\(\int \text {arctanh}(\tan (a+b x)) \, dx\) [315]

Optimal result

Integrand size = 7, antiderivative size = 79 \[ \int \text {arctanh}(\tan (a+b x)) \, dx=i x \arctan \left (e^{2 i (a+b x)}\right )+x \text {arctanh}(\tan (a+b x))-\frac {i \operatorname {PolyLog}\left (2,-i e^{2 i (a+b x)}\right )}{4 b}+\frac {i \operatorname {PolyLog}\left (2,i e^{2 i (a+b x)}\right )}{4 b} \] Output:

I*x*arctan(exp(2*I*(b*x+a)))+x*arctanh(tan(b*x+a))-1/4*I*polylog(2,-I*exp( 
2*I*(b*x+a)))/b+1/4*I*polylog(2,I*exp(2*I*(b*x+a)))/b
 

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.61 \[ \int \text {arctanh}(\tan (a+b x)) \, dx=x \text {arctanh}(\tan (a+b x))-\frac {(-4 a+\pi -4 b x) \left (\log \left (1-i e^{-2 i (a+b x)}\right )-\log \left (1+i e^{-2 i (a+b x)}\right )\right )-(-4 a+\pi ) \log \left (\cot \left (a+\frac {\pi }{4}+b x\right )\right )+2 i \left (\operatorname {PolyLog}\left (2,-i e^{-2 i (a+b x)}\right )-\operatorname {PolyLog}\left (2,i e^{-2 i (a+b x)}\right )\right )}{8 b} \] Input:

Integrate[ArcTanh[Tan[a + b*x]],x]
 

Output:

x*ArcTanh[Tan[a + b*x]] - ((-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)
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {6801, 3042, 4669, 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.

\(\displaystyle \int \text {arctanh}(\tan (a+b x)) \, dx\)

\(\Big \downarrow \) 6801

\(\displaystyle x \text {arctanh}(\tan (a+b x))-b \int x \sec (2 a+2 b x)dx\)

\(\Big \downarrow \) 3042

\(\displaystyle x \text {arctanh}(\tan (a+b x))-b \int x \csc \left (2 a+2 b x+\frac {\pi }{2}\right )dx\)

\(\Big \downarrow \) 4669

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

\(\displaystyle x \text {arctanh}(\tan (a+b x))-b \left (-\frac {i x \arctan \left (e^{2 i (a+b x)}\right )}{b}+\frac {i \operatorname {PolyLog}\left (2,-i e^{2 i (a+b x)}\right )}{4 b^2}-\frac {i \operatorname {PolyLog}\left (2,i e^{2 i (a+b x)}\right )}{4 b^2}\right )\)

Input:

Int[ArcTanh[Tan[a + b*x]],x]
 

Output:

x*ArcTanh[Tan[a + b*x]] - b*(((-I)*x*ArcTan[E^((2*I)*(a + b*x))])/b + ((I/ 
4)*PolyLog[2, (-I)*E^((2*I)*(a + b*x))])/b^2 - ((I/4)*PolyLog[2, I*E^((2*I 
)*(a + b*x))])/b^2)
 

Defintions of rubi rules used

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 3042

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

rule 4669

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] + (-Si 
mp[d*(m/f)   Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))], x], 
 x] + Simp[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 6801

Int[ArcTanh[Tan[(a_.) + (b_.)*(x_)]], x_Symbol] :> Simp[x*ArcTanh[Tan[a + b 
*x]], x] - Simp[b   Int[x*Sec[2*a + 2*b*x], x], x] /; FreeQ[{a, b}, x]
 

Maple [A] (verified)

