Optimal. Leaf size=94 \[ \frac {i \text {Li}_2\left (-\left ((i d+1) e^{2 i a+2 i b x}\right )\right )}{4 b}-\frac {1}{2} x \log \left (1+(1+i d) e^{2 i a+2 i b x}\right )+x \coth ^{-1}(d (-\tan (a+b x))+i d+1)+\frac {1}{2} i b x^2 \]
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Rubi [A] time = 0.16, 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 = {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 2184
Rule 2190
Rule 2279
Rule 2391
Rule 6256
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*}
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Mathematica [B] time = 3.24, size = 723, normalized size = 7.69 \[ x \coth ^{-1}(d (-\tan (a+b x))+i d+1)-\frac {x \sec (a+b x) (\cos (b x)+i \sin (b x)) (\sin (b x)+i \cos (b x)) \left (-\text {Li}_2\left (\frac {1}{2} (\cos (a)+i \sin (a)) (d \cos (a)+(i d+2) \sin (a)) (\tan (b x)-i)\right )+\text {Li}_2\left (\frac {\sec (b x) (d \cos (a)+(i d+2) \sin (a)) (\cos (a+b x)-i \sin (a+b x))}{2 (d-i)}\right )+\log (1-i \tan (b x)) \log \left (\frac {(\cos (a)-i \sin (a)) \sec (b x) (i d \sin (a+b x)+(d-2 i) \cos (a+b x))}{2 (d-i)}\right )-\log (1+i \tan (b x)) \log \left (\frac {\sec (b x) (-d \sin (a+b x)+(2+i d) \cos (a+b x))}{2 \cos (a)-2 i \sin (a)}\right )+\text {Li}_2(i \sin (2 b x)-\cos (2 b x))-2 i b x \log (2 \cos (b x) (\cos (b x)-i \sin (b x)))\right )}{(\tan (a+b x)-i) (i d \sin (a+b x)+(d-2 i) \cos (a+b x)) \left (-\frac {\sec ^2(b x) \log \left (\frac {\sec (b x) (-d \sin (a+b x)+(2+i d) \cos (a+b x))}{2 \cos (a)-2 i \sin (a)}\right )}{\tan (b x)-i}+\frac {\sec ^2(b x) \log \left (1-\frac {1}{2} (\cos (a)+i \sin (a)) (\tan (b x)-i) (d \cos (a)+(2+i d) \sin (a))\right )}{\tan (b x)-i}+\frac {\sec ^2(b x) \log \left (\frac {(\cos (a)-i \sin (a)) \sec (b x) (i d \sin (a+b x)+(d-2 i) \cos (a+b x))}{2 (d-i)}\right )}{\tan (b x)+i}+\frac {i \sec (b x) (d \cos (a)+(2+i d) \sin (a)) \log (1-i \tan (b x))}{i d \sin (a+b x)+(d-2 i) \cos (a+b x)}+\frac {\sec (b x) ((d-2 i) \sin (a)-i d \cos (a)) \log (1+i \tan (b x))}{i d \sin (a+b x)+(d-2 i) \cos (a+b x)}-(\tan (b x)-i) \log \left (1-\frac {\sec (b x) (d \cos (a)+(2+i d) \sin (a)) (\cos (a+b x)-i \sin (a+b x))}{2 (d-i)}\right )+2 i b x (\tan (b x)+i)\right )} \]
Warning: Unable to verify antiderivative.
