\(\int \cot ^{-1}(a+b f^{c+d x}) \, dx\) [76]

Optimal result

Integrand size = 12, antiderivative size = 196 \[ \int \cot ^{-1}\left (a+b f^{c+d x}\right ) \, dx=-\frac {\cot ^{-1}\left (a+b f^{c+d x}\right ) \log \left (\frac {2}{1-i \left (a+b f^{c+d x}\right )}\right )}{d \log (f)}+\frac {\cot ^{-1}\left (a+b f^{c+d x}\right ) \log \left (\frac {2 b f^{c+d x}}{(i-a) \left (1-i \left (a+b f^{c+d x}\right )\right )}\right )}{d \log (f)}-\frac {i \operatorname {PolyLog}\left (2,1-\frac {2}{1-i \left (a+b f^{c+d x}\right )}\right )}{2 d \log (f)}+\frac {i \operatorname {PolyLog}\left (2,1-\frac {2 b f^{c+d x}}{(i-a) \left (1-i \left (a+b f^{c+d x}\right )\right )}\right )}{2 d \log (f)} \] Output:

-arccot(a+b*f^(d*x+c))*ln(2/(1-I*(a+b*f^(d*x+c))))/d/ln(f)+arccot(a+b*f^(d 
*x+c))*ln(2*b*f^(d*x+c)/(I-a)/(1-I*(a+b*f^(d*x+c))))/d/ln(f)-1/2*I*polylog 
(2,1-2/(1-I*(a+b*f^(d*x+c))))/d/ln(f)+1/2*I*polylog(2,1-2*b*f^(d*x+c)/(I-a 
)/(1-I*(a+b*f^(d*x+c))))/d/ln(f)
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.85 \[ \int \cot ^{-1}\left (a+b f^{c+d x}\right ) \, dx=x \cot ^{-1}\left (a+b f^{c+d x}\right )+\frac {b \left (d x \log (f) \left (\log \left (1+\frac {b^2 f^{c+d x}}{a b-\sqrt {-b^2}}\right )-\log \left (1+\frac {b^2 f^{c+d x}}{a b+\sqrt {-b^2}}\right )\right )+\operatorname {PolyLog}\left (2,-\frac {b^2 f^{c+d x}}{a b-\sqrt {-b^2}}\right )-\operatorname {PolyLog}\left (2,-\frac {b^2 f^{c+d x}}{a b+\sqrt {-b^2}}\right )\right )}{2 \sqrt {-b^2} d \log (f)} \] Input:

Integrate[ArcCot[a + b*f^(c + d*x)],x]
 

Output:

x*ArcCot[a + b*f^(c + d*x)] + (b*(d*x*Log[f]*(Log[1 + (b^2*f^(c + d*x))/(a 
*b - Sqrt[-b^2])] - Log[1 + (b^2*f^(c + d*x))/(a*b + Sqrt[-b^2])]) + PolyL 
og[2, -((b^2*f^(c + d*x))/(a*b - Sqrt[-b^2]))] - PolyLog[2, -((b^2*f^(c + 
d*x))/(a*b + Sqrt[-b^2]))]))/(2*Sqrt[-b^2]*d*Log[f])
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.90, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {2720, 5571, 25, 27, 5382, 2849, 2752, 2897}

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.

\(\displaystyle \int \cot ^{-1}\left (a+b f^{c+d x}\right ) \, dx\)

\(\Big \downarrow \) 2720

\(\displaystyle \frac {\int f^{-c-d x} \cot ^{-1}\left (b f^{c+d x}+a\right )df^{c+d x}}{d \log (f)}\)

\(\Big \downarrow \) 5571

\(\displaystyle \frac {\int f^{-c-d x} \cot ^{-1}\left (b f^{c+d x}+a\right )d\left (b f^{c+d x}+a\right )}{b d \log (f)}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\int -f^{-c-d x} \cot ^{-1}\left (b f^{c+d x}+a\right )d\left (b f^{c+d x}+a\right )}{b d \log (f)}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int -\frac {f^{-c-d x} \cot ^{-1}\left (b f^{c+d x}+a\right )}{b}d\left (b f^{c+d x}+a\right )}{d \log (f)}\)

\(\Big \downarrow \) 5382

\(\displaystyle -\frac {\int \frac {\log \left (\frac {2}{1-i \left (b f^{c+d x}+a\right )}\right )}{f^{2 c+2 d x}+1}d\left (b f^{c+d x}+a\right )-\int \frac {\log \left (\frac {2 b f^{c+d x}}{(i-a) \left (1-i \left (b f^{c+d x}+a\right )\right )}\right )}{f^{2 c+2 d x}+1}d\left (b f^{c+d x}+a\right )+\log \left (\frac {2}{1-i \left (a+b f^{c+d x}\right )}\right ) \cot ^{-1}\left (a+b f^{c+d x}\right )-\log \left (\frac {2 b f^{c+d x}}{(-a+i) \left (1-i \left (a+b f^{c+d x}\right )\right )}\right ) \cot ^{-1}\left (a+b f^{c+d x}\right )}{d \log (f)}\)

