\(\int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))^3} \, dx\) [213]

Optimal result

Integrand size = 15, antiderivative size = 201 \[ \int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))^3} \, dx=-\frac {35 b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right )}{4 (b x-\text {arctanh}(\tanh (a+b x)))^{9/2}}+\frac {35 b}{4 \sqrt {x} (b x-\text {arctanh}(\tanh (a+b x)))^4}+\frac {35}{12 x^{3/2} (b x-\text {arctanh}(\tanh (a+b x)))^3}+\frac {7}{4 b x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))^2}+\frac {5}{4 b^2 x^{7/2} (b x-\text {arctanh}(\tanh (a+b x)))}-\frac {1}{2 b x^{5/2} \text {arctanh}(\tanh (a+b x))^2}+\frac {5}{4 b^2 x^{7/2} \text {arctanh}(\tanh (a+b x))} \] Output:

-35/4*b^(3/2)*arctanh(b^(1/2)*x^(1/2)/(b*x-arctanh(tanh(b*x+a)))^(1/2))/(b 
*x-arctanh(tanh(b*x+a)))^(9/2)+35/4*b/x^(1/2)/(b*x-arctanh(tanh(b*x+a)))^4 
+35/12/x^(3/2)/(b*x-arctanh(tanh(b*x+a)))^3+7/4/b/x^(5/2)/(b*x-arctanh(tan 
h(b*x+a)))^2+5/4/b^2/x^(7/2)/(b*x-arctanh(tanh(b*x+a)))-1/2/b/x^(5/2)/arct 
anh(tanh(b*x+a))^2+5/4/b^2/x^(7/2)/arctanh(tanh(b*x+a))
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))^3} \, dx=\frac {1}{12} \left (\frac {105 b^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {-b x+\text {arctanh}(\tanh (a+b x))}}\right )}{(-b x+\text {arctanh}(\tanh (a+b x)))^{9/2}}+\frac {80 b x-8 \text {arctanh}(\tanh (a+b x))}{x^{3/2} (-b x+\text {arctanh}(\tanh (a+b x)))^4}+\frac {33 b^2 \sqrt {x}}{\text {arctanh}(\tanh (a+b x)) (-b x+\text {arctanh}(\tanh (a+b x)))^4}+\frac {6 b^2 \sqrt {x}}{\text {arctanh}(\tanh (a+b x))^2 (-b x+\text {arctanh}(\tanh (a+b x)))^3}\right ) \] Input:

Integrate[1/(x^(5/2)*ArcTanh[Tanh[a + b*x]]^3),x]
 

Output:

((105*b^(3/2)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x] 
]]])/(-(b*x) + ArcTanh[Tanh[a + b*x]])^(9/2) + (80*b*x - 8*ArcTanh[Tanh[a 
+ b*x]])/(x^(3/2)*(-(b*x) + ArcTanh[Tanh[a + b*x]])^4) + (33*b^2*Sqrt[x])/ 
(ArcTanh[Tanh[a + b*x]]*(-(b*x) + ArcTanh[Tanh[a + b*x]])^4) + (6*b^2*Sqrt 
[x])/(ArcTanh[Tanh[a + b*x]]^2*(-(b*x) + ArcTanh[Tanh[a + b*x]])^3))/12
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.28, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {2599, 2599, 2594, 2594, 2594, 2594, 2593}

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.

