Integrand size = 15, antiderivative size = 76 \[ \int \coth ^{-1}(1-d-d \tanh (a+b x)) \, dx=\frac {b x^2}{2}+x \coth ^{-1}(1-d-d \tanh (a+b x))-\frac {1}{2} x \log \left (1+(1-d) e^{2 a+2 b x}\right )-\frac {\operatorname {PolyLog}\left (2,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{4 b} \]
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Time = 0.11 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6367, 2215, 2221, 2317, 2438} \[ \int \coth ^{-1}(1-d-d \tanh (a+b x)) \, dx=-\frac {\operatorname {PolyLog}\left (2,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{4 b}-\frac {1}{2} x \log \left ((1-d) e^{2 a+2 b x}+1\right )+x \coth ^{-1}(d (-\tanh (a+b x))-d+1)+\frac {b x^2}{2} \]
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Rule 2215
Rule 2221
Rule 2317
Rule 2438
Rule 6367
Rubi steps \begin{align*} \text {integral}& = x \coth ^{-1}(1-d-d \tanh (a+b x))+b \int \frac {x}{1+(1-d) e^{2 a+2 b x}} \, dx \\ & = \frac {b x^2}{2}+x \coth ^{-1}(1-d-d \tanh (a+b x))-(b (1-d)) \int \frac {e^{2 a+2 b x} x}{1+(1-d) e^{2 a+2 b x}} \, dx \\ & = \frac {b x^2}{2}+x \coth ^{-1}(1-d-d \tanh (a+b x))-\frac {1}{2} x \log \left (1+(1-d) e^{2 a+2 b x}\right )+\frac {1}{2} \int \log \left (1+(1-d) e^{2 a+2 b x}\right ) \, dx \\ & = \frac {b x^2}{2}+x \coth ^{-1}(1-d-d \tanh (a+b x))-\frac {1}{2} x \log \left (1+(1-d) e^{2 a+2 b x}\right )+\frac {\text {Subst}\left (\int \frac {\log (1+(1-d) x)}{x} \, dx,x,e^{2 a+2 b x}\right )}{4 b} \\ & = \frac {b x^2}{2}+x \coth ^{-1}(1-d-d \tanh (a+b x))-\frac {1}{2} x \log \left (1+(1-d) e^{2 a+2 b x}\right )-\frac {\operatorname {PolyLog}\left (2,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{4 b} \\ \end{align*}
Time = 0.09 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.87 \[ \int \coth ^{-1}(1-d-d \tanh (a+b x)) \, dx=x \coth ^{-1}(1-d-d \tanh (a+b x))+\frac {-2 b x \log \left (1-\frac {e^{-2 (a+b x)}}{-1+d}\right )+\operatorname {PolyLog}\left (2,\frac {e^{-2 (a+b x)}}{-1+d}\right )}{4 b} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(270\) vs. \(2(68)=136\).
Time = 1.63 (sec) , antiderivative size = 271, normalized size of antiderivative = 3.57
method | result | size |
derivativedivides | \(-\frac {-\frac {\operatorname {arccoth}\left (1-d -d \tanh \left (b x +a \right )\right ) d \ln \left (-d \tanh \left (b x +a \right )-d \right )}{2}+\frac {\operatorname {arccoth}\left (1-d -d \tanh \left (b x +a \right )\right ) d \ln \left (-d \tanh \left (b x +a \right )+d \right )}{2}-\frac {d^{2} \left (\frac {-\frac {\operatorname {dilog}\left (-\frac {d \tanh \left (b x +a \right )}{2}-\frac {d}{2}+1\right )}{2}-\frac {\ln \left (-d \tanh \left (b x +a \right )-d \right ) \ln \left (-\frac {d \tanh \left (b x +a \right )}{2}-\frac {d}{2}+1\right )}{2}+\frac {\ln \left (-d \tanh \left (b x +a \right )-d \right )^{2}}{4}}{d}-\frac {-\frac {\operatorname {dilog}\left (\frac {-d \tanh \left (b x +a \right )-d +2}{-2 d +2}\right )}{2}-\frac {\ln \left (-d \tanh \left (b x +a \right )+d \right ) \ln \left (\frac {-d \tanh \left (b x +a \right )-d +2}{-2 d +2}\right )}{2}+\frac {\operatorname {dilog}\left (-\frac {-d \tanh \left (b x +a \right )-d}{2 d}\right )}{2}+\frac {\ln \left (-d \tanh \left (b x +a \right )+d \right ) \ln \left (-\frac {-d \tanh \left (b x +a \right )-d}{2 d}\right )}{2}}{d}\right )}{2}}{b d}\) | \(271\) |
