Integrand size = 28, antiderivative size = 145 \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )^2}{x^2} \, dx=-\frac {a^2 A}{x}+a (2 A b+a C) x+a b B x^2+\frac {1}{3} \left (A \left (b^2+2 a c\right )+2 a b C\right ) x^3+\frac {1}{4} B \left (b^2+2 a c\right ) x^4+\frac {1}{5} \left (2 A b c+\left (b^2+2 a c\right ) C\right ) x^5+\frac {1}{3} b B c x^6+\frac {1}{7} c (A c+2 b C) x^7+\frac {1}{8} B c^2 x^8+\frac {1}{9} c^2 C x^9+a^2 B \log (x) \]
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Time = 0.08 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {1642} \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )^2}{x^2} \, dx=-\frac {a^2 A}{x}+a^2 B \log (x)+\frac {1}{5} x^5 \left (C \left (2 a c+b^2\right )+2 A b c\right )+\frac {1}{3} x^3 \left (A \left (2 a c+b^2\right )+2 a b C\right )+a x (a C+2 A b)+\frac {1}{4} B x^4 \left (2 a c+b^2\right )+a b B x^2+\frac {1}{7} c x^7 (A c+2 b C)+\frac {1}{3} b B c x^6+\frac {1}{8} B c^2 x^8+\frac {1}{9} c^2 C x^9 \]
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Rule 1642
Rubi steps \begin{align*} \text {integral}& = \int \left (a (2 A b+a C)+\frac {a^2 A}{x^2}+\frac {a^2 B}{x}+2 a b B x+\left (A \left (b^2+2 a c\right )+2 a b C\right ) x^2+B \left (b^2+2 a c\right ) x^3+\left (2 A b c+\left (b^2+2 a c\right ) C\right ) x^4+2 b B c x^5+c (A c+2 b C) x^6+B c^2 x^7+c^2 C x^8\right ) \, dx \\ & = -\frac {a^2 A}{x}+a (2 A b+a C) x+a b B x^2+\frac {1}{3} \left (A \left (b^2+2 a c\right )+2 a b C\right ) x^3+\frac {1}{4} B \left (b^2+2 a c\right ) x^4+\frac {1}{5} \left (2 A b c+\left (b^2+2 a c\right ) C\right ) x^5+\frac {1}{3} b B c x^6+\frac {1}{7} c (A c+2 b C) x^7+\frac {1}{8} B c^2 x^8+\frac {1}{9} c^2 C x^9+a^2 B \log (x) \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.00 \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )^2}{x^2} \, dx=-\frac {a^2 A}{x}+a (2 A b+a C) x+a b B x^2+\frac {1}{3} \left (A b^2+2 a A c+2 a b C\right ) x^3+\frac {1}{4} B \left (b^2+2 a c\right ) x^4+\frac {1}{5} \left (2 A b c+b^2 C+2 a c C\right ) x^5+\frac {1}{3} b B c x^6+\frac {1}{7} c (A c+2 b C) x^7+\frac {1}{8} B c^2 x^8+\frac {1}{9} c^2 C x^9+a^2 B \log (x) \]
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Time = 0.05 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.98
method | result | size |
norman | \(\frac {\left (\frac {1}{7} A \,c^{2}+\frac {2}{7} C b c \right ) x^{8}+\left (\frac {1}{2} B a c +\frac {1}{4} B \,b^{2}\right ) x^{5}+\left (\frac {2}{3} A a c +\frac {1}{3} A \,b^{2}+\frac {2}{3} a b C \right ) x^{4}+\left (\frac {2}{5} A b c +\frac {2}{5} a c C +\frac {1}{5} b^{2} C \right ) x^{6}+\left (2 A a b +C \,a^{2}\right ) x^{2}+B a b \,x^{3}-A \,a^{2}+\frac {B \,c^{2} x^{9}}{8}+\frac {c^{2} C \,x^{10}}{9}+\frac {b B c \,x^{7}}{3}}{x}+a^{2} B \ln \left (x \right )\) | \(142\) |
default | \(\frac {c^{2} C \,x^{9}}{9}+\frac {B \,c^{2} x^{8}}{8}+\frac {A \,c^{2} x^{7}}{7}+\frac {2 C b c \,x^{7}}{7}+\frac {b B c \,x^{6}}{3}+\frac {2 A b c \,x^{5}}{5}+\frac {2 C a c \,x^{5}}{5}+\frac {C \,b^{2} x^{5}}{5}+\frac {B a c \,x^{4}}{2}+\frac {B \,b^{2} x^{4}}{4}+\frac {2 A a c \,x^{3}}{3}+\frac {A \,b^{2} x^{3}}{3}+\frac {2 C a b \,x^{3}}{3}+a b B \,x^{2}+2 A a b x +C \,a^{2} x +a^{2} B \ln \left (x \right )-\frac {a^{2} A}{x}\) | \(147\) |
risch | \(\frac {c^{2} C \,x^{9}}{9}+\frac {B \,c^{2} x^{8}}{8}+\frac {A \,c^{2} x^{7}}{7}+\frac {2 C b c \,x^{7}}{7}+\frac {b B c \,x^{6}}{3}+\frac {2 A b c \,x^{5}}{5}+\frac {2 C a c \,x^{5}}{5}+\frac {C \,b^{2} x^{5}}{5}+\frac {B a c \,x^{4}}{2}+\frac {B \,b^{2} x^{4}}{4}+\frac {2 A a c \,x^{3}}{3}+\frac {A \,b^{2} x^{3}}{3}+\frac {2 C a b \,x^{3}}{3}+a b B \,x^{2}+2 A a b x +C \,a^{2} x +a^{2} B \ln \left (x \right )-\frac {a^{2} A}{x}\) | \(147\) |
