Optimal. Leaf size=94 \[ \frac {i \text {Li}_2\left ((1-i d) e^{2 i a+2 i b x}\right )}{4 b}-\frac {1}{2} x \log \left (1-(1-i d) e^{2 i a+2 i b x}\right )+x \coth ^{-1}(d (-\cot (a+b x))-i d+1)+\frac {1}{2} i b x^2 \]
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Rubi [A] time = 0.15, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {6258, 2184, 2190, 2279, 2391} \[ \frac {i \text {PolyLog}\left (2,(1-i d) e^{2 i a+2 i b x}\right )}{4 b}-\frac {1}{2} x \log \left (1-(1-i d) e^{2 i a+2 i b x}\right )+x \coth ^{-1}(d (-\cot (a+b x))-i d+1)+\frac {1}{2} i b x^2 \]
Antiderivative was successfully verified.
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Rule 2184
Rule 2190
Rule 2279
Rule 2391
Rule 6258
Rubi steps
\begin {align*} \int \coth ^{-1}(1-i d-d \cot (a+b x)) \, dx &=x \coth ^{-1}(1-i d-d \cot (a+b x))+(i b) \int \frac {x}{1+(-1+i d) e^{2 i a+2 i b x}} \, dx\\ &=\frac {1}{2} i b x^2+x \coth ^{-1}(1-i d-d \cot (a+b x))+(b (i+d)) \int \frac {e^{2 i a+2 i b x} x}{1+(-1+i d) e^{2 i a+2 i b x}} \, dx\\ &=\frac {1}{2} i b x^2+x \coth ^{-1}(1-i d-d \cot (a+b x))-\frac {1}{2} x \log \left (1-(1-i d) e^{2 i a+2 i b x}\right )+\frac {1}{2} \int \log \left (1+(-1+i d) e^{2 i a+2 i b x}\right ) \, dx\\ &=\frac {1}{2} i b x^2+x \coth ^{-1}(1-i d-d \cot (a+b x))-\frac {1}{2} x \log \left (1-(1-i d) e^{2 i a+2 i b x}\right )-\frac {i \operatorname {Subst}\left (\int \frac {\log (1+(-1+i d) x)}{x} \, dx,x,e^{2 i a+2 i b x}\right )}{4 b}\\ &=\frac {1}{2} i b x^2+x \coth ^{-1}(1-i d-d \cot (a+b x))-\frac {1}{2} x \log \left (1-(1-i d) e^{2 i a+2 i b x}\right )+\frac {i \text {Li}_2\left ((1-i d) e^{2 i a+2 i b x}\right )}{4 b}\\ \end {align*}
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Mathematica [B] time = 2.94, size = 605, normalized size = 6.44 \[ \frac {x \csc ^2(a+b x) (\cos (b x)-i \sin (b x)) (\cos (b x)+i \sin (b x)) \left (i \text {Li}_2\left (\frac {(\cos (a)-i \sin (a)) ((2-i d) \cos (a)+d \sin (a)) (\tan (b x)+i)}{2 (d+i)}\right )-i \text {Li}_2\left (\frac {1}{2} \sec (b x) ((2-i d) \cos (a)+d \sin (a)) (\cos (a+b x)+i \sin (a+b x))\right )+i \log (1-i \tan (b x)) \log \left (\frac {(\cos (a)-i \sin (a)) \sec (b x) (d \cos (a+b x)+i (d+2 i) \sin (a+b x))}{2 (d+i)}\right )-i \log (1+i \tan (b x)) \log \left (\frac {i \sec (b x) (d \cos (a+b x)+i (d+2 i) \sin (a+b x))}{2 \cos (a)-2 i \sin (a)}\right )+i \text {Li}_2(i \sin (2 b x)-\cos (2 b x))+2 b x \log (2 \cos (b x) (\cos (b x)-i \sin (b x)))\right )}{(\cot (a+b x)+i) (d \cot (a+b x)+i d-2) \left (\frac {\sec ^2(b x) \log \left (\frac {i \sec (b x) (d \cos (a+b x)+i (d+2 i) \sin (a+b x))}{2 \cos (a)-2 i \sin (a)}\right )}{1+i \tan (b x)}-\frac {\sec (b x) (i d \sin (a)+(d+2 i) \cos (a)) \log (1-i \tan (b x))}{d \cos (a+b x)+i (d+2 i) \sin (a+b x)}+\frac {\sec (b x) (i d \sin (a)+(d+2 i) \cos (a)) \log (1+i \tan (b x))}{d \cos (a+b x)+i (d+2 i) \sin (a+b x)}+i (\tan (b x)+i) \log \left (1-\frac {1}{2} \sec (b x) (d \sin (a)+(2-i d) \cos (a)) (\cos (a+b x)+i \sin (a+b x))\right )-2 b x (\tan (b x)+i)\right )}+x \coth ^{-1}(d (-\cot (a+b x))-i d+1) \]
Warning: Unable to verify antiderivative.