\(\Big \downarrow \) 2849

\(\displaystyle -\frac {i \int \frac {\log \left (\frac {2}{1-i \left (b f^{c+d x}+a\right )}\right )}{1-\frac {2}{1-i \left (b f^{c+d x}+a\right )}}d\frac {1}{1-i \left (b f^{c+d x}+a\right )}-\int \frac {\log \left (\frac {2 b f^{c+d x}}{(i-a) \left (1-i \left (b f^{c+d x}+a\right )\right )}\right )}{f^{2 c+2 d x}+1}d\left (b f^{c+d x}+a\right )+\log \left (\frac {2}{1-i \left (a+b f^{c+d x}\right )}\right ) \cot ^{-1}\left (a+b f^{c+d x}\right )-\log \left (\frac {2 b f^{c+d x}}{(-a+i) \left (1-i \left (a+b f^{c+d x}\right )\right )}\right ) \cot ^{-1}\left (a+b f^{c+d x}\right )}{d \log (f)}\)

\(\Big \downarrow \) 2752

\(\displaystyle -\frac {-\int \frac {\log \left (\frac {2 b f^{c+d x}}{(i-a) \left (1-i \left (b f^{c+d x}+a\right )\right )}\right )}{f^{2 c+2 d x}+1}d\left (b f^{c+d x}+a\right )+\frac {1}{2} i \operatorname {PolyLog}\left (2,1-\frac {2}{1-i \left (b f^{c+d x}+a\right )}\right )+\log \left (\frac {2}{1-i \left (a+b f^{c+d x}\right )}\right ) \cot ^{-1}\left (a+b f^{c+d x}\right )-\log \left (\frac {2 b f^{c+d x}}{(-a+i) \left (1-i \left (a+b f^{c+d x}\right )\right )}\right ) \cot ^{-1}\left (a+b f^{c+d x}\right )}{d \log (f)}\)

\(\Big \downarrow \) 2897

\(\displaystyle -\frac {\frac {1}{2} i \operatorname {PolyLog}\left (2,1-\frac {2}{1-i \left (b f^{c+d x}+a\right )}\right )-\frac {1}{2} i \operatorname {PolyLog}\left (2,1-\frac {2 b f^{c+d x}}{(i-a) \left (1-i \left (b f^{c+d x}+a\right )\right )}\right )+\log \left (\frac {2}{1-i \left (a+b f^{c+d x}\right )}\right ) \cot ^{-1}\left (a+b f^{c+d x}\right )-\log \left (\frac {2 b f^{c+d x}}{(-a+i) \left (1-i \left (a+b f^{c+d x}\right )\right )}\right ) \cot ^{-1}\left (a+b f^{c+d x}\right )}{d \log (f)}\)

Input:

Int[ArcCot[a + b*f^(c + d*x)],x]
 

Output:

-((ArcCot[a + b*f^(c + d*x)]*Log[2/(1 - I*(a + b*f^(c + d*x)))] - ArcCot[a 
 + b*f^(c + d*x)]*Log[(2*b*f^(c + d*x))/((I - a)*(1 - I*(a + b*f^(c + d*x) 
)))] + (I/2)*PolyLog[2, 1 - 2/(1 - I*(a + b*f^(c + d*x)))] - (I/2)*PolyLog 
[2, 1 - (2*b*f^(c + d*x))/((I - a)*(1 - I*(a + b*f^(c + d*x))))])/(d*Log[f 
]))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 2752

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLo 
g[2, 1 - c*x], x] /; FreeQ[{c, d, e}, x] && EqQ[e + c*d, 0]
 

rule 2849

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> Simp 
[-e/g   Subst[Int[Log[2*d*x]/(1 - 2*d*x), x], x, 1/(d + e*x)], x] /; FreeQ[ 
{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]
 

rule 2897

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[Pq^m*((1 - u)/ 
D[u, x])]}, Simp[C*PolyLog[2, 1 - u], x] /; FreeQ[C, x]] /; IntegerQ[m] && 
PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponents[u, 
 x][[2]], Expon[Pq, x]]
 

rule 5382

Int[((a_.) + ArcCot[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> Si 
mp[(-(a + b*ArcCot[c*x]))*(Log[2/(1 - I*c*x)]/e), x] + (Simp[(a + b*ArcCot[ 
c*x])*(Log[2*c*((d + e*x)/((c*d + I*e)*(1 - I*c*x)))]/e), x] - Simp[b*(c/e) 
   Int[Log[2/(1 - I*c*x)]/(1 + c^2*x^2), x], x] + Simp[b*(c/e)   Int[Log[2* 
c*((d + e*x)/((c*d + I*e)*(1 - I*c*x)))]/(1 + c^2*x^2), x], x]) /; FreeQ[{a 
, b, c, d, e}, x] && NeQ[c^2*d^2 + e^2, 0]
 

rule 5571

Int[((a_.) + ArcCot[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m 
_.), x_Symbol] :> Simp[1/d   Subst[Int[((d*e - c*f)/d + f*(x/d))^m*(a + b*A 
rcCot[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && I 
GtQ[p, 0]
 

