3.108 \(\int \tan ^{-1}(c-(i-c) \coth (a+b x)) \, dx\)

Optimal. Leaf size=82 \[ -\frac{i \text{PolyLog}\left (2,-i c e^{2 a+2 b x}\right )}{4 b}-\frac{1}{2} i x \log \left (1+i c e^{2 a+2 b x}\right )+x \tan ^{-1}(c-(-c+i) \coth (a+b x))+\frac{1}{2} i b x^2 \]

[Out]

(I/2)*b*x^2 + x*ArcTan[c - (I - c)*Coth[a + b*x]] - (I/2)*x*Log[1 + I*c*E^(2*a + 2*b*x)] - ((I/4)*PolyLog[2, (
-I)*c*E^(2*a + 2*b*x)])/b

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Rubi [A]  time = 0.120685, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.278, Rules used = {5189, 2184, 2190, 2279, 2391} \[ -\frac{i \text{PolyLog}\left (2,-i c e^{2 a+2 b x}\right )}{4 b}-\frac{1}{2} i x \log \left (1+i c e^{2 a+2 b x}\right )+x \tan ^{-1}(c-(-c+i) \coth (a+b x))+\frac{1}{2} i b x^2 \]

Antiderivative was successfully verified.

[In]

Int[ArcTan[c - (I - c)*Coth[a + b*x]],x]

[Out]

(I/2)*b*x^2 + x*ArcTan[c - (I - c)*Coth[a + b*x]] - (I/2)*x*Log[1 + I*c*E^(2*a + 2*b*x)] - ((I/4)*PolyLog[2, (
-I)*c*E^(2*a + 2*b*x)])/b

Rule 5189

Int[ArcTan[(c_.) + Coth[(a_.) + (b_.)*(x_)]*(d_.)], x_Symbol] :> Simp[x*ArcTan[c + d*Coth[a + b*x]], x] - Dist
[b, Int[x/(c - d - c*E^(2*a + 2*b*x)), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[(c - d)^2, -1]

Rule 2184

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[(c
+ d*x)^(m + 1)/(a*d*(m + 1)), x] - Dist[b/a, Int[((c + d*x)^m*(F^(g*(e + f*x)))^n)/(a + b*(F^(g*(e + f*x)))^n)
, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int \tan ^{-1}(c-(i-c) \coth (a+b x)) \, dx &=x \tan ^{-1}(c-(i-c) \coth (a+b x))-b \int \frac{x}{i-c e^{2 a+2 b x}} \, dx\\ &=\frac{1}{2} i b x^2+x \tan ^{-1}(c-(i-c) \coth (a+b x))+(i b c) \int \frac{e^{2 a+2 b x} x}{i-c e^{2 a+2 b x}} \, dx\\ &=\frac{1}{2} i b x^2+x \tan ^{-1}(c-(i-c) \coth (a+b x))-\frac{1}{2} i x \log \left (1+i c e^{2 a+2 b x}\right )+\frac{1}{2} i \int \log \left (1+i c e^{2 a+2 b x}\right ) \, dx\\ &=\frac{1}{2} i b x^2+x \tan ^{-1}(c-(i-c) \coth (a+b x))-\frac{1}{2} i x \log \left (1+i c e^{2 a+2 b x}\right )+\frac{i \operatorname{Subst}\left (\int \frac{\log (1+i c x)}{x} \, dx,x,e^{2 a+2 b x}\right )}{4 b}\\ &=\frac{1}{2} i b x^2+x \tan ^{-1}(c-(i-c) \coth (a+b x))-\frac{1}{2} i x \log \left (1+i c e^{2 a+2 b x}\right )-\frac{i \text{Li}_2\left (-i c e^{2 a+2 b x}\right )}{4 b}\\ \end{align*}

Mathematica [A]  time = 0.662361, size = 71, normalized size = 0.87 \[ x \tan ^{-1}(c+(c-i) \coth (a+b x))-\frac{i \left (2 b x \log \left (1-\frac{i e^{-2 (a+b x)}}{c}\right )-\text{PolyLog}\left (2,\frac{i e^{-2 (a+b x)}}{c}\right )\right )}{4 b} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTan[c - (I - c)*Coth[a + b*x]],x]

[Out]

x*ArcTan[c + (-I + c)*Coth[a + b*x]] - ((I/4)*(2*b*x*Log[1 - I/(c*E^(2*(a + b*x)))] - PolyLog[2, I/(c*E^(2*(a
+ b*x)))]))/b

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Maple [B]  time = 0.144, size = 1351, normalized size = 16.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctan(c-(I-c)*coth(b*x+a)),x)

[Out]

