\(\int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^6} \, dx\) [139]

Optimal result

Integrand size = 15, antiderivative size = 221 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^6} \, dx=\frac {3 b^5 \arctan \left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right )}{128 (b x-\text {arctanh}(\tanh (a+b x)))^{5/2}}+\frac {b^4}{128 x \text {arctanh}(\tanh (a+b x))^{3/2}}-\frac {b^5}{128 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{3/2}}-\frac {b^3}{64 x^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}+\frac {3 b^5}{128 (b x-\text {arctanh}(\tanh (a+b x)))^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}-\frac {b^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}{16 x^3}-\frac {b \text {arctanh}(\tanh (a+b x))^{3/2}}{8 x^4}-\frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{5 x^5} \] Output:

3/128*b^5*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)+1/128*b^4/x/arctanh(tanh(b*x+a))^(3/2 
)-1/128*b^5/(b*x-arctanh(tanh(b*x+a)))/arctanh(tanh(b*x+a))^(3/2)-1/64*b^3 
/x^2/arctanh(tanh(b*x+a))^(1/2)+3/128*b^5/(b*x-arctanh(tanh(b*x+a)))^2/arc 
tanh(tanh(b*x+a))^(1/2)-1/16*b^2*arctanh(tanh(b*x+a))^(1/2)/x^3-1/8*b*arct 
anh(tanh(b*x+a))^(3/2)/x^4-1/5*arctanh(tanh(b*x+a))^(5/2)/x^5
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.68 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^6} \, dx=\frac {1}{640} \left (-\frac {15 b^5 \text {arctanh}\left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {-b x+\text {arctanh}(\tanh (a+b x))}}\right )}{(-b x+\text {arctanh}(\tanh (a+b x)))^{5/2}}-\frac {\sqrt {\text {arctanh}(\tanh (a+b x))} \left (15 b^4 x^4+10 b^3 x^3 \text {arctanh}(\tanh (a+b x))+8 b^2 x^2 \text {arctanh}(\tanh (a+b x))^2-176 b x \text {arctanh}(\tanh (a+b x))^3+128 \text {arctanh}(\tanh (a+b x))^4\right )}{x^5 (-b x+\text {arctanh}(\tanh (a+b x)))^2}\right ) \] Input:

Integrate[ArcTanh[Tanh[a + b*x]]^(5/2)/x^6,x]
 

Output:

((-15*b^5*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]]]*(15*b^4*x^4 + 10*b^3*x^3*ArcTanh[Tanh[a + b*x]] + 8*b^2*x^2*Arc 
Tanh[Tanh[a + b*x]]^2 - 176*b*x*ArcTanh[Tanh[a + b*x]]^3 + 128*ArcTanh[Tan 
h[a + b*x]]^4))/(x^5*(-(b*x) + ArcTanh[Tanh[a + b*x]])^2))/640
 

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.10, 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, 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.

\(\displaystyle \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^6} \, dx\)

\(\Big \downarrow \) 2599

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

\(\Big \downarrow \) 2599

\(\displaystyle \frac {1}{2} b \left (\frac {3}{8} b \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^4}dx-\frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{4 x^4}\right )-\frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{5 x^5}\)

\(\Big \downarrow \) 2599

\(\displaystyle \frac {1}{2} b \left (\frac {3}{8} b \left (\frac {1}{6} b \int \frac {1}{x^3 \sqrt {\text {arctanh}(\tanh (a+b x))}}dx-\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{3 x^3}\right )-\frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{4 x^4}\right )-\frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{5 x^5}\)

\(\Big \downarrow \) 2599

\(\displaystyle \frac {1}{2} b \left (\frac {3}{8} b \left (\frac {1}{6} b \left (-\frac {1}{4} b \int \frac {1}{x^2 \text {arctanh}(\tanh (a+b x))^{3/2}}dx-\frac {1}{2 x^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}\right )-\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{3 x^3}\right )-\frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{4 x^4}\right )-\frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{5 x^5}\)

\(\Big \downarrow \) 2599

\(\displaystyle \frac {1}{2} b \left (\frac {3}{8} b \left (\frac {1}{6} b \left (-\frac {1}{4} b \left (-\frac {3}{2} b \int \frac {1}{x \text {arctanh}(\tanh (a+b x))^{5/2}}dx-\frac {1}{x \text {arctanh}(\tanh (a+b x))^{3/2}}\right )-\frac {1}{2 x^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}\right )-\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{3 x^3}\right )-\frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{4 x^4}\right )-\frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{5 x^5}\)

