Integrand size = 8, antiderivative size = 46 \[ \int \frac {\text {Shi}(b x)}{x^3} \, dx=-\frac {b \cosh (b x)}{4 x}-\frac {\sinh (b x)}{4 x^2}+\frac {1}{4} b^2 \text {Shi}(b x)-\frac {\text {Shi}(b x)}{2 x^2} \]
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Time = 0.05 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6667, 12, 3378, 3379} \[ \int \frac {\text {Shi}(b x)}{x^3} \, dx=\frac {1}{4} b^2 \text {Shi}(b x)-\frac {\text {Shi}(b x)}{2 x^2}-\frac {\sinh (b x)}{4 x^2}-\frac {b \cosh (b x)}{4 x} \]
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Rule 12
Rule 3378
Rule 3379
Rule 6667
Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Shi}(b x)}{2 x^2}+\frac {1}{2} b \int \frac {\sinh (b x)}{b x^3} \, dx \\ & = -\frac {\text {Shi}(b x)}{2 x^2}+\frac {1}{2} \int \frac {\sinh (b x)}{x^3} \, dx \\ & = -\frac {\sinh (b x)}{4 x^2}-\frac {\text {Shi}(b x)}{2 x^2}+\frac {1}{4} b \int \frac {\cosh (b x)}{x^2} \, dx \\ & = -\frac {b \cosh (b x)}{4 x}-\frac {\sinh (b x)}{4 x^2}-\frac {\text {Shi}(b x)}{2 x^2}+\frac {1}{4} b^2 \int \frac {\sinh (b x)}{x} \, dx \\ & = -\frac {b \cosh (b x)}{4 x}-\frac {\sinh (b x)}{4 x^2}+\frac {1}{4} b^2 \text {Shi}(b x)-\frac {\text {Shi}(b x)}{2 x^2} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00 \[ \int \frac {\text {Shi}(b x)}{x^3} \, dx=-\frac {b \cosh (b x)}{4 x}-\frac {\sinh (b x)}{4 x^2}+\frac {1}{4} b^2 \text {Shi}(b x)-\frac {\text {Shi}(b x)}{2 x^2} \]
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Time = 0.40 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.02
method | result | size |
parts | \(-\frac {\operatorname {Shi}\left (b x \right )}{2 x^{2}}+\frac {b^{2} \left (-\frac {\sinh \left (b x \right )}{2 b^{2} x^{2}}-\frac {\cosh \left (b x \right )}{2 b x}+\frac {\operatorname {Shi}\left (b x \right )}{2}\right )}{2}\) | \(47\) |
derivativedivides | \(b^{2} \left (-\frac {\operatorname {Shi}\left (b x \right )}{2 b^{2} x^{2}}-\frac {\sinh \left (b x \right )}{4 b^{2} x^{2}}-\frac {\cosh \left (b x \right )}{4 b x}+\frac {\operatorname {Shi}\left (b x \right )}{4}\right )\) | \(48\) |
default | \(b^{2} \left (-\frac {\operatorname {Shi}\left (b x \right )}{2 b^{2} x^{2}}-\frac {\sinh \left (b x \right )}{4 b^{2} x^{2}}-\frac {\cosh \left (b x \right )}{4 b x}+\frac {\operatorname {Shi}\left (b x \right )}{4}\right )\) | \(48\) |
meijerg | \(\frac {i \sqrt {\pi }\, b^{2} \left (\frac {4 i \cosh \left (b x \right )}{b x \sqrt {\pi }}+\frac {4 i \sinh \left (b x \right )}{b^{2} x^{2} \sqrt {\pi }}+\frac {4 i \left (-b^{2} x^{2}+2\right ) \operatorname {Shi}\left (b x \right )}{b^{2} x^{2} \sqrt {\pi }}\right )}{16}\) | \(69\) |
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\[ \int \frac {\text {Shi}(b x)}{x^3} \, dx=\int { \frac {{\rm Shi}\left (b x\right )}{x^{3}} \,d x } \]
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Time = 0.52 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.85 \[ \int \frac {\text {Shi}(b x)}{x^3} \, dx=\frac {b^{2} \operatorname {Shi}{\left (b x \right )}}{4} - \frac {b \cosh {\left (b x \right )}}{4 x} - \frac {\sinh {\left (b x \right )}}{4 x^{2}} - \frac {\operatorname {Shi}{\left (b x \right )}}{2 x^{2}} \]
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\[ \int \frac {\text {Shi}(b x)}{x^3} \, dx=\int { \frac {{\rm Shi}\left (b x\right )}{x^{3}} \,d x } \]
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\[ \int \frac {\text {Shi}(b x)}{x^3} \, dx=\int { \frac {{\rm Shi}\left (b x\right )}{x^{3}} \,d x } \]
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Timed out. \[ \int \frac {\text {Shi}(b x)}{x^3} \, dx=\frac {b^2\,\mathrm {sinhint}\left (b\,x\right )}{4}-\frac {\frac {\mathrm {sinhint}\left (b\,x\right )}{2}+\frac {\mathrm {sinh}\left (b\,x\right )}{4}+\frac {b\,x\,\mathrm {cosh}\left (b\,x\right )}{4}}{x^2} \]
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