\(\int \coth ^{-1}(c+d \tan (a+b x)) \, dx\) [238]

Optimal result

Integrand size = 11, antiderivative size = 194 \[ \int \coth ^{-1}(c+d \tan (a+b x)) \, dx=x \coth ^{-1}(c+d \tan (a+b x))+\frac {1}{2} x \log \left (1+\frac {(1-c+i d) e^{2 i a+2 i b x}}{1-c-i d}\right )-\frac {1}{2} x \log \left (1+\frac {(1+c-i d) e^{2 i a+2 i b x}}{1+c+i d}\right )-\frac {i \operatorname {PolyLog}\left (2,-\frac {(1-c+i d) e^{2 i a+2 i b x}}{1-c-i d}\right )}{4 b}+\frac {i \operatorname {PolyLog}\left (2,-\frac {(1+c-i d) e^{2 i a+2 i b x}}{1+c+i d}\right )}{4 b} \]

[Out]

x*arccoth(c+d*tan(b*x+a))+1/2*x*ln(1+(1-c+I*d)*exp(2*I*a+2*I*b*x)/(1-c-I*d))-1/2*x*ln(1+(1+c-I*d)*exp(2*I*a+2*
I*b*x)/(1+c+I*d))-1/4*I*polylog(2,-(1-c+I*d)*exp(2*I*a+2*I*b*x)/(1-c-I*d))/b+1/4*I*polylog(2,-(1+c-I*d)*exp(2*
I*a+2*I*b*x)/(1+c+I*d))/b

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {6395, 2221, 2317, 2438} \[ \int \coth ^{-1}(c+d \tan (a+b x)) \, dx=-\frac {i \operatorname {PolyLog}\left (2,-\frac {(-c+i d+1) e^{2 i a+2 i b x}}{-c-i d+1}\right )}{4 b}+\frac {i \operatorname {PolyLog}\left (2,-\frac {(c-i d+1) e^{2 i a+2 i b x}}{c+i d+1}\right )}{4 b}+\frac {1}{2} x \log \left (1+\frac {(-c+i d+1) e^{2 i a+2 i b x}}{-c-i d+1}\right )-\frac {1}{2} x \log \left (1+\frac {(c-i d+1) e^{2 i a+2 i b x}}{c+i d+1}\right )+x \coth ^{-1}(d \tan (a+b x)+c) \]

[In]

Int[ArcCoth[c + d*Tan[a + b*x]],x]

[Out]

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

Rule 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2317

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 2438

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]

Rule 6395

Int[ArcCoth[(c_.) + (d_.)*Tan[(a_.) + (b_.)*(x_)]], x_Symbol] :> Simp[x*ArcCoth[c + d*Tan[a + b*x]], x] + (-Di
st[I*b*(1 + c - I*d), Int[x*(E^(2*I*a + 2*I*b*x)/(1 + c + I*d + (1 + c - I*d)*E^(2*I*a + 2*I*b*x))), x], x] +
Dist[I*b*(1 - c + I*d), Int[x*(E^(2*I*a + 2*I*b*x)/(1 - c - I*d + (1 - c + I*d)*E^(2*I*a + 2*I*b*x))), x], x])
/; FreeQ[{a, b, c, d}, x] && NeQ[(c + I*d)^2, 1]

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

Mathematica [A] (warning: unable to verify)

Time = 0.35 (sec) , antiderivative size = 365, normalized size of antiderivative = 1.88 \[ \int \coth ^{-1}(c+d \tan (a+b x)) \, dx=x \left (\coth ^{-1}(c+d \tan (a+b x))+\frac {2 a \log (1-c-d \tan (a+b x))-i \log (1-i \tan (a+b x)) \log \left (\frac {-1+c+d \tan (a+b x)}{-1+c-i d}\right )+i \log (1+i \tan (a+b x)) \log \left (\frac {-1+c+d \tan (a+b x)}{-1+c+i d}\right )-2 a \log (1+c+d \tan (a+b x))+i \log (1-i \tan (a+b x)) \log \left (\frac {1+c+d \tan (a+b x)}{1+c-i d}\right )-i \log (1+i \tan (a+b x)) \log \left (\frac {1+c+d \tan (a+b x)}{1+c+i d}\right )+i \operatorname {PolyLog}\left (2,-\frac {d (-i+\tan (a+b x))}{-1+c+i d}\right )-i \operatorname {PolyLog}\left (2,-\frac {d (-i+\tan (a+b x))}{1+c+i d}\right )-i \operatorname {PolyLog}\left (2,-\frac {d (i+\tan (a+b x))}{-1+c-i d}\right )+i \operatorname {PolyLog}\left (2,-\frac {d (i+\tan (a+b x))}{1+c-i d}\right )}{4 a-2 i \log (1-i \tan (a+b x))+2 i \log (1+i \tan (a+b x))}\right ) \]