Time = 1.77 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.52

method	result	size
parts	\(x \,\operatorname {arctanh}\left (\tan \left (b x +a \right )\right )-\frac {-\frac {\left (b x +a \right ) \ln \left (i {\mathrm e}^{2 i \left (b x +a \right )}+1\right )}{2}+\frac {\left (b x +a \right ) \ln \left (-i {\mathrm e}^{2 i \left (b x +a \right )}+1\right )}{2}+\frac {i \operatorname {dilog}\left (i {\mathrm e}^{2 i \left (b x +a \right )}+1\right )}{4}-\frac {i \operatorname {dilog}\left (-i {\mathrm e}^{2 i \left (b x +a \right )}+1\right )}{4}-\frac {a \ln \left (\sec \left (2 b x +2 a \right )+\tan \left (2 b x +2 a \right )\right )}{2}}{b}\)	\(120\)
derivativedivides	\(\frac {\arctan \left (\tan \left (b x +a \right )\right ) \operatorname {arctanh}\left (\tan \left (b x +a \right )\right )+\frac {\arctan \left (\tan \left (b x +a \right )\right ) \ln \left (1+\frac {i \left (1+i \tan \left (b x +a \right )\right )^{2}}{1+\tan \left (b x +a \right )^{2}}\right )}{2}-\frac {\arctan \left (\tan \left (b x +a \right )\right ) \ln \left (1-\frac {i \left (1+i \tan \left (b x +a \right )\right )^{2}}{1+\tan \left (b x +a \right )^{2}}\right )}{2}-\frac {i \operatorname {dilog}\left (1+\frac {i \left (1+i \tan \left (b x +a \right )\right )^{2}}{1+\tan \left (b x +a \right )^{2}}\right )}{4}+\frac {i \operatorname {dilog}\left (1-\frac {i \left (1+i \tan \left (b x +a \right )\right )^{2}}{1+\tan \left (b x +a \right )^{2}}\right )}{4}}{b}\)	\(169\)
default	\(\frac {\arctan \left (\tan \left (b x +a \right )\right ) \operatorname {arctanh}\left (\tan \left (b x +a \right )\right )+\frac {\arctan \left (\tan \left (b x +a \right )\right ) \ln \left (1+\frac {i \left (1+i \tan \left (b x +a \right )\right )^{2}}{1+\tan \left (b x +a \right )^{2}}\right )}{2}-\frac {\arctan \left (\tan \left (b x +a \right )\right ) \ln \left (1-\frac {i \left (1+i \tan \left (b x +a \right )\right )^{2}}{1+\tan \left (b x +a \right )^{2}}\right )}{2}-\frac {i \operatorname {dilog}\left (1+\frac {i \left (1+i \tan \left (b x +a \right )\right )^{2}}{1+\tan \left (b x +a \right )^{2}}\right )}{4}+\frac {i \operatorname {dilog}\left (1-\frac {i \left (1+i \tan \left (b x +a \right )\right )^{2}}{1+\tan \left (b x +a \right )^{2}}\right )}{4}}{b}\)	\(169\)
risch	\(\text {Expression too large to display}\)	\(1161\)

Input:

int(arctanh(tan(b*x+a)),x,method=_RETURNVERBOSE)
 

Output:

x*arctanh(tan(b*x+a))-1/b*(-1/2*(b*x+a)*ln(I*exp(2*I*(b*x+a))+1)+1/2*(b*x+ 
a)*ln(-I*exp(2*I*(b*x+a))+1)+1/4*I*dilog(I*exp(2*I*(b*x+a))+1)-1/4*I*dilog 
(-I*exp(2*I*(b*x+a))+1)-1/2*a*ln(sec(2*b*x+2*a)+tan(2*b*x+2*a)))
 

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. 499 vs. \(2 (57) = 114\).

Time = 0.12 (sec) , antiderivative size = 499, normalized size of antiderivative = 6.32 \[ \int \text {arctanh}(\tan (a+b x)) \, dx =\text {Too large to display} \] Input:

integrate(arctanh(tan(b*x+a)),x, algorithm="fricas")
 

Output:

1/8*(4*b*x*log(-(tan(b*x + a) + 1)/(tan(b*x + a) - 1)) - 2*(b*x + a)*log(( 
(I + 1)*tan(b*x + a)^2 + 2*tan(b*x + a) - I + 1)/(tan(b*x + a)^2 + 1)) + 2 
*a*log(((I + 1)*tan(b*x + a)^2 + 2*I*tan(b*x + a) + I - 1)/(tan(b*x + a)^2 
 + 1)) - 2*a*log(((I + 1)*tan(b*x + a)^2 - 2*I*tan(b*x + a) + I - 1)/(tan( 
b*x + a)^2 + 1)) + 2*(b*x + a)*log(((I + 1)*tan(b*x + a)^2 - 2*tan(b*x + a 
) - I + 1)/(tan(b*x + a)^2 + 1)) - 2*(b*x + a)*log((-(I - 1)*tan(b*x + a)^ 
2 + 2*tan(b*x + a) + I + 1)/(tan(b*x + a)^2 + 1)) + 2*(b*x + a)*log((-(I - 
 1)*tan(b*x + a)^2 - 2*tan(b*x + a) + I + 1)/(tan(b*x + a)^2 + 1)) + 2*a*l 
og(((I - 1)*tan(b*x + a)^2 + 2*I*tan(b*x + a) + I + 1)/(tan(b*x + a)^2 + 1 
)) - 2*a*log(((I - 1)*tan(b*x + a)^2 - 2*I*tan(b*x + a) + I + 1)/(tan(b*x 
+ a)^2 + 1)) + I*dilog(-((I + 1)*tan(b*x + a)^2 + 2*tan(b*x + a) - I + 1)/ 
(tan(b*x + a)^2 + 1) + 1) + I*dilog(-((I + 1)*tan(b*x + a)^2 - 2*tan(b*x + 
 a) - I + 1)/(tan(b*x + a)^2 + 1) + 1) - I*dilog(-(-(I - 1)*tan(b*x + a)^2 
 + 2*tan(b*x + a) + I + 1)/(tan(b*x + a)^2 + 1) + 1) - I*dilog(-(-(I - 1)* 
tan(b*x + a)^2 - 2*tan(b*x + a) + I + 1)/(tan(b*x + a)^2 + 1) + 1))/b
 

