b*c*x^7 + 1/8*(2*C*b*c + A*c^2)*x^ 8 + 1/6*(C*b^2 + 2*(C*a + A*b)*c)*x^6 + 2/3*B*a*b*x^3 + 1/5*(B*b^2 + 2*B*a *c)*x^5 + 1/4*(2*C*a*b + A*b^2 + 2*A*a*c)*x^4 + B*a^2*x + A*a^2*log(x) + 1 /2*(C*a^2 + 2*A*a*b)*x^2
Time = 0.30 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.99 \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )^2}{x} \, dx=\frac {1}{10} \, C c^{2} x^{10} + \frac {1}{9} \, B c^{2} x^{9} + \frac {1}{4} \, C b c x^{8} + \frac {1}{8} \, A c^{2} x^{8} + \frac {2}{7} \, B b c x^{7} + \frac {1}{6} \, C b^{2} x^{6} + \frac {1}{3} \, C a c x^{6} + \frac {1}{3} \, A b c x^{6} + \frac {1}{5} \, B b^{2} x^{5} + \frac {2}{5} \, B a c x^{5} + \frac {1}{2} \, C a b x^{4} + \frac {1}{4} \, A b^{2} x^{4} + \frac {1}{2} \, A a c x^{4} + \frac {2}{3} \, B a b x^{3} + \frac {1}{2} \, C a^{2} x^{2} + A a b x^{2} + B a^{2} x + A a^{2} \log \left ({\left | x \right |}\right ) \]
1/10*C*c^2*x^10 + 1/9*B*c^2*x^9 + 1/4*C*b*c*x^8 + 1/8*A*c^2*x^8 + 2/7*B*b* c*x^7 + 1/6*C*b^2*x^6 + 1/3*C*a*c*x^6 + 1/3*A*b*c*x^6 + 1/5*B*b^2*x^5 + 2/ 5*B*a*c*x^5 + 1/2*C*a*b*x^4 + 1/4*A*b^2*x^4 + 1/2*A*a*c*x^4 + 2/3*B*a*b*x^ 3 + 1/2*C*a^2*x^2 + A*a*b*x^2 + B*a^2*x + A*a^2*log(abs(x))
Time = 7.86 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.90 \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )^2}{x} \, dx=x^2\,\left (\frac {C\,a^2}{2}+A\,b\,a\right )+x^8\,\left (\frac {A\,c^2}{8}+\frac {C\,b\,c}{4}\right )+x^4\,\left (\frac {A\,b^2}{4}+\frac {C\,a\,b}{2}+\frac {A\,a\,c}{2}\right )+x^6\,\left (\frac {C\,b^2}{6}+\frac {A\,c\,b}{3}+\frac {C\,a\,c}{3}\right )+\frac {B\,c^2\,x^9}{9}+\frac {C\,c^2\,x^{10}}{10}+A\,a^2\,\ln \left (x\right )+\frac {B\,x^5\,\left (b^2+2\,a\,c\right )}{5}+B\,a^2\,x+\frac {2\,B\,a\,b\,x^3}{3}+\frac {2\,B\,b\,c\,x^7}{7} \]