[In]

Integrate[ArcCoth[c + d*Tan[a + b*x]],x]

[Out]

x*(ArcCoth[c + d*Tan[a + b*x]] + (2*a*Log[1 - c - d*Tan[a + b*x]] - I*Log[1 - I*Tan[a + b*x]]*Log[(-1 + c + d*
Tan[a + b*x])/(-1 + c - I*d)] + I*Log[1 + I*Tan[a + b*x]]*Log[(-1 + c + d*Tan[a + b*x])/(-1 + c + I*d)] - 2*a*
Log[1 + c + d*Tan[a + b*x]] + I*Log[1 - I*Tan[a + b*x]]*Log[(1 + c + d*Tan[a + b*x])/(1 + c - I*d)] - I*Log[1
+ I*Tan[a + b*x]]*Log[(1 + c + d*Tan[a + b*x])/(1 + c + I*d)] + I*PolyLog[2, -((d*(-I + Tan[a + b*x]))/(-1 + c
+ I*d))] - I*PolyLog[2, -((d*(-I + Tan[a + b*x]))/(1 + c + I*d))] - I*PolyLog[2, -((d*(I + Tan[a + b*x]))/(-1
+ c - I*d))] + I*PolyLog[2, -((d*(I + Tan[a + b*x]))/(1 + c - I*d))])/(4*a - (2*I)*Log[1 - I*Tan[a + b*x]] +
(2*I)*Log[1 + I*Tan[a + b*x]]))

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 555 vs. \(2 (164 ) = 328\).

Time = 2.64 (sec) , antiderivative size = 556, normalized size of antiderivative = 2.87

method	result	size
derivativedivides	\(\frac {d \arctan \left (\tan \left (b x +a \right )\right ) \operatorname {arccoth}\left (c +d \tan \left (b x +a \right )\right )+d^{2} \left (-\frac {\arctan \left (-\frac {c +d \tan \left (b x +a \right )}{d}+\frac {c}{d}\right ) \ln \left (d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )+c -1\right )}{2 d}+\frac {\arctan \left (-\frac {c +d \tan \left (b x +a \right )}{d}+\frac {c}{d}\right ) \ln \left (d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )+c +1\right )}{2 d}+\frac {i \ln \left (d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )+c -1\right ) \left (\ln \left (\frac {i d -d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )}{i d +c -1}\right )-\ln \left (\frac {i d +d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )}{i d -c +1}\right )\right )}{4 d}+\frac {i \left (\operatorname {dilog}\left (\frac {i d -d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )}{i d +c -1}\right )-\operatorname {dilog}\left (\frac {i d +d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )}{i d -c +1}\right )\right )}{4 d}-\frac {i \ln \left (d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )+c +1\right ) \left (\ln \left (\frac {i d -d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )}{i d +c +1}\right )-\ln \left (\frac {i d +d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )}{i d -c -1}\right )\right )}{4 d}-\frac {i \left (\operatorname {dilog}\left (\frac {i d -d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )}{i d +c +1}\right )-\operatorname {dilog}\left (\frac {i d +d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )}{i d -c -1}\right )\right )}{4 d}\right )}{b d}\)	\(556\)
default	\(\frac {d \arctan \left (\tan \left (b x +a \right )\right ) \operatorname {arccoth}\left (c +d \tan \left (b x +a \right )\right )+d^{2} \left (-\frac {\arctan \left (-\frac {c +d \tan \left (b x +a \right )}{d}+\frac {c}{d}\right ) \ln \left (d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )+c -1\right )}{2 d}+\frac {\arctan \left (-\frac {c +d \tan \left (b x +a \right )}{d}+\frac {c}{d}\right ) \ln \left (d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )+c +1\right )}{2 d}+\frac {i \ln \left (d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )+c -1\right ) \left (\ln \left (\frac {i d -d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )}{i d +c -1}\right )-\ln \left (\frac {i d +d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )}{i d -c +1}\right )\right )}{4 d}+\frac {i \left (\operatorname {dilog}\left (\frac {i d -d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )}{i d +c -1}\right )-\operatorname {dilog}\left (\frac {i d +d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )}{i d -c +1}\right )\right )}{4 d}-\frac {i \ln \left (d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )+c +1\right ) \left (\ln \left (\frac {i d -d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )}{i d +c +1}\right )-\ln \left (\frac {i d +d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )}{i d -c -1}\right )\right )}{4 d}-\frac {i \left (\operatorname {dilog}\left (\frac {i d -d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )}{i d +c +1}\right )-\operatorname {dilog}\left (\frac {i d +d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )}{i d -c -1}\right )\right )}{4 d}\right )}{b d}\)	\(556\)
risch	\(\text {Expression too large to display}\)	\(3962\)

