\(\int (e+f x) \coth ^{-1}(\tan (a+b x)) \, dx\) [233]

Optimal result

Integrand size = 13, antiderivative size = 162 \[ \int (e+f x) \coth ^{-1}(\tan (a+b x)) \, dx=\frac {(e+f x)^2 \coth ^{-1}(\tan (a+b x))}{2 f}+\frac {i (e+f x)^2 \arctan \left (e^{2 i (a+b x)}\right )}{2 f}-\frac {i (e+f x) \operatorname {PolyLog}\left (2,-i e^{2 i (a+b x)}\right )}{4 b}+\frac {i (e+f x) \operatorname {PolyLog}\left (2,i e^{2 i (a+b x)}\right )}{4 b}+\frac {f \operatorname {PolyLog}\left (3,-i e^{2 i (a+b x)}\right )}{8 b^2}-\frac {f \operatorname {PolyLog}\left (3,i e^{2 i (a+b x)}\right )}{8 b^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {6387, 4266, 2611, 2320, 6724} \[ \int (e+f x) \coth ^{-1}(\tan (a+b x)) \, dx=\frac {i (e+f x)^2 \arctan \left (e^{2 i (a+b x)}\right )}{2 f}+\frac {f \operatorname {PolyLog}\left (3,-i e^{2 i (a+b x)}\right )}{8 b^2}-\frac {f \operatorname {PolyLog}\left (3,i e^{2 i (a+b x)}\right )}{8 b^2}-\frac {i (e+f x) \operatorname {PolyLog}\left (2,-i e^{2 i (a+b x)}\right )}{4 b}+\frac {i (e+f x) \operatorname {PolyLog}\left (2,i e^{2 i (a+b x)}\right )}{4 b}+\frac {(e+f x)^2 \coth ^{-1}(\tan (a+b x))}{2 f} \]

[In]

Int[(e + f*x)*ArcCoth[Tan[a + b*x]],x]

[Out]

((e + f*x)^2*ArcCoth[Tan[a + b*x]])/(2*f) + ((I/2)*(e + f*x)^2*ArcTan[E^((2*I)*(a + b*x))])/f - ((I/4)*(e + f*
x)*PolyLog[2, (-I)*E^((2*I)*(a + b*x))])/b + ((I/4)*(e + f*x)*PolyLog[2, I*E^((2*I)*(a + b*x))])/b + (f*PolyLo
g[3, (-I)*E^((2*I)*(a + b*x))])/(8*b^2) - (f*PolyLog[3, I*E^((2*I)*(a + b*x))])/(8*b^2)

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{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 2611

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] + Dist[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 4266

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

Rule 6387

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

Rule 6724

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> 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]

