\(\int (e+f x) \arctan (\coth (a+b x)) \, dx\) [95]

Optimal result

Integrand size = 13, antiderivative size = 159 \[ \int (e+f x) \arctan (\coth (a+b x)) \, dx=\frac {(e+f x)^2 \arctan \left (e^{2 a+2 b x}\right )}{2 f}+\frac {(e+f x)^2 \arctan (\coth (a+b x))}{2 f}-\frac {i (e+f x) \operatorname {PolyLog}\left (2,-i e^{2 a+2 b x}\right )}{4 b}+\frac {i (e+f x) \operatorname {PolyLog}\left (2,i e^{2 a+2 b x}\right )}{4 b}+\frac {i f \operatorname {PolyLog}\left (3,-i e^{2 a+2 b x}\right )}{8 b^2}-\frac {i f \operatorname {PolyLog}\left (3,i e^{2 a+2 b x}\right )}{8 b^2} \] Output:

1/2*(f*x+e)^2*arctan(exp(2*b*x+2*a))/f+1/2*(f*x+e)^2*arctan(coth(b*x+a))/f 
-1/4*I*(f*x+e)*polylog(2,-I*exp(2*b*x+2*a))/b+1/4*I*(f*x+e)*polylog(2,I*ex 
p(2*b*x+2*a))/b+1/8*I*f*polylog(3,-I*exp(2*b*x+2*a))/b^2-1/8*I*f*polylog(3 
,I*exp(2*b*x+2*a))/b^2
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.49 \[ \int (e+f x) \arctan (\coth (a+b x)) \, dx=e x \arctan (\coth (a+b x))+\frac {1}{2} f x^2 \arctan (\coth (a+b x))+\frac {i e \left (2 b x \left (\log \left (1-i e^{2 (a+b x)}\right )-\log \left (1+i e^{2 (a+b x)}\right )\right )-\operatorname {PolyLog}\left (2,-i e^{2 (a+b x)}\right )+\operatorname {PolyLog}\left (2,i e^{2 (a+b x)}\right )\right )}{4 b}+\frac {i f \left (2 b^2 x^2 \log \left (1-i e^{2 (a+b x)}\right )-2 b^2 x^2 \log \left (1+i e^{2 (a+b x)}\right )-2 b x \operatorname {PolyLog}\left (2,-i e^{2 (a+b x)}\right )+2 b x \operatorname {PolyLog}\left (2,i e^{2 (a+b x)}\right )+\operatorname {PolyLog}\left (3,-i e^{2 (a+b x)}\right )-\operatorname {PolyLog}\left (3,i e^{2 (a+b x)}\right )\right )}{8 b^2} \] Input:

Integrate[(e + f*x)*ArcTan[Coth[a + b*x]],x]
 

Output:

e*x*ArcTan[Coth[a + b*x]] + (f*x^2*ArcTan[Coth[a + b*x]])/2 + ((I/4)*e*(2* 
b*x*(Log[1 - I*E^(2*(a + b*x))] - Log[1 + I*E^(2*(a + b*x))]) - PolyLog[2, 
 (-I)*E^(2*(a + b*x))] + PolyLog[2, I*E^(2*(a + b*x))]))/b + ((I/8)*f*(2*b 
^2*x^2*Log[1 - I*E^(2*(a + b*x))] - 2*b^2*x^2*Log[1 + I*E^(2*(a + b*x))] - 
 2*b*x*PolyLog[2, (-I)*E^(2*(a + b*x))] + 2*b*x*PolyLog[2, I*E^(2*(a + b*x 
))] + PolyLog[3, (-I)*E^(2*(a + b*x))] - PolyLog[3, I*E^(2*(a + b*x))]))/b 
^2
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.10, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {5708, 3042, 4668, 3011, 2720, 7143}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int (e+f x) \arctan (\coth (a+b x)) \, dx\)

\(\Big \downarrow \) 5708

\(\displaystyle \frac {b \int (e+f x)^2 \text {sech}(2 a+2 b x)dx}{2 f}+\frac {(e+f x)^2 \arctan (\coth (a+b x))}{2 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(e+f x)^2 \arctan (\coth (a+b x))}{2 f}+\frac {b \int (e+f x)^2 \csc \left (2 i a+2 i b x+\frac {\pi }{2}\right )dx}{2 f}\)

\(\Big \downarrow \) 4668

\(\displaystyle \frac {(e+f x)^2 \arctan (\coth (a+b x))}{2 f}+\frac {b \left (-\frac {i f \int (e+f x) \log \left (1-i e^{2 a+2 b x}\right )dx}{b}+\frac {i f \int (e+f x) \log \left (1+i e^{2 a+2 b x}\right )dx}{b}+\frac {(e+f x)^2 \arctan \left (e^{2 a+2 b x}\right )}{b}\right )}{2 f}\)

