\(\int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^2} \, dx\) [135]

Optimal result

Integrand size = 15, antiderivative size = 110 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^2} \, dx=5 b \arctan \left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right ) (b x-\text {arctanh}(\tanh (a+b x)))^{3/2}-5 b (b x-\text {arctanh}(\tanh (a+b x))) \sqrt {\text {arctanh}(\tanh (a+b x))}+\frac {5}{3} b \text {arctanh}(\tanh (a+b x))^{3/2}-\frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x} \] Output:

5*b*arctan(arctanh(tanh(b*x+a))^(1/2)/(b*x-arctanh(tanh(b*x+a)))^(1/2))*(b 
*x-arctanh(tanh(b*x+a)))^(3/2)-5*b*(b*x-arctanh(tanh(b*x+a)))*arctanh(tanh 
(b*x+a))^(1/2)+5/3*b*arctanh(tanh(b*x+a))^(3/2)-arctanh(tanh(b*x+a))^(5/2) 
/x
 

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.96 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^2} \, dx=-5 b \text {arctanh}\left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {-b x+\text {arctanh}(\tanh (a+b x))}}\right ) (-b x+\text {arctanh}(\tanh (a+b x)))^{3/2}+\sqrt {\text {arctanh}(\tanh (a+b x))} \left (\frac {2 b^2 x}{3}+\frac {14}{3} b (-b x+\text {arctanh}(\tanh (a+b x)))-\frac {(-b x+\text {arctanh}(\tanh (a+b x)))^2}{x}\right ) \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.05, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {2599, 2590, 2590, 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.

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

\(\Big \downarrow \) 2599

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

\(\Big \downarrow \) 2590

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

\(\Big \downarrow \) 2590

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

\(\Big \downarrow \) 2592

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 2590

Int[(v_)^(n_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[ 
D[v, x]]}, Simp[v^n/(a*n), x] - Simp[(b*u - a*v)/a   Int[v^(n - 1)/u, x], x 
] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && GtQ[n, 0] && NeQ[n, 
 1]
 

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 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 = 193, normalized size of antiderivative = 1.75

method	result	size
default	\(2 b \left (\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{3}+2 \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}\, a +2 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right ) \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}+\frac {\left (-\frac {a^{2}}{2}-a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )-\frac {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{2}\right ) \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{b x}-\frac {5 \left (a^{2}+2 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )+\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}}\right )}{2 \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}}\right )\)	\(193\)

Input:

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

Output:

2*b*(1/3*arctanh(tanh(b*x+a))^(3/2)+2*arctanh(tanh(b*x+a))^(1/2)*a+2*(arct 
anh(tanh(b*x+a))-b*x-a)*arctanh(tanh(b*x+a))^(1/2)+(-1/2*a^2-a*(arctanh(ta 
nh(b*x+a))-b*x-a)-1/2*(arctanh(tanh(b*x+a))-b*x-a)^2)*arctanh(tanh(b*x+a)) 
^(1/2)/b/x-5/2*(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(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 = 123, normalized size of antiderivative = 1.12 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^2} \, dx=\left [\frac {15 \, a^{\frac {3}{2}} b x \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (2 \, b^{2} x^{2} + 14 \, a b x - 3 \, a^{2}\right )} \sqrt {b x + a}}{6 \, x}, \frac {15 \, \sqrt {-a} a b x \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x + a}}\right ) + {\left (2 \, b^{2} x^{2} + 14 \, a b x - 3 \, a^{2}\right )} \sqrt {b x + a}}{3 \, x}\right ] \] Input:

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

Output:

[1/6*(15*a^(3/2)*b*x*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(2*b 
^2*x^2 + 14*a*b*x - 3*a^2)*sqrt(b*x + a))/x, 1/3*(15*sqrt(-a)*a*b*x*arctan 
(sqrt(-a)/sqrt(b*x + a)) + (2*b^2*x^2 + 14*a*b*x - 3*a^2)*sqrt(b*x + a))/x 
]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.73 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^2} \, dx=\frac {1}{6} \, \sqrt {2} {\left (\frac {15 \, \sqrt {2} a^{2} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + 2 \, \sqrt {2} {\left (b x + a\right )}^{\frac {3}{2}} + 12 \, \sqrt {2} \sqrt {b x + a} a - \frac {3 \, \sqrt {2} \sqrt {b x + a} a^{2}}{b x}\right )} b \] Input:

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

Output:

1/6*sqrt(2)*(15*sqrt(2)*a^2*arctan(sqrt(b*x + a)/sqrt(-a))/sqrt(-a) + 2*sq 
rt(2)*(b*x + a)^(3/2) + 12*sqrt(2)*sqrt(b*x + a)*a - 3*sqrt(2)*sqrt(b*x + 
a)*a^2/(b*x))*b
 

Mupad [B] (verification not implemented)

Time = 7.47 (sec) , antiderivative size = 616, normalized size of antiderivative = 5.60 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(2*b^2*x*(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 - ((log((2*exp(2*a)*exp(2*b*x))/(e 
xp(2*a)*exp(2*b*x) + 1))/2 - log(2/(exp(2*a)*exp(2*b*x) + 1))/2)^(1/2)*(lo 
g(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)/(4*x) - ((3*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) - (4*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))/2 + b*x))/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))/b + (2^(1/ 
2)*b*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*exp(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)*16i)/(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)^(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)^(3/2)*5i)/8
 

Reduce [F]

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

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

Output:

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