\(\int \cot ^{-1}(c+d \tanh (a+b x)) \, dx\) [42]

Optimal result

Integrand size = 11, antiderivative size = 174 \[ \int \cot ^{-1}(c+d \tanh (a+b x)) \, dx=x \cot ^{-1}(c+d \tanh (a+b x))-\frac {1}{2} i x \log \left (1+\frac {(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )+\frac {1}{2} i x \log \left (1+\frac {(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )-\frac {i \operatorname {PolyLog}\left (2,-\frac {(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )}{4 b}+\frac {i \operatorname {PolyLog}\left (2,-\frac {(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )}{4 b} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 0.27 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.66 \[ \int \cot ^{-1}(c+d \tanh (a+b x)) \, dx=x \cot ^{-1}(c+d \tanh (a+b x))-\frac {4 a \sqrt {-d^2} \arctan \left (\frac {1+c^2-d^2+\left (1+c^2+2 c d+d^2\right ) e^{2 (a+b x)}}{2 d}\right )-2 d (a+b x) \log \left (1+\frac {2 \left (1+(c+d)^2\right ) e^{2 (a+b x)}}{2+2 c^2-2 d^2-4 \sqrt {-d^2}}\right )+2 d (a+b x) \log \left (1+\frac {\left (1+(c+d)^2\right ) e^{2 (a+b x)}}{1+c^2-d^2+2 \sqrt {-d^2}}\right )+d \operatorname {PolyLog}\left (2,-\frac {\left (1+c^2+2 c d+d^2\right ) e^{2 (a+b x)}}{1+c^2-d^2+2 \sqrt {-d^2}}\right )-d \operatorname {PolyLog}\left (2,\frac {\left (1+c^2+2 c d+d^2\right ) e^{2 (a+b x)}}{-1-c^2+d^2+2 \sqrt {-d^2}}\right )}{4 b \sqrt {-d^2}} \] Input:

Integrate[ArcCot[c + d*Tanh[a + b*x]],x]
 

Output:

x*ArcCot[c + d*Tanh[a + b*x]] - (4*a*Sqrt[-d^2]*ArcTan[(1 + c^2 - d^2 + (1 
 + c^2 + 2*c*d + d^2)*E^(2*(a + b*x)))/(2*d)] - 2*d*(a + b*x)*Log[1 + (2*( 
1 + (c + d)^2)*E^(2*(a + b*x)))/(2 + 2*c^2 - 2*d^2 - 4*Sqrt[-d^2])] + 2*d* 
(a + b*x)*Log[1 + ((1 + (c + d)^2)*E^(2*(a + b*x)))/(1 + c^2 - d^2 + 2*Sqr 
t[-d^2])] + d*PolyLog[2, -(((1 + c^2 + 2*c*d + d^2)*E^(2*(a + b*x)))/(1 + 
c^2 - d^2 + 2*Sqrt[-d^2]))] - d*PolyLog[2, ((1 + c^2 + 2*c*d + d^2)*E^(2*( 
a + b*x)))/(-1 - c^2 + d^2 + 2*Sqrt[-d^2])])/(4*b*Sqrt[-d^2])
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.36, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {5715, 2620, 2715, 2838}

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.

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

\(\Big \downarrow \) 5715

\(\displaystyle b (1+i (c+d)) \int \frac {e^{2 a+2 b x} x}{-c+(-c-d+i) e^{2 a+2 b x}+d+i}dx-b (1-i (c+d)) \int \frac {e^{2 a+2 b x} x}{c+(c+d+i) e^{2 a+2 b x}-d+i}dx+x \cot ^{-1}(d \tanh (a+b x)+c)\)

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 2715

\(\displaystyle b (1+i (c+d)) \left (\frac {x \log \left (1+\frac {(-c-d+i) e^{2 a+2 b x}}{-c+d+i}\right )}{2 b (-c-d+i)}-\frac {\int e^{-2 a-2 b x} \log \left (\frac {e^{2 a+2 b x} (-c-d+i)}{-c+d+i}+1\right )de^{2 a+2 b x}}{4 b^2 (-c-d+i)}\right )-b (1-i (c+d)) \left (\frac {x \log \left (1+\frac {(c+d+i) e^{2 a+2 b x}}{c-d+i}\right )}{2 b (c+d+i)}-\frac {\int e^{-2 a-2 b x} \log \left (\frac {e^{2 a+2 b x} (c+d+i)}{c-d+i}+1\right )de^{2 a+2 b x}}{4 b^2 (c+d+i)}\right )+x \cot ^{-1}(d \tanh (a+b x)+c)\)