[In]

int(arccoth(c+d*tan(b*x+a)),x,method=_RETURNVERBOSE)

[Out]

1/b/d*(d*arctan(tan(b*x+a))*arccoth(c+d*tan(b*x+a))+d^2*(-1/2*arctan(-(c+d*tan(b*x+a))/d+c/d)/d*ln(d*((c+d*tan
(b*x+a))/d-c/d)+c-1)+1/2*arctan(-(c+d*tan(b*x+a))/d+c/d)/d*ln(d*((c+d*tan(b*x+a))/d-c/d)+c+1)+1/4*I*ln(d*((c+d
*tan(b*x+a))/d-c/d)+c-1)*(ln((I*d-d*((c+d*tan(b*x+a))/d-c/d))/(I*d+c-1))-ln((I*d+d*((c+d*tan(b*x+a))/d-c/d))/(
1-c+I*d)))/d+1/4*I*(dilog((I*d-d*((c+d*tan(b*x+a))/d-c/d))/(I*d+c-1))-dilog((I*d+d*((c+d*tan(b*x+a))/d-c/d))/(
1-c+I*d)))/d-1/4*I*ln(d*((c+d*tan(b*x+a))/d-c/d)+c+1)*(ln((I*d-d*((c+d*tan(b*x+a))/d-c/d))/(1+c+I*d))-ln((I*d+
d*((c+d*tan(b*x+a))/d-c/d))/(I*d-c-1)))/d-1/4*I*(dilog((I*d-d*((c+d*tan(b*x+a))/d-c/d))/(1+c+I*d))-dilog((I*d+
d*((c+d*tan(b*x+a))/d-c/d))/(I*d-c-1)))/d))

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1184 vs. \(2 (136) = 272\).

Time = 0.32 (sec) , antiderivative size = 1184, normalized size of antiderivative = 6.10 \[ \int \coth ^{-1}(c+d \tan (a+b x)) \, dx=\text {Too large to display} \]

[In]

integrate(arccoth(c+d*tan(b*x+a)),x, algorithm="fricas")

[Out]

