3.1.100 \(\int \frac {1}{x^2 \text {arctanh}(\tanh (a+b x))^2} \, dx\) [100]

3.1.100.1 Optimal result

Integrand size = 13, antiderivative size = 102 \[ \int \frac {1}{x^2 \text {arctanh}(\tanh (a+b x))^2} \, dx=-\frac {2 b}{(b x-\text {arctanh}(\tanh (a+b x)))^2 \text {arctanh}(\tanh (a+b x))}+\frac {1}{x (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))}+\frac {2 b \log (x)}{(b x-\text {arctanh}(\tanh (a+b x)))^3}-\frac {2 b \log (\text {arctanh}(\tanh (a+b x)))}{(b x-\text {arctanh}(\tanh (a+b x)))^3} \]

output

-2*b/(b*x-arctanh(tanh(b*x+a)))^2/arctanh(tanh(b*x+a))+1/x/(b*x-arctanh(ta 
nh(b*x+a)))/arctanh(tanh(b*x+a))+2*b*ln(x)/(b*x-arctanh(tanh(b*x+a)))^3-2* 
b*ln(arctanh(tanh(b*x+a)))/(b*x-arctanh(tanh(b*x+a)))^3
 

3.1.100.2 Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.69 \[ \int \frac {1}{x^2 \text {arctanh}(\tanh (a+b x))^2} \, dx=\frac {-b^2 x^2+\text {arctanh}(\tanh (a+b x))^2+2 b x \text {arctanh}(\tanh (a+b x)) (\log (x)-\log (\text {arctanh}(\tanh (a+b x))))}{x (b x-\text {arctanh}(\tanh (a+b x)))^3 \text {arctanh}(\tanh (a+b x))} \]

input

Integrate[1/(x^2*ArcTanh[Tanh[a + b*x]]^2),x]
 

output

(-(b^2*x^2) + ArcTanh[Tanh[a + b*x]]^2 + 2*b*x*ArcTanh[Tanh[a + b*x]]*(Log 
[x] - Log[ArcTanh[Tanh[a + b*x]]]))/(x*(b*x - ArcTanh[Tanh[a + b*x]])^3*Ar 
cTanh[Tanh[a + b*x]])
 

3.1.100.3 Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.32, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {2602, 2594, 2591, 14, 2588, 14}

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.

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

\(\Big \downarrow \) 2602

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

\(\Big \downarrow \) 2594

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

\(\Big \downarrow \) 2591

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

\(\Big \downarrow \) 14

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

\(\Big \downarrow \) 2588

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

\(\Big \downarrow \) 14

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

input

Int[1/(x^2*ArcTanh[Tanh[a + b*x]]^2),x]
 

output

1/(x*(b*x - ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a + b*x]]) + (2*b*(-(1/(( 
b*x - ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a + b*x]])) - (-(Log[x]/(b*x - 
ArcTanh[Tanh[a + b*x]])) + Log[ArcTanh[Tanh[a + b*x]]]/(b*x - ArcTanh[Tanh 
[a + b*x]]))/(b*x - ArcTanh[Tanh[a + b*x]])))/(b*x - ArcTanh[Tanh[a + b*x] 
])
 

3.1.100.3.1 Defintions of rubi rules used

rule 14

Int[(a_.)/(x_), x_Symbol] :> Simp[a*Log[x], x] /; FreeQ[a, x]
 

rule 2588

Int[(u_)^(m_.), x_Symbol] :> With[{c = Simplify[D[u, x]]}, Simp[1/c   Subst 
[Int[x^m, x], x, u], x]] /; FreeQ[m, x] && PiecewiseLinearQ[u, x]
 

rule 2591

Int[1/((u_)*(v_)), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D 
[v, x]]}, Simp[b/(b*u - a*v)   Int[1/v, x], x] - Simp[a/(b*u - a*v)   Int[1 
/u, x], x] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x]
 

rule 2594

Int[(v_)^(n_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[ 
D[v, x]]}, Simp[v^(n + 1)/((n + 1)*(b*u - a*v)), x] - Simp[a*((n + 1)/((n + 
 1)*(b*u - a*v)))   Int[v^(n + 1)/u, x], x] /; NeQ[b*u - a*v, 0]] /; Piecew 
iseLinearQ[u, v, x] && LtQ[n, -1]
 

rule 2602

Int[(u_)^(m_)*(v_)^(n_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simp 
lify[D[v, x]]}, Simp[(-u^(m + 1))*(v^(n + 1)/((m + 1)*(b*u - a*v))), x] + S 
imp[b*((m + n + 2)/((m + 1)*(b*u - a*v)))   Int[u^(m + 1)*v^n, x], x] /; Ne 
Q[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && NeQ[m + n + 2, 0] && LtQ[m 
, -1]
 

3.1.100.4 Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.89

\[-\frac {b}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}+\frac {2 b \ln \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{3}}-\frac {1}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2} x}-\frac {2 b \ln \left (x \right )}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{3}}\]

input

int(1/x^2/arctanh(tanh(b*x+a))^2,x)
 

output

-1/(arctanh(tanh(b*x+a))-b*x)^2*b/arctanh(tanh(b*x+a))+2/(arctanh(tanh(b*x 
+a))-b*x)^3*b*ln(arctanh(tanh(b*x+a)))-1/(arctanh(tanh(b*x+a))-b*x)^2/x-2/ 
(arctanh(tanh(b*x+a))-b*x)^3*b*ln(x)
 