Maple [A] (verified)

Time = 0.82 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.83

method	result	size
derivativedivides	\(\frac {\ln \left (-b \,f^{d x +c}\right ) \operatorname {arccot}\left (a +b \,f^{d x +c}\right )+\frac {i \ln \left (-b \,f^{d x +c}\right ) \ln \left (\frac {i+b \,f^{d x +c}+a}{i+a}\right )}{2}-\frac {i \ln \left (-b \,f^{d x +c}\right ) \ln \left (\frac {i-b \,f^{d x +c}-a}{i-a}\right )}{2}+\frac {i \operatorname {dilog}\left (\frac {i+b \,f^{d x +c}+a}{i+a}\right )}{2}-\frac {i \operatorname {dilog}\left (\frac {i-b \,f^{d x +c}-a}{i-a}\right )}{2}}{d \ln \left (f \right )}\)	\(162\)
default	\(\frac {\ln \left (-b \,f^{d x +c}\right ) \operatorname {arccot}\left (a +b \,f^{d x +c}\right )+\frac {i \ln \left (-b \,f^{d x +c}\right ) \ln \left (\frac {i+b \,f^{d x +c}+a}{i+a}\right )}{2}-\frac {i \ln \left (-b \,f^{d x +c}\right ) \ln \left (\frac {i-b \,f^{d x +c}-a}{i-a}\right )}{2}+\frac {i \operatorname {dilog}\left (\frac {i+b \,f^{d x +c}+a}{i+a}\right )}{2}-\frac {i \operatorname {dilog}\left (\frac {i-b \,f^{d x +c}-a}{i-a}\right )}{2}}{d \ln \left (f \right )}\)	\(162\)
risch	\(\frac {i x \ln \left (1+i \left (a +b \,f^{d x +c}\right )\right )}{2}+\frac {\pi x}{2}-\frac {i \ln \left (-i b \,f^{d x +c}-a i+1\right ) \ln \left (-\frac {i f^{d x +c} b}{a i-1}\right )}{2 d \ln \left (f \right )}-\frac {i \operatorname {dilog}\left (-\frac {i f^{d x +c} b}{a i-1}\right )}{2 d \ln \left (f \right )}-\frac {i \operatorname {dilog}\left (\frac {b \,f^{d x} f^{c}+a -i}{a -i}\right )}{2 \ln \left (f \right ) d}-\frac {i \ln \left (\frac {b \,f^{d x} f^{c}+a -i}{a -i}\right ) x}{2}-\frac {i \ln \left (\frac {b \,f^{d x} f^{c}+a -i}{a -i}\right ) c}{2 d}+\frac {i c \ln \left (i b \,f^{d x} f^{c}+a i+1\right )}{2 d}\)	\(218\)

Input:

int(arccot(a+b*f^(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

1/d/ln(f)*(ln(-b*f^(d*x+c))*arccot(a+b*f^(d*x+c))+1/2*I*ln(-b*f^(d*x+c))*l 
n((I+b*f^(d*x+c)+a)/(I+a))-1/2*I*ln(-b*f^(d*x+c))*ln((I-b*f^(d*x+c)-a)/(I- 
a))+1/2*I*dilog((I+b*f^(d*x+c)+a)/(I+a))-1/2*I*dilog((I-b*f^(d*x+c)-a)/(I- 
a)))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.08 \[ \int \cot ^{-1}\left (a+b f^{c+d x}\right ) \, dx=\frac {2 \, d x \operatorname {arccot}\left (b f^{d x + c} + a\right ) \log \left (f\right ) - i \, c \log \left (b f^{d x + c} + a + i\right ) \log \left (f\right ) + i \, c \log \left (b f^{d x + c} + a - i\right ) \log \left (f\right ) + {\left (-i \, d x - i \, c\right )} \log \left (f\right ) \log \left (\frac {a^{2} + {\left (a b + i \, b\right )} f^{d x + c} + 1}{a^{2} + 1}\right ) + {\left (i \, d x + i \, c\right )} \log \left (f\right ) \log \left (\frac {a^{2} + {\left (a b - i \, b\right )} f^{d x + c} + 1}{a^{2} + 1}\right ) - i \, {\rm Li}_2\left (-\frac {a^{2} + {\left (a b + i \, b\right )} f^{d x + c} + 1}{a^{2} + 1} + 1\right ) + i \, {\rm Li}_2\left (-\frac {a^{2} + {\left (a b - i \, b\right )} f^{d x + c} + 1}{a^{2} + 1} + 1\right )}{2 \, d \log \left (f\right )} \] Input:

integrate(arccot(a+b*f^(d*x+c)),x, algorithm="fricas")
 