Rubi steps \begin{align*} \text {integral}& = \frac {(e+f x)^2 \coth ^{-1}(\tan (a+b x))}{2 f}-\frac {b \int (e+f x)^2 \sec (2 a+2 b x) \, dx}{2 f} \\ & = \frac {(e+f x)^2 \coth ^{-1}(\tan (a+b x))}{2 f}+\frac {i (e+f x)^2 \arctan \left (e^{2 i (a+b x)}\right )}{2 f}+\frac {1}{2} \int (e+f x) \log \left (1-i e^{i (2 a+2 b x)}\right ) \, dx-\frac {1}{2} \int (e+f x) \log \left (1+i e^{i (2 a+2 b x)}\right ) \, dx \\ & = \frac {(e+f x)^2 \coth ^{-1}(\tan (a+b x))}{2 f}+\frac {i (e+f x)^2 \arctan \left (e^{2 i (a+b x)}\right )}{2 f}-\frac {i (e+f x) \operatorname {PolyLog}\left (2,-i e^{2 i (a+b x)}\right )}{4 b}+\frac {i (e+f x) \operatorname {PolyLog}\left (2,i e^{2 i (a+b x)}\right )}{4 b}+\frac {(i f) \int \operatorname {PolyLog}\left (2,-i e^{i (2 a+2 b x)}\right ) \, dx}{4 b}-\frac {(i f) \int \operatorname {PolyLog}\left (2,i e^{i (2 a+2 b x)}\right ) \, dx}{4 b} \\ & = \frac {(e+f x)^2 \coth ^{-1}(\tan (a+b x))}{2 f}+\frac {i (e+f x)^2 \arctan \left (e^{2 i (a+b x)}\right )}{2 f}-\frac {i (e+f x) \operatorname {PolyLog}\left (2,-i e^{2 i (a+b x)}\right )}{4 b}+\frac {i (e+f x) \operatorname {PolyLog}\left (2,i e^{2 i (a+b x)}\right )}{4 b}+\frac {f \text {Subst}\left (\int \frac {\operatorname {PolyLog}(2,-i x)}{x} \, dx,x,e^{i (2 a+2 b x)}\right )}{8 b^2}-\frac {f \text {Subst}\left (\int \frac {\operatorname {PolyLog}(2,i x)}{x} \, dx,x,e^{i (2 a+2 b x)}\right )}{8 b^2} \\ & = \frac {(e+f x)^2 \coth ^{-1}(\tan (a+b x))}{2 f}+\frac {i (e+f x)^2 \arctan \left (e^{2 i (a+b x)}\right )}{2 f}-\frac {i (e+f x) \operatorname {PolyLog}\left (2,-i e^{2 i (a+b x)}\right )}{4 b}+\frac {i (e+f x) \operatorname {PolyLog}\left (2,i e^{2 i (a+b x)}\right )}{4 b}+\frac {f \operatorname {PolyLog}\left (3,-i e^{2 i (a+b x)}\right )}{8 b^2}-\frac {f \operatorname {PolyLog}\left (3,i e^{2 i (a+b x)}\right )}{8 b^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.82 \[ \int (e+f x) \coth ^{-1}(\tan (a+b x)) \, dx=e x \coth ^{-1}(\tan (a+b x))+\frac {1}{2} f x^2 \coth ^{-1}(\tan (a+b x))-\frac {e \left ((-4 a+\pi -4 b x) \left (\log \left (1-i e^{-2 i (a+b x)}\right )-\log \left (1+i e^{-2 i (a+b x)}\right )\right )-(-4 a+\pi ) \log \left (\cot \left (a+\frac {\pi }{4}+b x\right )\right )+2 i \left (\operatorname {PolyLog}\left (2,-i e^{-2 i (a+b x)}\right )-\operatorname {PolyLog}\left (2,i e^{-2 i (a+b x)}\right )\right )\right )}{8 b}+\frac {f \left (4 i b^2 x^2 \arctan (\cos (2 (a+b x))+i \sin (2 (a+b x)))+2 i b x \operatorname {PolyLog}(2,i \cos (2 (a+b x))-\sin (2 (a+b x)))-2 i b x \operatorname {PolyLog}(2,-i \cos (2 (a+b x))+\sin (2 (a+b x)))-\operatorname {PolyLog}(3,i \cos (2 (a+b x))-\sin (2 (a+b x)))+\operatorname {PolyLog}(3,-i \cos (2 (a+b x))+\sin (2 (a+b x)))\right )}{8 b^2} \]

[In]

Integrate[(e + f*x)*ArcCoth[Tan[a + b*x]],x]

[Out]

e*x*ArcCoth[Tan[a + b*x]] + (f*x^2*ArcCoth[Tan[a + b*x]])/2 - (e*((-4*a + Pi - 4*b*x)*(Log[1 - I/E^((2*I)*(a +
b*x))] - Log[1 + I/E^((2*I)*(a + b*x))]) - (-4*a + Pi)*Log[Cot[a + Pi/4 + b*x]] + (2*I)*(PolyLog[2, (-I)/E^((
2*I)*(a + b*x))] - PolyLog[2, I/E^((2*I)*(a + b*x))])))/(8*b) + (f*((4*I)*b^2*x^2*ArcTan[Cos[2*(a + b*x)] + I*
Sin[2*(a + b*x)]] + (2*I)*b*x*PolyLog[2, I*Cos[2*(a + b*x)] - Sin[2*(a + b*x)]] - (2*I)*b*x*PolyLog[2, (-I)*Co
s[2*(a + b*x)] + Sin[2*(a + b*x)]] - PolyLog[3, I*Cos[2*(a + b*x)] - Sin[2*(a + b*x)]] + PolyLog[3, (-I)*Cos[2
*(a + b*x)] + Sin[2*(a + b*x)]]))/(8*b^2)

Maple [C] (warning: unable to verify)

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

Time = 1.83 (sec) , antiderivative size = 1818, normalized size of antiderivative = 11.22

method	result	size
risch	\(\text {Expression too large to display}\)	\(1818\)

[In]

int((f*x+e)*arccoth(tan(b*x+a)),x,method=_RETURNVERBOSE)

[Out]

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

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

Time = 0.29 (sec) , antiderivative size = 834, normalized size of antiderivative = 5.15 \[ \int (e+f x) \coth ^{-1}(\tan (a+b x)) \, dx=\text {Too large to display} \]

[In]

integrate((f*x+e)*arccoth(tan(b*x+a)),x, algorithm="fricas")

[Out]

