Integrand size = 34, antiderivative size = 26 \[ \int \frac {\sinh \left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )}{1-a^2 x^2} \, dx=-\frac {\text {Shi}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )}{a} \]
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Time = 0.03 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {6813, 3379} \[ \int \frac {\sinh \left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )}{1-a^2 x^2} \, dx=-\frac {\text {Shi}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )}{a} \]
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Rule 3379
Rule 6813
Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int \frac {\sinh (x)}{x} \, dx,x,\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )}{a} \\ & = -\frac {\text {Shi}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )}{a} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {\sinh \left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )}{1-a^2 x^2} \, dx=-\frac {\text {Shi}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )}{a} \]
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\[\int \frac {\sinh \left (\frac {\sqrt {-a x +1}}{\sqrt {a x +1}}\right )}{-a^{2} x^{2}+1}d x\]
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\[ \int \frac {\sinh \left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )}{1-a^2 x^2} \, dx=\int { -\frac {\sinh \left (\frac {\sqrt {-a x + 1}}{\sqrt {a x + 1}}\right )}{a^{2} x^{2} - 1} \,d x } \]
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\[ \int \frac {\sinh \left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )}{1-a^2 x^2} \, dx=- \int \frac {\sinh {\left (\frac {\sqrt {- a x + 1}}{\sqrt {a x + 1}} \right )}}{a^{2} x^{2} - 1}\, dx \]
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\[ \int \frac {\sinh \left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )}{1-a^2 x^2} \, dx=\int { -\frac {\sinh \left (\frac {\sqrt {-a x + 1}}{\sqrt {a x + 1}}\right )}{a^{2} x^{2} - 1} \,d x } \]
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\[ \int \frac {\sinh \left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )}{1-a^2 x^2} \, dx=\int { -\frac {\sinh \left (\frac {\sqrt {-a x + 1}}{\sqrt {a x + 1}}\right )}{a^{2} x^{2} - 1} \,d x } \]
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Timed out. \[ \int \frac {\sinh \left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )}{1-a^2 x^2} \, dx=-\int \frac {\mathrm {sinh}\left (\frac {\sqrt {1-a\,x}}{\sqrt {a\,x+1}}\right )}{a^2\,x^2-1} \,d x \]
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