Integrand size = 22, antiderivative size = 152 \[ \int \frac {\left (a+b x^2\right )^2}{x^2 \left (c+d x^2\right )^3} \, dx=-\frac {a^2}{c x \left (c+d x^2\right )^2}-\frac {\left (b^2 c^2-2 a b c d+5 a^2 d^2\right ) x}{4 c^2 d \left (c+d x^2\right )^2}+\frac {\left (b^2 c^2+3 a d (2 b c-5 a d)\right ) x}{8 c^3 d \left (c+d x^2\right )}+\frac {\left (b^2 c^2+3 a d (2 b c-5 a d)\right ) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{8 c^{7/2} d^{3/2}} \]
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Time = 0.08 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.98, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {473, 393, 205, 211} \[ \int \frac {\left (a+b x^2\right )^2}{x^2 \left (c+d x^2\right )^3} \, dx=\frac {x \left (-\frac {5 a^2 d}{c}+2 a b-\frac {b^2 c}{d}\right )}{4 c \left (c+d x^2\right )^2}-\frac {a^2}{c x \left (c+d x^2\right )^2}+\frac {\left (3 a d (2 b c-5 a d)+b^2 c^2\right ) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{8 c^{7/2} d^{3/2}}+\frac {x \left (\frac {3 a (2 b c-5 a d)}{c^2}+\frac {b^2}{d}\right )}{8 c \left (c+d x^2\right )} \]
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Rule 205
Rule 211
Rule 393
Rule 473
Rubi steps \begin{align*} \text {integral}& = -\frac {a^2}{c x \left (c+d x^2\right )^2}+\frac {\int \frac {a (2 b c-5 a d)+b^2 c x^2}{\left (c+d x^2\right )^3} \, dx}{c} \\ & = -\frac {a^2}{c x \left (c+d x^2\right )^2}+\frac {\left (2 a b-\frac {b^2 c}{d}-\frac {5 a^2 d}{c}\right ) x}{4 c \left (c+d x^2\right )^2}+\frac {1}{4} \left (\frac {b^2}{d}+\frac {3 a (2 b c-5 a d)}{c^2}\right ) \int \frac {1}{\left (c+d x^2\right )^2} \, dx \\ & = -\frac {a^2}{c x \left (c+d x^2\right )^2}+\frac {\left (2 a b-\frac {b^2 c}{d}-\frac {5 a^2 d}{c}\right ) x}{4 c \left (c+d x^2\right )^2}+\frac {\left (\frac {b^2}{d}+\frac {3 a (2 b c-5 a d)}{c^2}\right ) x}{8 c \left (c+d x^2\right )}+\frac {\left (\frac {b^2}{d}+\frac {3 a (2 b c-5 a d)}{c^2}\right ) \int \frac {1}{c+d x^2} \, dx}{8 c} \\ & = -\frac {a^2}{c x \left (c+d x^2\right )^2}+\frac {\left (2 a b-\frac {b^2 c}{d}-\frac {5 a^2 d}{c}\right ) x}{4 c \left (c+d x^2\right )^2}+\frac {\left (\frac {b^2}{d}+\frac {3 a (2 b c-5 a d)}{c^2}\right ) x}{8 c \left (c+d x^2\right )}+\frac {\left (b^2 c^2+6 a b c d-15 a^2 d^2\right ) \tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{8 c^{7/2} d^{3/2}} \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.88 \[ \int \frac {\left (a+b x^2\right )^2}{x^2 \left (c+d x^2\right )^3} \, dx=-\frac {a^2}{c^3 x}-\frac {(b c-a d)^2 x}{4 c^2 d \left (c+d x^2\right )^2}+\frac {\left (b^2 c^2+6 a b c d-7 a^2 d^2\right ) x}{8 c^3 d \left (c+d x^2\right )}+\frac {\left (b^2 c^2+6 a b c d-15 a^2 d^2\right ) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{8 c^{7/2} d^{3/2}} \]
