\(\int \frac {1}{x^2 \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx\) [164]

Optimal result

Integrand size = 15, antiderivative size = 155 \[ \int \frac {1}{x^2 \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\frac {5 b \arctan \left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right )}{(b x-\text {arctanh}(\tanh (a+b x)))^{7/2}}-\frac {1}{x \text {arctanh}(\tanh (a+b x))^{5/2}}+\frac {b}{(b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{5/2}}-\frac {5 b}{3 (b x-\text {arctanh}(\tanh (a+b x)))^2 \text {arctanh}(\tanh (a+b x))^{3/2}}+\frac {5 b}{(b x-\text {arctanh}(\tanh (a+b x)))^3 \sqrt {\text {arctanh}(\tanh (a+b x))}} \] 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)))^(7/2)-1/x/arctanh(tanh(b*x+a))^(5/2)+b/(b*x-arcta 
nh(tanh(b*x+a)))/arctanh(tanh(b*x+a))^(5/2)-5/3*b/(b*x-arctanh(tanh(b*x+a) 
))^2/arctanh(tanh(b*x+a))^(3/2)+5*b/(b*x-arctanh(tanh(b*x+a)))^3/arctanh(t 
anh(b*x+a))^(1/2)
 

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.73 \[ \int \frac {1}{x^2 \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\frac {5 b \text {arctanh}\left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {-b x+\text {arctanh}(\tanh (a+b x))}}\right )}{(-b x+\text {arctanh}(\tanh (a+b x)))^{7/2}}+\frac {-2 b^2 x^2+14 b x \text {arctanh}(\tanh (a+b x))+3 \text {arctanh}(\tanh (a+b x))^2}{3 x (b x-\text {arctanh}(\tanh (a+b x)))^3 \text {arctanh}(\tanh (a+b x))^{3/2}} \] Input:

Integrate[1/(x^2*ArcTanh[Tanh[a + b*x]]^(5/2)),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]])^(7/2) + (-2*b^2*x^2 + 14*b*x*Arc 
Tanh[Tanh[a + b*x]] + 3*ArcTanh[Tanh[a + b*x]]^2)/(3*x*(b*x - ArcTanh[Tanh 
[a + b*x]])^3*ArcTanh[Tanh[a + b*x]]^(3/2))
 

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.26, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2599, 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.

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

\(\Big \downarrow \) 2599

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

\(\Big \downarrow \) 2594

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

\(\Big \downarrow \) 2594

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

\(\Big \downarrow \) 2594

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

\(\Big \downarrow \) 2592

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

Input:

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

Output:

(-5*b*(-((-(((-2*ArcTan[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[b*x - ArcTanh[Ta 
nh[a + b*x]]]])/(b*x - ArcTanh[Tanh[a + b*x]])^(3/2) - 2/((b*x - ArcTanh[T 
anh[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)))/( 
b*x - ArcTanh[Tanh[a + b*x]])) - 2/(5*(b*x - ArcTanh[Tanh[a + b*x]])*ArcTa 
nh[Tanh[a + b*x]]^(5/2))))/2 - 1/(x*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 = 0.21 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.84

method	result	size
default	\(2 b \left (-\frac {1}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}-\frac {2}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}-\frac {\frac {\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{2 b x}-\frac {5 \,\operatorname {arctanh}\left (\frac {\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}}\right )}{2 \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}}}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{3}}\right )\)	\(130\)

Input:

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

Output:

2*b*(-1/3/(arctanh(tanh(b*x+a))-b*x)^2/arctanh(tanh(b*x+a))^(3/2)-2/(arcta 
nh(tanh(b*x+a))-b*x)^3/arctanh(tanh(b*x+a))^(1/2)-1/(arctanh(tanh(b*x+a))- 
b*x)^3*(1/2*arctanh(tanh(b*x+a))^(1/2)/b/x-5/2/(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 = 218, normalized size of antiderivative = 1.41 \[ \int \frac {1}{x^2 \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\left [\frac {15 \, {\left (b^{3} x^{3} + 2 \, a b^{2} x^{2} + a^{2} b x\right )} \sqrt {a} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) - 2 \, {\left (15 \, a b^{2} x^{2} + 20 \, a^{2} b x + 3 \, a^{3}\right )} \sqrt {b x + a}}{6 \, {\left (a^{4} b^{2} x^{3} + 2 \, a^{5} b x^{2} + a^{6} x\right )}}, -\frac {15 \, {\left (b^{3} x^{3} + 2 \, a b^{2} x^{2} + a^{2} b x\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x + a}}\right ) + {\left (15 \, a b^{2} x^{2} + 20 \, a^{2} b x + 3 \, a^{3}\right )} \sqrt {b x + a}}{3 \, {\left (a^{4} b^{2} x^{3} + 2 \, a^{5} b x^{2} + a^{6} x\right )}}\right ] \] Input:

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

Output:

[1/6*(15*(b^3*x^3 + 2*a*b^2*x^2 + a^2*b*x)*sqrt(a)*log((b*x + 2*sqrt(b*x + 
 a)*sqrt(a) + 2*a)/x) - 2*(15*a*b^2*x^2 + 20*a^2*b*x + 3*a^3)*sqrt(b*x + a 
))/(a^4*b^2*x^3 + 2*a^5*b*x^2 + a^6*x), -1/3*(15*(b^3*x^3 + 2*a*b^2*x^2 + 
a^2*b*x)*sqrt(-a)*arctan(sqrt(-a)/sqrt(b*x + a)) + (15*a*b^2*x^2 + 20*a^2* 
b*x + 3*a^3)*sqrt(b*x + a))/(a^4*b^2*x^3 + 2*a^5*b*x^2 + a^6*x)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.42 \[ \int \frac {1}{x^2 \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=-\frac {5 \, b \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{3}} - \frac {2 \, {\left (6 \, {\left (b x + a\right )} b + a b\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3}} - \frac {\sqrt {b x + a}}{a^{3} x} \] Input:

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

Output:

-5*b*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^3) - 2/3*(6*(b*x + a)*b + 
a*b)/((b*x + a)^(3/2)*a^3) - sqrt(b*x + a)/(a^3*x)
 

Mupad [B] (verification not implemented)

Time = 8.82 (sec) , antiderivative size = 1230, normalized size of antiderivative = 7.94 \[ \int \frac {1}{x^2 \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

(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)*((2*a - log((2*ex 
p(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)^7 + 84*a^2*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*ex 
p(2*b*x) + 1)) + log(2/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^5 - 280*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)^4 + 560*a^4*(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 
 - 672*a^5*(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 - 128*a^7 - 14*a*(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)^6 + 448*a^6*(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))*1i)/(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)^(1/2)))*20i)/(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)^(7/2) - ( 
32*b*(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/...
 

Reduce [F]

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

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

Output:

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