Sympy [F]

\[ \int \text {arctanh}(\tan (a+b x)) \, dx=\int \operatorname {atanh}{\left (\tan {\left (a + b x \right )} \right )}\, dx \] Input:

integrate(atanh(tan(b*x+a)),x)
 

Output:

Integral(atanh(tan(a + b*x)), x)
 

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 182 vs. \(2 (57) = 114\).

Time = 0.14 (sec) , antiderivative size = 182, normalized size of antiderivative = 2.30 \[ \int \text {arctanh}(\tan (a+b x)) \, dx=\frac {4 \, {\left (b x + a\right )} \operatorname {artanh}\left (\tan \left (b x + a\right )\right ) + {\left (\arctan \left (\frac {1}{2} \, \tan \left (b x + a\right ) + \frac {1}{2}, \frac {1}{2} \, \tan \left (b x + a\right ) + \frac {1}{2}\right ) - \arctan \left (\frac {1}{2} \, \tan \left (b x + a\right ) - \frac {1}{2}, -\frac {1}{2} \, \tan \left (b x + a\right ) + \frac {1}{2}\right )\right )} \log \left (\tan \left (b x + a\right )^{2} + 1\right ) - {\left (b x + a\right )} \log \left (\frac {1}{2} \, \tan \left (b x + a\right )^{2} + \tan \left (b x + a\right ) + \frac {1}{2}\right ) + {\left (b x + a\right )} \log \left (\frac {1}{2} \, \tan \left (b x + a\right )^{2} - \tan \left (b x + a\right ) + \frac {1}{2}\right ) - i \, {\rm Li}_2\left (\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \tan \left (b x + a\right ) - \frac {1}{2} i + \frac {1}{2}\right ) + i \, {\rm Li}_2\left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \tan \left (b x + a\right ) + \frac {1}{2} i + \frac {1}{2}\right ) + i \, {\rm Li}_2\left (\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \tan \left (b x + a\right ) + \frac {1}{2} i + \frac {1}{2}\right ) - i \, {\rm Li}_2\left (-\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \tan \left (b x + a\right ) - \frac {1}{2} i + \frac {1}{2}\right )}{4 \, b} \] Input:

integrate(arctanh(tan(b*x+a)),x, algorithm="maxima")
 

Output:

1/4*(4*(b*x + a)*arctanh(tan(b*x + a)) + (arctan2(1/2*tan(b*x + a) + 1/2, 
1/2*tan(b*x + a) + 1/2) - arctan2(1/2*tan(b*x + a) - 1/2, -1/2*tan(b*x + a 
) + 1/2))*log(tan(b*x + a)^2 + 1) - (b*x + a)*log(1/2*tan(b*x + a)^2 + tan 
(b*x + a) + 1/2) + (b*x + a)*log(1/2*tan(b*x + a)^2 - tan(b*x + a) + 1/2) 
- I*dilog((1/2*I + 1/2)*tan(b*x + a) - 1/2*I + 1/2) + I*dilog(-(1/2*I - 1/ 
2)*tan(b*x + a) + 1/2*I + 1/2) + I*dilog((1/2*I - 1/2)*tan(b*x + a) + 1/2* 
I + 1/2) - I*dilog(-(1/2*I + 1/2)*tan(b*x + a) - 1/2*I + 1/2))/b
 

Giac [F]

\[ \int \text {arctanh}(\tan (a+b x)) \, dx=\int { \operatorname {artanh}\left (\tan \left (b x + a\right )\right ) \,d x } \] Input:

integrate(arctanh(tan(b*x+a)),x, algorithm="giac")
 

Output:

integrate(arctanh(tan(b*x + a)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \text {arctanh}(\tan (a+b x)) \, dx=\int \mathrm {atanh}\left (\mathrm {tan}\left (a+b\,x\right )\right ) \,d x \] Input:

int(atanh(tan(a + b*x)),x)
 

Output:

int(atanh(tan(a + b*x)), x)
 

Reduce [F]

\[ \int \text {arctanh}(\tan (a+b x)) \, dx=\int \mathit {atanh} \left (\tan \left (b x +a \right )\right )d x \] Input:

int(atanh(tan(b*x+a)),x)
 

Output:

int(atanh(tan(a + b*x)),x)