\(\int \text {arctanh}(1+d+d \tanh (a+b x)) \, dx\) [291]

Optimal result

Integrand size = 12, antiderivative size = 69 \[ \int \text {arctanh}(1+d+d \tanh (a+b x)) \, dx=\frac {b x^2}{2}+x \text {arctanh}(1+d+d \tanh (a+b x))-\frac {1}{2} x \log \left (1+(1+d) e^{2 a+2 b x}\right )-\frac {\operatorname {PolyLog}\left (2,-\left ((1+d) e^{2 a+2 b x}\right )\right )}{4 b} \] Output:

1/2*b*x^2+x*arctanh(1+d+d*tanh(b*x+a))-1/2*x*ln(1+(1+d)*exp(2*b*x+2*a))-1/ 
4*polylog(2,-(1+d)*exp(2*b*x+2*a))/b
 

Mathematica [A] (verified)

Time = 0.79 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.91 \[ \int \text {arctanh}(1+d+d \tanh (a+b x)) \, dx=x \text {arctanh}(1+d+d \tanh (a+b x))+\frac {-2 b x \log \left (1+\frac {e^{-2 (a+b x)}}{1+d}\right )+\operatorname {PolyLog}\left (2,-\frac {e^{-2 (a+b x)}}{1+d}\right )}{4 b} \] Input:

Integrate[ArcTanh[1 + d + d*Tanh[a + b*x]],x]
 

Output:

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

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.30, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {6785, 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.

\(\displaystyle \int \text {arctanh}(d \tanh (a+b x)+d+1) \, dx\)

\(\Big \downarrow \) 6785

\(\displaystyle b \int \frac {x}{e^{2 a+2 b x} (d+1)+1}dx+x \text {arctanh}(d \tanh (a+b x)+d+1)\)

\(\Big \downarrow \) 2615

\(\displaystyle b \left (\frac {x^2}{2}-(d+1) \int \frac {e^{2 a+2 b x} x}{e^{2 a+2 b x} (d+1)+1}dx\right )+x \text {arctanh}(d \tanh (a+b x)+d+1)\)

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

Input:

Int[ArcTanh[1 + d + d*Tanh[a + b*x]],x]
 

Output:

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

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 6785

Int[ArcTanh[(c_.) + (d_.)*Tanh[(a_.) + (b_.)*(x_)]], x_Symbol] :> Simp[x*Ar 
cTanh[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)

Leaf count of result is larger than twice the leaf count of optimal. \(254\) vs. \(2(61)=122\).

Time = 2.30 (sec) , antiderivative size = 255, normalized size of antiderivative = 3.70

