\(\int \frac {1}{x^3 \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx\) [165]

Optimal result

Integrand size = 15, antiderivative size = 224 \[ \int \frac {1}{x^3 \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\frac {35 b^2 \arctan \left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right )}{4 (b x-\text {arctanh}(\tanh (a+b x)))^{9/2}}+\frac {5 b}{4 x \text {arctanh}(\tanh (a+b x))^{7/2}}-\frac {5 b^2}{4 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{7/2}}-\frac {1}{2 x^2 \text {arctanh}(\tanh (a+b x))^{5/2}}+\frac {7 b^2}{4 (b x-\text {arctanh}(\tanh (a+b x)))^2 \text {arctanh}(\tanh (a+b x))^{5/2}}-\frac {35 b^2}{12 (b x-\text {arctanh}(\tanh (a+b x)))^3 \text {arctanh}(\tanh (a+b x))^{3/2}}+\frac {35 b^2}{4 (b x-\text {arctanh}(\tanh (a+b x)))^4 \sqrt {\text {arctanh}(\tanh (a+b x))}} \] Output:

35/4*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)))^(9/2)+5/4*b/x/arctanh(tanh(b*x+a))^(7/2)-5/4 
*b^2/(b*x-arctanh(tanh(b*x+a)))/arctanh(tanh(b*x+a))^(7/2)-1/2/x^2/arctanh 
(tanh(b*x+a))^(5/2)+7/4*b^2/(b*x-arctanh(tanh(b*x+a)))^2/arctanh(tanh(b*x+ 
a))^(5/2)-35/12*b^2/(b*x-arctanh(tanh(b*x+a)))^3/arctanh(tanh(b*x+a))^(3/2 
)+35/4*b^2/(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 = 0.59 \[ \int \frac {1}{x^3 \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=-\frac {35 b^2 \text {arctanh}\left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {-b x+\text {arctanh}(\tanh (a+b x))}}\right )}{4 (-b x+\text {arctanh}(\tanh (a+b x)))^{9/2}}+\frac {-8 b^3 x^3+80 b^2 x^2 \text {arctanh}(\tanh (a+b x))+39 b x \text {arctanh}(\tanh (a+b x))^2-6 \text {arctanh}(\tanh (a+b x))^3}{12 x^2 \text {arctanh}(\tanh (a+b x))^{3/2} (-b x+\text {arctanh}(\tanh (a+b x)))^4} \] Input:

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

Output:

(-35*b^2*ArcTanh[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[-(b*x) + ArcTanh[Tanh[a 
 + b*x]]]])/(4*(-(b*x) + ArcTanh[Tanh[a + b*x]])^(9/2)) + (-8*b^3*x^3 + 80 
*b^2*x^2*ArcTanh[Tanh[a + b*x]] + 39*b*x*ArcTanh[Tanh[a + b*x]]^2 - 6*ArcT 
anh[Tanh[a + b*x]]^3)/(12*x^2*ArcTanh[Tanh[a + b*x]]^(3/2)*(-(b*x) + ArcTa 
nh[Tanh[a + b*x]])^4)
 

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.20, 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, 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.

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

\(\Big \downarrow \) 2599

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

\(\Big \downarrow \) 2599

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

\(\Big \downarrow \) 2594

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

\(\Big \downarrow \) 2594

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

\(\Big \downarrow \) 2594

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

\(\Big \downarrow \) 2594

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

\(\Big \downarrow \) 2592

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

Input:

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

Output:

(-5*b*((-7*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)))/(b*x - ArcTanh[Tanh[a + b*x]])) - 2/(5*(b*x - ArcTanh[Tanh[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*A 
rcTanh[Tanh[a + b*x]]^(7/2))))/4 - 1/(2*x^2*ArcTanh[Tanh[a + b*x]]^(5/2))
 

