\(\int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x} \, dx\) [134]

Optimal result

Integrand size = 15, antiderivative size = 121 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x} \, dx=-2 \arctan \left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right ) (b x-\text {arctanh}(\tanh (a+b x)))^{5/2}+2 (b x-\text {arctanh}(\tanh (a+b x)))^2 \sqrt {\text {arctanh}(\tanh (a+b x))}-\frac {2}{3} (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{3/2}+\frac {2}{5} \text {arctanh}(\tanh (a+b x))^{5/2} \] Output:

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.82 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x} \, dx=\frac {2}{15} \left (15 b^2 x^2 \sqrt {\text {arctanh}(\tanh (a+b x))}-35 b x \text {arctanh}(\tanh (a+b x))^{3/2}+23 \text {arctanh}(\tanh (a+b x))^{5/2}-15 \text {arctanh}\left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {-b x+\text {arctanh}(\tanh (a+b x))}}\right ) (-b x+\text {arctanh}(\tanh (a+b x)))^{5/2}\right ) \] Input:

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

Output:

(2*(15*b^2*x^2*Sqrt[ArcTanh[Tanh[a + b*x]]] - 35*b*x*ArcTanh[Tanh[a + b*x] 
]^(3/2) + 23*ArcTanh[Tanh[a + b*x]]^(5/2) - 15*ArcTanh[Sqrt[ArcTanh[Tanh[a 
 + b*x]]]/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]]*(-(b*x) + ArcTanh[Tanh[a 
+ b*x]])^(5/2)))/15
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {2590, 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.

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

\(\Big \downarrow \) 2590

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

\(\Big \downarrow \) 2590

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

\(\Big \downarrow \) 2590

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

\(\Big \downarrow \) 2592

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

Input:

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

Output:

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

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(221\) vs. \(2(105)=210\).

Time = 0.20 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.83

method	result	size
default	\(\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{5}+\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}} a}{3}+\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}} \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )}{3}+2 \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}\, a^{2}+4 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right ) \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}+2 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2} \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}-\frac {2 \left (a^{3}+3 a^{2} \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )+3 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}+\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}\right ) \operatorname {arctanh}\left (\frac {\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}}\right )}{\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}}\)	\(222\)

Input:

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

Output:

2/5*arctanh(tanh(b*x+a))^(5/2)+2/3*arctanh(tanh(b*x+a))^(3/2)*a+2/3*arctan 
h(tanh(b*x+a))^(3/2)*(arctanh(tanh(b*x+a))-b*x-a)+2*arctanh(tanh(b*x+a))^( 
1/2)*a^2+4*a*(arctanh(tanh(b*x+a))-b*x-a)*arctanh(tanh(b*x+a))^(1/2)+2*(ar 
ctanh(tanh(b*x+a))-b*x-a)^2*arctanh(tanh(b*x+a))^(1/2)-2*(a^3+3*a^2*(arcta 
nh(tanh(b*x+a))-b*x-a)+3*a*(arctanh(tanh(b*x+a))-b*x-a)^2+(arctanh(tanh(b* 
x+a))-b*x-a)^3)/(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 = 111, normalized size of antiderivative = 0.92 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x} \, dx=\left [a^{\frac {5}{2}} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + \frac {2}{15} \, {\left (3 \, b^{2} x^{2} + 11 \, a b x + 23 \, a^{2}\right )} \sqrt {b x + a}, 2 \, \sqrt {-a} a^{2} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x + a}}\right ) + \frac {2}{15} \, {\left (3 \, b^{2} x^{2} + 11 \, a b x + 23 \, a^{2}\right )} \sqrt {b x + a}\right ] \] Input:

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

Output:

[a^(5/2)*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2/15*(3*b^2*x^2 + 
11*a*b*x + 23*a^2)*sqrt(b*x + a), 2*sqrt(-a)*a^2*arctan(sqrt(-a)/sqrt(b*x 
+ a)) + 2/15*(3*b^2*x^2 + 11*a*b*x + 23*a^2)*sqrt(b*x + a)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.60 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x} \, dx=\frac {1}{15} \, \sqrt {2} {\left (\frac {15 \, \sqrt {2} a^{3} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + 3 \, \sqrt {2} {\left (b x + a\right )}^{\frac {5}{2}} + 5 \, \sqrt {2} {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {2} \sqrt {b x + a} a^{2}\right )} \] Input:

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

Output:

1/15*sqrt(2)*(15*sqrt(2)*a^3*arctan(sqrt(b*x + a)/sqrt(-a))/sqrt(-a) + 3*s 
qrt(2)*(b*x + a)^(5/2) + 5*sqrt(2)*(b*x + a)^(3/2)*a + 15*sqrt(2)*sqrt(b*x 
 + a)*a^2)
 

Mupad [B] (verification not implemented)

Time = 7.14 (sec) , antiderivative size = 789, normalized size of antiderivative = 6.52 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x} \, dx=\text {Too large to display} \] Input:

int(atanh(tanh(a + b*x))^(5/2)/x,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)*((3*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)^2)/2 - (2*(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) - (8*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))/5)*(lo 
g(2/(exp(2*a)*exp(2*b*x) + 1))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*e 
xp(2*b*x) + 1))/2 + b*x))/(3*b)))/b + (2*b^2*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))/5 + (2^(1/2)*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)*(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*(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)^(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)^(5/2)*1i)/8 - (x* 
(3*b^2*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(ex 
p(2*a)*exp(2*b*x) + 1)) + 2*b*x) - (8*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))/...
 

Reduce [F]

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

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

Output:

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