Integrand size = 26, antiderivative size = 31 \[ \int \frac {x}{\sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{\sqrt {c}} \]
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Time = 0.03 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3, 2038, 634, 212} \[ \int \frac {x}{\sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{\sqrt {c}} \]
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Rule 3
Rule 212
Rule 634
Rule 2038
Rubi steps \begin{align*} \text {integral}& = \int \frac {x}{\sqrt {b x^2+c x^4}} \, dx \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {1}{\sqrt {b x+c x^2}} \, dx,x,x^2\right ) \\ & = \text {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x^2}{\sqrt {b x^2+c x^4}}\right ) \\ & = \frac {\tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{\sqrt {c}} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.68 \[ \int \frac {x}{\sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx=\frac {x \sqrt {b+c x^2} \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b+c x^2}}\right )}{\sqrt {c} \sqrt {x^2 \left (b+c x^2\right )}} \]
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Time = 0.17 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.35
method | result | size |
pseudoelliptic | \(\frac {-\ln \left (2\right )+\ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right )}{2 \sqrt {c}}\) | \(42\) |
default | \(\frac {x \sqrt {c \,x^{2}+b}\, \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+b}\right )}{\sqrt {c \,x^{4}+b \,x^{2}}\, \sqrt {c}}\) | \(44\) |
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Time = 0.29 (sec) , antiderivative size = 74, normalized size of antiderivative = 2.39 \[ \int \frac {x}{\sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx=\left [\frac {\log \left (-2 \, c x^{2} - b - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{2 \, \sqrt {c}}, -\frac {\sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right )}{c}\right ] \]
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\[ \int \frac {x}{\sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx=\int \frac {x}{\sqrt {x^{2} \left (b + c x^{2}\right )}}\, dx \]
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Time = 0.20 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \frac {x}{\sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx=\frac {\log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{2 \, \sqrt {c}} \]
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Time = 0.29 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.23 \[ \int \frac {x}{\sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx=\frac {\log \left ({\left | b \right |}\right ) \mathrm {sgn}\left (x\right )}{2 \, \sqrt {c}} - \frac {\log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right )}{\sqrt {c} \mathrm {sgn}\left (x\right )} \]
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Time = 13.43 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.06 \[ \int \frac {x}{\sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx=\frac {\ln \left (\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}+\sqrt {c\,x^4+b\,x^2}\right )}{2\,\sqrt {c}} \]
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