parallelrisch | \(\frac {280 c^{2} C \,x^{10}+315 B \,c^{2} x^{9}+360 A \,c^{2} x^{8}+720 C b c \,x^{8}+840 b B c \,x^{7}+1008 A b c \,x^{6}+1008 C a c \,x^{6}+504 C \,b^{2} x^{6}+1260 B a c \,x^{5}+630 B \,b^{2} x^{5}+1680 A a c \,x^{4}+840 A \,b^{2} x^{4}+1680 C a b \,x^{4}+2520 B a b \,x^{3}+5040 A a b \,x^{2}+2520 B \,a^{2} \ln \left (x \right ) x +2520 C \,a^{2} x^{2}-2520 A \,a^{2}}{2520 x}\) | \(157\) |
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Time = 0.26 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.00 \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )^2}{x^2} \, dx=\frac {280 \, C c^{2} x^{10} + 315 \, B c^{2} x^{9} + 840 \, B b c x^{7} + 360 \, {\left (2 \, C b c + A c^{2}\right )} x^{8} + 504 \, {\left (C b^{2} + 2 \, {\left (C a + A b\right )} c\right )} x^{6} + 2520 \, B a b x^{3} + 630 \, {\left (B b^{2} + 2 \, B a c\right )} x^{5} + 840 \, {\left (2 \, C a b + A b^{2} + 2 \, A a c\right )} x^{4} + 2520 \, B a^{2} x \log \left (x\right ) - 2520 \, A a^{2} + 2520 \, {\left (C a^{2} + 2 \, A a b\right )} x^{2}}{2520 \, x} \]
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Time = 0.14 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.08 \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )^2}{x^2} \, dx=- \frac {A a^{2}}{x} + B a^{2} \log {\left (x \right )} + B a b x^{2} + \frac {B b c x^{6}}{3} + \frac {B c^{2} x^{8}}{8} + \frac {C c^{2} x^{9}}{9} + x^{7} \left (\frac {A c^{2}}{7} + \frac {2 C b c}{7}\right ) + x^{5} \cdot \left (\frac {2 A b c}{5} + \frac {2 C a c}{5} + \frac {C b^{2}}{5}\right ) + x^{4} \left (\frac {B a c}{2} + \frac {B b^{2}}{4}\right ) + x^{3} \cdot \left (\frac {2 A a c}{3} + \frac {A b^{2}}{3} + \frac {2 C a b}{3}\right ) + x \left (2 A a b + C a^{2}\right ) \]
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Time = 0.18 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.94 \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )^2}{x^2} \, dx=\frac {1}{9} \, C c^{2} x^{9} + \frac {1}{8} \, B c^{2} x^{8} + \frac {1}{3} \, B b c x^{6} + \frac {1}{7} \, {\left (2 \, C b c + A c^{2}\right )} x^{7} + \frac {1}{5} \, {\left (C b^{2} + 2 \, {\left (C a + A b\right )} c\right )} x^{5} + B a b x^{2} + \frac {1}{4} \, {\left (B b^{2} + 2 \, B a c\right )} x^{4} + \frac {1}{3} \, {\left (2 \, C a b + A b^{2} + 2 \, A a c\right )} x^{3} + B a^{2} \log \left (x\right ) - \frac {A a^{2}}{x} + {\left (C a^{2} + 2 \, A a b\right )} x \]
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Time = 0.30 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.01 \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )^2}{x^2} \, dx=\frac {1}{9} \, C c^{2} x^{9} + \frac {1}{8} \, B c^{2} x^{8} + \frac {2}{7} \, C b c x^{7} + \frac {1}{7} \, A c^{2} x^{7} + \frac {1}{3} \, B b c x^{6} + \frac {1}{5} \, C b^{2} x^{5} + \frac {2}{5} \, C a c x^{5} + \frac {2}{5} \, A b c x^{5} + \frac {1}{4} \, B b^{2} x^{4} + \frac {1}{2} \, B a c x^{4} + \frac {2}{3} \, C a b x^{3} + \frac {1}{3} \, A b^{2} x^{3} + \frac {2}{3} \, A a c x^{3} + B a b x^{2} + C a^{2} x + 2 \, A a b x + B a^{2} \log \left ({\left | x \right |}\right ) - \frac {A a^{2}}{x} \]
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Time = 0.06 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.93 \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )^2}{x^2} \, dx=x^7\,\left (\frac {A\,c^2}{7}+\frac {2\,C\,b\,c}{7}\right )+x^3\,\left (\frac {A\,b^2}{3}+\frac {2\,C\,a\,b}{3}+\frac {2\,A\,a\,c}{3}\right )+x^5\,\left (\frac {C\,b^2}{5}+\frac {2\,A\,c\,b}{5}+\frac {2\,C\,a\,c}{5}\right )+x\,\left (C\,a^2+2\,A\,b\,a\right )-\frac {A\,a^2}{x}+\frac {B\,c^2\,x^8}{8}+\frac {C\,c^2\,x^9}{9}+B\,a^2\,\ln \left (x\right )+\frac {B\,x^4\,\left (b^2+2\,a\,c\right )}{4}+B\,a\,b\,x^2+\frac {B\,b\,c\,x^6}{3} \]
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