default | \(-\frac {-\frac {\operatorname {arccoth}\left (1-d -d \tanh \left (b x +a \right )\right ) d \ln \left (-d \tanh \left (b x +a \right )-d \right )}{2}+\frac {\operatorname {arccoth}\left (1-d -d \tanh \left (b x +a \right )\right ) d \ln \left (-d \tanh \left (b x +a \right )+d \right )}{2}-\frac {d^{2} \left (\frac {-\frac {\operatorname {dilog}\left (-\frac {d \tanh \left (b x +a \right )}{2}-\frac {d}{2}+1\right )}{2}-\frac {\ln \left (-d \tanh \left (b x +a \right )-d \right ) \ln \left (-\frac {d \tanh \left (b x +a \right )}{2}-\frac {d}{2}+1\right )}{2}+\frac {\ln \left (-d \tanh \left (b x +a \right )-d \right )^{2}}{4}}{d}-\frac {-\frac {\operatorname {dilog}\left (\frac {-d \tanh \left (b x +a \right )-d +2}{-2 d +2}\right )}{2}-\frac {\ln \left (-d \tanh \left (b x +a \right )+d \right ) \ln \left (\frac {-d \tanh \left (b x +a \right )-d +2}{-2 d +2}\right )}{2}+\frac {\operatorname {dilog}\left (-\frac {-d \tanh \left (b x +a \right )-d}{2 d}\right )}{2}+\frac {\ln \left (-d \tanh \left (b x +a \right )+d \right ) \ln \left (-\frac {-d \tanh \left (b x +a \right )-d}{2 d}\right )}{2}}{d}\right )}{2}}{b d}\) | \(271\) |
risch | \(\text {Expression too large to display}\) | \(1124\) |
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Leaf count of result is larger than twice the leaf count of optimal. 227 vs. \(2 (63) = 126\).
Time = 0.29 (sec) , antiderivative size = 227, normalized size of antiderivative = 2.99 \[ \int \coth ^{-1}(1-d-d \tanh (a+b x)) \, dx=\frac {b^{2} x^{2} - b x \log \left (\frac {d \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )}{{\left (d - 2\right )} \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )}\right ) + a \log \left (2 \, {\left (d - 1\right )} \cosh \left (b x + a\right ) + 2 \, {\left (d - 1\right )} \sinh \left (b x + a\right ) + 2 \, \sqrt {d - 1}\right ) + a \log \left (2 \, {\left (d - 1\right )} \cosh \left (b x + a\right ) + 2 \, {\left (d - 1\right )} \sinh \left (b x + a\right ) - 2 \, \sqrt {d - 1}\right ) - {\left (b x + a\right )} \log \left (\sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) - {\left (b x + a\right )} \log \left (-\sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) - {\rm Li}_2\left (\sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - {\rm Li}_2\left (-\sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right )}{2 \, b} \]
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\[ \int \coth ^{-1}(1-d-d \tanh (a+b x)) \, dx=- \int \operatorname {acoth}{\left (d \tanh {\left (a + b x \right )} + d - 1 \right )}\, dx \]
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Time = 0.76 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.96 \[ \int \coth ^{-1}(1-d-d \tanh (a+b x)) \, dx=\frac {1}{4} \, b d {\left (\frac {2 \, x^{2}}{d} - \frac {2 \, b x \log \left (-{\left (d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + {\rm Li}_2\left ({\left (d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}\right )}{b^{2} d}\right )} - x \operatorname {arcoth}\left (d \tanh \left (b x + a\right ) + d - 1\right ) \]
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\[ \int \coth ^{-1}(1-d-d \tanh (a+b x)) \, dx=\int { \operatorname {arcoth}\left (-d \tanh \left (b x + a\right ) - d + 1\right ) \,d x } \]
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Timed out. \[ \int \coth ^{-1}(1-d-d \tanh (a+b x)) \, dx=\int -\mathrm {acoth}\left (d+d\,\mathrm {tanh}\left (a+b\,x\right )-1\right ) \,d x \]
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