\(\int \frac {1}{x^4 \text {arctanh}(\tanh (a+b x))^{3/2}} \, dx\) [157]

Optimal result

Integrand size = 15, antiderivative size = 245 \[ \int \frac {1}{x^4 \text {arctanh}(\tanh (a+b x))^{3/2}} \, dx=-\frac {35 b^3 \arctan \left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right )}{8 (b x-\text {arctanh}(\tanh (a+b x)))^{9/2}}-\frac {5 b^2}{8 x \text {arctanh}(\tanh (a+b x))^{7/2}}+\frac {5 b^3}{8 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{7/2}}+\frac {b}{4 x^2 \text {arctanh}(\tanh (a+b x))^{5/2}}-\frac {7 b^3}{8 (b x-\text {arctanh}(\tanh (a+b x)))^2 \text {arctanh}(\tanh (a+b x))^{5/2}}-\frac {1}{3 x^3 \text {arctanh}(\tanh (a+b x))^{3/2}}+\frac {35 b^3}{24 (b x-\text {arctanh}(\tanh (a+b x)))^3 \text {arctanh}(\tanh (a+b x))^{3/2}}-\frac {35 b^3}{8 (b x-\text {arctanh}(\tanh (a+b x)))^4 \sqrt {\text {arctanh}(\tanh (a+b x))}} \] Output:

-35/8*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)))^(9/2)-5/8*b^2/x/arctanh(tanh(b*x+a))^(7/2)+ 
5/8*b^3/(b*x-arctanh(tanh(b*x+a)))/arctanh(tanh(b*x+a))^(7/2)+1/4*b/x^2/ar 
ctanh(tanh(b*x+a))^(5/2)-7/8*b^3/(b*x-arctanh(tanh(b*x+a)))^2/arctanh(tanh 
(b*x+a))^(5/2)-1/3/x^3/arctanh(tanh(b*x+a))^(3/2)+35/24*b^3/(b*x-arctanh(t 
anh(b*x+a)))^3/arctanh(tanh(b*x+a))^(3/2)-35/8*b^3/(b*x-arctanh(tanh(b*x+a 
)))^4/arctanh(tanh(b*x+a))^(1/2)
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.54 \[ \int \frac {1}{x^4 \text {arctanh}(\tanh (a+b x))^{3/2}} \, dx=\frac {35 b^3 \text {arctanh}\left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {-b x+\text {arctanh}(\tanh (a+b x))}}\right )}{8 (-b x+\text {arctanh}(\tanh (a+b x)))^{9/2}}-\frac {48 b^3 x^3+87 b^2 x^2 \text {arctanh}(\tanh (a+b x))-38 b x \text {arctanh}(\tanh (a+b x))^2+8 \text {arctanh}(\tanh (a+b x))^3}{24 x^3 \sqrt {\text {arctanh}(\tanh (a+b x))} (-b x+\text {arctanh}(\tanh (a+b x)))^4} \] Input:

Integrate[1/(x^4*ArcTanh[Tanh[a + b*x]]^(3/2)),x]
 

Output:

(35*b^3*ArcTanh[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[-(b*x) + ArcTanh[Tanh[a 
+ b*x]]]])/(8*(-(b*x) + ArcTanh[Tanh[a + b*x]])^(9/2)) - (48*b^3*x^3 + 87* 
b^2*x^2*ArcTanh[Tanh[a + b*x]] - 38*b*x*ArcTanh[Tanh[a + b*x]]^2 + 8*ArcTa 
nh[Tanh[a + b*x]]^3)/(24*x^3*Sqrt[ArcTanh[Tanh[a + b*x]]]*(-(b*x) + ArcTan 
h[Tanh[a + b*x]])^4)
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.19, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {2599, 2599, 2599, 2594, 2594, 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.