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 = 157, normalized size of antiderivative = 0.70

method	result	size
default	\(2 b^{2} \left (\frac {3}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{4} \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}+\frac {1}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}+\frac {\frac {\frac {11 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{8}+\left (-\frac {13 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}{8}+\frac {13 b x}{8}\right ) \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{b^{2} x^{2}}-\frac {35 \,\operatorname {arctanh}\left (\frac {\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}}\right )}{8 \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}}}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{4}}\right )\)	\(157\)

Input:

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

Output:

2*b^2*(3/(arctanh(tanh(b*x+a))-b*x)^4/arctanh(tanh(b*x+a))^(1/2)+1/3/(arct 
anh(tanh(b*x+a))-b*x)^3/arctanh(tanh(b*x+a))^(3/2)+1/(arctanh(tanh(b*x+a)) 
-b*x)^4*((11/8*arctanh(tanh(b*x+a))^(3/2)+(-13/8*arctanh(tanh(b*x+a))+13/8 
*b*x)*arctanh(tanh(b*x+a))^(1/2))/b^2/x^2-35/8/(arctanh(tanh(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 = 252, normalized size of antiderivative = 1.12 \[ \int \frac {1}{x^3 \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\left [\frac {105 \, {\left (b^{4} x^{4} + 2 \, a b^{3} x^{3} + a^{2} b^{2} x^{2}\right )} \sqrt {a} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (105 \, a b^{3} x^{3} + 140 \, a^{2} b^{2} x^{2} + 21 \, a^{3} b x - 6 \, a^{4}\right )} \sqrt {b x + a}}{24 \, {\left (a^{5} b^{2} x^{4} + 2 \, a^{6} b x^{3} + a^{7} x^{2}\right )}}, \frac {105 \, {\left (b^{4} x^{4} + 2 \, a b^{3} x^{3} + a^{2} b^{2} x^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x + a}}\right ) + {\left (105 \, a b^{3} x^{3} + 140 \, a^{2} b^{2} x^{2} + 21 \, a^{3} b x - 6 \, a^{4}\right )} \sqrt {b x + a}}{12 \, {\left (a^{5} b^{2} x^{4} + 2 \, a^{6} b x^{3} + a^{7} x^{2}\right )}}\right ] \] Input:

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

Output:

[1/24*(105*(b^4*x^4 + 2*a*b^3*x^3 + a^2*b^2*x^2)*sqrt(a)*log((b*x - 2*sqrt 
(b*x + a)*sqrt(a) + 2*a)/x) + 2*(105*a*b^3*x^3 + 140*a^2*b^2*x^2 + 21*a^3* 
b*x - 6*a^4)*sqrt(b*x + a))/(a^5*b^2*x^4 + 2*a^6*b*x^3 + a^7*x^2), 1/12*(1 
05*(b^4*x^4 + 2*a*b^3*x^3 + a^2*b^2*x^2)*sqrt(-a)*arctan(sqrt(-a)/sqrt(b*x 
 + a)) + (105*a*b^3*x^3 + 140*a^2*b^2*x^2 + 21*a^3*b*x - 6*a^4)*sqrt(b*x + 
 a))/(a^5*b^2*x^4 + 2*a^6*b*x^3 + a^7*x^2)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.42 \[ \int \frac {1}{x^3 \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\frac {35 \, b^{2} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{4 \, \sqrt {-a} a^{4}} + \frac {2 \, {\left (9 \, {\left (b x + a\right )} b^{2} + a b^{2}\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{4}} + \frac {11 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{2} - 13 \, \sqrt {b x + a} a b^{2}}{4 \, a^{4} b^{2} x^{2}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 9.22 (sec) , antiderivative size = 1514, normalized size of antiderivative = 6.76 \[ \int \frac {1}{x^3 \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

((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)*((4*(2*b*(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) - 7*b* 
(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*(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)*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)) + (56*b^2*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)*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)*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))^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)*((140*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)) + (2^(1/2)*b^2*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)*(lo...
 

Reduce [F]

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

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

Output:

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