\(\int \arctan (c-(i-c) \tanh (a+b x)) \, dx\) [91]

Optimal result

Integrand size = 18, antiderivative size = 82 \[ \int \arctan (c-(i-c) \tanh (a+b x)) \, dx=\frac {1}{2} i b x^2+x \arctan (c-(i-c) \tanh (a+b x))-\frac {1}{2} i x \log \left (1-i c e^{2 a+2 b x}\right )-\frac {i \operatorname {PolyLog}\left (2,i c e^{2 a+2 b x}\right )}{4 b} \] Output:

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

Mathematica [A] (verified)

Time = 1.47 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.87 \[ \int \arctan (c-(i-c) \tanh (a+b x)) \, dx=x \arctan (c+(-i+c) \tanh (a+b x))-\frac {i \left (2 b x \log \left (1+\frac {i e^{-2 (a+b x)}}{c}\right )-\operatorname {PolyLog}\left (2,-\frac {i e^{-2 (a+b x)}}{c}\right )\right )}{4 b} \] Input:

Integrate[ArcTan[c - (I - c)*Tanh[a + b*x]],x]
 

Output:

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

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.17, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {5710, 2615, 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.

\(\displaystyle \int \arctan (c-(-c+i) \tanh (a+b x)) \, dx\)

\(\Big \downarrow \) 5710

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

\(\Big \downarrow \) 2615

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

Input:

Int[ArcTan[c - (I - c)*Tanh[a + b*x]],x]
 

Output:

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

Defintions of rubi rules used

rule 2615

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

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 5710

Int[ArcTan[(c_.) + (d_.)*Tanh[(a_.) + (b_.)*(x_)]], x_Symbol] :> Simp[x*Arc 
Tan[c + d*Tanh[a + b*x]], x] - Simp[b   Int[x/(c - d + c*E^(2*a + 2*b*x)), 
x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[(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. 515 vs. \(2 (68 ) = 136\).

Time = 0.70 (sec) , antiderivative size = 516, normalized size of antiderivative = 6.29

method	result	size
derivativedivides	\(\frac {-\frac {\arctan \left (c +\tanh \left (b x +a \right ) \left (-i+c \right )\right ) \ln \left (\tanh \left (b x +a \right ) \left (-i+c \right )-c +i\right )}{2 i-2 c}-\frac {2 i \arctan \left (c +\tanh \left (b x +a \right ) \left (-i+c \right )\right ) \ln \left (\tanh \left (b x +a \right ) \left (-i+c \right )-c +i\right ) c}{2 i-2 c}+\frac {\arctan \left (c +\tanh \left (b x +a \right ) \left (-i+c \right )\right ) \ln \left (\tanh \left (b x +a \right ) \left (-i+c \right )-c +i\right ) c^{2}}{2 i-2 c}+\frac {\arctan \left (c +\tanh \left (b x +a \right ) \left (-i+c \right )\right ) \ln \left (-i+\tanh \left (b x +a \right ) \left (-i+c \right )+c \right )}{2 i-2 c}+\frac {2 i \arctan \left (c +\tanh \left (b x +a \right ) \left (-i+c \right )\right ) \ln \left (-i+\tanh \left (b x +a \right ) \left (-i+c \right )+c \right ) c}{2 i-2 c}-\frac {\arctan \left (c +\tanh \left (b x +a \right ) \left (-i+c \right )\right ) \ln \left (-i+\tanh \left (b x +a \right ) \left (-i+c \right )+c \right ) c^{2}}{2 i-2 c}+\left (i-c \right )^{2} \left (\frac {\frac {i \left (\operatorname {dilog}\left (-\frac {i \left (\tanh \left (b x +a \right ) \left (-i+c \right )+c +i\right )}{2}\right )+\ln \left (-i+\tanh \left (b x +a \right ) \left (-i+c \right )+c \right ) \ln \left (-\frac {i \left (\tanh \left (b x +a \right ) \left (-i+c \right )+c +i\right )}{2}\right )\right )}{2}-\frac {i \ln \left (-i+\tanh \left (b x +a \right ) \left (-i+c \right )+c \right )^{2}}{4}}{2 i-2 c}-\frac {-\frac {i \left (\operatorname {dilog}\left (\frac {-i+\tanh \left (b x +a \right ) \left (-i+c \right )+c}{-2 i+2 c}\right )+\ln \left (\tanh \left (b x +a \right ) \left (-i+c \right )-c +i\right ) \ln \left (\frac {-i+\tanh \left (b x +a \right ) \left (-i+c \right )+c}{-2 i+2 c}\right )\right )}{2}+\frac {i \left (\operatorname {dilog}\left (\frac {\tanh \left (b x +a \right ) \left (-i+c \right )+c +i}{2 c}\right )+\ln \left (\tanh \left (b x +a \right ) \left (-i+c \right )-c +i\right ) \ln \left (\frac {\tanh \left (b x +a \right ) \left (-i+c \right )+c +i}{2 c}\right )\right )}{2}}{2 \left (i-c \right )}\right )}{b \left (-i+c \right )}\)	\(516\)
default	\(\frac {-\frac {\arctan \left (c +\tanh \left (b x +a \right ) \left (-i+c \right )\right ) \ln \left (\tanh \left (b x +a \right ) \left (-i+c \right )-c +i\right )}{2 i-2 c}-\frac {2 i \arctan \left (c +\tanh \left (b x +a \right ) \left (-i+c \right )\right ) \ln \left (\tanh \left (b x +a \right ) \left (-i+c \right )-c +i\right ) c}{2 i-2 c}+\frac {\arctan \left (c +\tanh \left (b x +a \right ) \left (-i+c \right )\right ) \ln \left (\tanh \left (b x +a \right ) \left (-i+c \right )-c +i\right ) c^{2}}{2 i-2 c}+\frac {\arctan \left (c +\tanh \left (b x +a \right ) \left (-i+c \right )\right ) \ln \left (-i+\tanh \left (b x +a \right ) \left (-i+c \right )+c \right )}{2 i-2 c}+\frac {2 i \arctan \left (c +\tanh \left (b x +a \right ) \left (-i+c \right )\right ) \ln \left (-i+\tanh \left (b x +a \right ) \left (-i+c \right )+c \right ) c}{2 i-2 c}-\frac {\arctan \left (c +\tanh \left (b x +a \right ) \left (-i+c \right )\right ) \ln \left (-i+\tanh \left (b x +a \right ) \left (-i+c \right )+c \right ) c^{2}}{2 i-2 c}+\left (i-c \right )^{2} \left (\frac {\frac {i \left (\operatorname {dilog}\left (-\frac {i \left (\tanh \left (b x +a \right ) \left (-i+c \right )+c +i\right )}{2}\right )+\ln \left (-i+\tanh \left (b x +a \right ) \left (-i+c \right )+c \right ) \ln \left (-\frac {i \left (\tanh \left (b x +a \right ) \left (-i+c \right )+c +i\right )}{2}\right )\right )}{2}-\frac {i \ln \left (-i+\tanh \left (b x +a \right ) \left (-i+c \right )+c \right )^{2}}{4}}{2 i-2 c}-\frac {-\frac {i \left (\operatorname {dilog}\left (\frac {-i+\tanh \left (b x +a \right ) \left (-i+c \right )+c}{-2 i+2 c}\right )+\ln \left (\tanh \left (b x +a \right ) \left (-i+c \right )-c +i\right ) \ln \left (\frac {-i+\tanh \left (b x +a \right ) \left (-i+c \right )+c}{-2 i+2 c}\right )\right )}{2}+\frac {i \left (\operatorname {dilog}\left (\frac {\tanh \left (b x +a \right ) \left (-i+c \right )+c +i}{2 c}\right )+\ln \left (\tanh \left (b x +a \right ) \left (-i+c \right )-c +i\right ) \ln \left (\frac {\tanh \left (b x +a \right ) \left (-i+c \right )+c +i}{2 c}\right )\right )}{2}}{2 \left (i-c \right )}\right )}{b \left (-i+c \right )}\)	\(516\)
risch	\(\text {Expression too large to display}\)	\(1230\)

