Integrand size = 19, antiderivative size = 55 \[ \int \frac {x^5}{\left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {x^2}{c \sqrt {b x^2+c x^4}}+\frac {\text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{c^{3/2}} \]
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Time = 0.05 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {2043, 666, 634, 212} \[ \int \frac {x^5}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{c^{3/2}}-\frac {x^2}{c \sqrt {b x^2+c x^4}} \]
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Rule 212
Rule 634
Rule 666
Rule 2043
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {x^2}{\left (b x+c x^2\right )^{3/2}} \, dx,x,x^2\right ) \\ & = -\frac {x^2}{c \sqrt {b x^2+c x^4}}+\frac {\text {Subst}\left (\int \frac {1}{\sqrt {b x+c x^2}} \, dx,x,x^2\right )}{2 c} \\ & = -\frac {x^2}{c \sqrt {b x^2+c x^4}}+\frac {\text {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x^2}{\sqrt {b x^2+c x^4}}\right )}{c} \\ & = -\frac {x^2}{c \sqrt {b x^2+c x^4}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{c^{3/2}} \\ \end{align*}
Time = 0.10 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.33 \[ \int \frac {x^5}{\left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {x \left (\sqrt {c} x+2 \sqrt {b+c x^2} \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b}-\sqrt {b+c x^2}}\right )\right )}{c^{3/2} \sqrt {x^2 \left (b+c x^2\right )}} \]
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Time = 0.08 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.15
method | result | size |
default | \(-\frac {x^{3} \left (c \,x^{2}+b \right ) \left (x \,c^{\frac {3}{2}}-\ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+b}\right ) c \sqrt {c \,x^{2}+b}\right )}{\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} c^{\frac {5}{2}}}\) | \(63\) |
pseudoelliptic | \(\frac {-2 x^{2} \sqrt {c}+\sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right )-\sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \ln \left (2\right )}{2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, c^{\frac {3}{2}}}\) | \(90\) |
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Time = 0.26 (sec) , antiderivative size = 150, normalized size of antiderivative = 2.73 \[ \int \frac {x^5}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\left [\frac {{\left (c x^{2} + b\right )} \sqrt {c} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - 2 \, \sqrt {c x^{4} + b x^{2}} c}{2 \, {\left (c^{3} x^{2} + b c^{2}\right )}}, -\frac {{\left (c x^{2} + b\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) + \sqrt {c x^{4} + b x^{2}} c}{c^{3} x^{2} + b c^{2}}\right ] \]
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\[ \int \frac {x^5}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {x^{5}}{\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}\, dx \]
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Time = 0.29 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.98 \[ \int \frac {x^5}{\left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {x^{2}}{\sqrt {c x^{4} + b x^{2}} c} + \frac {\log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{2 \, c^{\frac {3}{2}}} \]
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Time = 0.29 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.04 \[ \int \frac {x^5}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {\log \left ({\left | b \right |}\right ) \mathrm {sgn}\left (x\right )}{2 \, c^{\frac {3}{2}}} - \frac {x}{\sqrt {c x^{2} + b} c \mathrm {sgn}\left (x\right )} - \frac {\log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right )}{c^{\frac {3}{2}} \mathrm {sgn}\left (x\right )} \]
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Time = 13.22 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00 \[ \int \frac {x^5}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {\ln \left (\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}+\sqrt {c\,x^4+b\,x^2}\right )}{2\,c^{3/2}}-\frac {x^2}{c\,\sqrt {c\,x^4+b\,x^2}} \]
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