3.3.96 \(\int \text {arctanh}(1-d-d \tanh (a+b x)) \, dx\) [296]

3.3.96.1 Optimal result

Integrand size = 15, antiderivative size = 76 \[ \int \text {arctanh}(1-d-d \tanh (a+b x)) \, dx=\frac {b x^2}{2}+x \text {arctanh}(1-d-d \tanh (a+b x))-\frac {1}{2} x \log \left (1+(1-d) e^{2 a+2 b x}\right )-\frac {\operatorname {PolyLog}\left (2,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{4 b} \]

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
 

3.3.96.2 Mathematica [A] (verified)

Time = 0.83 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.87 \[ \int \text {arctanh}(1-d-d \tanh (a+b x)) \, dx=x \text {arctanh}(1-d-d \tanh (a+b x))+\frac {-2 b x \log \left (1-\frac {e^{-2 (a+b x)}}{-1+d}\right )+\operatorname {PolyLog}\left (2,\frac {e^{-2 (a+b x)}}{-1+d}\right )}{4 b} \]

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)
 

3.3.96.3 Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.36, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, 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

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

\(\Big \downarrow \) 2615

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

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))))
 

3.3.96.3.1 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]
 

3.3.96.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(271\) vs. \(2(66)=132\).

Time = 0.86 (sec) , antiderivative size = 272, normalized size of antiderivative = 3.58

method	result	size
derivativedivides	\(-\frac {\frac {\operatorname {arctanh}\left (-1+d +d \tanh \left (b x +a \right )\right ) d \ln \left (d +d \tanh \left (b x +a \right )\right )}{2}-\frac {\operatorname {arctanh}\left (-1+d +d \tanh \left (b x +a \right )\right ) d \ln \left (-d \tanh \left (b x +a \right )+d \right )}{2}-\frac {d^{2} \left (\frac {\frac {\operatorname {dilog}\left (\frac {-d \tanh \left (b x +a \right )-d +2}{-2 d +2}\right )}{2}+\frac {\ln \left (-d \tanh \left (b x +a \right )+d \right ) \ln \left (\frac {-d \tanh \left (b x +a \right )-d +2}{-2 d +2}\right )}{2}-\frac {\operatorname {dilog}\left (-\frac {-d \tanh \left (b x +a \right )-d}{2 d}\right )}{2}-\frac {\ln \left (-d \tanh \left (b x +a \right )+d \right ) \ln \left (-\frac {-d \tanh \left (b x +a \right )-d}{2 d}\right )}{2}}{d}-\frac {\frac {\left (\ln \left (d +d \tanh \left (b x +a \right )\right )-\ln \left (\frac {d \tanh \left (b x +a \right )}{2}+\frac {d}{2}\right )\right ) \ln \left (-\frac {d \tanh \left (b x +a \right )}{2}-\frac {d}{2}+1\right )}{2}-\frac {\operatorname {dilog}\left (\frac {d \tanh \left (b x +a \right )}{2}+\frac {d}{2}\right )}{2}-\frac {\ln \left (d +d \tanh \left (b x +a \right )\right )^{2}}{4}}{d}\right )}{2}}{b d}\)	\(272\)
default	\(-\frac {\frac {\operatorname {arctanh}\left (-1+d +d \tanh \left (b x +a \right )\right ) d \ln \left (d +d \tanh \left (b x +a \right )\right )}{2}-\frac {\operatorname {arctanh}\left (-1+d +d \tanh \left (b x +a \right )\right ) d \ln \left (-d \tanh \left (b x +a \right )+d \right )}{2}-\frac {d^{2} \left (\frac {\frac {\operatorname {dilog}\left (\frac {-d \tanh \left (b x +a \right )-d +2}{-2 d +2}\right )}{2}+\frac {\ln \left (-d \tanh \left (b x +a \right )+d \right ) \ln \left (\frac {-d \tanh \left (b x +a \right )-d +2}{-2 d +2}\right )}{2}-\frac {\operatorname {dilog}\left (-\frac {-d \tanh \left (b x +a \right )-d}{2 d}\right )}{2}-\frac {\ln \left (-d \tanh \left (b x +a \right )+d \right ) \ln \left (-\frac {-d \tanh \left (b x +a \right )-d}{2 d}\right )}{2}}{d}-\frac {\frac {\left (\ln \left (d +d \tanh \left (b x +a \right )\right )-\ln \left (\frac {d \tanh \left (b x +a \right )}{2}+\frac {d}{2}\right )\right ) \ln \left (-\frac {d \tanh \left (b x +a \right )}{2}-\frac {d}{2}+1\right )}{2}-\frac {\operatorname {dilog}\left (\frac {d \tanh \left (b x +a \right )}{2}+\frac {d}{2}\right )}{2}-\frac {\ln \left (d +d \tanh \left (b x +a \right )\right )^{2}}{4}}{d}\right )}{2}}{b d}\)	\(272\)
risch	\(\text {Expression too large to display}\)	\(1095\)

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*tan 
h(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*(ln(d+d*tanh(b*x+a))-ln(1/2*d*tanh(b* 
x+a)+1/2*d))*ln(-1/2*d*tanh(b*x+a)-1/2*d+1)-1/2*dilog(1/2*d*tanh(b*x+a)+1/ 
2*d)-1/4*ln(d+d*tanh(b*x+a))^2)))
 

3.3.96.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 228 vs. \(2 (61) = 122\).

Time = 0.28 (sec) , antiderivative size = 228, normalized size of antiderivative = 3.00 \[ \int \text {arctanh}(1-d-d \tanh (a+b x)) \, dx=\frac {b^{2} x^{2} - b x \log \left (-\frac {d \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )}{{\left (d - 2\right )} \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )}\right ) + a \log \left (2 \, {\left (d - 1\right )} \cosh \left (b x + a\right ) + 2 \, {\left (d - 1\right )} \sinh \left (b x + a\right ) + 2 \, \sqrt {d - 1}\right ) + a \log \left (2 \, {\left (d - 1\right )} \cosh \left (b x + a\right ) + 2 \, {\left (d - 1\right )} \sinh \left (b x + a\right ) - 2 \, \sqrt {d - 1}\right ) - {\left (b x + a\right )} \log \left (\sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) - {\left (b x + a\right )} \log \left (-\sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) - {\rm Li}_2\left (\sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - {\rm Li}_2\left (-\sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right )}{2 \, b} \]

input

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

output

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

3.3.96.6 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)
 

3.3.96.7 Maxima [A] (verification not implemented)

Time = 0.73 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.96 \[ \int \text {arctanh}(1-d-d \tanh (a+b x)) \, dx=\frac {1}{4} \, b d {\left (\frac {2 \, x^{2}}{d} - \frac {2 \, b x \log \left (-{\left (d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + {\rm Li}_2\left ({\left (d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}\right )}{b^{2} d}\right )} - x \operatorname {artanh}\left (d \tanh \left (b x + a\right ) + d - 1\right ) \]

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)
 

3.3.96.8 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)
 

3.3.96.9 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)