3.2.9 \(\int \frac {1}{x^2 \text {arctanh}(\tanh (a+b x))^3} \, dx\) [109]

3.2.9.1 Optimal result

Integrand size = 13, antiderivative size = 131 \[ \int \frac {1}{x^2 \text {arctanh}(\tanh (a+b x))^3} \, dx=-\frac {3 b}{2 (b x-\text {arctanh}(\tanh (a+b x)))^2 \text {arctanh}(\tanh (a+b x))^2}+\frac {1}{x (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^2}+\frac {3 b}{(b x-\text {arctanh}(\tanh (a+b x)))^3 \text {arctanh}(\tanh (a+b x))}-\frac {3 b \log (x)}{(b x-\text {arctanh}(\tanh (a+b x)))^4}+\frac {3 b \log (\text {arctanh}(\tanh (a+b x)))}{(b x-\text {arctanh}(\tanh (a+b x)))^4} \]

output

-3/2*b/(b*x-arctanh(tanh(b*x+a)))^2/arctanh(tanh(b*x+a))^2+1/x/(b*x-arctan 
h(tanh(b*x+a)))/arctanh(tanh(b*x+a))^2+3*b/(b*x-arctanh(tanh(b*x+a)))^3/ar 
ctanh(tanh(b*x+a))-3*b*ln(x)/(b*x-arctanh(tanh(b*x+a)))^4+3*b*ln(arctanh(t 
anh(b*x+a)))/(b*x-arctanh(tanh(b*x+a)))^4
 

3.2.9.2 Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.71 \[ \int \frac {1}{x^2 \text {arctanh}(\tanh (a+b x))^3} \, dx=-\frac {b^3 x^3-6 b^2 x^2 \text {arctanh}(\tanh (a+b x))+2 \text {arctanh}(\tanh (a+b x))^3+3 b x \text {arctanh}(\tanh (a+b x))^2 (1+2 \log (x)-2 \log (\text {arctanh}(\tanh (a+b x))))}{2 x \text {arctanh}(\tanh (a+b x))^2 (-b x+\text {arctanh}(\tanh (a+b x)))^4} \]

input

Integrate[1/(x^2*ArcTanh[Tanh[a + b*x]]^3),x]
 

output

-1/2*(b^3*x^3 - 6*b^2*x^2*ArcTanh[Tanh[a + b*x]] + 2*ArcTanh[Tanh[a + b*x] 
]^3 + 3*b*x*ArcTanh[Tanh[a + b*x]]^2*(1 + 2*Log[x] - 2*Log[ArcTanh[Tanh[a 
+ b*x]]]))/(x*ArcTanh[Tanh[a + b*x]]^2*(-(b*x) + ArcTanh[Tanh[a + b*x]])^4 
)
 

3.2.9.3 Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.38, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {2602, 2594, 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.

\(\displaystyle \int \frac {1}{x^2 \text {arctanh}(\tanh (a+b x))^3} \, dx\)

\(\Big \downarrow \) 2602

\(\displaystyle \frac {3 b \int \frac {1}{x \text {arctanh}(\tanh (a+b x))^3}dx}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {1}{x (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^2}\)

\(\Big \downarrow \) 2594

\(\displaystyle \frac {3 b \left (-\frac {\int \frac {1}{x \text {arctanh}(\tanh (a+b x))^2}dx}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {1}{2 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^2}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {1}{x (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^2}\)

\(\Big \downarrow \) 2594

\(\displaystyle \frac {3 b \left (-\frac {-\frac {\int \frac {1}{x \text {arctanh}(\tanh (a+b x))}dx}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {1}{(b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {1}{2 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^2}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {1}{x (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^2}\)

\(\Big \downarrow \) 2591

\(\displaystyle \frac {3 b \left (-\frac {-\frac {\frac {b \int \frac {1}{\text {arctanh}(\tanh (a+b x))}dx}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {\int \frac {1}{x}dx}{b x-\text {arctanh}(\tanh (a+b x))}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {1}{(b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {1}{2 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^2}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {1}{x (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^2}\)

\(\Big \downarrow \) 14

\(\displaystyle \frac {3 b \left (-\frac {-\frac {\frac {b \int \frac {1}{\text {arctanh}(\tanh (a+b x))}dx}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {\log (x)}{b x-\text {arctanh}(\tanh (a+b x))}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {1}{(b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {1}{2 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^2}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {1}{x (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^2}\)

\(\Big \downarrow \) 2588

\(\displaystyle \frac {3 b \left (-\frac {-\frac {\frac {\int \frac {1}{\text {arctanh}(\tanh (a+b x))}d\text {arctanh}(\tanh (a+b x))}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {\log (x)}{b x-\text {arctanh}(\tanh (a+b x))}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {1}{(b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {1}{2 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^2}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {1}{x (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^2}\)

\(\Big \downarrow \) 14

\(\displaystyle \frac {1}{x (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^2}+\frac {3 b \left (-\frac {1}{2 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^2}-\frac {-\frac {1}{(b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))}-\frac {\frac {\log (\text {arctanh}(\tanh (a+b x)))}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {\log (x)}{b x-\text {arctanh}(\tanh (a+b x))}}{b x-\text {arctanh}(\tanh (a+b x))}}{b x-\text {arctanh}(\tanh (a+b x))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}\)

input

Int[1/(x^2*ArcTanh[Tanh[a + b*x]]^3),x]
 

output

1/(x*(b*x - ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a + b*x]]^2) + (3*b*(-1/2 
*1/((b*x - ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a + b*x]]^2) - (-(1/((b*x 
- ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a + b*x]])) - (-(Log[x]/(b*x - ArcT 
anh[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]])))/ 
(b*x - ArcTanh[Tanh[a + b*x]])
 

