\(\int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^3} \, dx\) [118]

Optimal result

Integrand size = 15, antiderivative size = 125 \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^3} \, dx=\frac {b^2 \arctan \left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right )}{4 (b x-\text {arctanh}(\tanh (a+b x)))^{3/2}}-\frac {b}{4 x \sqrt {\text {arctanh}(\tanh (a+b x))}}+\frac {b^2}{4 (b x-\text {arctanh}(\tanh (a+b x))) \sqrt {\text {arctanh}(\tanh (a+b x))}}-\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{2 x^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.71 \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^3} \, dx=\frac {1}{4} \left (\frac {\left (-2+\frac {b x}{b x-\text {arctanh}(\tanh (a+b x))}\right ) \sqrt {\text {arctanh}(\tanh (a+b x))}}{x^2}+\frac {b^2 \text {arctanh}\left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {-b x+\text {arctanh}(\tanh (a+b x))}}\right )}{(-b x+\text {arctanh}(\tanh (a+b x)))^{3/2}}\right ) \] Input:

Integrate[Sqrt[ArcTanh[Tanh[a + b*x]]]/x^3,x]
 

Output:

(((-2 + (b*x)/(b*x - ArcTanh[Tanh[a + b*x]]))*Sqrt[ArcTanh[Tanh[a + b*x]]] 
)/x^2 + (b^2*ArcTanh[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[-(b*x) + ArcTanh[Ta 
nh[a + b*x]]]])/(-(b*x) + ArcTanh[Tanh[a + b*x]])^(3/2))/4
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {2599, 2599, 2594, 2592}

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.

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

\(\Big \downarrow \) 2599

\(\displaystyle \frac {1}{4} b \int \frac {1}{x^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}dx-\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{2 x^2}\)

\(\Big \downarrow \) 2599

\(\displaystyle \frac {1}{4} b \left (-\frac {1}{2} b \int \frac {1}{x \text {arctanh}(\tanh (a+b x))^{3/2}}dx-\frac {1}{x \sqrt {\text {arctanh}(\tanh (a+b x))}}\right )-\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{2 x^2}\)

\(\Big \downarrow \) 2594

\(\displaystyle \frac {1}{4} b \left (-\frac {1}{2} b \left (-\frac {\int \frac {1}{x \sqrt {\text {arctanh}(\tanh (a+b x))}}dx}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{(b x-\text {arctanh}(\tanh (a+b x))) \sqrt {\text {arctanh}(\tanh (a+b x))}}\right )-\frac {1}{x \sqrt {\text {arctanh}(\tanh (a+b x))}}\right )-\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{2 x^2}\)

\(\Big \downarrow \) 2592

\(\displaystyle \frac {1}{4} b \left (-\frac {1}{2} b \left (-\frac {2 \arctan \left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right )}{(b x-\text {arctanh}(\tanh (a+b x)))^{3/2}}-\frac {2}{(b x-\text {arctanh}(\tanh (a+b x))) \sqrt {\text {arctanh}(\tanh (a+b x))}}\right )-\frac {1}{x \sqrt {\text {arctanh}(\tanh (a+b x))}}\right )-\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{2 x^2}\)

Input:

Int[Sqrt[ArcTanh[Tanh[a + b*x]]]/x^3,x]
 

Output:

(b*(-1/2*(b*((-2*ArcTan[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[b*x - ArcTanh[Ta 
nh[a + b*x]]]])/(b*x - ArcTanh[Tanh[a + b*x]])^(3/2) - 2/((b*x - ArcTanh[T 
anh[a + b*x]])*Sqrt[ArcTanh[Tanh[a + b*x]]]))) - 1/(x*Sqrt[ArcTanh[Tanh[a 
+ b*x]]])))/4 - Sqrt[ArcTanh[Tanh[a + b*x]]]/(2*x^2)
 

Defintions of rubi rules used

rule 2592

Int[1/((u_)*Sqrt[v_]), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simpli 
fy[D[v, x]]}, Simp[2*(ArcTan[Sqrt[v]/Rt[(b*u - a*v)/a, 2]]/(a*Rt[(b*u - a*v 
)/a, 2])), x] /; NeQ[b*u - a*v, 0] && PosQ[(b*u - a*v)/a]] /; PiecewiseLine 
arQ[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 2599

Int[(u_)^(m_)*(v_)^(n_.), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Sim 
plify[D[v, x]]}, Simp[u^(m + 1)*(v^n/(a*(m + 1))), x] - Simp[b*(n/(a*(m + 1 
)))   Int[u^(m + 1)*v^(n - 1), x], x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m, n} 
, x] && PiecewiseLinearQ[u, v, x] && NeQ[m, -1] && ((LtQ[m, -1] && GtQ[n, 0 
] &&  !(ILtQ[m + n, -2] && (FractionQ[m] || GeQ[2*n + m + 1, 0]))) || (IGtQ 
[n, 0] && IGtQ[m, 0] && LeQ[n, m]) || (IGtQ[n, 0] &&  !IntegerQ[m]) || (ILt 
Q[m, 0] &&  !IntegerQ[n]))
 

