Integrand size = 19, antiderivative size = 74 \[ \int \frac {1}{x \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {1}{b x^2 \sqrt {b x^2+c x^4}}-\frac {4 \sqrt {b x^2+c x^4}}{3 b^2 x^4}+\frac {8 c \sqrt {b x^2+c x^4}}{3 b^3 x^2} \]
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Time = 0.08 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2040, 2041, 2039} \[ \int \frac {1}{x \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {8 c \sqrt {b x^2+c x^4}}{3 b^3 x^2}-\frac {4 \sqrt {b x^2+c x^4}}{3 b^2 x^4}+\frac {1}{b x^2 \sqrt {b x^2+c x^4}} \]
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Rule 2039
Rule 2040
Rule 2041
Rubi steps \begin{align*} \text {integral}& = \frac {1}{b x^2 \sqrt {b x^2+c x^4}}+\frac {4 \int \frac {1}{x^3 \sqrt {b x^2+c x^4}} \, dx}{b} \\ & = \frac {1}{b x^2 \sqrt {b x^2+c x^4}}-\frac {4 \sqrt {b x^2+c x^4}}{3 b^2 x^4}-\frac {(8 c) \int \frac {1}{x \sqrt {b x^2+c x^4}} \, dx}{3 b^2} \\ & = \frac {1}{b x^2 \sqrt {b x^2+c x^4}}-\frac {4 \sqrt {b x^2+c x^4}}{3 b^2 x^4}+\frac {8 c \sqrt {b x^2+c x^4}}{3 b^3 x^2} \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.62 \[ \int \frac {1}{x \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {-b^2+4 b c x^2+8 c^2 x^4}{3 b^3 x^2 \sqrt {x^2 \left (b+c x^2\right )}} \]
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Time = 0.10 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.55
method | result | size |
pseudoelliptic | \(-\frac {-8 c^{2} x^{4}-4 b c \,x^{2}+b^{2}}{3 b^{3} x^{2} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\) | \(41\) |
gosper | \(-\frac {\left (c \,x^{2}+b \right ) \left (-8 c^{2} x^{4}-4 b c \,x^{2}+b^{2}\right )}{3 b^{3} \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}\) | \(45\) |
default | \(-\frac {\left (c \,x^{2}+b \right ) \left (-8 c^{2} x^{4}-4 b c \,x^{2}+b^{2}\right )}{3 b^{3} \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}\) | \(45\) |
trager | \(-\frac {\left (-8 c^{2} x^{4}-4 b c \,x^{2}+b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{3 \left (c \,x^{2}+b \right ) b^{3} x^{4}}\) | \(50\) |
risch | \(-\frac {\left (c \,x^{2}+b \right ) \left (-5 c \,x^{2}+b \right )}{3 b^{3} x^{2} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}+\frac {x^{2} c^{2}}{b^{3} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\) | \(61\) |
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Time = 0.24 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.73 \[ \int \frac {1}{x \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {{\left (8 \, c^{2} x^{4} + 4 \, b c x^{2} - b^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{3 \, {\left (b^{3} c x^{6} + b^{4} x^{4}\right )}} \]
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\[ \int \frac {1}{x \left (b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {1}{x \left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}\, dx \]
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none
Time = 0.26 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.88 \[ \int \frac {1}{x \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {8 \, c^{2} x^{2}}{3 \, \sqrt {c x^{4} + b x^{2}} b^{3}} + \frac {4 \, c}{3 \, \sqrt {c x^{4} + b x^{2}} b^{2}} - \frac {1}{3 \, \sqrt {c x^{4} + b x^{2}} b x^{2}} \]
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Time = 0.29 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.54 \[ \int \frac {1}{x \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {c^{2} x}{\sqrt {c x^{2} + b} b^{3} \mathrm {sgn}\left (x\right )} - \frac {2 \, {\left (3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} c^{\frac {3}{2}} - 12 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} b c^{\frac {3}{2}} + 5 \, b^{2} c^{\frac {3}{2}}\right )}}{3 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b\right )}^{3} b^{2} \mathrm {sgn}\left (x\right )} \]
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Time = 12.98 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.69 \[ \int \frac {1}{x \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {\sqrt {c\,x^4+b\,x^2}\,\left (-b^2+4\,b\,c\,x^2+8\,c^2\,x^4\right )}{3\,b^3\,x^4\,\left (c\,x^2+b\right )} \]
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