-1/16*(2*(-I*b*f*x - I*b*e)*dilog(-((I + 1)*tan(b*x + a)^2 + 2*tan(b*x + a) - I + 1)/(tan(b*x + a)^2 + 1) + 1)
+ 2*(-I*b*f*x - I*b*e)*dilog(-((I + 1)*tan(b*x + a)^2 - 2*tan(b*x + a) - I + 1)/(tan(b*x + a)^2 + 1) + 1) + 2
*(I*b*f*x + I*b*e)*dilog(-(-(I - 1)*tan(b*x + a)^2 + 2*tan(b*x + a) + I + 1)/(tan(b*x + a)^2 + 1) + 1) + 2*(I*
b*f*x + I*b*e)*dilog(-(-(I - 1)*tan(b*x + a)^2 - 2*tan(b*x + a) + I + 1)/(tan(b*x + a)^2 + 1) + 1) + 2*(b^2*f*
x^2 + 2*b^2*e*x + 2*a*b*e - a^2*f)*log(((I + 1)*tan(b*x + a)^2 + 2*tan(b*x + a) - I + 1)/(tan(b*x + a)^2 + 1))
- 2*(2*a*b*e - a^2*f)*log(((I + 1)*tan(b*x + a)^2 + 2*I*tan(b*x + a) + I - 1)/(tan(b*x + a)^2 + 1)) + 2*(2*a*
b*e - a^2*f)*log(((I + 1)*tan(b*x + a)^2 - 2*I*tan(b*x + a) + I - 1)/(tan(b*x + a)^2 + 1)) - 2*(b^2*f*x^2 + 2*
b^2*e*x + 2*a*b*e - a^2*f)*log(((I + 1)*tan(b*x + a)^2 - 2*tan(b*x + a) - I + 1)/(tan(b*x + a)^2 + 1)) + 2*(b^
2*f*x^2 + 2*b^2*e*x + 2*a*b*e - a^2*f)*log((-(I - 1)*tan(b*x + a)^2 + 2*tan(b*x + a) + I + 1)/(tan(b*x + a)^2
+ 1)) - 2*(b^2*f*x^2 + 2*b^2*e*x + 2*a*b*e - a^2*f)*log((-(I - 1)*tan(b*x + a)^2 - 2*tan(b*x + a) + I + 1)/(ta
n(b*x + a)^2 + 1)) - 2*(2*a*b*e - a^2*f)*log(((I - 1)*tan(b*x + a)^2 + 2*I*tan(b*x + a) + I + 1)/(tan(b*x + a)
^2 + 1)) + 2*(2*a*b*e - a^2*f)*log(((I - 1)*tan(b*x + a)^2 - 2*I*tan(b*x + a) + I + 1)/(tan(b*x + a)^2 + 1)) -
4*(b^2*f*x^2 + 2*b^2*e*x)*log((tan(b*x + a) + 1)/(tan(b*x + a) - 1)) - f*polylog(3, (I*tan(b*x + a)^2 + 2*tan
(b*x + a) - I)/(tan(b*x + a)^2 + 1)) + f*polylog(3, (I*tan(b*x + a)^2 - 2*tan(b*x + a) - I)/(tan(b*x + a)^2 +
1)) - f*polylog(3, (-I*tan(b*x + a)^2 + 2*tan(b*x + a) + I)/(tan(b*x + a)^2 + 1)) + f*polylog(3, (-I*tan(b*x +
a)^2 - 2*tan(b*x + a) + I)/(tan(b*x + a)^2 + 1)))/b^2

Sympy [F]

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

[In]

integrate((f*x+e)*acoth(tan(b*x+a)),x)

[Out]

Integral((e + f*x)*acoth(tan(a + b*x)), x)

Maxima [F]

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

[In]

integrate((f*x+e)*arccoth(tan(b*x+a)),x, algorithm="maxima")

[Out]

1/8*(f*x^2 + 2*e*x)*log(2*cos(2*b*x + 2*a)^2 + 2*sin(2*b*x + 2*a)^2 + 4*sin(2*b*x + 2*a) + 2) - 1/8*(f*x^2 + 2
*e*x)*log(2*cos(2*b*x + 2*a)^2 + 2*sin(2*b*x + 2*a)^2 - 4*sin(2*b*x + 2*a) + 2) - integrate(((b*f*x^2 + 2*b*e*
x)*cos(4*b*x + 4*a)*cos(2*b*x + 2*a) + (b*f*x^2 + 2*b*e*x)*sin(4*b*x + 4*a)*sin(2*b*x + 2*a) + (b*f*x^2 + 2*b*
e*x)*cos(2*b*x + 2*a))/(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1), x)

Giac [F]

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

[In]

integrate((f*x+e)*arccoth(tan(b*x+a)),x, algorithm="giac")

[Out]

integrate((f*x + e)*arccoth(tan(b*x + a)), x)

Mupad [F(-1)]

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

[In]

int(acoth(tan(a + b*x))*(e + f*x),x)

[Out]

int(acoth(tan(a + b*x))*(e + f*x), x)