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fricas [B] time = 0.48, size = 222, normalized size = 2.36 \[ \frac {i \, b^{2} x^{2} - b x \log \left (\frac {d e^{\left (2 i \, b x + 2 i \, a\right )}}{{\left (d - i\right )} e^{\left (2 i \, b x + 2 i \, a\right )} - i}\right ) - i \, a^{2} - {\left (b x + a\right )} \log \left (\frac {1}{2} \, \sqrt {-4 i \, d - 4} e^{\left (i \, b x + i \, a\right )} + 1\right ) - {\left (b x + a\right )} \log \left (-\frac {1}{2} \, \sqrt {-4 i \, d - 4} e^{\left (i \, b x + i \, a\right )} + 1\right ) + a \log \left (\frac {{\left (2 \, d - 2 i\right )} e^{\left (i \, b x + i \, a\right )} + i \, \sqrt {-4 i \, d - 4}}{2 \, d - 2 i}\right ) + a \log \left (\frac {{\left (2 \, d - 2 i\right )} e^{\left (i \, b x + i \, a\right )} - i \, \sqrt {-4 i \, d - 4}}{2 \, d - 2 i}\right ) + i \, {\rm Li}_2\left (\frac {1}{2} \, \sqrt {-4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right ) + i \, {\rm Li}_2\left (-\frac {1}{2} \, \sqrt {-4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right )}{2 \, b} \]
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 \[ \int \operatorname {arcoth}\left (-d \tan \left (b x + a\right ) + i \, d + 1\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.51, size = 297, normalized size = 3.16 \[ \frac {i \mathrm {arccoth}\left (1+i d -d \tan \left (b x +a \right )\right ) \ln \left (i d +d \tan \left (b x +a \right )\right )}{2 b}-\frac {i \mathrm {arccoth}\left (1+i d -d \tan \left (b x +a \right )\right ) \ln \left (i d -d \tan \left (b x +a \right )\right )}{2 b}-\frac {i \ln \left (i d -d \tan \left (b x +a \right )\right )^{2}}{8 b}+\frac {i \dilog \left (1+\frac {i d}{2}-\frac {d \tan \left (b x +a \right )}{2}\right )}{4 b}+\frac {i \ln \left (i d -d \tan \left (b x +a \right )\right ) \ln \left (1+\frac {i d}{2}-\frac {d \tan \left (b x +a \right )}{2}\right )}{4 b}-\frac {i \dilog \left (\frac {-2-i d +d \tan \left (b x +a \right )}{-2 i d -2}\right )}{4 b}-\frac {i \ln \left (i d +d \tan \left (b x +a \right )\right ) \ln \left (\frac {-2-i d +d \tan \left (b x +a \right )}{-2 i d -2}\right )}{4 b}+\frac {i \dilog \left (\frac {i \left (-i d +d \tan \left (b x +a \right )\right )}{2 d}\right )}{4 b}+\frac {i \ln \left (i d +d \tan \left (b x +a \right )\right ) \ln \left (\frac {i \left (-i d +d \tan \left (b x +a \right )\right )}{2 d}\right )}{4 b} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.42, size = 263, normalized size = 2.80 \[ -\frac {4 \, {\left (b x + a\right )} d {\left (\frac {\log \left (d \tan \left (b x + a\right ) - i \, d - 2\right )}{d} - \frac {\log \left (\tan \left (b x + a\right ) - i\right )}{d}\right )} + d {\left (-\frac {2 i \, {\left (\log \left (d \tan \left (b x + a\right ) - i \, d - 2\right ) \log \left (-\frac {i \, d \tan \left (b x + a\right ) + d - 2 i}{2 \, d - 2 i} + 1\right ) + {\rm Li}_2\left (\frac {i \, d \tan \left (b x + a\right ) + d - 2 i}{2 \, d - 2 i}\right )\right )}}{d} + \frac {2 i \, \log \left (d \tan \left (b x + a\right ) - i \, d - 2\right ) \log \left (\tan \left (b x + a\right ) - i\right ) - i \, \log \left (\tan \left (b x + a\right ) - i\right )^{2}}{d} - \frac {2 i \, {\left (\log \left (-\frac {1}{2} \, d \tan \left (b x + a\right ) + \frac {1}{2} i \, d + 1\right ) \log \left (\tan \left (b x + a\right ) - i\right ) + {\rm Li}_2\left (\frac {1}{2} \, d \tan \left (b x + a\right ) - \frac {1}{2} i \, d\right )\right )}}{d} + \frac {2 i \, {\left (\log \left (\tan \left (b x + a\right ) - i\right ) \log \left (-\frac {1}{2} i \, \tan \left (b x + a\right ) + \frac {1}{2}\right ) + {\rm Li}_2\left (\frac {1}{2} i \, \tan \left (b x + a\right ) + \frac {1}{2}\right )\right )}}{d}\right )} + 8 \, {\left (b x + a\right )} \operatorname {arcoth}\left (d \tan \left (b x + a\right ) - i \, d - 1\right )}{8 \, b} \]
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 \[ \int \mathrm {acoth}\left (1-d\,\mathrm {tan}\left (a+b\,x\right )+d\,1{}\mathrm {i}\right ) \,d x \]
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 \[ \int \operatorname {acoth}{\left (- d \tan {\left (a + b x \right )} + i d + 1 \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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