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

\(\Big \downarrow \) 2599

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

\(\Big \downarrow \) 2599

\(\displaystyle -\frac {1}{2} b \left (-\frac {5}{4} b \int \frac {1}{x^2 \text {arctanh}(\tanh (a+b x))^{7/2}}dx-\frac {1}{2 x^2 \text {arctanh}(\tanh (a+b x))^{5/2}}\right )-\frac {1}{3 x^3 \text {arctanh}(\tanh (a+b x))^{3/2}}\)

\(\Big \downarrow \) 2599

\(\displaystyle -\frac {1}{2} b \left (-\frac {5}{4} b \left (-\frac {7}{2} b \int \frac {1}{x \text {arctanh}(\tanh (a+b x))^{9/2}}dx-\frac {1}{x \text {arctanh}(\tanh (a+b x))^{7/2}}\right )-\frac {1}{2 x^2 \text {arctanh}(\tanh (a+b x))^{5/2}}\right )-\frac {1}{3 x^3 \text {arctanh}(\tanh (a+b x))^{3/2}}\)

\(\Big \downarrow \) 2594

\(\displaystyle -\frac {1}{2} b \left (-\frac {5}{4} b \left (-\frac {7}{2} b \left (-\frac {\int \frac {1}{x \text {arctanh}(\tanh (a+b x))^{7/2}}dx}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{7 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{7/2}}\right )-\frac {1}{x \text {arctanh}(\tanh (a+b x))^{7/2}}\right )-\frac {1}{2 x^2 \text {arctanh}(\tanh (a+b x))^{5/2}}\right )-\frac {1}{3 x^3 \text {arctanh}(\tanh (a+b x))^{3/2}}\)

\(\Big \downarrow \) 2594

\(\displaystyle -\frac {1}{2} b \left (-\frac {5}{4} b \left (-\frac {7}{2} b \left (-\frac {-\frac {\int \frac {1}{x \text {arctanh}(\tanh (a+b x))^{5/2}}dx}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{5 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{5/2}}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{7 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{7/2}}\right )-\frac {1}{x \text {arctanh}(\tanh (a+b x))^{7/2}}\right )-\frac {1}{2 x^2 \text {arctanh}(\tanh (a+b x))^{5/2}}\right )-\frac {1}{3 x^3 \text {arctanh}(\tanh (a+b x))^{3/2}}\)

\(\Big \downarrow \) 2594

\(\displaystyle -\frac {1}{2} b \left (-\frac {5}{4} b \left (-\frac {7}{2} b \left (-\frac {-\frac {-\frac {\int \frac {1}{x \text {arctanh}(\tanh (a+b x))^{3/2}}dx}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{3 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{3/2}}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{5 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{5/2}}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{7 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{7/2}}\right )-\frac {1}{x \text {arctanh}(\tanh (a+b x))^{7/2}}\right )-\frac {1}{2 x^2 \text {arctanh}(\tanh (a+b x))^{5/2}}\right )-\frac {1}{3 x^3 \text {arctanh}(\tanh (a+b x))^{3/2}}\)

\(\Big \downarrow \) 2594

\(\displaystyle -\frac {1}{2} b \left (-\frac {5}{4} b \left (-\frac {7}{2} b \left (-\frac {-\frac {-\frac {-\frac {\int \frac {1}{x \sqrt {\text {arctanh}(\tanh (a+b x))}}dx}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{(b x-\text {arctanh}(\tanh (a+b x))) \sqrt {\text {arctanh}(\tanh (a+b x))}}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{3 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{3/2}}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{5 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{5/2}}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{7 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{7/2}}\right )-\frac {1}{x \text {arctanh}(\tanh (a+b x))^{7/2}}\right )-\frac {1}{2 x^2 \text {arctanh}(\tanh (a+b x))^{5/2}}\right )-\frac {1}{3 x^3 \text {arctanh}(\tanh (a+b x))^{3/2}}\)

