∫ tan−1(c + (i + c) coth(a + bx))dx

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

Optimal. Leaf size=79 \[ \frac{i \text{PolyLog}\left (2,i c e^{2 a+2 b x}\right )}{4 b}+\frac{1}{2} i x \log \left (1-i c e^{2 a+2 b x}\right )+x \tan ^{-1}(c+(c+i) \coth (a+b x))-\frac{1}{2} i b x^2 \]

[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 = 0.122423, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, 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 = 0.660659, size = 71, normalized size = 0.9 \[ \frac{i \left (2 b x \log \left (1+\frac{i e^{-2 (a+b x)}}{c}\right )-\text{PolyLog}\left (2,-\frac{i e^{-2 (a+b x)}}{c}\right )\right )}{4 b}+x \tan ^{-1}(c+(c+i) \coth (a+b x)) \]

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 = 0.152, size = 1381, normalized size = 17.5 \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/(I+c)/b*arctan(c+(I+c)*coth(b*x+a))/(2*I+2*c)*ln(c+(I+c)*coth(b*x+a)+I)*c^2-1/(I+c)/b*arctan(c+(I+c)*coth(b*

x+a))/(2*I+2*c)*ln(c-(I+c)*coth(b*x+a)+I)*c^2+1/4*I/(I+c)^2/b*ln(-1/2*I*(-c-(I+c)*coth(b*x+a)+I))*ln(c+(I+c)*c

oth(b*x+a)+I)*c^2-1/4*I/(I+c)^2/b*ln(c-(I+c)*coth(b*x+a)+I)*ln(-1/2*(-c-(I+c)*coth(b*x+a)+I)/c)*c^2+1/4*I/(I+c

)^2/b*ln(c-(I+c)*coth(b*x+a)+I)*ln((-c-(I+c)*coth(b*x+a)-I)/(-2*I-2*c))*c^2-1/4*I/(I+c)^2/b*ln(-1/2*I*(c+(I+c)

*coth(b*x+a)+I))*ln(-1/2*I*(-c-(I+c)*coth(b*x+a)+I))*c^2+1/2/(I+c)^2/b*ln(c-(I+c)*coth(b*x+a)+I)*ln(-1/2*(-c-(

I+c)*coth(b*x+a)+I)/c)*c-1/4*I/(I+c)^2/b*dilog((-c-(I+c)*coth(b*x+a)-I)/(-2*I-2*c))+1/8*I/(I+c)^2/b*ln(c+(I+c)

*coth(b*x+a)+I)^2+1/4*I/(I+c)^2/b*dilog(-1/2*I*(c+(I+c)*coth(b*x+a)+I))+1/4/(I+c)^2/b*ln(c+(I+c)*coth(b*x+a)+I

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

ln(c+(I+c)*coth(b*x+a)+I)*c-2*I/(I+c)/b*arctan(c+(I+c)*coth(b*x+a))/(2*I+2*c)*ln(c-(I+c)*coth(b*x+a)+I)*c-1/2/

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

x+a))/(2*I+2*c)*ln(c-(I+c)*coth(b*x+a)+I)-1/4*I/(I+c)^2/b*ln(-1/2*I*(-c-(I+c)*coth(b*x+a)+I))*ln(c+(I+c)*coth(

b*x+a)+I)+1/4*I/(I+c)^2/b*ln(c-(I+c)*coth(b*x+a)+I)*ln(-1/2*(-c-(I+c)*coth(b*x+a)+I)/c)-1/4*I/(I+c)^2/b*ln(c-(

I+c)*coth(b*x+a)+I)*ln((-c-(I+c)*coth(b*x+a)-I)/(-2*I-2*c))-1/8*I/(I+c)^2/b*ln(c+(I+c)*coth(b*x+a)+I)^2*c^2-1/

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

*c^2+1/4*I/(I+c)^2/b*dilog((-c-(I+c)*coth(b*x+a)-I)/(-2*I-2*c))*c^2+1/4*I/(I+c)^2/b*ln(-1/2*I*(c+(I+c)*coth(b*

x+a)+I))*ln(-1/2*I*(-c-(I+c)*coth(b*x+a)+I))-1/(I+c)/b*arctan(c+(I+c)*coth(b*x+a))/(2*I+2*c)*ln(c+(I+c)*coth(b

*x+a)+I)+1/2/(I+c)^2/b*ln(-1/2*I*(c+(I+c)*coth(b*x+a)+I))*ln(-1/2*I*(-c-(I+c)*coth(b*x+a)+I))*c-1/2/(I+c)^2/b*

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

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

b*x+a)+I)/c)

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Maxima [A] time = 5.92393, size = 108, normalized size = 1.37 \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 = 2.24685, size = 510, normalized size = 6.46 \begin{align*} \frac{-i \, b^{2} x^{2} + i \, b x \log \left (-\frac{{\left (c + i\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c e^{\left (2 \, b x + 2 \, a\right )} + 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 + I)*e^(2*b*x + 2*a)/(c*e^(2*b*x + 2*a) + 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)

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