Output:

1/2*(2*d*x*arccot(b*f^(d*x + c) + a)*log(f) - I*c*log(b*f^(d*x + c) + a + 
I)*log(f) + I*c*log(b*f^(d*x + c) + a - I)*log(f) + (-I*d*x - I*c)*log(f)* 
log((a^2 + (a*b + I*b)*f^(d*x + c) + 1)/(a^2 + 1)) + (I*d*x + I*c)*log(f)* 
log((a^2 + (a*b - I*b)*f^(d*x + c) + 1)/(a^2 + 1)) - I*dilog(-(a^2 + (a*b 
+ I*b)*f^(d*x + c) + 1)/(a^2 + 1) + 1) + I*dilog(-(a^2 + (a*b - I*b)*f^(d* 
x + c) + 1)/(a^2 + 1) + 1))/(d*log(f))
 

Sympy [F]

\[ \int \cot ^{-1}\left (a+b f^{c+d x}\right ) \, dx=\int \operatorname {acot}{\left (a + b f^{c + d x} \right )}\, dx \] Input:

integrate(acot(a+b*f**(d*x+c)),x)
 

Output:

Integral(acot(a + b*f**(c + d*x)), x)
 

Maxima [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.96 \[ \int \cot ^{-1}\left (a+b f^{c+d x}\right ) \, dx=\frac {{\left (d x + c\right )} \operatorname {arccot}\left (b f^{d x + c} + a\right )}{d} + \frac {2 \, {\left (d x + c\right )} \arctan \left (\frac {b^{2} f^{d x + c} + a b}{b}\right ) \log \left (f\right ) + {\left (\pi - \arctan \left (\frac {1}{a}\right )\right )} \log \left (b^{2} f^{2 \, d x + 2 \, c} + 2 \, a b f^{d x + c} + a^{2} + 1\right ) - \arctan \left (b f^{d x + c} + a\right ) \log \left (\frac {b^{2} f^{2 \, d x + 2 \, c}}{a^{2} + 1}\right ) + i \, {\rm Li}_2\left (\frac {i \, b f^{d x + c} + i \, a + 1}{i \, a + 1}\right ) - i \, {\rm Li}_2\left (\frac {i \, b f^{d x + c} + i \, a - 1}{i \, a - 1}\right )}{2 \, d \log \left (f\right )} \] Input:

integrate(arccot(a+b*f^(d*x+c)),x, algorithm="maxima")
 

Output:

(d*x + c)*arccot(b*f^(d*x + c) + a)/d + 1/2*(2*(d*x + c)*arctan((b^2*f^(d* 
x + c) + a*b)/b)*log(f) + (pi - arctan(1/a))*log(b^2*f^(2*d*x + 2*c) + 2*a 
*b*f^(d*x + c) + a^2 + 1) - arctan(b*f^(d*x + c) + a)*log(b^2*f^(2*d*x + 2 
*c)/(a^2 + 1)) + I*dilog((I*b*f^(d*x + c) + I*a + 1)/(I*a + 1)) - I*dilog( 
(I*b*f^(d*x + c) + I*a - 1)/(I*a - 1)))/(d*log(f))
 

Giac [F]

\[ \int \cot ^{-1}\left (a+b f^{c+d x}\right ) \, dx=\int { \operatorname {arccot}\left (b f^{d x + c} + a\right ) \,d x } \] Input:

integrate(arccot(a+b*f^(d*x+c)),x, algorithm="giac")
 

Output:

integrate(arccot(b*f^(d*x + c) + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \cot ^{-1}\left (a+b f^{c+d x}\right ) \, dx=\int \mathrm {acot}\left (a+b\,f^{c+d\,x}\right ) \,d x \] Input:

int(acot(a + b*f^(c + d*x)),x)
 

Output:

int(acot(a + b*f^(c + d*x)), x)
 

Reduce [F]

\[ \int \cot ^{-1}\left (a+b f^{c+d x}\right ) \, dx=\int \mathit {acot} \left (f^{d x +c} b +a \right )d x \] Input:

int(acot(a+b*f^(d*x+c)),x)
 

Output:

int(acot(f**(c + d*x)*b + a),x)