Integrand size = 20, antiderivative size = 48 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )}{x^2} \, dx=-\frac {a^2 c}{x}+a (2 b c+a d) x+\frac {1}{3} b (b c+2 a d) x^3+\frac {1}{5} b^2 d x^5 \]
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Time = 0.02 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {459} \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )}{x^2} \, dx=-\frac {a^2 c}{x}+\frac {1}{3} b x^3 (2 a d+b c)+a x (a d+2 b c)+\frac {1}{5} b^2 d x^5 \]
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Rule 459
Rubi steps \begin{align*} \text {integral}& = \int \left (a (2 b c+a d)+\frac {a^2 c}{x^2}+b (b c+2 a d) x^2+b^2 d x^4\right ) \, dx \\ & = -\frac {a^2 c}{x}+a (2 b c+a d) x+\frac {1}{3} b (b c+2 a d) x^3+\frac {1}{5} b^2 d x^5 \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )}{x^2} \, dx=-\frac {a^2 c}{x}+a (2 b c+a d) x+\frac {1}{3} b (b c+2 a d) x^3+\frac {1}{5} b^2 d x^5 \]
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Time = 2.56 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.02
method | result | size |
default | \(\frac {b^{2} d \,x^{5}}{5}+\frac {2 x^{3} a b d}{3}+\frac {b^{2} c \,x^{3}}{3}+a^{2} d x +2 a b c x -\frac {a^{2} c}{x}\) | \(49\) |
risch | \(\frac {b^{2} d \,x^{5}}{5}+\frac {2 x^{3} a b d}{3}+\frac {b^{2} c \,x^{3}}{3}+a^{2} d x +2 a b c x -\frac {a^{2} c}{x}\) | \(49\) |
norman | \(\frac {\frac {b^{2} d \,x^{6}}{5}+\left (\frac {2}{3} a b d +\frac {1}{3} b^{2} c \right ) x^{4}+\left (a^{2} d +2 a b c \right ) x^{2}-a^{2} c}{x}\) | \(52\) |
gosper | \(-\frac {-3 b^{2} d \,x^{6}-10 a b d \,x^{4}-5 b^{2} c \,x^{4}-15 a^{2} d \,x^{2}-30 a b c \,x^{2}+15 a^{2} c}{15 x}\) | \(56\) |
parallelrisch | \(\frac {3 b^{2} d \,x^{6}+10 a b d \,x^{4}+5 b^{2} c \,x^{4}+15 a^{2} d \,x^{2}+30 a b c \,x^{2}-15 a^{2} c}{15 x}\) | \(56\) |
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Time = 0.28 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.10 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )}{x^2} \, dx=\frac {3 \, b^{2} d x^{6} + 5 \, {\left (b^{2} c + 2 \, a b d\right )} x^{4} - 15 \, a^{2} c + 15 \, {\left (2 \, a b c + a^{2} d\right )} x^{2}}{15 \, x} \]
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Time = 0.06 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )}{x^2} \, dx=- \frac {a^{2} c}{x} + \frac {b^{2} d x^{5}}{5} + x^{3} \cdot \left (\frac {2 a b d}{3} + \frac {b^{2} c}{3}\right ) + x \left (a^{2} d + 2 a b c\right ) \]
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Time = 0.27 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )}{x^2} \, dx=\frac {1}{5} \, b^{2} d x^{5} + \frac {1}{3} \, {\left (b^{2} c + 2 \, a b d\right )} x^{3} - \frac {a^{2} c}{x} + {\left (2 \, a b c + a^{2} d\right )} x \]
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Time = 0.28 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )}{x^2} \, dx=\frac {1}{5} \, b^{2} d x^{5} + \frac {1}{3} \, b^{2} c x^{3} + \frac {2}{3} \, a b d x^{3} + 2 \, a b c x + a^{2} d x - \frac {a^{2} c}{x} \]
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Time = 0.05 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )}{x^2} \, dx=x\,\left (d\,a^2+2\,b\,c\,a\right )+x^3\,\left (\frac {c\,b^2}{3}+\frac {2\,a\,d\,b}{3}\right )-\frac {a^2\,c}{x}+\frac {b^2\,d\,x^5}{5} \]
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