Optimal. Leaf size=157 \[ \frac{\sin (3) \text{CosIntegral}(3-3 \tanh (a+b x))}{8 b}+\frac{\sin (3) \text{CosIntegral}(3 \tanh (a+b x)+3)}{8 b}-\frac{3 \sin (1) \text{CosIntegral}(1-\tanh (a+b x))}{8 b}-\frac{3 \sin (1) \text{CosIntegral}(\tanh (a+b x)+1)}{8 b}-\frac{\cos (3) \text{Si}(3-3 \tanh (a+b x))}{8 b}+\frac{3 \cos (1) \text{Si}(1-\tanh (a+b x))}{8 b}+\frac{3 \cos (1) \text{Si}(\tanh (a+b x)+1)}{8 b}-\frac{\cos (3) \text{Si}(3 \tanh (a+b x)+3)}{8 b} \]
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Rubi [A] time = 0.39518, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 19, number of rules used = 5, integrand size = 9, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.556, Rules used = {6725, 3312, 3303, 3299, 3302} \[ \frac{\sin (3) \text{CosIntegral}(3-3 \tanh (a+b x))}{8 b}+\frac{\sin (3) \text{CosIntegral}(3 \tanh (a+b x)+3)}{8 b}-\frac{3 \sin (1) \text{CosIntegral}(1-\tanh (a+b x))}{8 b}-\frac{3 \sin (1) \text{CosIntegral}(\tanh (a+b x)+1)}{8 b}-\frac{\cos (3) \text{Si}(3-3 \tanh (a+b x))}{8 b}+\frac{3 \cos (1) \text{Si}(1-\tanh (a+b x))}{8 b}+\frac{3 \cos (1) \text{Si}(\tanh (a+b x)+1)}{8 b}-\frac{\cos (3) \text{Si}(3 \tanh (a+b x)+3)}{8 b} \]
Antiderivative was successfully verified.
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Rule 6725
Rule 3312
Rule 3303
Rule 3299
Rule 3302
Rubi steps
\begin{align*} \int \sin ^3(\tanh (a+b x)) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\sin ^3(x)}{1-x^2} \, dx,x,\tanh (a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \left (-\frac{\sin ^3(x)}{2 (-1+x)}+\frac{\sin ^3(x)}{2 (1+x)}\right ) \, dx,x,\tanh (a+b x)\right )}{b}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{\sin ^3(x)}{-1+x} \, dx,x,\tanh (a+b x)\right )}{2 b}+\frac{\operatorname{Subst}\left (\int \frac{\sin ^3(x)}{1+x} \, dx,x,\tanh (a+b x)\right )}{2 b}\\ &=-\frac{\operatorname{Subst}\left (\int \left (\frac{3 \sin (x)}{4 (-1+x)}-\frac{\sin (3 x)}{4 (-1+x)}\right ) \, dx,x,\tanh (a+b x)\right )}{2 b}+\frac{\operatorname{Subst}\left (\int \left (\frac{3 \sin (x)}{4 (1+x)}-\frac{\sin (3 x)}{4 (1+x)}\right ) \, dx,x,\tanh (a+b x)\right )}{2 b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\sin (3 x)}{-1+x} \, dx,x,\tanh (a+b x)\right )}{8 b}-\frac{\operatorname{Subst}\left (\int \frac{\sin (3 x)}{1+x} \, dx,x,\tanh (a+b x)\right )}{8 b}-\frac{3 \operatorname{Subst}\left (\int \frac{\sin (x)}{-1+x} \, dx,x,\tanh (a+b x)\right )}{8 b}+\frac{3 \operatorname{Subst}\left (\int \frac{\sin (x)}{1+x} \, dx,x,\tanh (a+b x)\right )}{8 b}\\ &=\frac{(3 \cos (1)) \operatorname{Subst}\left (\int \frac{\sin (1-x)}{-1+x} \, dx,x,\tanh (a+b x)\right )}{8 b}+\frac{(3 \cos (1)) \operatorname{Subst}\left (\int \frac{\sin (1+x)}{1+x} \, dx,x,\tanh (a+b x)\right )}{8 b}-\frac{\cos (3) \operatorname{Subst}\left (\int \frac{\sin (3-3 x)}{-1+x} \, dx,x,\tanh (a+b x)\right )}{8 b}-\frac{\cos (3) \operatorname{Subst}\left (\int \frac{\sin (3+3 x)}{1+x} \, dx,x,\tanh (a+b x)\right )}{8 b}-\frac{(3 \sin (1)) \operatorname{Subst}\left (\int \frac{\cos (1-x)}{-1+x} \, dx,x,\tanh (a+b x)\right )}{8 b}-\frac{(3 \sin (1)) \operatorname{Subst}\left (\int \frac{\cos (1+x)}{1+x} \, dx,x,\tanh (a+b x)\right )}{8 b}+\frac{\sin (3) \operatorname{Subst}\left (\int \frac{\cos (3-3 x)}{-1+x} \, dx,x,\tanh (a+b x)\right )}{8 b}+\frac{\sin (3) \operatorname{Subst}\left (\int \frac{\cos (3+3 x)}{1+x} \, dx,x,\tanh (a+b x)\right )}{8 b}\\ &=-\frac{3 \text{Ci}(1-\tanh (a+b x)) \sin (1)}{8 b}-\frac{3 \text{Ci}(1+\tanh (a+b x)) \sin (1)}{8 b}+\frac{\text{Ci}(3-3 \tanh (a+b x)) \sin (3)}{8 b}+\frac{\text{Ci}(3+3 \tanh (a+b x)) \sin (3)}{8 b}-\frac{\cos (3) \text{Si}(3-3 \tanh (a+b x))}{8 b}+\frac{3 \cos (1) \text{Si}(1-\tanh (a+b x))}{8 b}+\frac{3 \cos (1) \text{Si}(1+\tanh (a+b x))}{8 b}-\frac{\cos (3) \text{Si}(3+3 \tanh (a+b x))}{8 b}\\ \end{align*}
