Optimal. Leaf size=93 \[ \frac{i \text{PolyLog}\left (2,-(1-i d) e^{2 i a+2 i b x}\right )}{4 b}-\frac{1}{2} x \log \left (1+(1-i d) e^{2 i a+2 i b x}\right )+x \coth ^{-1}(d \tan (a+b x)-i d+1)+\frac{1}{2} i b x^2 \]
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Rubi [A] time = 0.154592, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {6256, 2184, 2190, 2279, 2391} \[ \frac{i \text{PolyLog}\left (2,-(1-i d) e^{2 i a+2 i b x}\right )}{4 b}-\frac{1}{2} x \log \left (1+(1-i d) e^{2 i a+2 i b x}\right )+x \coth ^{-1}(d \tan (a+b x)-i d+1)+\frac{1}{2} i b x^2 \]
Antiderivative was successfully verified.
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Rule 6256
Rule 2184
Rule 2190
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int \coth ^{-1}(1-i d+d \tan (a+b x)) \, dx &=x \coth ^{-1}(1-i d+d \tan (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 \tan (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 \tan (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 \tan (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 \tan (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*}
Mathematica [B] time = 2.85825, size = 766, normalized size = 8.24 \[ \frac{x \sec ^2(a+b x) (\cos (b x)+i \sin (b x)) (\sin (b x)+i \cos (b x)) (d \sin (a+b x)+(2-i d) \cos (a+b x)) \left (\text{PolyLog}\left (2,-\frac{1}{2} (\cos (a)+i \sin (a)) (\tan (b x)-i) (d \cos (a)+i (d+2 i) \sin (a))\right )-\text{PolyLog}\left (2,\frac{\sec (b x) (d \cos (a)+i (d+2 i) \sin (a)) (\cos (a+b x)-i \sin (a+b x))}{2 (d+i)}\right )-\text{PolyLog}(2,-\cos (2 b x)+i \sin (2 b x))-\log (1-i \tan (b x)) \log \left (\frac{(\cos (a)-i \sin (a)) \sec (b x) (i d \sin (a+b x)+(d+2 i) \cos (a+b x))}{2 (d+i)}\right )+\log (1+i \tan (b x)) \log \left (\frac{\sec (b x) (d \sin (a+b x)+(2-i d) \cos (a+b x))}{2 \cos (a)-2 i \sin (a)}\right )+2 i b x \log (2 \cos (b x) (\cos (b x)-i \sin (b x)))\right )}{(\tan (a+b x)-i) (d \tan (a+b x)-i d+2) (i d \sin (a+b x)+(d+2 i) \cos (a+b x)) \left (\frac{\sec ^2(b x) \log \left (\frac{\sec (b x) (d \sin (a+b x)+(2-i d) \cos (a+b x))}{2 \cos (a)-2 i \sin (a)}\right )}{\tan (b x)-i}-\frac{\sec ^2(b x) \log \left (1+\frac{1}{2} (\cos (a)+i \sin (a)) (\tan (b x)-i) (d \cos (a)+i (d+2 i) \sin (a))\right )}{\tan (b x)-i}-\frac{\sec ^2(b x) \log \left (\frac{(\cos (a)-i \sin (a)) \sec (b x) (i d \sin (a+b x)+(d+2 i) \cos (a+b x))}{2 (d+i)}\right )}{\tan (b x)+i}+\frac{i \sec (b x) (d \cos (a)+i (d+2 i) \sin (a)) \log (1+i \tan (b x))}{i d \sin (a+b x)+(d+2 i) \cos (a+b x)}+\frac{\sec (b x) ((d+2 i) \sin (a)-i d \cos (a)) \log (1-i \tan (b x))}{i d \sin (a+b x)+(d+2 i) \cos (a+b x)}+(\tan (b x)-i) \log \left (\frac{(\cos (a)-i \sin (a)) \sec (b x) (i d \sin (a+b x)+(d+2 i) \cos (a+b x))}{2 (d+i)}\right )+2 b x (1-i \tan (b x))\right )}+x \coth ^{-1}(d \tan (a+b x)-i d+1) \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 0.175, size = 292, normalized size = 3.1 \begin{align*}{\frac{{\frac{i}{2}}{\rm arccoth} \left (1-id+d\tan \left ( bx+a \right ) \right )\ln \left ( id+d\tan \left ( bx+a \right ) \right ) }{b}}-{\frac{{\frac{i}{2}}{\rm arccoth} \left (1-id+d\tan \left ( bx+a \right ) \right )\ln \left ( -id+d\tan \left ( bx+a \right ) \right ) }{b}}-{\frac{{\frac{i}{8}} \left ( \ln \left ( -id+d\tan \left ( bx+a \right ) \right ) \right ) ^{2}}{b}}+{\frac{{\frac{i}{4}}}{b}{\it dilog} \left ( 1-{\frac{i}{2}}d+{\frac{d\tan \left ( bx+a \right ) }{2}} \right ) }+{\frac{{\frac{i}{4}}\ln \left ( -id+d\tan \left ( bx+a \right ) \right ) }{b}\ln \left ( 1-{\frac{i}{2}}d+{\frac{d\tan \left ( bx+a \right ) }{2}} \right ) }+{\frac{{\frac{i}{4}}}{b}{\it dilog} \left ({\frac{{\frac{i}{2}} \left ( -id+d\tan \left ( bx+a \right ) \right ) }{d}} \right ) }+{\frac{{\frac{i}{4}}\ln \left ( id+d\tan \left ( bx+a \right ) \right ) }{b}\ln \left ({\frac{{\frac{i}{2}} \left ( -id+d\tan \left ( bx+a \right ) \right ) }{d}} \right ) }-{\frac{{\frac{i}{4}}}{b}{\it dilog} \left ({\frac{2-id+d\tan \left ( bx+a \right ) }{-2\,id+2}} \right ) }-{\frac{{\frac{i}{4}}\ln \left ( id+d\tan \left ( bx+a \right ) \right ) }{b}\ln \left ({\frac{2-id+d\tan \left ( bx+a \right ) }{-2\,id+2}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 1.55359, size = 358, normalized size = 3.85 \begin{align*} -\frac{4 \,{\left (b x + a\right )} d{\left (\frac{\log \left (d \tan \left (b x + a\right ) - i \, d + 2\right )}{d} - \frac{\log \left (\tan \left (b x + a\right ) - i\right )}{d}\right )} - d{\left (\frac{2 i \,{\left (\log \left (d \tan \left (b x + a\right ) - i \, d + 2\right ) \log \left (-\frac{i \, d \tan \left (b x + a\right ) + d + 2 i}{2 \, d + 2 i} + 1\right ) +{\rm Li}_2\left (\frac{i \, d \tan \left (b x + a\right ) + d + 2 i}{2 \, d + 2 i}\right )\right )}}{d} - \frac{2 i \, \log \left (d \tan \left (b x + a\right ) - i \, d + 2\right ) \log \left (\tan \left (b x + a\right ) - i\right ) - i \, \log \left (\tan \left (b x + a\right ) - i\right )^{2}}{d} + \frac{2 i \,{\left (\log \left (\frac{1}{2} \, d \tan \left (b x + a\right ) - \frac{1}{2} i \, d + 1\right ) \log \left (\tan \left (b x + a\right ) - i\right ) +{\rm Li}_2\left (-\frac{1}{2} \, d \tan \left (b x + a\right ) + \frac{1}{2} i \, d\right )\right )}}{d} - \frac{2 i \,{\left (\log \left (\tan \left (b x + a\right ) - i\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 (d \tan \left (b x + a\right ) - i \, d + 1\right )}{8 \, b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.71042, size = 603, normalized size = 6.48 \begin{align*} \frac{i \, b^{2} x^{2} + b x \log \left (\frac{{\left ({\left (d + i\right )} e^{\left (2 i \, b x + 2 i \, a\right )} + i\right )} e^{\left (-2 i \, b x - 2 i \, a\right )}}{d}\right ) - i \, a^{2} -{\left (b x + a\right )} \log \left (\frac{1}{2} \, \sqrt{4 i \, d - 4} e^{\left (i \, b x + i \, a\right )} + 1\right ) -{\left (b x + a\right )} \log \left (-\frac{1}{2} \, \sqrt{4 i \, d - 4} e^{\left (i \, b x + i \, a\right )} + 1\right ) + a \log \left (\frac{{\left (2 \, d + 2 i\right )} e^{\left (i \, b x + i \, a\right )} + i \, \sqrt{4 i \, d - 4}}{2 \, d + 2 i}\right ) + a \log \left (\frac{{\left (2 \, d + 2 i\right )} e^{\left (i \, b x + i \, a\right )} - i \, \sqrt{4 i \, d - 4}}{2 \, d + 2 i}\right ) + i \,{\rm Li}_2\left (\frac{1}{2} \, \sqrt{4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right ) + i \,{\rm Li}_2\left (-\frac{1}{2} \, \sqrt{4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right )}{2 \, b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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 \operatorname{arcoth}\left (d \tan \left (b x + a\right ) - i \, d + 1\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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