Input:

int(arctan(c-(I-c)*tanh(b*x+a)),x,method=_RETURNVERBOSE)
 

Output:

1/b/(-I+c)*(-arctan(c+tanh(b*x+a)*(-I+c))/(2*I-2*c)*ln(tanh(b*x+a)*(-I+c)- 
c+I)-2*I*arctan(c+tanh(b*x+a)*(-I+c))/(2*I-2*c)*ln(tanh(b*x+a)*(-I+c)-c+I) 
*c+arctan(c+tanh(b*x+a)*(-I+c))/(2*I-2*c)*ln(tanh(b*x+a)*(-I+c)-c+I)*c^2+a 
rctan(c+tanh(b*x+a)*(-I+c))/(2*I-2*c)*ln(-I+tanh(b*x+a)*(-I+c)+c)+2*I*arct 
an(c+tanh(b*x+a)*(-I+c))/(2*I-2*c)*ln(-I+tanh(b*x+a)*(-I+c)+c)*c-arctan(c+ 
tanh(b*x+a)*(-I+c))/(2*I-2*c)*ln(-I+tanh(b*x+a)*(-I+c)+c)*c^2+(I-c)^2*(1/2 
/(I-c)*(1/2*I*(dilog(-1/2*I*(tanh(b*x+a)*(-I+c)+c+I))+ln(-I+tanh(b*x+a)*(- 
I+c)+c)*ln(-1/2*I*(tanh(b*x+a)*(-I+c)+c+I)))-1/4*I*ln(-I+tanh(b*x+a)*(-I+c 
)+c)^2)-1/2/(I-c)*(-1/2*I*(dilog((-I+tanh(b*x+a)*(-I+c)+c)/(-2*I+2*c))+ln( 
tanh(b*x+a)*(-I+c)-c+I)*ln((-I+tanh(b*x+a)*(-I+c)+c)/(-2*I+2*c)))+1/2*I*(d 
ilog(1/2*(tanh(b*x+a)*(-I+c)+c+I)/c)+ln(tanh(b*x+a)*(-I+c)-c+I)*ln(1/2*(ta 
nh(b*x+a)*(-I+c)+c+I)/c)))))
 

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. 187 vs. \(2 (58) = 116\).