1/8*(4*b*x*log((d*tan(b*x + a) + c + 1)/(d*tan(b*x + a) + c - 1)) - 2*(b*x + a)*log(-2*((I*(c + 1)*d - d^2)*ta
n(b*x + a)^2 - c^2 - I*(c + 1)*d + (I*c^2 - 2*(c + 1)*d - I*d^2 + 2*I*c + I)*tan(b*x + a) - 2*c - 1)/((c^2 + d
^2 + 2*c + 1)*tan(b*x + a)^2 + c^2 + d^2 + 2*c + 1)) - 2*(b*x + a)*log(-2*((-I*(c + 1)*d - d^2)*tan(b*x + a)^2
- c^2 + I*(c + 1)*d + (-I*c^2 - 2*(c + 1)*d + I*d^2 - 2*I*c - I)*tan(b*x + a) - 2*c - 1)/((c^2 + d^2 + 2*c +
1)*tan(b*x + a)^2 + c^2 + d^2 + 2*c + 1)) + 2*(b*x + a)*log(-2*((I*(c - 1)*d - d^2)*tan(b*x + a)^2 - c^2 - I*(
c - 1)*d + (I*c^2 - 2*(c - 1)*d - I*d^2 - 2*I*c + I)*tan(b*x + a) + 2*c - 1)/((c^2 + d^2 - 2*c + 1)*tan(b*x +
a)^2 + c^2 + d^2 - 2*c + 1)) + 2*(b*x + a)*log(-2*((-I*(c - 1)*d - d^2)*tan(b*x + a)^2 - c^2 + I*(c - 1)*d + (
-I*c^2 - 2*(c - 1)*d + I*d^2 + 2*I*c - I)*tan(b*x + a) + 2*c - 1)/((c^2 + d^2 - 2*c + 1)*tan(b*x + a)^2 + c^2
+ d^2 - 2*c + 1)) + 2*a*log(((I*(c + 1)*d + d^2)*tan(b*x + a)^2 - c^2 + I*(c + 1)*d + (I*c^2 + I*d^2 + 2*I*c +
I)*tan(b*x + a) - 2*c - 1)/(tan(b*x + a)^2 + 1)) + 2*a*log(((I*(c + 1)*d - d^2)*tan(b*x + a)^2 + c^2 + I*(c +
1)*d + (I*c^2 + I*d^2 + 2*I*c + I)*tan(b*x + a) + 2*c + 1)/(tan(b*x + a)^2 + 1)) - 2*a*log(((I*(c - 1)*d + d^
2)*tan(b*x + a)^2 - c^2 + I*(c - 1)*d + (I*c^2 + I*d^2 - 2*I*c + I)*tan(b*x + a) + 2*c - 1)/(tan(b*x + a)^2 +
1)) - 2*a*log(((I*(c - 1)*d - d^2)*tan(b*x + a)^2 + c^2 + I*(c - 1)*d + (I*c^2 + I*d^2 - 2*I*c + I)*tan(b*x +
a) - 2*c + 1)/(tan(b*x + a)^2 + 1)) - I*dilog(2*((I*(c + 1)*d - d^2)*tan(b*x + a)^2 - c^2 - I*(c + 1)*d + (I*c
^2 - 2*(c + 1)*d - I*d^2 + 2*I*c + I)*tan(b*x + a) - 2*c - 1)/((c^2 + d^2 + 2*c + 1)*tan(b*x + a)^2 + c^2 + d^
2 + 2*c + 1) + 1) + I*dilog(2*((-I*(c + 1)*d - d^2)*tan(b*x + a)^2 - c^2 + I*(c + 1)*d + (-I*c^2 - 2*(c + 1)*d
+ I*d^2 - 2*I*c - I)*tan(b*x + a) - 2*c - 1)/((c^2 + d^2 + 2*c + 1)*tan(b*x + a)^2 + c^2 + d^2 + 2*c + 1) + 1
) + I*dilog(2*((I*(c - 1)*d - d^2)*tan(b*x + a)^2 - c^2 - I*(c - 1)*d + (I*c^2 - 2*(c - 1)*d - I*d^2 - 2*I*c +
I)*tan(b*x + a) + 2*c - 1)/((c^2 + d^2 - 2*c + 1)*tan(b*x + a)^2 + c^2 + d^2 - 2*c + 1) + 1) - I*dilog(2*((-I
*(c - 1)*d - d^2)*tan(b*x + a)^2 - c^2 + I*(c - 1)*d + (-I*c^2 - 2*(c - 1)*d + I*d^2 + 2*I*c - I)*tan(b*x + a)
+ 2*c - 1)/((c^2 + d^2 - 2*c + 1)*tan(b*x + a)^2 + c^2 + d^2 - 2*c + 1) + 1))/b

Sympy [F]

\[ \int \coth ^{-1}(c+d \tan (a+b x)) \, dx=\int \operatorname {acoth}{\left (c + d \tan {\left (a + b x \right )} \right )}\, dx \]

[In]

integrate(acoth(c+d*tan(b*x+a)),x)

[Out]

Integral(acoth(c + d*tan(a + b*x)), x)

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 372 vs. \(2 (136) = 272\).