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Time = 2.78 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.84
method | result | size |
default | \(-\frac {a^{2}}{c^{3} x}-\frac {\frac {\left (\frac {7}{8} a^{2} d^{2}-\frac {3}{4} a b c d -\frac {1}{8} b^{2} c^{2}\right ) x^{3}+\frac {c \left (9 a^{2} d^{2}-10 a b c d +b^{2} c^{2}\right ) x}{8 d}}{\left (d \,x^{2}+c \right )^{2}}+\frac {\left (15 a^{2} d^{2}-6 a b c d -b^{2} c^{2}\right ) \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{8 d \sqrt {c d}}}{c^{3}}\) | \(128\) |
risch | \(\frac {-\frac {\left (15 a^{2} d^{2}-6 a b c d -b^{2} c^{2}\right ) x^{4}}{8 c^{3}}-\frac {\left (25 a^{2} d^{2}-10 a b c d +b^{2} c^{2}\right ) x^{2}}{8 c^{2} d}-\frac {a^{2}}{c}}{x \left (d \,x^{2}+c \right )^{2}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (c^{7} d^{3} \textit {\_Z}^{2}+225 a^{4} d^{4}-180 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}+12 a \,b^{3} c^{3} d +b^{4} c^{4}\right )}{\sum }\textit {\_R} \ln \left (\left (3 \textit {\_R}^{2} c^{7} d^{3}+450 a^{4} d^{4}-360 a^{3} b c \,d^{3}+12 a^{2} b^{2} c^{2} d^{2}+24 a \,b^{3} c^{3} d +2 b^{4} c^{4}\right ) x +\left (15 d^{3} a^{2} c^{4}-6 d^{2} a b \,c^{5}-b^{2} c^{6} d \right ) \textit {\_R} \right )\right )}{16}\) | \(255\) |
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Time = 0.31 (sec) , antiderivative size = 475, normalized size of antiderivative = 3.12 \[ \int \frac {\left (a+b x^2\right )^2}{x^2 \left (c+d x^2\right )^3} \, dx=\left [-\frac {16 \, a^{2} c^{3} d^{2} - 2 \, {\left (b^{2} c^{3} d^{2} + 6 \, a b c^{2} d^{3} - 15 \, a^{2} c d^{4}\right )} x^{4} + 2 \, {\left (b^{2} c^{4} d - 10 \, a b c^{3} d^{2} + 25 \, a^{2} c^{2} d^{3}\right )} x^{2} - {\left ({\left (b^{2} c^{2} d^{2} + 6 \, a b c d^{3} - 15 \, a^{2} d^{4}\right )} x^{5} + 2 \, {\left (b^{2} c^{3} d + 6 \, a b c^{2} d^{2} - 15 \, a^{2} c d^{3}\right )} x^{3} + {\left (b^{2} c^{4} + 6 \, a b c^{3} d - 15 \, a^{2} c^{2} d^{2}\right )} x\right )} \sqrt {-c d} \log \left (\frac {d x^{2} + 2 \, \sqrt {-c d} x - c}{d x^{2} + c}\right )}{16 \, {\left (c^{4} d^{4} x^{5} + 2 \, c^{5} d^{3} x^{3} + c^{6} d^{2} x\right )}}, -\frac {8 \, a^{2} c^{3} d^{2} - {\left (b^{2} c^{3} d^{2} + 6 \, a b c^{2} d^{3} - 15 \, a^{2} c d^{4}\right )} x^{4} + {\left (b^{2} c^{4} d - 10 \, a b c^{3} d^{2} + 25 \, a^{2} c^{2} d^{3}\right )} x^{2} - {\left ({\left (b^{2} c^{2} d^{2} + 6 \, a b c d^{3} - 15 \, a^{2} d^{4}\right )} x^{5} + 2 \, {\left (b^{2} c^{3} d + 6 \, a b c^{2} d^{2} - 15 \, a^{2} c d^{3}\right )} x^{3} + {\left (b^{2} c^{4} + 6 \, a b c^{3} d - 15 \, a^{2} c^{2} d^{2}\right )} x\right )} \sqrt {c d} \arctan \left (\frac {\sqrt {c d} x}{c}\right )}{8 \, {\left (c^{4} d^{4} x^{5} + 2 \, c^{5} d^{3} x^{3} + c^{6} d^{2} x\right )}}\right ] \]