\(\Big \downarrow \) 3011

\(\displaystyle \frac {(e+f x)^2 \arctan (\coth (a+b x))}{2 f}+\frac {b \left (\frac {i f \left (\frac {f \int \operatorname {PolyLog}\left (2,-i e^{2 a+2 b x}\right )dx}{2 b}-\frac {(e+f x) \operatorname {PolyLog}\left (2,-i e^{2 a+2 b x}\right )}{2 b}\right )}{b}-\frac {i f \left (\frac {f \int \operatorname {PolyLog}\left (2,i e^{2 a+2 b x}\right )dx}{2 b}-\frac {(e+f x) \operatorname {PolyLog}\left (2,i e^{2 a+2 b x}\right )}{2 b}\right )}{b}+\frac {(e+f x)^2 \arctan \left (e^{2 a+2 b x}\right )}{b}\right )}{2 f}\)

\(\Big \downarrow \) 2720

\(\displaystyle \frac {(e+f x)^2 \arctan (\coth (a+b x))}{2 f}+\frac {b \left (\frac {i f \left (\frac {f \int e^{-2 a-2 b x} \operatorname {PolyLog}\left (2,-i e^{2 a+2 b x}\right )de^{2 a+2 b x}}{4 b^2}-\frac {(e+f x) \operatorname {PolyLog}\left (2,-i e^{2 a+2 b x}\right )}{2 b}\right )}{b}-\frac {i f \left (\frac {f \int e^{-2 a-2 b x} \operatorname {PolyLog}\left (2,i e^{2 a+2 b x}\right )de^{2 a+2 b x}}{4 b^2}-\frac {(e+f x) \operatorname {PolyLog}\left (2,i e^{2 a+2 b x}\right )}{2 b}\right )}{b}+\frac {(e+f x)^2 \arctan \left (e^{2 a+2 b x}\right )}{b}\right )}{2 f}\)

\(\Big \downarrow \) 7143

\(\displaystyle \frac {(e+f x)^2 \arctan (\coth (a+b x))}{2 f}+\frac {b \left (\frac {(e+f x)^2 \arctan \left (e^{2 a+2 b x}\right )}{b}+\frac {i f \left (\frac {f \operatorname {PolyLog}\left (3,-i e^{2 a+2 b x}\right )}{4 b^2}-\frac {(e+f x) \operatorname {PolyLog}\left (2,-i e^{2 a+2 b x}\right )}{2 b}\right )}{b}-\frac {i f \left (\frac {f \operatorname {PolyLog}\left (3,i e^{2 a+2 b x}\right )}{4 b^2}-\frac {(e+f x) \operatorname {PolyLog}\left (2,i e^{2 a+2 b x}\right )}{2 b}\right )}{b}\right )}{2 f}\)

Input:

Int[(e + f*x)*ArcTan[Coth[a + b*x]],x]
 

Output:

((e + f*x)^2*ArcTan[Coth[a + b*x]])/(2*f) + (b*(((e + f*x)^2*ArcTan[E^(2*a 
 + 2*b*x)])/b + (I*f*(-1/2*((e + f*x)*PolyLog[2, (-I)*E^(2*a + 2*b*x)])/b 
+ (f*PolyLog[3, (-I)*E^(2*a + 2*b*x)])/(4*b^2)))/b - (I*f*(-1/2*((e + f*x) 
*PolyLog[2, I*E^(2*a + 2*b*x)])/b + (f*PolyLog[3, I*E^(2*a + 2*b*x)])/(4*b 
^2)))/b))/(2*f)
 

Defintions of rubi rules used

rule 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) 
*(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
 

rule 3011

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) 
*(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + 
b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F]))   Int[(f + g*x)^( 
m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e 
, f, g, n}, x] && GtQ[m, 0]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4668

Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_ 
))^(m_.), x_Symbol] :> Simp[-2*(c + d*x)^m*(ArcTanh[E^((-I)*e + f*fz*x)/E^( 
I*k*Pi)]/(f*fz*I)), x] + (-Simp[d*(m/(f*fz*I))   Int[(c + d*x)^(m - 1)*Log[ 
1 - E^((-I)*e + f*fz*x)/E^(I*k*Pi)], x], x] + Simp[d*(m/(f*fz*I))   Int[(c 
+ d*x)^(m - 1)*Log[1 + E^((-I)*e + f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c 
, d, e, f, fz}, x] && IntegerQ[2*k] && IGtQ[m, 0]
 

rule 5708

Int[ArcTan[Coth[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] 
:> Simp[(e + f*x)^(m + 1)*(ArcTan[Coth[a + b*x]]/(f*(m + 1))), x] + Simp[b/ 
(f*(m + 1))   Int[(e + f*x)^(m + 1)*Sech[2*a + 2*b*x], x], x] /; FreeQ[{a, 
b, e, f}, x] && IGtQ[m, 0]
 

rule 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d 
, e, n, p}, x] && EqQ[b*d, a*e]
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.60 (sec) , antiderivative size = 1777, normalized size of antiderivative = 11.18

method	result	size
risch	\(\text {Expression too large to display}\)	\(1777\)