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

\(\Big \downarrow \) 2599

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

\(\Big \downarrow \) 2599

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

\(\Big \downarrow \) 2594

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

\(\Big \downarrow \) 2594

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

\(\Big \downarrow \) 2594

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

\(\Big \downarrow \) 2594

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

\(\Big \downarrow \) 2593

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

Input:

Int[1/(x^(5/2)*ArcTanh[Tanh[a + b*x]]^3),x]
 

Output:

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

Defintions of rubi rules used

rule 2593

Int[1/((u_)*Sqrt[v_]), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simpli 
fy[D[v, x]]}, Simp[-2*(ArcTanh[Sqrt[v]/Rt[-(b*u - a*v)/a, 2]]/(a*Rt[-(b*u - 
 a*v)/a, 2])), x] /; NeQ[b*u - a*v, 0] && NegQ[(b*u - a*v)/a]] /; Piecewise 
LinearQ[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 = 1.59 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.68

method	result	size
derivativedivides	\(\frac {2 b^{2} \left (\frac {\frac {11 b \,x^{\frac {3}{2}}}{8}+\left (\frac {13 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}{8}-\frac {13 b x}{8}\right ) \sqrt {x}}{\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2}}+\frac {35 \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{8 \sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{4}}-\frac {2}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{3} x^{\frac {3}{2}}}+\frac {6 b}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{4} \sqrt {x}}\)	\(136\)
default	\(\frac {2 b^{2} \left (\frac {\frac {11 b \,x^{\frac {3}{2}}}{8}+\left (\frac {13 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}{8}-\frac {13 b x}{8}\right ) \sqrt {x}}{\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2}}+\frac {35 \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{8 \sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{4}}-\frac {2}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{3} x^{\frac {3}{2}}}+\frac {6 b}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{4} \sqrt {x}}\)	\(136\)
risch	\(\text {Expression too large to display}\)	\(12319\)

Input:

int(1/x^(5/2)/arctanh(tanh(b*x+a))^3,x,method=_RETURNVERBOSE)
 

Output:

2/(arctanh(tanh(b*x+a))-b*x)^4*b^2*((11/8*b*x^(3/2)+(13/8*arctanh(tanh(b*x 
+a))-13/8*b*x)*x^(1/2))/arctanh(tanh(b*x+a))^2+35/8/((arctanh(tanh(b*x+a)) 
-b*x)*b)^(1/2)*arctan(b*x^(1/2)/((arctanh(tanh(b*x+a))-b*x)*b)^(1/2)))-2/3 
/(arctanh(tanh(b*x+a))-b*x)^3/x^(3/2)+6/(arctanh(tanh(b*x+a))-b*x)^4*b/x^( 
1/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.22 \[ \int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))^3} \, dx=\left [\frac {105 \, {\left (b^{3} x^{4} + 2 \, a b^{2} x^{3} + a^{2} b x^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x + 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) + 2 \, {\left (105 \, b^{3} x^{3} + 175 \, a b^{2} x^{2} + 56 \, a^{2} b x - 8 \, a^{3}\right )} \sqrt {x}}{24 \, {\left (a^{4} b^{2} x^{4} + 2 \, a^{5} b x^{3} + a^{6} x^{2}\right )}}, \frac {105 \, {\left (b^{3} x^{4} + 2 \, a b^{2} x^{3} + a^{2} b x^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\sqrt {x} \sqrt {\frac {b}{a}}\right ) + {\left (105 \, b^{3} x^{3} + 175 \, a b^{2} x^{2} + 56 \, a^{2} b x - 8 \, a^{3}\right )} \sqrt {x}}{12 \, {\left (a^{4} b^{2} x^{4} + 2 \, a^{5} b x^{3} + a^{6} x^{2}\right )}}\right ] \] Input:

integrate(1/x^(5/2)/arctanh(tanh(b*x+a))^3,x, algorithm="fricas")
 

Output:

[1/24*(105*(b^3*x^4 + 2*a*b^2*x^3 + a^2*b*x^2)*sqrt(-b/a)*log((b*x + 2*a*s 
qrt(x)*sqrt(-b/a) - a)/(b*x + a)) + 2*(105*b^3*x^3 + 175*a*b^2*x^2 + 56*a^ 
2*b*x - 8*a^3)*sqrt(x))/(a^4*b^2*x^4 + 2*a^5*b*x^3 + a^6*x^2), 1/12*(105*( 
b^3*x^4 + 2*a*b^2*x^3 + a^2*b*x^2)*sqrt(b/a)*arctan(sqrt(x)*sqrt(b/a)) + ( 
105*b^3*x^3 + 175*a*b^2*x^2 + 56*a^2*b*x - 8*a^3)*sqrt(x))/(a^4*b^2*x^4 + 
2*a^5*b*x^3 + a^6*x^2)]
 