Time = 0.10 (sec) , antiderivative size = 187, normalized size of antiderivative = 2.28 \[ \int \arctan (c-(i-c) \tanh (a+b x)) \, dx=\frac {i \, b^{2} x^{2} + i \, b x \log \left (-\frac {{\left (c e^{\left (2 \, b x + 2 \, a\right )} + i\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{c - i}\right ) - i \, a^{2} + {\left (-i \, b x - i \, a\right )} \log \left (\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (b x + a\right )} + 1\right ) + {\left (-i \, b x - i \, a\right )} \log \left (-\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (b x + a\right )} + 1\right ) + i \, a \log \left (\frac {2 \, c e^{\left (b x + a\right )} + i \, \sqrt {4 i \, c}}{2 \, c}\right ) + i \, a \log \left (\frac {2 \, c e^{\left (b x + a\right )} - i \, \sqrt {4 i \, c}}{2 \, c}\right ) - i \, {\rm Li}_2\left (\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (b x + a\right )}\right ) - i \, {\rm Li}_2\left (-\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (b x + a\right )}\right )}{2 \, b} \] Input:

integrate(arctan(c-(I-c)*tanh(b*x+a)),x, algorithm="fricas")
 

Output:

1/2*(I*b^2*x^2 + I*b*x*log(-(c*e^(2*b*x + 2*a) + I)*e^(-2*b*x - 2*a)/(c - 
I)) - I*a^2 + (-I*b*x - I*a)*log(1/2*sqrt(4*I*c)*e^(b*x + a) + 1) + (-I*b* 
x - I*a)*log(-1/2*sqrt(4*I*c)*e^(b*x + a) + 1) + I*a*log(1/2*(2*c*e^(b*x + 
 a) + I*sqrt(4*I*c))/c) + I*a*log(1/2*(2*c*e^(b*x + a) - I*sqrt(4*I*c))/c) 
 - I*dilog(1/2*sqrt(4*I*c)*e^(b*x + a)) - I*dilog(-1/2*sqrt(4*I*c)*e^(b*x 
+ a)))/b
 

Sympy [F(-2)]

Exception generated. \[ \int \arctan (c-(i-c) \tanh (a+b x)) \, dx=\text {Exception raised: CoercionFailed} \] Input:

integrate(atan(c-(I-c)*tanh(b*x+a)),x)
 

Output:

Exception raised: CoercionFailed >> Cannot convert _t0**2 + exp(2*a) of ty 
pe <class 'sympy.core.add.Add'> to QQ_I[b,_t0,exp(a)]
 

Maxima [A] (verification not implemented)

Time = 1.37 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.98 \[ \int \arctan (c-(i-c) \tanh (a+b x)) \, dx=-2 \, b {\left (c - i\right )} {\left (\frac {2 \, x^{2}}{2 i \, c + 2} - \frac {2 \, b x \log \left (-i \, c e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + {\rm Li}_2\left (i \, c e^{\left (2 \, b x + 2 \, a\right )}\right )}{-2 \, b^{2} {\left (-i \, c - 1\right )}}\right )} + x \arctan \left ({\left (c - i\right )} \tanh \left (b x + a\right ) + c\right ) \] Input:

integrate(arctan(c-(I-c)*tanh(b*x+a)),x, algorithm="maxima")
 

Output:

-2*b*(c - I)*(2*x^2/(2*I*c + 2) - (2*b*x*log(-I*c*e^(2*b*x + 2*a) + 1) + d 
ilog(I*c*e^(2*b*x + 2*a)))/(b^2*(2*I*c + 2))) + x*arctan((c - I)*tanh(b*x 
+ a) + c)
 

Giac [F]

\[ \int \arctan (c-(i-c) \tanh (a+b x)) \, dx=\int { \arctan \left ({\left (c - i\right )} \tanh \left (b x + a\right ) + c\right ) \,d x } \] Input:

integrate(arctan(c-(I-c)*tanh(b*x+a)),x, algorithm="giac")
 

Output:

integrate(arctan((c - I)*tanh(b*x + a) + c), x)
 

Mupad [F(-1)]

Timed out. \[ \int \arctan (c-(i-c) \tanh (a+b x)) \, dx=\int \mathrm {atan}\left (c+\mathrm {tanh}\left (a+b\,x\right )\,\left (c-\mathrm {i}\right )\right ) \,d x \] Input:

int(atan(c + tanh(a + b*x)*(c - 1i)),x)
 

Output:

int(atan(c + tanh(a + b*x)*(c - 1i)), x)
 

Reduce [F]

\[ \int \arctan (c-(i-c) \tanh (a+b x)) \, dx=\int \mathit {atan} \left (\tanh \left (b x +a \right ) c -\tanh \left (b x +a \right ) i +c \right )d x \] Input:

int(atan(c-(I-c)*tanh(b*x+a)),x)
 

Output:

int(atan(tanh(a + b*x)*c - tanh(a + b*x)*i + c),x)