\(\Big \downarrow \) 2594

\(\displaystyle \frac {1}{2} b \left (\frac {3}{8} b \left (\frac {1}{6} b \left (-\frac {1}{4} b \left (-\frac {3}{2} b \left (-\frac {\int \frac {1}{x \text {arctanh}(\tanh (a+b x))^{3/2}}dx}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{3 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{3/2}}\right )-\frac {1}{x \text {arctanh}(\tanh (a+b x))^{3/2}}\right )-\frac {1}{2 x^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}\right )-\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{3 x^3}\right )-\frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{4 x^4}\right )-\frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{5 x^5}\)

\(\Big \downarrow \) 2594

\(\displaystyle \frac {1}{2} b \left (\frac {3}{8} b \left (\frac {1}{6} b \left (-\frac {1}{4} b \left (-\frac {3}{2} b \left (-\frac {-\frac {\int \frac {1}{x \sqrt {\text {arctanh}(\tanh (a+b x))}}dx}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{(b x-\text {arctanh}(\tanh (a+b x))) \sqrt {\text {arctanh}(\tanh (a+b x))}}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{3 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{3/2}}\right )-\frac {1}{x \text {arctanh}(\tanh (a+b x))^{3/2}}\right )-\frac {1}{2 x^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}\right )-\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{3 x^3}\right )-\frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{4 x^4}\right )-\frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{5 x^5}\)

\(\Big \downarrow \) 2592

\(\displaystyle \frac {1}{2} b \left (\frac {3}{8} b \left (\frac {1}{6} b \left (-\frac {1}{4} b \left (-\frac {3}{2} b \left (-\frac {-\frac {2 \arctan \left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right )}{(b x-\text {arctanh}(\tanh (a+b x)))^{3/2}}-\frac {2}{(b x-\text {arctanh}(\tanh (a+b x))) \sqrt {\text {arctanh}(\tanh (a+b x))}}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{3 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{3/2}}\right )-\frac {1}{x \text {arctanh}(\tanh (a+b x))^{3/2}}\right )-\frac {1}{2 x^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}\right )-\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{3 x^3}\right )-\frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{4 x^4}\right )-\frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{5 x^5}\)

Input:

Int[ArcTanh[Tanh[a + b*x]]^(5/2)/x^6,x]
 

Output:

-1/5*ArcTanh[Tanh[a + b*x]]^(5/2)/x^5 + (b*((3*b*((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[Tanh[a + b*x]])) - 2/(3*(b* 
x - ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a + b*x]]^(3/2))))/2 - 1/(x*ArcTa 
nh[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)))/8 - ArcTanh[Tanh[a + b*x]]^(3/2)/(4 
*x^4)))/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 = 1.37 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.19

method	result	size
default	\(2 b^{5} \left (\frac {\frac {3 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {9}{2}}}{256 \left (a^{2}+2 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )+\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}\right )}-\frac {7 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{128 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )}-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{10}+\left (\frac {7 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}{128}-\frac {7 b x}{128}\right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}+\left (-\frac {3 a^{2}}{256}-\frac {3 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )}{128}-\frac {3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{256}\right ) \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{b^{5} x^{5}}-\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}}\right )}{256 \left (a^{2}+2 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )+\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}\right ) \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}}\right )\)	\(262\)

Input:

int(arctanh(tanh(b*x+a))^(5/2)/x^6,x,method=_RETURNVERBOSE)
 

Output:

2*b^5*((3/256/(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))^(9/2)-7/128/(arctanh(tanh(b*x+a))-b*x)*arct 
anh(tanh(b*x+a))^(7/2)-1/10*arctanh(tanh(b*x+a))^(5/2)+(7/128*arctanh(tanh 
(b*x+a))-7/128*b*x)*arctanh(tanh(b*x+a))^(3/2)+(-3/256*a^2-3/128*a*(arctan 
h(tanh(b*x+a))-b*x-a)-3/256*(arctanh(tanh(b*x+a))-b*x-a)^2)*arctanh(tanh(b 
*x+a))^(1/2))/b^5/x^5-3/256/(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))-b*x)^(1/2)*arctanh(arctanh(t 
anh(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 = 186, normalized size of antiderivative = 0.84 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^6} \, dx=\left [\frac {15 \, \sqrt {a} b^{5} x^{5} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (15 \, a b^{4} x^{4} - 10 \, a^{2} b^{3} x^{3} - 248 \, a^{3} b^{2} x^{2} - 336 \, a^{4} b x - 128 \, a^{5}\right )} \sqrt {b x + a}}{1280 \, a^{3} x^{5}}, \frac {15 \, \sqrt {-a} b^{5} x^{5} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x + a}}\right ) + {\left (15 \, a b^{4} x^{4} - 10 \, a^{2} b^{3} x^{3} - 248 \, a^{3} b^{2} x^{2} - 336 \, a^{4} b x - 128 \, a^{5}\right )} \sqrt {b x + a}}{640 \, a^{3} x^{5}}\right ] \] Input:

integrate(arctanh(tanh(b*x+a))^(5/2)/x^6,x, algorithm="fricas")
 