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fricas [A] time = 0.71, size = 121, normalized size = 1.29 \[ \frac {2 i \, b^{2} x^{2} - 2 \, b x \log \left (\frac {d e^{\left (2 i \, b x + 2 i \, a\right )}}{{\left (d + i\right )} e^{\left (2 i \, b x + 2 i \, a\right )} - i}\right ) - 2 i \, a^{2} - 2 \, {\left (b x + a\right )} \log \left ({\left (i \, d - 1\right )} e^{\left (2 i \, b x + 2 i \, a\right )} + 1\right ) + 2 \, a \log \left (\frac {{\left (d + i\right )} e^{\left (2 i \, b x + 2 i \, a\right )} - i}{d + i}\right ) + i \, {\rm Li}_2\left (-{\left (i \, d - 1\right )} e^{\left (2 i \, b x + 2 i \, a\right )}\right )}{4 \, b} \]
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 \[ \int \operatorname {arcoth}\left (-d \cot \left (b x + a\right ) - i \, d + 1\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.70, size = 304, normalized size = 3.23 \[ \frac {i \mathrm {arccoth}\left (1-i d -d \cot \left (b x +a \right )\right ) \ln \left (i d -d \cot \left (b x +a \right )\right )}{2 b}-\frac {i \mathrm {arccoth}\left (1-i d -d \cot \left (b x +a \right )\right ) \ln \left (-i d -d \cot \left (b x +a \right )\right )}{2 b}-\frac {i \ln \left (-i d -d \cot \left (b x +a \right )\right )^{2}}{8 b}+\frac {i \dilog \left (1-\frac {i d}{2}-\frac {d \cot \left (b x +a \right )}{2}\right )}{4 b}+\frac {i \ln \left (-i d -d \cot \left (b x +a \right )\right ) \ln \left (1-\frac {i d}{2}-\frac {d \cot \left (b x +a \right )}{2}\right )}{4 b}-\frac {i \dilog \left (\frac {2-i d -d \cot \left (b x +a \right )}{-2 i d +2}\right )}{4 b}-\frac {i \ln \left (i d -d \cot \left (b x +a \right )\right ) \ln \left (\frac {2-i d -d \cot \left (b x +a \right )}{-2 i d +2}\right )}{4 b}+\frac {i \dilog \left (\frac {i \left (-i d -d \cot \left (b x +a \right )\right )}{2 d}\right )}{4 b}+\frac {i \ln \left (i d -d \cot \left (b x +a \right )\right ) \ln \left (\frac {i \left (-i d -d \cot \left (b x +a \right )\right )}{2 d}\right )}{4 b} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.45, size = 288, normalized size = 3.06 \[ -\frac {4 \, {\left (b x + a\right )} d {\left (\frac {\log \left ({\left (i \, d - 2\right )} \tan \left (b x + a\right ) + d\right )}{d} - \frac {\log \left (i \, \tan \left (b x + a\right ) + 1\right )}{d}\right )} - d {\left (\frac {2 i \, {\left (\log \left ({\left (i \, d - 2\right )} \tan \left (b x + a\right ) + d\right ) \log \left (\frac {{\left (d + 2 i\right )} \tan \left (b x + a\right ) - i \, d}{2 i \, d - 2} + 1\right ) + {\rm Li}_2\left (-\frac {{\left (d + 2 i\right )} \tan \left (b x + a\right ) - i \, d}{2 i \, d - 2}\right )\right )}}{d} + \frac {2 i \, {\left (\log \left (-\frac {1}{2} \, {\left (d + 2 i\right )} \tan \left (b x + a\right ) + \frac {1}{2} i \, d\right ) \log \left (i \, \tan \left (b x + a\right ) + 1\right ) + {\rm Li}_2\left (\frac {1}{2} \, {\left (d + 2 i\right )} \tan \left (b x + a\right ) - \frac {1}{2} i \, d + 1\right )\right )}}{d} - \frac {2 i \, \log \left ({\left (i \, d - 2\right )} \tan \left (b x + a\right ) + d\right ) \log \left (i \, \tan \left (b x + a\right ) + 1\right ) - i \, \log \left (i \, \tan \left (b x + a\right ) + 1\right )^{2}}{d} - \frac {2 i \, {\left (\log \left (i \, \tan \left (b x + a\right ) + 1\right ) \log \left (-\frac {1}{2} i \, \tan \left (b x + a\right ) + \frac {1}{2}\right ) + {\rm Li}_2\left (\frac {1}{2} i \, \tan \left (b x + a\right ) + \frac {1}{2}\right )\right )}}{d}\right )} + 8 \, {\left (b x + a\right )} \operatorname {arcoth}\left (i \, d + \frac {d}{\tan \left (b x + a\right )} - 1\right )}{8 \, b} \]
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 \[ \int -\mathrm {acoth}\left (d\,\mathrm {cot}\left (a+b\,x\right )-1+d\,1{}\mathrm {i}\right ) \,d x \]
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 \[ \int \operatorname {acoth}{\left (- d \cot {\left (a + b x \right )} - i d + 1 \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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