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Time = 0.64 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.47 \[ \int \frac {\left (a+b x^2\right )^2}{x^2 \left (c+d x^2\right )^3} \, dx=\frac {\sqrt {- \frac {1}{c^{7} d^{3}}} \cdot \left (15 a^{2} d^{2} - 6 a b c d - b^{2} c^{2}\right ) \log {\left (- c^{4} d \sqrt {- \frac {1}{c^{7} d^{3}}} + x \right )}}{16} - \frac {\sqrt {- \frac {1}{c^{7} d^{3}}} \cdot \left (15 a^{2} d^{2} - 6 a b c d - b^{2} c^{2}\right ) \log {\left (c^{4} d \sqrt {- \frac {1}{c^{7} d^{3}}} + x \right )}}{16} + \frac {- 8 a^{2} c^{2} d + x^{4} \left (- 15 a^{2} d^{3} + 6 a b c d^{2} + b^{2} c^{2} d\right ) + x^{2} \left (- 25 a^{2} c d^{2} + 10 a b c^{2} d - b^{2} c^{3}\right )}{8 c^{5} d x + 16 c^{4} d^{2} x^{3} + 8 c^{3} d^{3} x^{5}} \]
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Time = 0.27 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.96 \[ \int \frac {\left (a+b x^2\right )^2}{x^2 \left (c+d x^2\right )^3} \, dx=-\frac {8 \, a^{2} c^{2} d - {\left (b^{2} c^{2} d + 6 \, a b c d^{2} - 15 \, a^{2} d^{3}\right )} x^{4} + {\left (b^{2} c^{3} - 10 \, a b c^{2} d + 25 \, a^{2} c d^{2}\right )} x^{2}}{8 \, {\left (c^{3} d^{3} x^{5} + 2 \, c^{4} d^{2} x^{3} + c^{5} d x\right )}} + \frac {{\left (b^{2} c^{2} + 6 \, a b c d - 15 \, a^{2} d^{2}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{8 \, \sqrt {c d} c^{3} d} \]
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Time = 0.28 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.89 \[ \int \frac {\left (a+b x^2\right )^2}{x^2 \left (c+d x^2\right )^3} \, dx=-\frac {a^{2}}{c^{3} x} + \frac {{\left (b^{2} c^{2} + 6 \, a b c d - 15 \, a^{2} d^{2}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{8 \, \sqrt {c d} c^{3} d} + \frac {b^{2} c^{2} d x^{3} + 6 \, a b c d^{2} x^{3} - 7 \, a^{2} d^{3} x^{3} - b^{2} c^{3} x + 10 \, a b c^{2} d x - 9 \, a^{2} c d^{2} x}{8 \, {\left (d x^{2} + c\right )}^{2} c^{3} d} \]
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Time = 5.08 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.89 \[ \int \frac {\left (a+b x^2\right )^2}{x^2 \left (c+d x^2\right )^3} \, dx=\frac {\mathrm {atan}\left (\frac {\sqrt {d}\,x}{\sqrt {c}}\right )\,\left (-15\,a^2\,d^2+6\,a\,b\,c\,d+b^2\,c^2\right )}{8\,c^{7/2}\,d^{3/2}}-\frac {\frac {a^2}{c}-\frac {x^4\,\left (-15\,a^2\,d^2+6\,a\,b\,c\,d+b^2\,c^2\right )}{8\,c^3}+\frac {x^2\,\left (25\,a^2\,d^2-10\,a\,b\,c\,d+b^2\,c^2\right )}{8\,c^2\,d}}{c^2\,x+2\,c\,d\,x^3+d^2\,x^5} \]
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