Integrand size = 19, antiderivative size = 109 \[ \int \frac {1}{x^2 \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {1}{b x^3 \sqrt {b x^2+c x^4}}-\frac {5 \sqrt {b x^2+c x^4}}{4 b^2 x^5}+\frac {15 c \sqrt {b x^2+c x^4}}{8 b^3 x^3}-\frac {15 c^2 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{8 b^{7/2}} \]
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Time = 0.09 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {2048, 2050, 2033, 212} \[ \int \frac {1}{x^2 \left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {15 c^2 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{8 b^{7/2}}+\frac {15 c \sqrt {b x^2+c x^4}}{8 b^3 x^3}-\frac {5 \sqrt {b x^2+c x^4}}{4 b^2 x^5}+\frac {1}{b x^3 \sqrt {b x^2+c x^4}} \]
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Rule 212
Rule 2033
Rule 2048
Rule 2050
Rubi steps \begin{align*} \text {integral}& = \frac {1}{b x^3 \sqrt {b x^2+c x^4}}+\frac {5 \int \frac {1}{x^4 \sqrt {b x^2+c x^4}} \, dx}{b} \\ & = \frac {1}{b x^3 \sqrt {b x^2+c x^4}}-\frac {5 \sqrt {b x^2+c x^4}}{4 b^2 x^5}-\frac {(15 c) \int \frac {1}{x^2 \sqrt {b x^2+c x^4}} \, dx}{4 b^2} \\ & = \frac {1}{b x^3 \sqrt {b x^2+c x^4}}-\frac {5 \sqrt {b x^2+c x^4}}{4 b^2 x^5}+\frac {15 c \sqrt {b x^2+c x^4}}{8 b^3 x^3}+\frac {\left (15 c^2\right ) \int \frac {1}{\sqrt {b x^2+c x^4}} \, dx}{8 b^3} \\ & = \frac {1}{b x^3 \sqrt {b x^2+c x^4}}-\frac {5 \sqrt {b x^2+c x^4}}{4 b^2 x^5}+\frac {15 c \sqrt {b x^2+c x^4}}{8 b^3 x^3}-\frac {\left (15 c^2\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {b x^2+c x^4}}\right )}{8 b^3} \\ & = \frac {1}{b x^3 \sqrt {b x^2+c x^4}}-\frac {5 \sqrt {b x^2+c x^4}}{4 b^2 x^5}+\frac {15 c \sqrt {b x^2+c x^4}}{8 b^3 x^3}-\frac {15 c^2 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{8 b^{7/2}} \\ \end{align*}
Time = 0.11 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.84 \[ \int \frac {1}{x^2 \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {\sqrt {b} \left (-2 b^2+5 b c x^2+15 c^2 x^4\right )-15 c^2 x^4 \sqrt {b+c x^2} \text {arctanh}\left (\frac {\sqrt {b+c x^2}}{\sqrt {b}}\right )}{8 b^{7/2} x^3 \sqrt {x^2 \left (b+c x^2\right )}} \]
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Time = 0.25 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.86
method | result | size |
default | \(-\frac {\left (c \,x^{2}+b \right ) \left (-15 b^{\frac {3}{2}} c^{2} x^{4}+15 \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) \sqrt {c \,x^{2}+b}\, b \,c^{2} x^{4}-5 b^{\frac {5}{2}} c \,x^{2}+2 b^{\frac {7}{2}}\right )}{8 x \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} b^{\frac {9}{2}}}\) | \(94\) |
risch | \(-\frac {\left (c \,x^{2}+b \right ) \left (-7 c \,x^{2}+2 b \right )}{8 b^{3} x^{3} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}+\frac {\left (\frac {c^{2}}{b^{3} \sqrt {c \,x^{2}+b}}-\frac {15 c^{2} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right )}{8 b^{\frac {7}{2}}}\right ) x \sqrt {c \,x^{2}+b}}{\sqrt {x^{2} \left (c \,x^{2}+b \right )}}\) | \(112\) |
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Time = 0.27 (sec) , antiderivative size = 229, normalized size of antiderivative = 2.10 \[ \int \frac {1}{x^2 \left (b x^2+c x^4\right )^{3/2}} \, dx=\left [\frac {15 \, {\left (c^{3} x^{7} + b c^{2} x^{5}\right )} \sqrt {b} \log \left (-\frac {c x^{3} + 2 \, b x - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {b}}{x^{3}}\right ) + 2 \, {\left (15 \, b c^{2} x^{4} + 5 \, b^{2} c x^{2} - 2 \, b^{3}\right )} \sqrt {c x^{4} + b x^{2}}}{16 \, {\left (b^{4} c x^{7} + b^{5} x^{5}\right )}}, \frac {15 \, {\left (c^{3} x^{7} + b c^{2} x^{5}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-b}}{c x^{3} + b x}\right ) + {\left (15 \, b c^{2} x^{4} + 5 \, b^{2} c x^{2} - 2 \, b^{3}\right )} \sqrt {c x^{4} + b x^{2}}}{8 \, {\left (b^{4} c x^{7} + b^{5} x^{5}\right )}}\right ] \]
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\[ \int \frac {1}{x^2 \left (b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {1}{x^{2} \left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}\, dx \]
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\[ \int \frac {1}{x^2 \left (b x^2+c x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} x^{2}} \,d x } \]
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Time = 0.29 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.91 \[ \int \frac {1}{x^2 \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {15 \, c^{2} \arctan \left (\frac {\sqrt {c x^{2} + b}}{\sqrt {-b}}\right )}{8 \, \sqrt {-b} b^{3} \mathrm {sgn}\left (x\right )} + \frac {c^{2}}{\sqrt {c x^{2} + b} b^{3} \mathrm {sgn}\left (x\right )} + \frac {7 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} c^{2} - 9 \, \sqrt {c x^{2} + b} b c^{2}}{8 \, b^{3} c^{2} x^{4} \mathrm {sgn}\left (x\right )} \]
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Time = 13.77 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.40 \[ \int \frac {1}{x^2 \left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {{\left (\frac {b}{c\,x^2}+1\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (\frac {3}{2},\frac {7}{2};\ \frac {9}{2};\ -\frac {b}{c\,x^2}\right )}{7\,x\,{\left (c\,x^4+b\,x^2\right )}^{3/2}} \]
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