\(\Big \downarrow \) 2592

\(\displaystyle -\frac {1}{2} b \left (-\frac {5}{4} b \left (-\frac {7}{2} b \left (-\frac {-\frac {-\frac {-\frac {2 \arctan \left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right )}{(b x-\text {arctanh}(\tanh (a+b x)))^{3/2}}-\frac {2}{(b x-\text {arctanh}(\tanh (a+b x))) \sqrt {\text {arctanh}(\tanh (a+b x))}}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{3 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{3/2}}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{5 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{5/2}}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{7 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{7/2}}\right )-\frac {1}{x \text {arctanh}(\tanh (a+b x))^{7/2}}\right )-\frac {1}{2 x^2 \text {arctanh}(\tanh (a+b x))^{5/2}}\right )-\frac {1}{3 x^3 \text {arctanh}(\tanh (a+b x))^{3/2}}\)

Input:

Int[1/(x^4*ArcTanh[Tanh[a + b*x]]^(3/2)),x]
 

Output:

-1/2*(b*((-5*b*((-7*b*(-((-((-(((-2*ArcTan[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sq 
rt[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 - Ar 
cTanh[Tanh[a + b*x]])) - 2/(3*(b*x - ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[ 
a + b*x]]^(3/2)))/(b*x - ArcTanh[Tanh[a + b*x]])) - 2/(5*(b*x - ArcTanh[Ta 
nh[a + b*x]])*ArcTanh[Tanh[a + b*x]]^(5/2)))/(b*x - ArcTanh[Tanh[a + b*x]] 
)) - 2/(7*(b*x - ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a + b*x]]^(7/2))))/2 
 - 1/(x*ArcTanh[Tanh[a + b*x]]^(7/2))))/4 - 1/(2*x^2*ArcTanh[Tanh[a + b*x] 
]^(5/2)))) - 1/(3*x^3*ArcTanh[Tanh[a + b*x]]^(3/2))
 

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 = 0.21 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.76

method	result	size
default	\(2 b^{3} \left (-\frac {1}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{4} \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}-\frac {\frac {\frac {19 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{16}+\left (-\frac {17 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}{6}+\frac {17 b x}{6}\right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}+\left (\frac {29 a^{2}}{16}+\frac {29 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )}{8}+\frac {29 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{16}\right ) \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{b^{3} x^{3}}-\frac {35 \,\operatorname {arctanh}\left (\frac {\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}}\right )}{16 \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}}}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{4}}\right )\)	\(186\)

Input:

int(1/x^4/arctanh(tanh(b*x+a))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

2*b^3*(-1/(arctanh(tanh(b*x+a))-b*x)^4/arctanh(tanh(b*x+a))^(1/2)-1/(arcta 
nh(tanh(b*x+a))-b*x)^4*((19/16*arctanh(tanh(b*x+a))^(5/2)+(-17/6*arctanh(t 
anh(b*x+a))+17/6*b*x)*arctanh(tanh(b*x+a))^(3/2)+(29/16*a^2+29/8*a*(arctan 
h(tanh(b*x+a))-b*x-a)+29/16*(arctanh(tanh(b*x+a))-b*x-a)^2)*arctanh(tanh(b 
*x+a))^(1/2))/b^3/x^3-35/16/(arctanh(tanh(b*x+a))-b*x)^(1/2)*arctanh(arcta 
nh(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 = 208, normalized size of antiderivative = 0.85 \[ \int \frac {1}{x^4 \text {arctanh}(\tanh (a+b x))^{3/2}} \, dx=\left [\frac {105 \, {\left (b^{4} x^{4} + a b^{3} x^{3}\right )} \sqrt {a} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) - 2 \, {\left (105 \, a b^{3} x^{3} + 35 \, a^{2} b^{2} x^{2} - 14 \, a^{3} b x + 8 \, a^{4}\right )} \sqrt {b x + a}}{48 \, {\left (a^{5} b x^{4} + a^{6} x^{3}\right )}}, -\frac {105 \, {\left (b^{4} x^{4} + a b^{3} x^{3}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x + a}}\right ) + {\left (105 \, a b^{3} x^{3} + 35 \, a^{2} b^{2} x^{2} - 14 \, a^{3} b x + 8 \, a^{4}\right )} \sqrt {b x + a}}{24 \, {\left (a^{5} b x^{4} + a^{6} x^{3}\right )}}\right ] \] Input:

integrate(1/x^4/arctanh(tanh(b*x+a))^(3/2),x, algorithm="fricas")
 