3.2.9.3.1 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]
 

3.2.9.4 Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.89

\[-\frac {b}{2 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2}}+\frac {3 b \ln \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{4}}-\frac {2 b}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}-\frac {1}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{3} x}-\frac {3 b \ln \left (x \right )}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{4}}\]

input

int(1/x^2/arctanh(tanh(b*x+a))^3,x)
 

output

-1/2/(arctanh(tanh(b*x+a))-b*x)^2/arctanh(tanh(b*x+a))^2*b+3/(arctanh(tanh 
(b*x+a))-b*x)^4*b*ln(arctanh(tanh(b*x+a)))-2/(arctanh(tanh(b*x+a))-b*x)^3* 
b/arctanh(tanh(b*x+a))-1/(arctanh(tanh(b*x+a))-b*x)^3/x-3/(arctanh(tanh(b* 
x+a))-b*x)^4*b*ln(x)
 

3.2.9.5 Fricas [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x^2 \text {arctanh}(\tanh (a+b x))^3} \, dx=-\frac {6 \, a b^{2} x^{2} + 9 \, a^{2} b x + 2 \, a^{3} - 6 \, {\left (b^{3} x^{3} + 2 \, a b^{2} x^{2} + a^{2} b x\right )} \log \left (b x + a\right ) + 6 \, {\left (b^{3} x^{3} + 2 \, a b^{2} x^{2} + a^{2} b x\right )} \log \left (x\right )}{2 \, {\left (a^{4} b^{2} x^{3} + 2 \, a^{5} b x^{2} + a^{6} x\right )}} \]

input

integrate(1/x^2/arctanh(tanh(b*x+a))^3,x, algorithm="fricas")
 

output

-1/2*(6*a*b^2*x^2 + 9*a^2*b*x + 2*a^3 - 6*(b^3*x^3 + 2*a*b^2*x^2 + a^2*b*x 
)*log(b*x + a) + 6*(b^3*x^3 + 2*a*b^2*x^2 + a^2*b*x)*log(x))/(a^4*b^2*x^3 
+ 2*a^5*b*x^2 + a^6*x)
 

3.2.9.6 Sympy [F]

\[ \int \frac {1}{x^2 \text {arctanh}(\tanh (a+b x))^3} \, dx=\int \frac {1}{x^{2} \operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]

input

integrate(1/x**2/atanh(tanh(b*x+a))**3,x)
 

output

Integral(1/(x**2*atanh(tanh(a + b*x))**3), x)
 

3.2.9.7 Maxima [A] (verification not implemented)

Time = 0.77 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.53 \[ \int \frac {1}{x^2 \text {arctanh}(\tanh (a+b x))^3} \, dx=-\frac {6 \, b^{2} x^{2} + 9 \, a b x + 2 \, a^{2}}{2 \, {\left (a^{3} b^{2} x^{3} + 2 \, a^{4} b x^{2} + a^{5} x\right )}} + \frac {3 \, b \log \left (b x + a\right )}{a^{4}} - \frac {3 \, b \log \left (x\right )}{a^{4}} \]

input

integrate(1/x^2/arctanh(tanh(b*x+a))^3,x, algorithm="maxima")
 

output

-1/2*(6*b^2*x^2 + 9*a*b*x + 2*a^2)/(a^3*b^2*x^3 + 2*a^4*b*x^2 + a^5*x) + 3 
*b*log(b*x + a)/a^4 - 3*b*log(x)/a^4
 

3.2.9.8 Giac [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.46 \[ \int \frac {1}{x^2 \text {arctanh}(\tanh (a+b x))^3} \, dx=\frac {3 \, b \log \left ({\left | b x + a \right |}\right )}{a^{4}} - \frac {3 \, b \log \left ({\left | x \right |}\right )}{a^{4}} - \frac {6 \, a b^{2} x^{2} + 9 \, a^{2} b x + 2 \, a^{3}}{2 \, {\left (b x + a\right )}^{2} a^{4} x} \]

input

integrate(1/x^2/arctanh(tanh(b*x+a))^3,x, algorithm="giac")
 

output

3*b*log(abs(b*x + a))/a^4 - 3*b*log(abs(x))/a^4 - 1/2*(6*a*b^2*x^2 + 9*a^2 
*b*x + 2*a^3)/((b*x + a)^2*a^4*x)
 

3.2.9.9 Mupad [B] (verification not implemented)

Time = 5.81 (sec) , antiderivative size = 804, normalized size of antiderivative = 6.14 \[ \int \frac {1}{x^2 \text {arctanh}(\tanh (a+b x))^3} \, dx=\text {Too large to display} \]

input

int(1/(x^2*atanh(tanh(a + b*x))^3),x)
 

output

-(24*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)) - 24*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 - 8*log(1/(exp(2*a)*exp(2*b*x) + 1)) 
^3 + 8*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))^3 + 32*b^3*x^3 
 + 24*b*x*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))^2 + 96*b^2* 
x^2*log(1/(exp(2*a)*exp(2*b*x) + 1)) - 96*b^2*x^2*log((exp(2*a)*exp(2*b*x) 
)/(exp(2*a)*exp(2*b*x) + 1)) + 24*b*x*log(1/(exp(2*a)*exp(2*b*x) + 1))^2 - 
 b*x*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))^2*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)*ex 
p(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x))*96i - b*x*log(1/(exp(2*a)*e 
xp(2*b*x) + 1))^2*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))*96i - 48*b*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)) + b*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))*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))/(e 
xp(2*a)*exp(2*b*x) + 1)) + 2*b*x))*192i)/(x*(log(1/(exp(2*a)*exp(2*b*x)...