Integrand size = 20, antiderivative size = 51 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )}{x^3} \, dx=-\frac {a^2 c}{2 x^2}+\frac {1}{2} b (b c+2 a d) x^2+\frac {1}{4} b^2 d x^4+a (2 b c+a d) \log (x) \]
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Time = 0.03 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {457, 77} \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )}{x^3} \, dx=-\frac {a^2 c}{2 x^2}+\frac {1}{2} b x^2 (2 a d+b c)+a \log (x) (a d+2 b c)+\frac {1}{4} b^2 d x^4 \]
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Rule 77
Rule 457
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {(a+b x)^2 (c+d x)}{x^2} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (b (b c+2 a d)+\frac {a^2 c}{x^2}+\frac {a (2 b c+a d)}{x}+b^2 d x\right ) \, dx,x,x^2\right ) \\ & = -\frac {a^2 c}{2 x^2}+\frac {1}{2} b (b c+2 a d) x^2+\frac {1}{4} b^2 d x^4+a (2 b c+a d) \log (x) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.96 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )}{x^3} \, dx=\frac {1}{4} \left (-\frac {2 a^2 c}{x^2}+2 b (b c+2 a d) x^2+b^2 d x^4+4 a (2 b c+a d) \log (x)\right ) \]
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Time = 2.60 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.94
method | result | size |
default | \(\frac {b^{2} d \,x^{4}}{4}+x^{2} a b d +\frac {b^{2} c \,x^{2}}{2}+a \left (a d +2 b c \right ) \ln \left (x \right )-\frac {a^{2} c}{2 x^{2}}\) | \(48\) |
norman | \(\frac {\left (a b d +\frac {1}{2} b^{2} c \right ) x^{4}-\frac {a^{2} c}{2}+\frac {b^{2} d \,x^{6}}{4}}{x^{2}}+\left (a^{2} d +2 a b c \right ) \ln \left (x \right )\) | \(51\) |
parallelrisch | \(\frac {b^{2} d \,x^{6}+4 a b d \,x^{4}+2 b^{2} c \,x^{4}+4 \ln \left (x \right ) x^{2} a^{2} d +8 \ln \left (x \right ) x^{2} a b c -2 a^{2} c}{4 x^{2}}\) | \(59\) |
risch | \(\frac {b^{2} d \,x^{4}}{4}+x^{2} a b d +\frac {b^{2} c \,x^{2}}{2}+a^{2} d +a b c +\frac {b^{2} c^{2}}{4 d}-\frac {a^{2} c}{2 x^{2}}+\ln \left (x \right ) a^{2} d +2 \ln \left (x \right ) a b c\) | \(70\) |
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Time = 0.33 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.06 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )}{x^3} \, dx=\frac {b^{2} d x^{6} + 2 \, {\left (b^{2} c + 2 \, a b d\right )} x^{4} + 4 \, {\left (2 \, a b c + a^{2} d\right )} x^{2} \log \left (x\right ) - 2 \, a^{2} c}{4 \, x^{2}} \]
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Time = 0.12 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.94 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )}{x^3} \, dx=- \frac {a^{2} c}{2 x^{2}} + a \left (a d + 2 b c\right ) \log {\left (x \right )} + \frac {b^{2} d x^{4}}{4} + x^{2} \left (a b d + \frac {b^{2} c}{2}\right ) \]
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Time = 0.24 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.02 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )}{x^3} \, dx=\frac {1}{4} \, b^{2} d x^{4} + \frac {1}{2} \, {\left (b^{2} c + 2 \, a b d\right )} x^{2} + \frac {1}{2} \, {\left (2 \, a b c + a^{2} d\right )} \log \left (x^{2}\right ) - \frac {a^{2} c}{2 \, x^{2}} \]
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Time = 0.29 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.37 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )}{x^3} \, dx=\frac {1}{4} \, b^{2} d x^{4} + \frac {1}{2} \, b^{2} c x^{2} + a b d x^{2} + \frac {1}{2} \, {\left (2 \, a b c + a^{2} d\right )} \log \left (x^{2}\right ) - \frac {2 \, a b c x^{2} + a^{2} d x^{2} + a^{2} c}{2 \, x^{2}} \]
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Time = 0.05 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.94 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )}{x^3} \, dx=x^2\,\left (\frac {c\,b^2}{2}+a\,d\,b\right )+\ln \left (x\right )\,\left (d\,a^2+2\,b\,c\,a\right )-\frac {a^2\,c}{2\,x^2}+\frac {b^2\,d\,x^4}{4} \]
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