\(\Big \downarrow \) 2838

\(\displaystyle b (1+i (c+d)) \left (\frac {\operatorname {PolyLog}\left (2,-\frac {(-c-d+i) e^{2 a+2 b x}}{-c+d+i}\right )}{4 b^2 (-c-d+i)}+\frac {x \log \left (1+\frac {(-c-d+i) e^{2 a+2 b x}}{-c+d+i}\right )}{2 b (-c-d+i)}\right )-b (1-i (c+d)) \left (\frac {\operatorname {PolyLog}\left (2,-\frac {(c+d+i) e^{2 a+2 b x}}{c-d+i}\right )}{4 b^2 (c+d+i)}+\frac {x \log \left (1+\frac {(c+d+i) e^{2 a+2 b x}}{c-d+i}\right )}{2 b (c+d+i)}\right )+x \cot ^{-1}(d \tanh (a+b x)+c)\)

Input:

Int[ArcCot[c + d*Tanh[a + b*x]],x]
 

Output:

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

Defintions of rubi rules used

rule 2620

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] - Si 
mp[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 2715

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[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 2838

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 5715

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

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 351 vs. \(2 (150 ) = 300\).

Time = 2.27 (sec) , antiderivative size = 352, normalized size of antiderivative = 2.02

method	result	size
derivativedivides	\(\frac {\frac {\operatorname {arccot}\left (c +d \tanh \left (b x +a \right )\right ) d \ln \left (-d \tanh \left (b x +a \right )-d \right )}{2}-\frac {\operatorname {arccot}\left (c +d \tanh \left (b x +a \right )\right ) d \ln \left (-d \tanh \left (b x +a \right )+d \right )}{2}-\frac {d^{2} \left (\frac {\frac {i \ln \left (-d \tanh \left (b x +a \right )+d \right ) \ln \left (\frac {i+d \tanh \left (b x +a \right )+c}{i+c +d}\right )}{2}-\frac {i \ln \left (-d \tanh \left (b x +a \right )+d \right ) \ln \left (\frac {i-d \tanh \left (b x +a \right )-c}{i-c -d}\right )}{2}+\frac {i \operatorname {dilog}\left (\frac {i+d \tanh \left (b x +a \right )+c}{i+c +d}\right )}{2}-\frac {i \operatorname {dilog}\left (\frac {i-d \tanh \left (b x +a \right )-c}{i-c -d}\right )}{2}}{d}-\frac {\frac {i \ln \left (-d \tanh \left (b x +a \right )-d \right ) \ln \left (\frac {i+d \tanh \left (b x +a \right )+c}{i+c -d}\right )}{2}-\frac {i \ln \left (-d \tanh \left (b x +a \right )-d \right ) \ln \left (\frac {i-d \tanh \left (b x +a \right )-c}{i-c +d}\right )}{2}+\frac {i \operatorname {dilog}\left (\frac {i+d \tanh \left (b x +a \right )+c}{i+c -d}\right )}{2}-\frac {i \operatorname {dilog}\left (\frac {i-d \tanh \left (b x +a \right )-c}{i-c +d}\right )}{2}}{d}\right )}{2}}{b d}\)	\(352\)
default	\(\frac {\frac {\operatorname {arccot}\left (c +d \tanh \left (b x +a \right )\right ) d \ln \left (-d \tanh \left (b x +a \right )-d \right )}{2}-\frac {\operatorname {arccot}\left (c +d \tanh \left (b x +a \right )\right ) d \ln \left (-d \tanh \left (b x +a \right )+d \right )}{2}-\frac {d^{2} \left (\frac {\frac {i \ln \left (-d \tanh \left (b x +a \right )+d \right ) \ln \left (\frac {i+d \tanh \left (b x +a \right )+c}{i+c +d}\right )}{2}-\frac {i \ln \left (-d \tanh \left (b x +a \right )+d \right ) \ln \left (\frac {i-d \tanh \left (b x +a \right )-c}{i-c -d}\right )}{2}+\frac {i \operatorname {dilog}\left (\frac {i+d \tanh \left (b x +a \right )+c}{i+c +d}\right )}{2}-\frac {i \operatorname {dilog}\left (\frac {i-d \tanh \left (b x +a \right )-c}{i-c -d}\right )}{2}}{d}-\frac {\frac {i \ln \left (-d \tanh \left (b x +a \right )-d \right ) \ln \left (\frac {i+d \tanh \left (b x +a \right )+c}{i+c -d}\right )}{2}-\frac {i \ln \left (-d \tanh \left (b x +a \right )-d \right ) \ln \left (\frac {i-d \tanh \left (b x +a \right )-c}{i-c +d}\right )}{2}+\frac {i \operatorname {dilog}\left (\frac {i+d \tanh \left (b x +a \right )+c}{i+c -d}\right )}{2}-\frac {i \operatorname {dilog}\left (\frac {i-d \tanh \left (b x +a \right )-c}{i-c +d}\right )}{2}}{d}\right )}{2}}{b d}\)	\(352\)
risch	\(\text {Expression too large to display}\)	\(4201\)

Input:

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

Output:

1/b/d*(1/2*arccot(c+d*tanh(b*x+a))*d*ln(-d*tanh(b*x+a)-d)-1/2*arccot(c+d*t 
anh(b*x+a))*d*ln(-d*tanh(b*x+a)+d)-1/2*d^2*(1/d*(1/2*I*ln(-d*tanh(b*x+a)+d 
)*ln((I+d*tanh(b*x+a)+c)/(I+c+d))-1/2*I*ln(-d*tanh(b*x+a)+d)*ln((I-d*tanh( 
b*x+a)-c)/(I-c-d))+1/2*I*dilog((I+d*tanh(b*x+a)+c)/(I+c+d))-1/2*I*dilog((I 
-d*tanh(b*x+a)-c)/(I-c-d)))-1/d*(1/2*I*ln(-d*tanh(b*x+a)-d)*ln((I+d*tanh(b 
*x+a)+c)/(I+c-d))-1/2*I*ln(-d*tanh(b*x+a)-d)*ln((I-d*tanh(b*x+a)-c)/(I-c+d 
))+1/2*I*dilog((I+d*tanh(b*x+a)+c)/(I+c-d))-1/2*I*dilog((I-d*tanh(b*x+a)-c 
)/(I-c+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. 825 vs. \(2 (128) = 256\).

Time = 0.24 (sec) , antiderivative size = 825, normalized size of antiderivative = 4.74 \[ \int \cot ^{-1}(c+d \tanh (a+b x)) \, dx =\text {Too large to display} \] Input:

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

Output:

1/2*(2*b*x*arctan(cosh(b*x + a)/(c*cosh(b*x + a) + d*sinh(b*x + a))) + I*a 
*log(2*(c^2 + 2*c*d + d^2 + 1)*cosh(b*x + a) + 2*(c^2 + 2*c*d + d^2 + 1)*s 
inh(b*x + a) + 2*(c^2 - d^2 - 2*I*d + 1)*sqrt(-(c^2 - d^2 + 2*I*d + 1)/(c^ 
2 - 2*c*d + d^2 + 1))) + I*a*log(2*(c^2 + 2*c*d + d^2 + 1)*cosh(b*x + a) + 
 2*(c^2 + 2*c*d + d^2 + 1)*sinh(b*x + a) - 2*(c^2 - d^2 - 2*I*d + 1)*sqrt( 
-(c^2 - d^2 + 2*I*d + 1)/(c^2 - 2*c*d + d^2 + 1))) - I*a*log(2*(c^2 + 2*c* 
d + d^2 + 1)*cosh(b*x + a) + 2*(c^2 + 2*c*d + d^2 + 1)*sinh(b*x + a) + 2*( 
c^2 - d^2 + 2*I*d + 1)*sqrt(-(c^2 - d^2 - 2*I*d + 1)/(c^2 - 2*c*d + d^2 + 
1))) - I*a*log(2*(c^2 + 2*c*d + d^2 + 1)*cosh(b*x + a) + 2*(c^2 + 2*c*d + 
d^2 + 1)*sinh(b*x + a) - 2*(c^2 - d^2 + 2*I*d + 1)*sqrt(-(c^2 - d^2 - 2*I* 
d + 1)/(c^2 - 2*c*d + d^2 + 1))) + (-I*b*x - I*a)*log(sqrt(-(c^2 - d^2 + 2 
*I*d + 1)/(c^2 - 2*c*d + d^2 + 1))*(cosh(b*x + a) + sinh(b*x + a)) + 1) + 
(-I*b*x - I*a)*log(-sqrt(-(c^2 - d^2 + 2*I*d + 1)/(c^2 - 2*c*d + d^2 + 1)) 
*(cosh(b*x + a) + sinh(b*x + a)) + 1) + (I*b*x + I*a)*log(sqrt(-(c^2 - d^2 
 - 2*I*d + 1)/(c^2 - 2*c*d + d^2 + 1))*(cosh(b*x + a) + sinh(b*x + a)) + 1 
) + (I*b*x + I*a)*log(-sqrt(-(c^2 - d^2 - 2*I*d + 1)/(c^2 - 2*c*d + d^2 + 
1))*(cosh(b*x + a) + sinh(b*x + a)) + 1) - I*dilog(sqrt(-(c^2 - d^2 + 2*I* 
d + 1)/(c^2 - 2*c*d + d^2 + 1))*(cosh(b*x + a) + sinh(b*x + a))) - I*dilog 
(-sqrt(-(c^2 - d^2 + 2*I*d + 1)/(c^2 - 2*c*d + d^2 + 1))*(cosh(b*x + a) + 
sinh(b*x + a))) + I*dilog(sqrt(-(c^2 - d^2 - 2*I*d + 1)/(c^2 - 2*c*d + ...
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

4*b*d*integrate(x*e^(2*b*x + 2*a)/(c^2 - 2*c*d + d^2 + (c^2*e^(4*a) + 2*c* 
d*e^(4*a) + d^2*e^(4*a) + e^(4*a))*e^(4*b*x) + 2*(c^2*e^(2*a) - d^2*e^(2*a 
) + e^(2*a))*e^(2*b*x) + 1), x) + x*arctan2(e^(2*b*x + 2*a) + 1, (c*e^(2*a 
) + d*e^(2*a))*e^(2*b*x) + c - d)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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