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.74

method	result	size
default	\(2 b^{2} \left (\frac {-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{8 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )}-\frac {\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{8}}{b^{2} x^{2}}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}}\right )}{8 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{\frac {3}{2}}}\right )\)	\(92\)

Input:

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

Output:

2*b^2*((-1/8/(arctanh(tanh(b*x+a))-b*x)*arctanh(tanh(b*x+a))^(3/2)-1/8*arc 
tanh(tanh(b*x+a))^(1/2))/b^2/x^2+1/8/(arctanh(tanh(b*x+a))-b*x)^(3/2)*arct 
anh(arctanh(tanh(b*x+a))^(1/2)/(arctanh(tanh(b*x+a))-b*x)^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.93 \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^3} \, dx=\left [\frac {\sqrt {a} b^{2} x^{2} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) - 2 \, {\left (a b x + 2 \, a^{2}\right )} \sqrt {b x + a}}{8 \, a^{2} x^{2}}, -\frac {\sqrt {-a} b^{2} x^{2} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x + a}}\right ) + {\left (a b x + 2 \, a^{2}\right )} \sqrt {b x + a}}{4 \, a^{2} x^{2}}\right ] \] Input:

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

Output:

[1/8*(sqrt(a)*b^2*x^2*log((b*x + 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) - 2*(a* 
b*x + 2*a^2)*sqrt(b*x + a))/(a^2*x^2), -1/4*(sqrt(-a)*b^2*x^2*arctan(sqrt( 
-a)/sqrt(b*x + a)) + (a*b*x + 2*a^2)*sqrt(b*x + a))/(a^2*x^2)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

\[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^3} \, dx=\int { \frac {\sqrt {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )}}{x^{3}} \,d x } \] Input:

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

Output:

integrate(sqrt(arctanh(tanh(b*x + a)))/x^3, x)
 

Giac [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.60 \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^3} \, dx=-\frac {\sqrt {2} {\left (\frac {\sqrt {2} b^{3} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} + \frac {\sqrt {2} {\left ({\left (b x + a\right )}^{\frac {3}{2}} b^{3} + \sqrt {b x + a} a b^{3}\right )}}{a b^{2} x^{2}}\right )}}{8 \, b} \] Input:

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

Output:

-1/8*sqrt(2)*(sqrt(2)*b^3*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a) + sq 
rt(2)*((b*x + a)^(3/2)*b^3 + sqrt(b*x + a)*a*b^3)/(a*b^2*x^2))/b
 

Mupad [B] (verification not implemented)

Time = 7.99 (sec) , antiderivative size = 741, normalized size of antiderivative = 5.93 \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^3} \, dx =\text {Too large to display} \] Input:

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

Output:

(b*(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(exp( 
2*a)*exp(2*b*x) + 1))/2)^(1/2))/(2*x*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - l 
og((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)) - ((log((2 
*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(exp(2*a)*exp(2 
*b*x) + 1))/2)^(1/2)*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*e 
xp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x))/(x^2*(2*log(2/(exp(2*a)*ex 
p(2*b*x) + 1)) - 2*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) 
+ 4*b*x)) + (2^(1/2)*b^2*log(((2*2^(1/2)*a + (log((2*exp(2*a)*exp(2*b*x))/ 
(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(exp(2*a)*exp(2*b*x) + 1))/2)^(1/2)*( 
log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*e 
xp(2*b*x) + 1)) + 2*b*x)^(1/2)*2i - 2^(1/2)*(2*a - log((2*exp(2*a)*exp(2*b 
*x))/(exp(2*a)*exp(2*b*x) + 1)) + log(2/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x 
) + 2^(1/2)*b*x)*((2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) 
+ 1)) + log(2/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^3 - 8*a^3 - 6*a*(2*a - l 
og((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + log(2/(exp(2*a)*ex 
p(2*b*x) + 1)) + 2*b*x)^2 + 12*a^2*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp 
(2*a)*exp(2*b*x) + 1)) + log(2/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x))*4i)/(x 
*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a) 
*exp(2*b*x) + 1)) + 2*b*x)^(1/2)))*1i)/(4*(log(2/(exp(2*a)*exp(2*b*x) + 1) 
) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^(3/...
 

Reduce [F]

\[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^3} \, dx=\frac {-2 \sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}+\left (\int \frac {\sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}}{\mathit {atanh} \left (\tanh \left (b x +a \right )\right ) x^{2}}d x \right ) b \,x^{2}}{4 x^{2}} \] Input:

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

Output:

( - 2*sqrt(atanh(tanh(a + b*x))) + int(sqrt(atanh(tanh(a + b*x)))/(atanh(t 
anh(a + b*x))*x**2),x)*b*x**2)/(4*x**2)