1/4*I/b/(c-I)/(I-c)*dilog(1/2*((c-I)*coth(b*x+a)+c+I)/c)-1/4*I/b/(c-I)/(I-c)*dilog(((c-I)*coth(b*x+a)+c-I)/(-2
*I+2*c))-1/4*I/b/(c-I)/(I-c)*dilog(-1/2*I*((c-I)*coth(b*x+a)+c+I))+1/8*I/b/(c-I)/(I-c)*ln((c-I)*coth(b*x+a)+c-
I)^2+1/b/(c-I)*arctan((c-I)*coth(b*x+a)+c)/(2*I-2*c)*ln((c-I)*coth(b*x+a)+c-I)+1/2/b/(c-I)/(I-c)*dilog(-1/2*I*
((c-I)*coth(b*x+a)+c+I))*c-1/4/b/(c-I)/(I-c)*ln((c-I)*coth(b*x+a)+c-I)^2*c-1/2/b/(c-I)/(I-c)*dilog(1/2*((c-I)*
coth(b*x+a)+c+I)/c)*c+1/2/b/(c-I)/(I-c)*dilog(((c-I)*coth(b*x+a)+c-I)/(-2*I+2*c))*c-1/b/(c-I)*arctan((c-I)*cot
h(b*x+a)+c)/(2*I-2*c)*ln((c-I)*coth(b*x+a)-c+I)+1/4*I/b/(c-I)/(I-c)*ln((c-I)*coth(b*x+a)-c+I)*ln(1/2*((c-I)*co
th(b*x+a)+c+I)/c)+1/4*I/b/(c-I)/(I-c)*dilog(((c-I)*coth(b*x+a)+c-I)/(-2*I+2*c))*c^2+1/b/(c-I)*arctan((c-I)*cot
h(b*x+a)+c)/(2*I-2*c)*ln((c-I)*coth(b*x+a)-c+I)*c^2-1/2/b/(c-I)/(I-c)*ln((c-I)*coth(b*x+a)-c+I)*ln(1/2*((c-I)*
coth(b*x+a)+c+I)/c)*c+1/2/b/(c-I)/(I-c)*ln((c-I)*coth(b*x+a)-c+I)*ln(((c-I)*coth(b*x+a)+c-I)/(-2*I+2*c))*c+1/2
/b/(c-I)/(I-c)*ln(-1/2*I*((c-I)*coth(b*x+a)+c+I))*ln((c-I)*coth(b*x+a)+c-I)*c-1/4*I/b/(c-I)/(I-c)*ln((c-I)*cot
h(b*x+a)-c+I)*ln(((c-I)*coth(b*x+a)+c-I)/(-2*I+2*c))-1/b/(c-I)*arctan((c-I)*coth(b*x+a)+c)/(2*I-2*c)*ln((c-I)*
coth(b*x+a)+c-I)*c^2-1/4*I/b/(c-I)/(I-c)*ln(-1/2*I*((c-I)*coth(b*x+a)+c+I))*ln((c-I)*coth(b*x+a)+c-I)+1/4*I/b/
(c-I)/(I-c)*dilog(-1/2*I*((c-I)*coth(b*x+a)+c+I))*c^2-1/8*I/b/(c-I)/(I-c)*ln((c-I)*coth(b*x+a)+c-I)^2*c^2-1/4*
I/b/(c-I)/(I-c)*dilog(1/2*((c-I)*coth(b*x+a)+c+I)/c)*c^2+1/4*I/b/(c-I)/(I-c)*ln((c-I)*coth(b*x+a)-c+I)*ln(((c-
I)*coth(b*x+a)+c-I)/(-2*I+2*c))*c^2+2*I/b/(c-I)*arctan((c-I)*coth(b*x+a)+c)/(2*I-2*c)*ln((c-I)*coth(b*x+a)+c-I
)*c-2*I/b/(c-I)*arctan((c-I)*coth(b*x+a)+c)/(2*I-2*c)*ln((c-I)*coth(b*x+a)-c+I)*c+1/4*I/b/(c-I)/(I-c)*ln(-1/2*
I*((c-I)*coth(b*x+a)+c+I))*ln((c-I)*coth(b*x+a)+c-I)*c^2-1/4*I/b/(c-I)/(I-c)*ln((c-I)*coth(b*x+a)-c+I)*ln(1/2*
((c-I)*coth(b*x+a)+c+I)/c)*c^2

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Maxima [A]  time = 5.82855, size = 108, normalized size = 1.32 \begin{align*} -2 \, b{\left (c - i\right )}{\left (\frac{2 \, x^{2}}{2 i \, c + 2} - \frac{2 \, b x \log \left (i \, c e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) +{\rm Li}_2\left (-i \, c e^{\left (2 \, b x + 2 \, a\right )}\right )}{-2 \, b^{2}{\left (-i \, c - 1\right )}}\right )} + x \arctan \left ({\left (c - i\right )} \coth \left (b x + a\right ) + c\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(c-(I-c)*coth(b*x+a)),x, algorithm="maxima")