Time = 0.36 (sec) , antiderivative size = 372, normalized size of antiderivative = 1.92 \[ \int \coth ^{-1}(c+d \tan (a+b x)) \, dx=\frac {4 \, {\left (b x + a\right )} \operatorname {arcoth}\left (d \tan \left (b x + a\right ) + c\right ) + {\left (\arctan \left (\frac {d^{2} \tan \left (b x + a\right ) + {\left (c + 1\right )} d}{c^{2} + d^{2} + 2 \, c + 1}, \frac {{\left (c + 1\right )} d \tan \left (b x + a\right ) + c^{2} + 2 \, c + 1}{c^{2} + d^{2} + 2 \, c + 1}\right ) - \arctan \left (\frac {d^{2} \tan \left (b x + a\right ) + {\left (c - 1\right )} d}{c^{2} + d^{2} - 2 \, c + 1}, \frac {{\left (c - 1\right )} d \tan \left (b x + a\right ) + c^{2} - 2 \, c + 1}{c^{2} + d^{2} - 2 \, c + 1}\right )\right )} \log \left (\tan \left (b x + a\right )^{2} + 1\right ) - {\left (b x + a\right )} \log \left (\frac {d^{2} \tan \left (b x + a\right )^{2} + 2 \, {\left (c + 1\right )} d \tan \left (b x + a\right ) + c^{2} + 2 \, c + 1}{c^{2} + d^{2} + 2 \, c + 1}\right ) + {\left (b x + a\right )} \log \left (\frac {d^{2} \tan \left (b x + a\right )^{2} + 2 \, {\left (c - 1\right )} d \tan \left (b x + a\right ) + c^{2} - 2 \, c + 1}{c^{2} + d^{2} - 2 \, c + 1}\right ) - i \, {\rm Li}_2\left (-\frac {i \, d \tan \left (b x + a\right ) - d}{i \, c + d + i}\right ) + i \, {\rm Li}_2\left (-\frac {i \, d \tan \left (b x + a\right ) - d}{i \, c + d - i}\right ) - i \, {\rm Li}_2\left (\frac {i \, d \tan \left (b x + a\right ) + d}{-i \, c + d + i}\right ) + i \, {\rm Li}_2\left (\frac {i \, d \tan \left (b x + a\right ) + d}{-i \, c + d - i}\right )}{4 \, b} \]

[In]

integrate(arccoth(c+d*tan(b*x+a)),x, algorithm="maxima")

[Out]

1/4*(4*(b*x + a)*arccoth(d*tan(b*x + a) + c) + (arctan2((d^2*tan(b*x + a) + (c + 1)*d)/(c^2 + d^2 + 2*c + 1),
((c + 1)*d*tan(b*x + a) + c^2 + 2*c + 1)/(c^2 + d^2 + 2*c + 1)) - arctan2((d^2*tan(b*x + a) + (c - 1)*d)/(c^2
+ d^2 - 2*c + 1), ((c - 1)*d*tan(b*x + a) + c^2 - 2*c + 1)/(c^2 + d^2 - 2*c + 1)))*log(tan(b*x + a)^2 + 1) - (
b*x + a)*log((d^2*tan(b*x + a)^2 + 2*(c + 1)*d*tan(b*x + a) + c^2 + 2*c + 1)/(c^2 + d^2 + 2*c + 1)) + (b*x + a
)*log((d^2*tan(b*x + a)^2 + 2*(c - 1)*d*tan(b*x + a) + c^2 - 2*c + 1)/(c^2 + d^2 - 2*c + 1)) - I*dilog(-(I*d*t
an(b*x + a) - d)/(I*c + d + I)) + I*dilog(-(I*d*tan(b*x + a) - d)/(I*c + d - I)) - I*dilog((I*d*tan(b*x + a) +
d)/(-I*c + d + I)) + I*dilog((I*d*tan(b*x + a) + d)/(-I*c + d - I)))/b

Giac [F]

\[ \int \coth ^{-1}(c+d \tan (a+b x)) \, dx=\int { \operatorname {arcoth}\left (d \tan \left (b x + a\right ) + c\right ) \,d x } \]

[In]

integrate(arccoth(c+d*tan(b*x+a)),x, algorithm="giac")

[Out]

integrate(arccoth(d*tan(b*x + a) + c), x)

Mupad [F(-1)]

Timed out. \[ \int \coth ^{-1}(c+d \tan (a+b x)) \, dx=\int \mathrm {acoth}\left (c+d\,\mathrm {tan}\left (a+b\,x\right )\right ) \,d x \]

[In]

int(acoth(c + d*tan(a + b*x)),x)

[Out]

int(acoth(c + d*tan(a + b*x)), x)