Input:

int((f*x+e)*arctan(coth(b*x+a)),x,method=_RETURNVERBOSE)
 

Output:

1/8*I*f*polylog(3,-I*exp(2*b*x+2*a))/b^2+1/4*I/b^2*f*a^2*ln(exp(2*b*x+2*a) 
+I)-1/2*I/b*e*a*ln(exp(2*b*x+2*a)+I)-1/8*I*f*polylog(3,I*exp(2*b*x+2*a))/b 
^2+1/2*I*ln(exp(2*b*x+2*a)-I)*e*x+1/4*I*ln(exp(2*b*x+2*a)-I)*f*x^2-1/2*I*f 
/b^2*a^2*ln(((-I)^(1/2)-exp(b*x+a))/(-I)^(1/2))-1/2*I*f/b^2*a^2*ln(((-I)^( 
1/2)+exp(b*x+a))/(-I)^(1/2))-1/2*I*f/b^2*a*dilog(((-I)^(1/2)-exp(b*x+a))/( 
-I)^(1/2))-1/2*I*f/b^2*a*dilog(((-I)^(1/2)+exp(b*x+a))/(-I)^(1/2))+1/2*I*e 
*ln(((-I)^(1/2)-exp(b*x+a))/(-I)^(1/2))*x+1/2*I*e*ln(((-I)^(1/2)+exp(b*x+a 
))/(-I)^(1/2))*x+1/2*I*e/b*dilog(((-I)^(1/2)-exp(b*x+a))/(-I)^(1/2))+1/2*I 
*e/b*dilog(((-I)^(1/2)+exp(b*x+a))/(-I)^(1/2))-1/2*I*f/b*ln(1+I*exp(2*b*x+ 
2*a))*a*x+1/2*I*f/b*a*ln(1+exp(b*x+a)*(-1)^(3/4))*x+1/2*I*(-1/2*f*x^2-e*x) 
*ln(exp(2*b*x+2*a)+I)+1/4*Pi*(csgn(I/(exp(2*b*x+2*a)-1))*csgn(I*(exp(2*b*x 
+2*a)-I))*csgn(I*(exp(2*b*x+2*a)-I)/(exp(2*b*x+2*a)-1))-csgn(I/(exp(2*b*x+ 
2*a)-1))*csgn(I*(exp(2*b*x+2*a)+I))*csgn(I*(exp(2*b*x+2*a)+I)/(exp(2*b*x+2 
*a)-1))-csgn(I/(exp(2*b*x+2*a)-1))*csgn(I*(exp(2*b*x+2*a)-I)/(exp(2*b*x+2* 
a)-1))^2+csgn(I/(exp(2*b*x+2*a)-1))*csgn(I*(exp(2*b*x+2*a)+I)/(exp(2*b*x+2 
*a)-1))^2-csgn(I*(exp(2*b*x+2*a)-I))*csgn(I*(exp(2*b*x+2*a)-I)/(exp(2*b*x+ 
2*a)-1))^2+csgn(I*(exp(2*b*x+2*a)+I))*csgn(I*(exp(2*b*x+2*a)+I)/(exp(2*b*x 
+2*a)-1))^2+csgn(I*(exp(2*b*x+2*a)-I)/(exp(2*b*x+2*a)-1))*csgn((1+I)*(exp( 
2*b*x+2*a)-I)/(exp(2*b*x+2*a)-1))-csgn((1+I)*(exp(2*b*x+2*a)-I)/(exp(2*b*x 
+2*a)-1))^2-csgn(I*(exp(2*b*x+2*a)+I)/(exp(2*b*x+2*a)-1))*csgn((1-I)*(e...
 

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. 600 vs. \(2 (130) = 260\).

Time = 0.16 (sec) , antiderivative size = 600, normalized size of antiderivative = 3.77 \[ \int (e+f x) \arctan (\coth (a+b x)) \, dx =\text {Too large to display} \] Input:

integrate((f*x+e)*arctan(coth(b*x+a)),x, algorithm="fricas")
 

Output:

1/4*(2*(b^2*f*x^2 + 2*b^2*e*x)*arctan(cosh(b*x + a)/sinh(b*x + a)) - 2*(-I 
*b*f*x - I*b*e)*dilog(1/2*sqrt(4*I)*(cosh(b*x + a) + sinh(b*x + a))) - 2*( 
-I*b*f*x - I*b*e)*dilog(-1/2*sqrt(4*I)*(cosh(b*x + a) + sinh(b*x + a))) - 
2*(I*b*f*x + I*b*e)*dilog(1/2*sqrt(-4*I)*(cosh(b*x + a) + sinh(b*x + a))) 
- 2*(I*b*f*x + I*b*e)*dilog(-1/2*sqrt(-4*I)*(cosh(b*x + a) + sinh(b*x + a) 
)) + (I*b^2*f*x^2 + 2*I*b^2*e*x + 2*I*a*b*e - I*a^2*f)*log(1/2*sqrt(4*I)*( 
cosh(b*x + a) + sinh(b*x + a)) + 1) + (I*b^2*f*x^2 + 2*I*b^2*e*x + 2*I*a*b 
*e - I*a^2*f)*log(-1/2*sqrt(4*I)*(cosh(b*x + a) + sinh(b*x + a)) + 1) + (- 
I*b^2*f*x^2 - 2*I*b^2*e*x - 2*I*a*b*e + I*a^2*f)*log(1/2*sqrt(-4*I)*(cosh( 
b*x + a) + sinh(b*x + a)) + 1) + (-I*b^2*f*x^2 - 2*I*b^2*e*x - 2*I*a*b*e + 
 I*a^2*f)*log(-1/2*sqrt(-4*I)*(cosh(b*x + a) + sinh(b*x + a)) + 1) + (-2*I 
*a*b*e + I*a^2*f)*log(I*sqrt(4*I) + 2*cosh(b*x + a) + 2*sinh(b*x + a)) + ( 
-2*I*a*b*e + I*a^2*f)*log(-I*sqrt(4*I) + 2*cosh(b*x + a) + 2*sinh(b*x + a) 
) + (2*I*a*b*e - I*a^2*f)*log(I*sqrt(-4*I) + 2*cosh(b*x + a) + 2*sinh(b*x 
+ a)) + (2*I*a*b*e - I*a^2*f)*log(-I*sqrt(-4*I) + 2*cosh(b*x + a) + 2*sinh 
(b*x + a)) - 2*I*f*polylog(3, 1/2*sqrt(4*I)*(cosh(b*x + a) + sinh(b*x + a) 
)) - 2*I*f*polylog(3, -1/2*sqrt(4*I)*(cosh(b*x + a) + sinh(b*x + a))) + 2* 
I*f*polylog(3, 1/2*sqrt(-4*I)*(cosh(b*x + a) + sinh(b*x + a))) + 2*I*f*pol 
ylog(3, -1/2*sqrt(-4*I)*(cosh(b*x + a) + sinh(b*x + a))))/b^2
 

Sympy [F]

\[ \int (e+f x) \arctan (\coth (a+b x)) \, dx=\int \left (e + f x\right ) \operatorname {atan}{\left (\coth {\left (a + b x \right )} \right )}\, dx \] Input:

integrate((f*x+e)*atan(coth(b*x+a)),x)
 

Output:

Integral((e + f*x)*atan(coth(a + b*x)), x)
 

Maxima [F]

\[ \int (e+f x) \arctan (\coth (a+b x)) \, dx=\int { {\left (f x + e\right )} \arctan \left (\coth \left (b x + a\right )\right ) \,d x } \] Input:

integrate((f*x+e)*arctan(coth(b*x+a)),x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

1/2*(f*x^2 + 2*e*x)*arctan2(e^(2*b*x + 2*a) + 1, e^(2*b*x + 2*a) - 1) + in 
tegrate((b*f*x^2*e^(2*a) + 2*b*e*x*e^(2*a))*e^(2*b*x)/(e^(4*b*x + 4*a) + 1 
), x)
 

Giac [F]

\[ \int (e+f x) \arctan (\coth (a+b x)) \, dx=\int { {\left (f x + e\right )} \arctan \left (\coth \left (b x + a\right )\right ) \,d x } \] Input:

integrate((f*x+e)*arctan(coth(b*x+a)),x, algorithm="giac")
 

Output:

sage0*x
 

Mupad [F(-1)]

Timed out. \[ \int (e+f x) \arctan (\coth (a+b x)) \, dx=\int \mathrm {atan}\left (\mathrm {coth}\left (a+b\,x\right )\right )\,\left (e+f\,x\right ) \,d x \] Input:

int(atan(coth(a + b*x))*(e + f*x),x)
 

Output:

int(atan(coth(a + b*x))*(e + f*x), x)
 

Reduce [F]

\[ \int (e+f x) \arctan (\coth (a+b x)) \, dx=\left (\int \mathit {atan} \left (\coth \left (b x +a \right )\right )d x \right ) e +\left (\int \mathit {atan} \left (\coth \left (b x +a \right )\right ) x d x \right ) f \] Input:

int((f*x+e)*atan(coth(b*x+a)),x)
 

Output:

int(atan(coth(a + b*x)),x)*e + int(atan(coth(a + b*x))*x,x)*f