Sympy [F]

\[ \int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))^3} \, dx=\int \frac {1}{x^{\frac {5}{2}} \operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \] Input:

integrate(1/x**(5/2)/atanh(tanh(b*x+a))**3,x)
 

Output:

Integral(1/(x**(5/2)*atanh(tanh(a + b*x))**3), x)
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.43 \[ \int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))^3} \, dx=\frac {105 \, b^{3} x^{3} + 175 \, a b^{2} x^{2} + 56 \, a^{2} b x - 8 \, a^{3}}{12 \, {\left (a^{4} b^{2} x^{\frac {7}{2}} + 2 \, a^{5} b x^{\frac {5}{2}} + a^{6} x^{\frac {3}{2}}\right )}} + \frac {35 \, b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{4}} \] Input:

integrate(1/x^(5/2)/arctanh(tanh(b*x+a))^3,x, algorithm="maxima")
 

Output:

1/12*(105*b^3*x^3 + 175*a*b^2*x^2 + 56*a^2*b*x - 8*a^3)/(a^4*b^2*x^(7/2) + 
 2*a^5*b*x^(5/2) + a^6*x^(3/2)) + 35/4*b^2*arctan(b*sqrt(x)/sqrt(a*b))/(sq 
rt(a*b)*a^4)
 

Giac [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.35 \[ \int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))^3} \, dx=\frac {35 \, b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{4}} + \frac {2 \, {\left (9 \, b x - a\right )}}{3 \, a^{4} x^{\frac {3}{2}}} + \frac {11 \, b^{3} x^{\frac {3}{2}} + 13 \, a b^{2} \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} a^{4}} \] Input:

integrate(1/x^(5/2)/arctanh(tanh(b*x+a))^3,x, algorithm="giac")
 

Output:

35/4*b^2*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^4) + 2/3*(9*b*x - a)/(a^ 
4*x^(3/2)) + 1/4*(11*b^3*x^(3/2) + 13*a*b^2*sqrt(x))/((b*x + a)^2*a^4)
 

Mupad [B] (verification not implemented)

Time = 4.70 (sec) , antiderivative size = 1362, normalized size of antiderivative = 6.78 \[ \int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))^3} \, dx=\text {Too large to display} \] Input:

int(1/(x^(5/2)*atanh(tanh(a + b*x))^3),x)
 

Output:

(x^(1/2)*((2*(2*b*(3*log(2/(exp(2*a)*exp(2*b*x) + 1)) - 3*log((2*exp(2*a)* 
exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 6*b*x) - 14*b*(log(2/(exp(2*a)*ex 
p(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*(2*a*b - b*(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))*(log(2/(exp(2*a)*e 
xp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 
 2*b*x)) + (56*b^2*x)/(3*(2*a*b - b*(2*a - log((2*exp(2*a)*exp(2*b*x))/(ex 
p(2*a)*exp(2*b*x) + 1)) + log(2/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*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*b*x^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 - (x^(1/ 
2)*((280*b)/(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)^3) - (280*b^2*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))/(2*b*x^2 - 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)) + (70*2^(1/2)*b^(3/2 
)*log((b^(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^(1/2)*(log(2/(exp(2*a)*e 
xp(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*b^(1/2)*x^(1/2)*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*...
 

Reduce [F]

\[ \int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))^3} \, dx=\int \frac {1}{\sqrt {x}\, \mathit {atanh} \left (\tanh \left (b x +a \right )\right )^{3} x^{2}}d x \] Input:

int(1/x^(5/2)/atanh(tanh(b*x+a))^3,x)
 

Output:

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