method	result	size
derivativedivides	\(\frac {\frac {\operatorname {arctanh}\left (1+d +d \tanh \left (b x +a \right )\right ) d \ln \left (d +d \tanh \left (b x +a \right )\right )}{2}-\frac {\operatorname {arctanh}\left (1+d +d \tanh \left (b x +a \right )\right ) d \ln \left (-d \tanh \left (b x +a \right )+d \right )}{2}-\frac {d^{2} \left (\frac {-\frac {\operatorname {dilog}\left (\frac {-d \tanh \left (b x +a \right )-d -2}{-2 d -2}\right )}{2}-\frac {\ln \left (-d \tanh \left (b x +a \right )+d \right ) \ln \left (\frac {-d \tanh \left (b x +a \right )-d -2}{-2 d -2}\right )}{2}+\frac {\operatorname {dilog}\left (-\frac {-d \tanh \left (b x +a \right )-d}{2 d}\right )}{2}+\frac {\ln \left (-d \tanh \left (b x +a \right )+d \right ) \ln \left (-\frac {-d \tanh \left (b x +a \right )-d}{2 d}\right )}{2}}{d}-\frac {-\frac {\operatorname {dilog}\left (\frac {d \tanh \left (b x +a \right )}{2}+\frac {d}{2}+1\right )}{2}-\frac {\ln \left (d +d \tanh \left (b x +a \right )\right ) \ln \left (\frac {d \tanh \left (b x +a \right )}{2}+\frac {d}{2}+1\right )}{2}+\frac {\ln \left (d +d \tanh \left (b x +a \right )\right )^{2}}{4}}{d}\right )}{2}}{b d}\)	\(255\)
default	\(\frac {\frac {\operatorname {arctanh}\left (1+d +d \tanh \left (b x +a \right )\right ) d \ln \left (d +d \tanh \left (b x +a \right )\right )}{2}-\frac {\operatorname {arctanh}\left (1+d +d \tanh \left (b x +a \right )\right ) d \ln \left (-d \tanh \left (b x +a \right )+d \right )}{2}-\frac {d^{2} \left (\frac {-\frac {\operatorname {dilog}\left (\frac {-d \tanh \left (b x +a \right )-d -2}{-2 d -2}\right )}{2}-\frac {\ln \left (-d \tanh \left (b x +a \right )+d \right ) \ln \left (\frac {-d \tanh \left (b x +a \right )-d -2}{-2 d -2}\right )}{2}+\frac {\operatorname {dilog}\left (-\frac {-d \tanh \left (b x +a \right )-d}{2 d}\right )}{2}+\frac {\ln \left (-d \tanh \left (b x +a \right )+d \right ) \ln \left (-\frac {-d \tanh \left (b x +a \right )-d}{2 d}\right )}{2}}{d}-\frac {-\frac {\operatorname {dilog}\left (\frac {d \tanh \left (b x +a \right )}{2}+\frac {d}{2}+1\right )}{2}-\frac {\ln \left (d +d \tanh \left (b x +a \right )\right ) \ln \left (\frac {d \tanh \left (b x +a \right )}{2}+\frac {d}{2}+1\right )}{2}+\frac {\ln \left (d +d \tanh \left (b x +a \right )\right )^{2}}{4}}{d}\right )}{2}}{b d}\)	\(255\)
risch	\(\text {Expression too large to display}\)	\(1083\)

Input:

int(arctanh(1+d+d*tanh(b*x+a)),x,method=_RETURNVERBOSE)
 

Output:

1/b/d*(1/2*arctanh(1+d+d*tanh(b*x+a))*d*ln(d+d*tanh(b*x+a))-1/2*arctanh(1+ 
d+d*tanh(b*x+a))*d*ln(-d*tanh(b*x+a)+d)-1/2*d^2*(1/d*(-1/2*dilog((-d*tanh( 
b*x+a)-d-2)/(-2*d-2))-1/2*ln(-d*tanh(b*x+a)+d)*ln((-d*tanh(b*x+a)-d-2)/(-2 
*d-2))+1/2*dilog(-1/2*(-d*tanh(b*x+a)-d)/d)+1/2*ln(-d*tanh(b*x+a)+d)*ln(-1 
/2*(-d*tanh(b*x+a)-d)/d))-1/d*(-1/2*dilog(1/2*d*tanh(b*x+a)+1/2*d+1)-1/2*l 
n(d+d*tanh(b*x+a))*ln(1/2*d*tanh(b*x+a)+1/2*d+1)+1/4*ln(d+d*tanh(b*x+a))^2 
)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 239 vs. \(2 (60) = 120\).

Time = 0.09 (sec) , antiderivative size = 239, normalized size of antiderivative = 3.46 \[ \int \text {arctanh}(1+d+d \tanh (a+b x)) \, dx=\frac {b^{2} x^{2} + b x \log \left (-\frac {{\left (d + 2\right )} \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )}{d \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )}\right ) + a \log \left (2 \, {\left (d + 1\right )} \cosh \left (b x + a\right ) + 2 \, {\left (d + 1\right )} \sinh \left (b x + a\right ) + \sqrt {-4 \, d - 4}\right ) + a \log \left (2 \, {\left (d + 1\right )} \cosh \left (b x + a\right ) + 2 \, {\left (d + 1\right )} \sinh \left (b x + a\right ) - \sqrt {-4 \, d - 4}\right ) - {\left (b x + a\right )} \log \left (\frac {1}{2} \, \sqrt {-4 \, d - 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) - {\left (b x + a\right )} \log \left (-\frac {1}{2} \, \sqrt {-4 \, d - 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) - {\rm Li}_2\left (\frac {1}{2} \, \sqrt {-4 \, d - 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - {\rm Li}_2\left (-\frac {1}{2} \, \sqrt {-4 \, d - 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right )}{2 \, b} \] Input:

integrate(arctanh(1+d+d*tanh(b*x+a)),x, algorithm="fricas")
 

