\(\int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^4} \, dx\) [119]

Optimal result

Integrand size = 15, antiderivative size = 179 \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^4} \, dx=\frac {b^3 \arctan \left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right )}{8 (b x-\text {arctanh}(\tanh (a+b x)))^{5/2}}+\frac {b^2}{24 x \text {arctanh}(\tanh (a+b x))^{3/2}}-\frac {b^3}{24 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{3/2}}-\frac {b}{12 x^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}+\frac {b^3}{8 (b x-\text {arctanh}(\tanh (a+b x)))^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}-\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{3 x^3} \] Output:

1/8*b^3*arctan(arctanh(tanh(b*x+a))^(1/2)/(b*x-arctanh(tanh(b*x+a)))^(1/2) 
)/(b*x-arctanh(tanh(b*x+a)))^(5/2)+1/24*b^2/x/arctanh(tanh(b*x+a))^(3/2)-1 
/24*b^3/(b*x-arctanh(tanh(b*x+a)))/arctanh(tanh(b*x+a))^(3/2)-1/12*b/x^2/a 
rctanh(tanh(b*x+a))^(1/2)+1/8*b^3/(b*x-arctanh(tanh(b*x+a)))^2/arctanh(tan 
h(b*x+a))^(1/2)-1/3*arctanh(tanh(b*x+a))^(1/2)/x^3
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.64 \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^4} \, dx=\frac {1}{24} \left (-\frac {3 b^3 \text {arctanh}\left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {-b x+\text {arctanh}(\tanh (a+b x))}}\right )}{(-b x+\text {arctanh}(\tanh (a+b x)))^{5/2}}+\frac {\sqrt {\text {arctanh}(\tanh (a+b x))} \left (-3 b^2 x^2+14 b x \text {arctanh}(\tanh (a+b x))-8 \text {arctanh}(\tanh (a+b x))^2\right )}{x^3 (-b x+\text {arctanh}(\tanh (a+b x)))^2}\right ) \] Input:

Integrate[Sqrt[ArcTanh[Tanh[a + b*x]]]/x^4,x]
 

Output:

((-3*b^3*ArcTanh[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[-(b*x) + ArcTanh[Tanh[a 
 + b*x]]]])/(-(b*x) + ArcTanh[Tanh[a + b*x]])^(5/2) + (Sqrt[ArcTanh[Tanh[a 
 + b*x]]]*(-3*b^2*x^2 + 14*b*x*ArcTanh[Tanh[a + b*x]] - 8*ArcTanh[Tanh[a + 
 b*x]]^2))/(x^3*(-(b*x) + ArcTanh[Tanh[a + b*x]])^2))/24
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2599, 2599, 2599, 2594, 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.

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

\(\Big \downarrow \) 2599

\(\displaystyle \frac {1}{6} b \int \frac {1}{x^3 \sqrt {\text {arctanh}(\tanh (a+b x))}}dx-\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{3 x^3}\)

\(\Big \downarrow \) 2599

\(\displaystyle \frac {1}{6} b \left (-\frac {1}{4} b \int \frac {1}{x^2 \text {arctanh}(\tanh (a+b x))^{3/2}}dx-\frac {1}{2 x^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}\right )-\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{3 x^3}\)

\(\Big \downarrow \) 2599

\(\displaystyle \frac {1}{6} b \left (-\frac {1}{4} b \left (-\frac {3}{2} b \int \frac {1}{x \text {arctanh}(\tanh (a+b x))^{5/2}}dx-\frac {1}{x \text {arctanh}(\tanh (a+b x))^{3/2}}\right )-\frac {1}{2 x^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}\right )-\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{3 x^3}\)

\(\Big \downarrow \) 2594

\(\displaystyle \frac {1}{6} b \left (-\frac {1}{4} b \left (-\frac {3}{2} b \left (-\frac {\int \frac {1}{x \text {arctanh}(\tanh (a+b x))^{3/2}}dx}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{3 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{3/2}}\right )-\frac {1}{x \text {arctanh}(\tanh (a+b x))^{3/2}}\right )-\frac {1}{2 x^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}\right )-\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{3 x^3}\)

\(\Big \downarrow \) 2594

\(\displaystyle \frac {1}{6} b \left (-\frac {1}{4} b \left (-\frac {3}{2} b \left (-\frac {-\frac {\int \frac {1}{x \sqrt {\text {arctanh}(\tanh (a+b x))}}dx}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{(b x-\text {arctanh}(\tanh (a+b x))) \sqrt {\text {arctanh}(\tanh (a+b x))}}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{3 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{3/2}}\right )-\frac {1}{x \text {arctanh}(\tanh (a+b x))^{3/2}}\right )-\frac {1}{2 x^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}\right )-\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{3 x^3}\)

\(\Big \downarrow \) 2592

\(\displaystyle \frac {1}{6} b \left (-\frac {1}{4} b \left (-\frac {3}{2} b \left (-\frac {-\frac {2 \arctan \left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right )}{(b x-\text {arctanh}(\tanh (a+b x)))^{3/2}}-\frac {2}{(b x-\text {arctanh}(\tanh (a+b x))) \sqrt {\text {arctanh}(\tanh (a+b x))}}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{3 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{3/2}}\right )-\frac {1}{x \text {arctanh}(\tanh (a+b x))^{3/2}}\right )-\frac {1}{2 x^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}\right )-\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{3 x^3}\)

Input:

Int[Sqrt[ArcTanh[Tanh[a + b*x]]]/x^4,x]
 

Output:

(b*(-1/4*(b*((-3*b*(-(((-2*ArcTan[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[b*x - 
ArcTanh[Tanh[a + b*x]]]])/(b*x - ArcTanh[Tanh[a + b*x]])^(3/2) - 2/((b*x - 
 ArcTanh[Tanh[a + b*x]])*Sqrt[ArcTanh[Tanh[a + b*x]]]))/(b*x - ArcTanh[Tan 
h[a + b*x]])) - 2/(3*(b*x - ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a + b*x]] 
^(3/2))))/2 - 1/(x*ArcTanh[Tanh[a + b*x]]^(3/2)))) - 1/(2*x^2*Sqrt[ArcTanh 
[Tanh[a + b*x]]])))/6 - Sqrt[ArcTanh[Tanh[a + b*x]]]/(3*x^3)
 

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 = 0.21 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.03

method	result	size
default	\(2 b^{3} \left (\frac {\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{16 a^{2}+32 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )+16 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{6 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )}-\frac {\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{16}}{b^{3} x^{3}}-\frac {\operatorname {arctanh}\left (\frac {\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}}\right )}{16 \left (a^{2}+2 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )+\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}\right ) \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}}\right )\)	\(185\)

Input:

int(arctanh(tanh(b*x+a))^(1/2)/x^4,x,method=_RETURNVERBOSE)
 

Output:

2*b^3*((1/16/(a^2+2*a*(arctanh(tanh(b*x+a))-b*x-a)+(arctanh(tanh(b*x+a))-b 
*x-a)^2)*arctanh(tanh(b*x+a))^(5/2)-1/6/(arctanh(tanh(b*x+a))-b*x)*arctanh 
(tanh(b*x+a))^(3/2)-1/16*arctanh(tanh(b*x+a))^(1/2))/b^3/x^3-1/16/(a^2+2*a 
*(arctanh(tanh(b*x+a))-b*x-a)+(arctanh(tanh(b*x+a))-b*x-a)^2)/(arctanh(tan 
h(b*x+a))-b*x)^(1/2)*arctanh(arctanh(tanh(b*x+a))^(1/2)/(arctanh(tanh(b*x+ 
a))-b*x)^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.79 \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^4} \, dx=\left [\frac {3 \, \sqrt {a} b^{3} x^{3} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (3 \, a b^{2} x^{2} - 2 \, a^{2} b x - 8 \, a^{3}\right )} \sqrt {b x + a}}{48 \, a^{3} x^{3}}, \frac {3 \, \sqrt {-a} b^{3} x^{3} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x + a}}\right ) + {\left (3 \, a b^{2} x^{2} - 2 \, a^{2} b x - 8 \, a^{3}\right )} \sqrt {b x + a}}{24 \, a^{3} x^{3}}\right ] \] Input:

integrate(arctanh(tanh(b*x+a))^(1/2)/x^4,x, algorithm="fricas")
 

Output:

[1/48*(3*sqrt(a)*b^3*x^3*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2* 
(3*a*b^2*x^2 - 2*a^2*b*x - 8*a^3)*sqrt(b*x + a))/(a^3*x^3), 1/24*(3*sqrt(- 
a)*b^3*x^3*arctan(sqrt(-a)/sqrt(b*x + a)) + (3*a*b^2*x^2 - 2*a^2*b*x - 8*a 
^3)*sqrt(b*x + a))/(a^3*x^3)]
 

Sympy [F]

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

integrate(atanh(tanh(b*x+a))**(1/2)/x**4,x)
 

Output:

Integral(sqrt(atanh(tanh(a + b*x)))/x**4, x)
 

Maxima [F]

\[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^4} \, dx=\int { \frac {\sqrt {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )}}{x^{4}} \,d x } \] Input:

integrate(arctanh(tanh(b*x+a))^(1/2)/x^4,x, algorithm="maxima")
 

Output:

integrate(sqrt(arctanh(tanh(b*x + a)))/x^4, x)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.45 \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^4} \, dx=\frac {1}{48} \, \sqrt {2} b^{3} {\left (\frac {3 \, \sqrt {2} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {\sqrt {2} {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 8 \, {\left (b x + a\right )}^{\frac {3}{2}} a - 3 \, \sqrt {b x + a} a^{2}\right )}}{a^{2} b^{3} x^{3}}\right )} \] Input:

integrate(arctanh(tanh(b*x+a))^(1/2)/x^4,x, algorithm="giac")
 

Output:

1/48*sqrt(2)*b^3*(3*sqrt(2)*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^2) 
+ sqrt(2)*(3*(b*x + a)^(5/2) - 8*(b*x + a)^(3/2)*a - 3*sqrt(b*x + a)*a^2)/ 
(a^2*b^3*x^3))
 

Mupad [B] (verification not implemented)

Time = 7.57 (sec) , antiderivative size = 964, normalized size of antiderivative = 5.39 \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^4} \, dx=\text {Too large to display} \] Input:

int(atanh(tanh(a + b*x))^(1/2)/x^4,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))/(3*x^2*(2*log(2/(exp(2*a)*exp(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)) + (b 
^2*(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)^2) + (2^(1/ 
2)*b^3*log((((log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - l 
og(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)*exp(2*b*x) + 1)) + 2*b*x)^(1/2)*2 
i - 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)*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)^5 + 40*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)^3 - 80*a^3*(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 - 32*a^5 - 10*a*(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)^ 
4 + 80*a^4*(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...
 

Reduce [F]

\[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^4} \, dx=\frac {-2 \sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}+\left (\int \frac {\sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}}{\mathit {atanh} \left (\tanh \left (b x +a \right )\right ) x^{3}}d x \right ) b \,x^{3}}{6 x^{3}} \] Input:

int(atanh(tanh(b*x+a))^(1/2)/x^4,x)
 

Output:

( - 2*sqrt(atanh(tanh(a + b*x))) + int(sqrt(atanh(tanh(a + b*x)))/(atanh(t 
anh(a + b*x))*x**3),x)*b*x**3)/(6*x**3)