3.1.100.5 Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.62 \[ \int \frac {1}{x^2 \text {arctanh}(\tanh (a+b x))^2} \, dx=-\frac {2 \, a b x + a^{2} - 2 \, {\left (b^{2} x^{2} + a b x\right )} \log \left (b x + a\right ) + 2 \, {\left (b^{2} x^{2} + a b x\right )} \log \left (x\right )}{a^{3} b x^{2} + a^{4} x} \]

input

integrate(1/x^2/arctanh(tanh(b*x+a))^2,x, algorithm="fricas")
 

output

-(2*a*b*x + a^2 - 2*(b^2*x^2 + a*b*x)*log(b*x + a) + 2*(b^2*x^2 + a*b*x)*l 
og(x))/(a^3*b*x^2 + a^4*x)
 

3.1.100.6 Sympy [F]

\[ \int \frac {1}{x^2 \text {arctanh}(\tanh (a+b x))^2} \, dx=\int \frac {1}{x^{2} \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]

input

integrate(1/x**2/atanh(tanh(b*x+a))**2,x)
 

output

Integral(1/(x**2*atanh(tanh(a + b*x))**2), x)
 

3.1.100.7 Maxima [A] (verification not implemented)

Time = 0.51 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.44 \[ \int \frac {1}{x^2 \text {arctanh}(\tanh (a+b x))^2} \, dx=-\frac {2 \, b x + a}{a^{2} b x^{2} + a^{3} x} + \frac {2 \, b \log \left (b x + a\right )}{a^{3}} - \frac {2 \, b \log \left (x\right )}{a^{3}} \]

input

integrate(1/x^2/arctanh(tanh(b*x+a))^2,x, algorithm="maxima")
 

output

-(2*b*x + a)/(a^2*b*x^2 + a^3*x) + 2*b*log(b*x + a)/a^3 - 2*b*log(x)/a^3
 

3.1.100.8 Giac [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.44 \[ \int \frac {1}{x^2 \text {arctanh}(\tanh (a+b x))^2} \, dx=\frac {2 \, b \log \left ({\left | b x + a \right |}\right )}{a^{3}} - \frac {2 \, b \log \left ({\left | x \right |}\right )}{a^{3}} - \frac {2 \, b x + a}{{\left (b x^{2} + a x\right )} a^{2}} \]

input

integrate(1/x^2/arctanh(tanh(b*x+a))^2,x, algorithm="giac")
 

output

2*b*log(abs(b*x + a))/a^3 - 2*b*log(abs(x))/a^3 - (2*b*x + a)/((b*x^2 + a* 
x)*a^2)
 

3.1.100.9 Mupad [B] (verification not implemented)

Time = 5.99 (sec) , antiderivative size = 432, normalized size of antiderivative = 4.24 \[ \int \frac {1}{x^2 \text {arctanh}(\tanh (a+b x))^2} \, dx=-\frac {4\,{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2-\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\left (8\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+b\,x\,\mathrm {atan}\left (\frac {-\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,1{}\mathrm {i}+\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,1{}\mathrm {i}+b\,x\,2{}\mathrm {i}}{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x}\right )\,32{}\mathrm {i}\right )+4\,{\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2-16\,b^2\,x^2+b\,x\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\mathrm {atan}\left (\frac {-\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,1{}\mathrm {i}+\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,1{}\mathrm {i}+b\,x\,2{}\mathrm {i}}{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x}\right )\,32{}\mathrm {i}}{x\,\left (\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\right )\,{\left (\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3} \]

input

int(1/(x^2*atanh(tanh(a + b*x))^2),x)
 

output

-(4*log(1/(exp(2*a)*exp(2*b*x) + 1))^2 - log((exp(2*a)*exp(2*b*x))/(exp(2* 
a)*exp(2*b*x) + 1))*(8*log(1/(exp(2*a)*exp(2*b*x) + 1)) + b*x*atan((log((e 
xp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))*1i - log(1/(exp(2*a)*exp(2* 
b*x) + 1))*1i + b*x*2i)/(log(1/(exp(2*a)*exp(2*b*x) + 1)) - log((exp(2*a)* 
exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x))*32i) + 4*log((exp(2*a)*ex 
p(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))^2 - 16*b^2*x^2 + b*x*log(1/(exp(2*a)* 
exp(2*b*x) + 1))*atan((log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1) 
)*1i - log(1/(exp(2*a)*exp(2*b*x) + 1))*1i + b*x*2i)/(log(1/(exp(2*a)*exp( 
2*b*x) + 1)) - log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b* 
x))*32i)/(x*(log(1/(exp(2*a)*exp(2*b*x) + 1)) - log((exp(2*a)*exp(2*b*x))/ 
(exp(2*a)*exp(2*b*x) + 1)))*(log(1/(exp(2*a)*exp(2*b*x) + 1)) - log((exp(2 
*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^3)