Output:

1/2*(b^2*x^2 + b*x*log(-((d + 2)*cosh(b*x + a) + d*sinh(b*x + a))/(d*cosh( 
b*x + a) + d*sinh(b*x + a))) + a*log(2*(d + 1)*cosh(b*x + a) + 2*(d + 1)*s 
inh(b*x + a) + sqrt(-4*d - 4)) + a*log(2*(d + 1)*cosh(b*x + a) + 2*(d + 1) 
*sinh(b*x + a) - sqrt(-4*d - 4)) - (b*x + a)*log(1/2*sqrt(-4*d - 4)*(cosh( 
b*x + a) + sinh(b*x + a)) + 1) - (b*x + a)*log(-1/2*sqrt(-4*d - 4)*(cosh(b 
*x + a) + sinh(b*x + a)) + 1) - dilog(1/2*sqrt(-4*d - 4)*(cosh(b*x + a) + 
sinh(b*x + a))) - dilog(-1/2*sqrt(-4*d - 4)*(cosh(b*x + a) + sinh(b*x + a) 
)))/b
 

Sympy [F]

\[ \int \text {arctanh}(1+d+d \tanh (a+b x)) \, dx=\int \operatorname {atanh}{\left (d \tanh {\left (a + b x \right )} + d + 1 \right )}\, dx \] Input:

integrate(atanh(1+d+d*tanh(b*x+a)),x)
 

Output:

Integral(atanh(d*tanh(a + b*x) + d + 1), x)
 

Maxima [A] (verification not implemented)

Time = 0.59 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.04 \[ \int \text {arctanh}(1+d+d \tanh (a+b x)) \, dx=\frac {1}{4} \, b d {\left (\frac {2 \, x^{2}}{d} - \frac {2 \, b x \log \left ({\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + {\rm Li}_2\left (-{\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}\right )}{b^{2} d}\right )} + x \operatorname {artanh}\left (d \tanh \left (b x + a\right ) + d + 1\right ) \] Input:

integrate(arctanh(1+d+d*tanh(b*x+a)),x, algorithm="maxima")
 

Output:

1/4*b*d*(2*x^2/d - (2*b*x*log((d + 1)*e^(2*b*x + 2*a) + 1) + dilog(-(d + 1 
)*e^(2*b*x + 2*a)))/(b^2*d)) + x*arctanh(d*tanh(b*x + a) + d + 1)
 

Giac [F]

\[ \int \text {arctanh}(1+d+d \tanh (a+b x)) \, dx=\int { \operatorname {artanh}\left (d \tanh \left (b x + a\right ) + d + 1\right ) \,d x } \] Input:

integrate(arctanh(1+d+d*tanh(b*x+a)),x, algorithm="giac")
 

Output:

integrate(arctanh(d*tanh(b*x + a) + d + 1), x)
 

Mupad [F(-1)]

Timed out. \[ \int \text {arctanh}(1+d+d \tanh (a+b x)) \, dx=\int \mathrm {atanh}\left (d+d\,\mathrm {tanh}\left (a+b\,x\right )+1\right ) \,d x \] Input:

int(atanh(d + d*tanh(a + b*x) + 1),x)
 

Output:

int(atanh(d + d*tanh(a + b*x) + 1), x)
 

Reduce [F]

\[ \int \text {arctanh}(1+d+d \tanh (a+b x)) \, dx=\int \mathit {atanh} \left (\tanh \left (b x +a \right ) d +d +1\right )d x \] Input:

int(atanh(1+d+d*tanh(b*x+a)),x)
 

Output:

int(atanh(tanh(a + b*x)*d + d + 1),x)