[Out]

-2*b*(c - I)*(2*x^2/(2*I*c + 2) - (2*b*x*log(I*c*e^(2*b*x + 2*a) + 1) + dilog(-I*c*e^(2*b*x + 2*a)))/(b^2*(2*I
*c + 2))) + x*arctan((c - I)*coth(b*x + a) + c)

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Fricas [B]  time = 2.2388, size = 521, normalized size = 6.35 \begin{align*} \frac{i \, b^{2} x^{2} + i \, b x \log \left (-\frac{{\left (c e^{\left (2 \, b x + 2 \, a\right )} - i\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{c - i}\right ) - i \, a^{2} +{\left (-i \, b x - i \, a\right )} \log \left (\frac{1}{2} \, \sqrt{-4 i \, c} e^{\left (b x + a\right )} + 1\right ) +{\left (-i \, b x - i \, a\right )} \log \left (-\frac{1}{2} \, \sqrt{-4 i \, c} e^{\left (b x + a\right )} + 1\right ) + i \, a \log \left (\frac{2 \, c e^{\left (b x + a\right )} + i \, \sqrt{-4 i \, c}}{2 \, c}\right ) + i \, a \log \left (\frac{2 \, c e^{\left (b x + a\right )} - i \, \sqrt{-4 i \, c}}{2 \, c}\right ) - i \,{\rm Li}_2\left (\frac{1}{2} \, \sqrt{-4 i \, c} e^{\left (b x + a\right )}\right ) - i \,{\rm Li}_2\left (-\frac{1}{2} \, \sqrt{-4 i \, c} e^{\left (b x + a\right )}\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(c-(I-c)*coth(b*x+a)),x, algorithm="fricas")

[Out]

1/2*(I*b^2*x^2 + I*b*x*log(-(c*e^(2*b*x + 2*a) - I)*e^(-2*b*x - 2*a)/(c - I)) - I*a^2 + (-I*b*x - I*a)*log(1/2
*sqrt(-4*I*c)*e^(b*x + a) + 1) + (-I*b*x - I*a)*log(-1/2*sqrt(-4*I*c)*e^(b*x + a) + 1) + I*a*log(1/2*(2*c*e^(b
*x + a) + I*sqrt(-4*I*c))/c) + I*a*log(1/2*(2*c*e^(b*x + a) - I*sqrt(-4*I*c))/c) - I*dilog(1/2*sqrt(-4*I*c)*e^
(b*x + a)) - I*dilog(-1/2*sqrt(-4*I*c)*e^(b*x + a)))/b

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} b \int \frac{x}{c e^{2 a} e^{2 b x} - i}\, dx + \frac{i x \log{\left (- i c - \frac{i c}{e^{2 a} e^{2 b x} - 1} - \frac{i c e^{a} e^{b x}}{e^{a} e^{b x} - e^{- a} e^{- b x}} + 1 - \frac{1}{e^{2 a} e^{2 b x} - 1} - \frac{e^{a} e^{b x}}{e^{a} e^{b x} - e^{- a} e^{- b x}} \right )}}{2} - \frac{i x \log{\left (i c + \frac{i c}{e^{2 a} e^{2 b x} - 1} + \frac{i c e^{a} e^{b x}}{e^{a} e^{b x} - e^{- a} e^{- b x}} + 1 + \frac{1}{e^{2 a} e^{2 b x} - 1} + \frac{e^{a} e^{b x}}{e^{a} e^{b x} - e^{- a} e^{- b x}} \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atan(c-(I-c)*coth(b*x+a)),x)

[Out]

b*Integral(x/(c*exp(2*a)*exp(2*b*x) - I), x) + I*x*log(-I*c - I*c/(exp(2*a)*exp(2*b*x) - 1) - I*c*exp(a)*exp(b
*x)/(exp(a)*exp(b*x) - exp(-a)*exp(-b*x)) + 1 - 1/(exp(2*a)*exp(2*b*x) - 1) - exp(a)*exp(b*x)/(exp(a)*exp(b*x)
- exp(-a)*exp(-b*x)))/2 - I*x*log(I*c + I*c/(exp(2*a)*exp(2*b*x) - 1) + I*c*exp(a)*exp(b*x)/(exp(a)*exp(b*x)
- exp(-a)*exp(-b*x)) + 1 + 1/(exp(2*a)*exp(2*b*x) - 1) + exp(a)*exp(b*x)/(exp(a)*exp(b*x) - exp(-a)*exp(-b*x))
)/2

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \arctan \left ({\left (c - i\right )} \coth \left (b x + a\right ) + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(c-(I-c)*coth(b*x+a)),x, algorithm="giac")

[Out]

integrate(arctan((c - I)*coth(b*x + a) + c), x)