Output:

[1/48*(105*(b^4*x^4 + a*b^3*x^3)*sqrt(a)*log((b*x + 2*sqrt(b*x + a)*sqrt(a 
) + 2*a)/x) - 2*(105*a*b^3*x^3 + 35*a^2*b^2*x^2 - 14*a^3*b*x + 8*a^4)*sqrt 
(b*x + a))/(a^5*b*x^4 + a^6*x^3), -1/24*(105*(b^4*x^4 + a*b^3*x^3)*sqrt(-a 
)*arctan(sqrt(-a)/sqrt(b*x + a)) + (105*a*b^3*x^3 + 35*a^2*b^2*x^2 - 14*a^ 
3*b*x + 8*a^4)*sqrt(b*x + a))/(a^5*b*x^4 + a^6*x^3)]
 

Sympy [F]

\[ \int \frac {1}{x^4 \text {arctanh}(\tanh (a+b x))^{3/2}} \, dx=\int \frac {1}{x^{4} \operatorname {atanh}^{\frac {3}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \] Input:

integrate(1/x**4/atanh(tanh(b*x+a))**(3/2),x)
 

Output:

Integral(1/(x**4*atanh(tanh(a + b*x))**(3/2)), x)
 

Maxima [F]

\[ \int \frac {1}{x^4 \text {arctanh}(\tanh (a+b x))^{3/2}} \, dx=\int { \frac {1}{x^{4} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{\frac {3}{2}}} \,d x } \] Input:

integrate(1/x^4/arctanh(tanh(b*x+a))^(3/2),x, algorithm="maxima")
 

Output:

integrate(1/(x^4*arctanh(tanh(b*x + a))^(3/2)), x)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.39 \[ \int \frac {1}{x^4 \text {arctanh}(\tanh (a+b x))^{3/2}} \, dx=-\frac {35 \, b^{3} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{8 \, \sqrt {-a} a^{4}} - \frac {2 \, b^{3}}{\sqrt {b x + a} a^{4}} - \frac {57 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{3} - 136 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{3} + 87 \, \sqrt {b x + a} a^{2} b^{3}}{24 \, a^{4} b^{3} x^{3}} \] Input:

integrate(1/x^4/arctanh(tanh(b*x+a))^(3/2),x, algorithm="giac")
 

Output:

-35/8*b^3*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^4) - 2*b^3/(sqrt(b*x 
+ a)*a^4) - 1/24*(57*(b*x + a)^(5/2)*b^3 - 136*(b*x + a)^(3/2)*a*b^3 + 87* 
sqrt(b*x + a)*a^2*b^3)/(a^4*b^3*x^3)
 

Mupad [B] (verification not implemented)

Time = 7.35 (sec) , antiderivative size = 1258, normalized size of antiderivative = 5.13 \[ \int \frac {1}{x^4 \text {arctanh}(\tanh (a+b x))^{3/2}} \, dx=\text {Too large to display} \] Input:

int(1/(x^4*atanh(tanh(a + b*x))^(3/2)),x)
 

Output:

(((38*b^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)^3 - (140*b^3*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)^4)*(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))/(x*(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)))) - (4*(log((2*ex 
p(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^3*(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) + (2^(1/2)*b^3*log( 
(((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*e 
xp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^(1/2)*2i + 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) 
^9 + 144*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)^7 - 672*a^3*(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)^6 + 2016*a^4*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*e 
xp(2*b*x) + 1)) + log(2/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^5 - 4032*a^...
 

Reduce [F]

\[ \int \frac {1}{x^4 \text {arctanh}(\tanh (a+b x))^{3/2}} \, dx=\int \frac {\sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}}{\mathit {atanh} \left (\tanh \left (b x +a \right )\right )^{2} x^{4}}d x \] Input:

int(1/x^4/atanh(tanh(b*x+a))^(3/2),x)
 

Output:

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