Optimal. Leaf size=115 \[ \frac {\cos (2) \text {CosIntegral}(2-2 \tanh (a+b x))}{4 b}-\frac {\cos (2) \text {CosIntegral}(2+2 \tanh (a+b x))}{4 b}-\frac {\log (1-\tanh (a+b x))}{4 b}+\frac {\log (1+\tanh (a+b x))}{4 b}+\frac {\sin (2) \text {Si}(2-2 \tanh (a+b x))}{4 b}-\frac {\sin (2) \text {Si}(2+2 \tanh (a+b x))}{4 b} \]
[Out]
________________________________________________________________________________________
Rubi [A]
time = 0.19, 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 = {6857, 3393,
3384, 3380, 3383} \begin {gather*} \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} \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
Rule 3380
Rule 3383
Rule 3384
Rule 3393
Rule 6857
Rubi steps
\begin {align*} \int \sin ^2(\tanh (a+b x)) \, dx &=\frac {\text {Subst}\left (\int \frac {\sin ^2(x)}{1-x^2} \, dx,x,\tanh (a+b x)\right )}{b}\\ &=\frac {\text {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 {\text {Subst}\left (\int \frac {\sin ^2(x)}{-1+x} \, dx,x,\tanh (a+b x)\right )}{2 b}+\frac {\text {Subst}\left (\int \frac {\sin ^2(x)}{1+x} \, dx,x,\tanh (a+b x)\right )}{2 b}\\ &=-\frac {\text {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 {\text {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 {\text {Subst}\left (\int \frac {\cos (2 x)}{-1+x} \, dx,x,\tanh (a+b x)\right )}{4 b}-\frac {\text {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) \text {Subst}\left (\int \frac {\cos (2-2 x)}{-1+x} \, dx,x,\tanh (a+b x)\right )}{4 b}-\frac {\cos (2) \text {Subst}\left (\int \frac {\cos (2+2 x)}{1+x} \, dx,x,\tanh (a+b x)\right )}{4 b}+\frac {\sin (2) \text {Subst}\left (\int \frac {\sin (2-2 x)}{-1+x} \, dx,x,\tanh (a+b x)\right )}{4 b}-\frac {\sin (2) \text {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*}
________________________________________________________________________________________
Mathematica [A]
time = 0.12, size = 88, normalized size = 0.77 \begin {gather*} \frac {\cos (2) \text {CosIntegral}(2-2 \tanh (a+b x))-\cos (2) \text {CosIntegral}(2 (1+\tanh (a+b x)))-\log (1-\tanh (a+b x))+\log (1+\tanh (a+b x))+\sin (2) \text {Si}(2-2 \tanh (a+b x))-\sin (2) \text {Si}(2 (1+\tanh (a+b x)))}{4 b} \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
Maple [A]
time = 1.03, size = 88, normalized size = 0.77
method | result | size |
derivativedivides | \(\frac {-\frac {\ln \left (-1+\tanh \left (b x +a \right )\right )}{4}+\frac {\ln \left (\tanh \left (b x +a \right )+1\right )}{4}-\frac {\sinIntegral \left (2+2 \tanh \left (b x +a \right )\right ) \sin \left (2\right )}{4}-\frac {\cosineIntegral \left (2+2 \tanh \left (b x +a \right )\right ) \cos \left (2\right )}{4}-\frac {\sinIntegral \left (-2+2 \tanh \left (b x +a \right )\right ) \sin \left (2\right )}{4}+\frac {\cosineIntegral \left (-2+2 \tanh \left (b x +a \right )\right ) \cos \left (2\right )}{4}}{b}\) | \(88\) |
default | \(\frac {-\frac {\ln \left (-1+\tanh \left (b x +a \right )\right )}{4}+\frac {\ln \left (\tanh \left (b x +a \right )+1\right )}{4}-\frac {\sinIntegral \left (2+2 \tanh \left (b x +a \right )\right ) \sin \left (2\right )}{4}-\frac {\cosineIntegral \left (2+2 \tanh \left (b x +a \right )\right ) \cos \left (2\right )}{4}-\frac {\sinIntegral \left (-2+2 \tanh \left (b x +a \right )\right ) \sin \left (2\right )}{4}+\frac {\cosineIntegral \left (-2+2 \tanh \left (b x +a \right )\right ) \cos \left (2\right )}{4}}{b}\) | \(88\) |
risch | \(\frac {x}{2}-\frac {{\mathrm e}^{-2 i} \expIntegral \left (1, -\frac {4 i}{{\mathrm e}^{2 b x +2 a}+1}\right )}{8 b}+\frac {{\mathrm e}^{2 i} \expIntegral \left (1, -\frac {4 i}{{\mathrm e}^{2 b x +2 a}+1}+4 i\right )}{8 b}-\frac {{\mathrm e}^{2 i} \expIntegral \left (1, \frac {4 i}{{\mathrm e}^{2 b x +2 a}+1}\right )}{8 b}+\frac {{\mathrm e}^{-2 i} \expIntegral \left (1, \frac {4 i}{{\mathrm e}^{2 b x +2 a}+1}-4 i\right )}{8 b}\) | \(115\) |
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Fricas [A]
time = 0.37, size = 155, normalized size = 1.35 \begin {gather*} \frac {4 \, b x - \cos \left (2\right ) \operatorname {Ci}\left (\frac {4 \, e^{\left (2 \, b x + 2 \, a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right ) - \cos \left (2\right ) \operatorname {Ci}\left (-\frac {4 \, e^{\left (2 \, b x + 2 \, a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right ) + \cos \left (2\right ) \operatorname {Ci}\left (\frac {4}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right ) + \cos \left (2\right ) \operatorname {Ci}\left (-\frac {4}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right ) - 2 \, \sin \left (2\right ) \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 \left (2\right ) \operatorname {Si}\left (\frac {4}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right )}{8 \, b} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sin ^{2}{\left (\tanh {\left (a + b x \right )} \right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\sin \left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2 \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________