Output:

[1/1280*(15*sqrt(a)*b^5*x^5*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 
 2*(15*a*b^4*x^4 - 10*a^2*b^3*x^3 - 248*a^3*b^2*x^2 - 336*a^4*b*x - 128*a^ 
5)*sqrt(b*x + a))/(a^3*x^5), 1/640*(15*sqrt(-a)*b^5*x^5*arctan(sqrt(-a)/sq 
rt(b*x + a)) + (15*a*b^4*x^4 - 10*a^2*b^3*x^3 - 248*a^3*b^2*x^2 - 336*a^4* 
b*x - 128*a^5)*sqrt(b*x + a))/(a^3*x^5)]
 

Sympy [F]

\[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^6} \, dx=\int \frac {\operatorname {atanh}^{\frac {5}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}{x^{6}}\, dx \] Input:

integrate(atanh(tanh(b*x+a))**(5/2)/x**6,x)
 

Output:

Integral(atanh(tanh(a + b*x))**(5/2)/x**6, x)
 

Maxima [F]

\[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^6} \, dx=\int { \frac {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{\frac {5}{2}}}{x^{6}} \,d x } \] Input:

integrate(arctanh(tanh(b*x+a))^(5/2)/x^6,x, algorithm="maxima")
 

Output:

integrate(arctanh(tanh(b*x + a))^(5/2)/x^6, x)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.48 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^6} \, dx=\frac {1}{1280} \, \sqrt {2} b^{5} {\left (\frac {15 \, \sqrt {2} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {\sqrt {2} {\left (15 \, {\left (b x + a\right )}^{\frac {9}{2}} - 70 \, {\left (b x + a\right )}^{\frac {7}{2}} a - 128 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} + 70 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} - 15 \, \sqrt {b x + a} a^{4}\right )}}{a^{2} b^{5} x^{5}}\right )} \] Input:

integrate(arctanh(tanh(b*x+a))^(5/2)/x^6,x, algorithm="giac")
 

Output:

1/1280*sqrt(2)*b^5*(15*sqrt(2)*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^ 
2) + sqrt(2)*(15*(b*x + a)^(9/2) - 70*(b*x + a)^(7/2)*a - 128*(b*x + a)^(5 
/2)*a^2 + 70*(b*x + a)^(3/2)*a^3 - 15*sqrt(b*x + a)*a^4)/(a^2*b^5*x^5))
 

Mupad [B] (verification not implemented)

Time = 8.65 (sec) , antiderivative size = 1292, normalized size of antiderivative = 5.85 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^6} \, dx=\text {Too large to display} \] Input:

int(atanh(tanh(a + b*x))^(5/2)/x^6,x)
 

Output:

(3*b^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))/(32*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)^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)*(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)/(4*x^5*(5*log(2/( 
exp(2*a)*exp(2*b*x) + 1)) - 5*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2* 
b*x) + 1)) + 10*b*x)) + (b^3*(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))/(16*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)) + (2^(1/2)*b^5*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)*(l 
og(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*ex 
p(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)^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 - 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 - 32*a^5 - 10*...
 

Reduce [F]

\[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^6} \, dx=\frac {-32 \sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}\, \mathit {atanh} \left (\tanh \left (b x +a \right )\right )^{2}-20 \sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}\, \mathit {atanh} \left (\tanh \left (b x +a \right )\right ) b x -10 \sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}\, b^{2} x^{2}+5 \left (\int \frac {\sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}}{\mathit {atanh} \left (\tanh \left (b x +a \right )\right ) x^{3}}d x \right ) b^{3} x^{5}}{160 x^{5}} \] Input:

int(atanh(tanh(b*x+a))^(5/2)/x^6,x)
 

Output:

( - 32*sqrt(atanh(tanh(a + b*x)))*atanh(tanh(a + b*x))**2 - 20*sqrt(atanh( 
tanh(a + b*x)))*atanh(tanh(a + b*x))*b*x - 10*sqrt(atanh(tanh(a + b*x)))*b 
**2*x**2 + 5*int(sqrt(atanh(tanh(a + b*x)))/(atanh(tanh(a + b*x))*x**3),x) 
*b**3*x**5)/(160*x**5)