Integrand size = 22, antiderivative size = 88 \[ \int \frac {e^{\text {sech}^{-1}(c x)} x^4}{1-c^2 x^2} \, dx=-\frac {x^2}{2 c^3}-\frac {2 \sqrt {1-c x}}{3 c^5 \sqrt {\frac {1}{1+c x}}}-\frac {x^2 \sqrt {1-c x}}{3 c^3 \sqrt {\frac {1}{1+c x}}}-\frac {\log \left (1-c^2 x^2\right )}{2 c^5} \]
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Time = 0.13 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {6476, 1972, 102, 12, 75, 272, 45} \[ \int \frac {e^{\text {sech}^{-1}(c x)} x^4}{1-c^2 x^2} \, dx=-\frac {2 \sqrt {1-c x}}{3 c^5 \sqrt {\frac {1}{c x+1}}}-\frac {x^2 \sqrt {1-c x}}{3 c^3 \sqrt {\frac {1}{c x+1}}}-\frac {x^2}{2 c^3}-\frac {\log \left (1-c^2 x^2\right )}{2 c^5} \]
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Rule 12
Rule 45
Rule 75
Rule 102
Rule 272
Rule 1972
Rule 6476
Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {x^3 \sqrt {\frac {1}{1+c x}}}{\sqrt {1-c x}} \, dx}{c}+\frac {\int \frac {x^3}{1-c^2 x^2} \, dx}{c} \\ & = \frac {\text {Subst}\left (\int \frac {x}{1-c^2 x} \, dx,x,x^2\right )}{2 c}+\frac {\left (\sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {x^3}{\sqrt {1-c x} \sqrt {1+c x}} \, dx}{c} \\ & = -\frac {x^2 \sqrt {1-c x}}{3 c^3 \sqrt {\frac {1}{1+c x}}}+\frac {\text {Subst}\left (\int \left (-\frac {1}{c^2}-\frac {1}{c^2 \left (-1+c^2 x\right )}\right ) \, dx,x,x^2\right )}{2 c}-\frac {\left (\sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int -\frac {2 x}{\sqrt {1-c x} \sqrt {1+c x}} \, dx}{3 c^3} \\ & = -\frac {x^2}{2 c^3}-\frac {x^2 \sqrt {1-c x}}{3 c^3 \sqrt {\frac {1}{1+c x}}}-\frac {\log \left (1-c^2 x^2\right )}{2 c^5}+\frac {\left (2 \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {x}{\sqrt {1-c x} \sqrt {1+c x}} \, dx}{3 c^3} \\ & = -\frac {x^2}{2 c^3}-\frac {2 \sqrt {1-c x}}{3 c^5 \sqrt {\frac {1}{1+c x}}}-\frac {x^2 \sqrt {1-c x}}{3 c^3 \sqrt {\frac {1}{1+c x}}}-\frac {\log \left (1-c^2 x^2\right )}{2 c^5} \\ \end{align*}
Time = 0.22 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.78 \[ \int \frac {e^{\text {sech}^{-1}(c x)} x^4}{1-c^2 x^2} \, dx=-\frac {3 c^2 x^2+2 \sqrt {\frac {1-c x}{1+c x}} \left (2+2 c x+c^2 x^2+c^3 x^3\right )+3 \log \left (1-c^2 x^2\right )}{6 c^5} \]
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Time = 0.69 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.84
method | result | size |
default | \(-\frac {\sqrt {-\frac {c x -1}{c x}}\, x \sqrt {\frac {c x +1}{c x}}\, \left (c^{2} x^{2}+2\right )}{3 c^{4}}+\frac {-\frac {x^{2}}{2 c^{2}}-\frac {\ln \left (c^{2} x^{2}-1\right )}{2 c^{4}}}{c}\) | \(74\) |
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Time = 0.24 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.78 \[ \int \frac {e^{\text {sech}^{-1}(c x)} x^4}{1-c^2 x^2} \, dx=-\frac {3 \, c^{2} x^{2} + 2 \, {\left (c^{3} x^{3} + 2 \, c x\right )} \sqrt {\frac {c x + 1}{c x}} \sqrt {-\frac {c x - 1}{c x}} + 3 \, \log \left (c^{2} x^{2} - 1\right )}{6 \, c^{5}} \]
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\[ \int \frac {e^{\text {sech}^{-1}(c x)} x^4}{1-c^2 x^2} \, dx=- \frac {\int \frac {x^{3}}{c^{2} x^{2} - 1}\, dx + \int \frac {c x^{4} \sqrt {-1 + \frac {1}{c x}} \sqrt {1 + \frac {1}{c x}}}{c^{2} x^{2} - 1}\, dx}{c} \]
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\[ \int \frac {e^{\text {sech}^{-1}(c x)} x^4}{1-c^2 x^2} \, dx=\int { -\frac {x^{4} {\left (\sqrt {\frac {1}{c x} + 1} \sqrt {\frac {1}{c x} - 1} + \frac {1}{c x}\right )}}{c^{2} x^{2} - 1} \,d x } \]
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\[ \int \frac {e^{\text {sech}^{-1}(c x)} x^4}{1-c^2 x^2} \, dx=\int { -\frac {x^{4} {\left (\sqrt {\frac {1}{c x} + 1} \sqrt {\frac {1}{c x} - 1} + \frac {1}{c x}\right )}}{c^{2} x^{2} - 1} \,d x } \]
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Time = 5.42 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.86 \[ \int \frac {e^{\text {sech}^{-1}(c x)} x^4}{1-c^2 x^2} \, dx=-\frac {\ln \left (c^2\,x^2-1\right )+c^2\,x^2}{2\,c^5}-x^3\,\sqrt {\frac {1}{c\,x}-1}\,\left (\frac {\sqrt {\frac {1}{c\,x}+1}}{3\,c^2}+\frac {2\,\sqrt {\frac {1}{c\,x}+1}}{3\,c^4\,x^2}\right ) \]
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