Integrand size = 25, antiderivative size = 46 \[ \int \frac {e^{c (a+b x)}}{\sqrt {\sinh ^2(a c+b c x)}} \, dx=\frac {\log \left (1-e^{2 c (a+b x)}\right ) \sinh (a c+b c x)}{b c \sqrt {\sinh ^2(a c+b c x)}} \]
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Time = 0.08 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {6852, 2320, 12, 266} \[ \int \frac {e^{c (a+b x)}}{\sqrt {\sinh ^2(a c+b c x)}} \, dx=\frac {\log \left (1-e^{2 c (a+b x)}\right ) \sinh (a c+b c x)}{b c \sqrt {\sinh ^2(a c+b c x)}} \]
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Rule 12
Rule 266
Rule 2320
Rule 6852
Rubi steps \begin{align*} \text {integral}& = \frac {\sinh (a c+b c x) \int e^{c (a+b x)} \text {csch}(a c+b c x) \, dx}{\sqrt {\sinh ^2(a c+b c x)}} \\ & = \frac {\sinh (a c+b c x) \text {Subst}\left (\int \frac {2 x}{-1+x^2} \, dx,x,e^{c (a+b x)}\right )}{b c \sqrt {\sinh ^2(a c+b c x)}} \\ & = \frac {(2 \sinh (a c+b c x)) \text {Subst}\left (\int \frac {x}{-1+x^2} \, dx,x,e^{c (a+b x)}\right )}{b c \sqrt {\sinh ^2(a c+b c x)}} \\ & = \frac {\log \left (1-e^{2 c (a+b x)}\right ) \sinh (a c+b c x)}{b c \sqrt {\sinh ^2(a c+b c x)}} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.96 \[ \int \frac {e^{c (a+b x)}}{\sqrt {\sinh ^2(a c+b c x)}} \, dx=\frac {\log \left (1-e^{2 c (a+b x)}\right ) \sinh (c (a+b x))}{b c \sqrt {\sinh ^2(c (a+b x))}} \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.64 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.63
method | result | size |
default | \(\operatorname {csgn}\left (\sinh \left (c \left (b x +a \right )\right )\right ) \left (x +\frac {\ln \left (\sinh \left (c \left (b x +a \right )\right )\right )}{c b}\right )\) | \(29\) |
risch | \(\frac {\ln \left ({\mathrm e}^{2 b c x}-{\mathrm e}^{-2 a c}\right ) \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) {\mathrm e}^{-c \left (b x +a \right )}}{c b \sqrt {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2} {\mathrm e}^{-2 c \left (b x +a \right )}}}\) | \(68\) |
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Time = 0.26 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.91 \[ \int \frac {e^{c (a+b x)}}{\sqrt {\sinh ^2(a c+b c x)}} \, dx=\frac {\log \left (\frac {2 \, \sinh \left (b c x + a c\right )}{\cosh \left (b c x + a c\right ) - \sinh \left (b c x + a c\right )}\right )}{b c} \]
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\[ \int \frac {e^{c (a+b x)}}{\sqrt {\sinh ^2(a c+b c x)}} \, dx=e^{a c} \int \frac {e^{b c x}}{\sqrt {\sinh ^{2}{\left (a c + b c x \right )}}}\, dx \]
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Time = 0.30 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.85 \[ \int \frac {e^{c (a+b x)}}{\sqrt {\sinh ^2(a c+b c x)}} \, dx=\frac {\log \left (e^{\left (b c x + a c\right )} + 1\right )}{b c} + \frac {\log \left (e^{\left (b c x + a c\right )} - 1\right )}{b c} \]
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Time = 0.28 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.85 \[ \int \frac {e^{c (a+b x)}}{\sqrt {\sinh ^2(a c+b c x)}} \, dx=\frac {\log \left (e^{\left (b c x\right )} + e^{\left (-a c\right )}\right ) \mathrm {sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right ) + \log \left ({\left | e^{\left (b c x\right )} - e^{\left (-a c\right )} \right |}\right ) \mathrm {sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right )}{b c} \]
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Timed out. \[ \int \frac {e^{c (a+b x)}}{\sqrt {\sinh ^2(a c+b c x)}} \, dx=\int \frac {{\mathrm {e}}^{c\,\left (a+b\,x\right )}}{\sqrt {{\mathrm {sinh}\left (a\,c+b\,c\,x\right )}^2}} \,d x \]
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