Mathematica [A] time = 0.214449, size = 124, normalized size = 0.79 \[ \frac{2 \sin (3) \text{CosIntegral}(3-3 \tanh (a+b x))+2 \sin (3) \text{CosIntegral}(3 \tanh (a+b x)+3)-6 \sin (1) \text{CosIntegral}(1-\tanh (a+b x))-6 \sin (1) \text{CosIntegral}(\tanh (a+b x)+1)-2 \cos (3) \text{Si}(3-3 \tanh (a+b x))+6 \cos (1) \text{Si}(1-\tanh (a+b x))+6 \cos (1) \text{Si}(\tanh (a+b x)+1)-2 \cos (3) \text{Si}(3 \tanh (a+b x)+3)}{16 b} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.026, size = 118, normalized size = 0.8 \begin{align*}{\frac{1}{b} \left ( -{\frac{{\it Si} \left ( 3+3\,\tanh \left ( bx+a \right ) \right ) \cos \left ( 3 \right ) }{8}}+{\frac{{\it Ci} \left ( 3+3\,\tanh \left ( bx+a \right ) \right ) \sin \left ( 3 \right ) }{8}}+{\frac{{\it Si} \left ( -3+3\,\tanh \left ( bx+a \right ) \right ) \cos \left ( 3 \right ) }{8}}+{\frac{{\it Ci} \left ( -3+3\,\tanh \left ( bx+a \right ) \right ) \sin \left ( 3 \right ) }{8}}+{\frac{3\,{\it Si} \left ( 1+\tanh \left ( bx+a \right ) \right ) \cos \left ( 1 \right ) }{8}}-{\frac{3\,{\it Ci} \left ( 1+\tanh \left ( bx+a \right ) \right ) \sin \left ( 1 \right ) }{8}}-{\frac{3\,{\it Si} \left ( -1+\tanh \left ( bx+a \right ) \right ) \cos \left ( 1 \right ) }{8}}-{\frac{3\,{\it Ci} \left ( -1+\tanh \left ( bx+a \right ) \right ) \sin \left ( 1 \right ) }{8}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sin \left (\tanh \left (b x + a\right )\right )^{3}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.34124, size = 929, normalized size = 5.92 \begin{align*} \frac{\operatorname{Ci}\left (\frac{6 \, e^{\left (2 \, b x + 2 \, a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right ) \sin \left (3\right ) + \operatorname{Ci}\left (-\frac{6 \, e^{\left (2 \, b x + 2 \, a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right ) \sin \left (3\right ) + \operatorname{Ci}\left (\frac{6}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right ) \sin \left (3\right ) + \operatorname{Ci}\left (-\frac{6}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right ) \sin \left (3\right ) - 3 \, \operatorname{Ci}\left (\frac{2 \, e^{\left (2 \, b x + 2 \, a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right ) \sin \left (1\right ) - 3 \, \operatorname{Ci}\left (-\frac{2 \, e^{\left (2 \, b x + 2 \, a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right ) \sin \left (1\right ) - 3 \, \operatorname{Ci}\left (\frac{2}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right ) \sin \left (1\right ) - 3 \, \operatorname{Ci}\left (-\frac{2}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right ) \sin \left (1\right ) - 2 \, \cos \left (3\right ) \operatorname{Si}\left (\frac{6 \, e^{\left (2 \, b x + 2 \, a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right ) + 6 \, \cos \left (1\right ) \operatorname{Si}\left (\frac{2 \, e^{\left (2 \, b x + 2 \, a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right ) - 2 \, \cos \left (3\right ) \operatorname{Si}\left (\frac{6}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right ) + 6 \, \cos \left (1\right ) \operatorname{Si}\left (\frac{2}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right )}{16 \, b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sin ^{3}{\left (\tanh{\left (a + b x \right )} \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sin \left (\tanh \left (b x + a\right )\right )^{3}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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