Optimal. Leaf size=115 \[ \frac {\cos (2) \text {Ci}(2-2 \tanh (a+b x))}{4 b}-\frac {\cos (2) \text {Ci}(2 \tanh (a+b x)+2)}{4 b}+\frac {\sin (2) \text {Si}(2-2 \tanh (a+b x))}{4 b}-\frac {\sin (2) \text {Si}(2 \tanh (a+b x)+2)}{4 b}-\frac {\log (1-\tanh (a+b x))}{4 b}+\frac {\log (\tanh (a+b x)+1)}{4 b} \]
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Rubi [A] time = 0.27, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 13, 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} \[ \frac {\cos (2) \text {CosIntegral}(2-2 \tanh (a+b x))}{4 b}-\frac {\cos (2) \text {CosIntegral}(2 \tanh (a+b x)+2)}{4 b}+\frac {\sin (2) \text {Si}(2-2 \tanh (a+b x))}{4 b}-\frac {\sin (2) \text {Si}(2 \tanh (a+b x)+2)}{4 b}-\frac {\log (1-\tanh (a+b x))}{4 b}+\frac {\log (\tanh (a+b x)+1)}{4 b} \]
Antiderivative was successfully verified.
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Rule 3299
Rule 3302
Rule 3303
Rule 3312
Rule 6725
Rubi steps
\begin {align*} \int \sin ^2(\tanh (a+b x)) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\sin ^2(x)}{1-x^2} \, dx,x,\tanh (a+b x)\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int \left (-\frac {\sin ^2(x)}{2 (-1+x)}+\frac {\sin ^2(x)}{2 (1+x)}\right ) \, dx,x,\tanh (a+b x)\right )}{b}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {\sin ^2(x)}{-1+x} \, dx,x,\tanh (a+b x)\right )}{2 b}+\frac {\operatorname {Subst}\left (\int \frac {\sin ^2(x)}{1+x} \, dx,x,\tanh (a+b x)\right )}{2 b}\\ &=-\frac {\operatorname {Subst}\left (\int \left (\frac {1}{2 (-1+x)}-\frac {\cos (2 x)}{2 (-1+x)}\right ) \, dx,x,\tanh (a+b x)\right )}{2 b}+\frac {\operatorname {Subst}\left (\int \left (\frac {1}{2 (1+x)}-\frac {\cos (2 x)}{2 (1+x)}\right ) \, dx,x,\tanh (a+b x)\right )}{2 b}\\ &=-\frac {\log (1-\tanh (a+b x))}{4 b}+\frac {\log (1+\tanh (a+b x))}{4 b}+\frac {\operatorname {Subst}\left (\int \frac {\cos (2 x)}{-1+x} \, dx,x,\tanh (a+b x)\right )}{4 b}-\frac {\operatorname {Subst}\left (\int \frac {\cos (2 x)}{1+x} \, dx,x,\tanh (a+b x)\right )}{4 b}\\ &=-\frac {\log (1-\tanh (a+b x))}{4 b}+\frac {\log (1+\tanh (a+b x))}{4 b}+\frac {\cos (2) \operatorname {Subst}\left (\int \frac {\cos (2-2 x)}{-1+x} \, dx,x,\tanh (a+b x)\right )}{4 b}-\frac {\cos (2) \operatorname {Subst}\left (\int \frac {\cos (2+2 x)}{1+x} \, dx,x,\tanh (a+b x)\right )}{4 b}+\frac {\sin (2) \operatorname {Subst}\left (\int \frac {\sin (2-2 x)}{-1+x} \, dx,x,\tanh (a+b x)\right )}{4 b}-\frac {\sin (2) \operatorname {Subst}\left (\int \frac {\sin (2+2 x)}{1+x} \, dx,x,\tanh (a+b x)\right )}{4 b}\\ &=\frac {\cos (2) \text {Ci}(2-2 \tanh (a+b x))}{4 b}-\frac {\cos (2) \text {Ci}(2+2 \tanh (a+b x))}{4 b}-\frac {\log (1-\tanh (a+b x))}{4 b}+\frac {\log (1+\tanh (a+b x))}{4 b}+\frac {\sin (2) \text {Si}(2-2 \tanh (a+b x))}{4 b}-\frac {\sin (2) \text {Si}(2+2 \tanh (a+b x))}{4 b}\\ \end {align*}
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Mathematica [A] time = 0.15, size = 88, normalized size = 0.77 \[ \frac {\cos (2) \text {Ci}(2-2 \tanh (a+b x))-\cos (2) \text {Ci}(2 (\tanh (a+b x)+1))+\sin (2) \text {Si}(2-2 \tanh (a+b x))-\sin (2) \text {Si}(2 (\tanh (a+b x)+1))-\log (1-\tanh (a+b x))+\log (\tanh (a+b x)+1)}{4 b} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.54, size = 155, normalized size = 1.35 \[ \frac {4 \, b x - \cos \relax (2) \operatorname {Ci}\left (\frac {4 \, e^{\left (2 \, b x + 2 \, a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right ) - \cos \relax (2) \operatorname {Ci}\left (-\frac {4 \, e^{\left (2 \, b x + 2 \, a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right ) + \cos \relax (2) \operatorname {Ci}\left (\frac {4}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right ) + \cos \relax (2) \operatorname {Ci}\left (-\frac {4}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right ) - 2 \, \sin \relax (2) \operatorname {Si}\left (\frac {4 \, e^{\left (2 \, b x + 2 \, a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right ) + 2 \, \sin \relax (2) \operatorname {Si}\left (\frac {4}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right )}{8 \, b} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \sin \left (\tanh \left (b x + a\right )\right )^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.09, size = 102, normalized size = 0.89 \[ -\frac {\ln \left (-1+\tanh \left (b x +a \right )\right )}{4 b}+\frac {\ln \left (1+\tanh \left (b x +a \right )\right )}{4 b}-\frac {\Si \left (2+2 \tanh \left (b x +a \right )\right ) \sin \relax (2)}{4 b}-\frac {\Ci \left (2+2 \tanh \left (b x +a \right )\right ) \cos \relax (2)}{4 b}-\frac {\Si \left (-2+2 \tanh \left (b x +a \right )\right ) \sin \relax (2)}{4 b}+\frac {\Ci \left (-2+2 \tanh \left (b x +a \right )\right ) \cos \relax (2)}{4 b} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, x - \frac {1}{2} \, \int \cos \left (\frac {2 \, {\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )}}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int {\sin \left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2 \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \sin ^{2}{\left (\tanh {\left (a + b x \right )} \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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