\(\int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^5} \, dx\) [138]

Optimal result

Integrand size = 15, antiderivative size = 167 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^5} \, dx=\frac {5 b^4 \arctan \left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right )}{64 (b x-\text {arctanh}(\tanh (a+b x)))^{3/2}}-\frac {5 b^3}{64 x \sqrt {\text {arctanh}(\tanh (a+b x))}}+\frac {5 b^4}{64 (b x-\text {arctanh}(\tanh (a+b x))) \sqrt {\text {arctanh}(\tanh (a+b x))}}-\frac {5 b^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}{32 x^2}-\frac {5 b \text {arctanh}(\tanh (a+b x))^{3/2}}{24 x^3}-\frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{4 x^4} \] Output:

5/64*b^4*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)-5/64*b^3/x/arctanh(tanh(b*x+a))^(1/2)+ 
5/64*b^4/(b*x-arctanh(tanh(b*x+a)))/arctanh(tanh(b*x+a))^(1/2)-5/32*b^2*ar 
ctanh(tanh(b*x+a))^(1/2)/x^2-5/24*b*arctanh(tanh(b*x+a))^(3/2)/x^3-1/4*arc 
tanh(tanh(b*x+a))^(5/2)/x^4
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.80 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^5} \, dx=\frac {5 b^4 \text {arctanh}\left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {-b x+\text {arctanh}(\tanh (a+b x))}}\right )}{64 (-b x+\text {arctanh}(\tanh (a+b x)))^{3/2}}-\frac {\sqrt {\text {arctanh}(\tanh (a+b x))} \left (15 b^3 x^3+10 b^2 x^2 \text {arctanh}(\tanh (a+b x))+8 b x \text {arctanh}(\tanh (a+b x))^2-48 \text {arctanh}(\tanh (a+b x))^3\right )}{192 x^4 (b x-\text {arctanh}(\tanh (a+b x)))} \] Input:

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

Output:

(5*b^4*ArcTanh[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[-(b*x) + ArcTanh[Tanh[a + 
 b*x]]]])/(64*(-(b*x) + ArcTanh[Tanh[a + b*x]])^(3/2)) - (Sqrt[ArcTanh[Tan 
h[a + b*x]]]*(15*b^3*x^3 + 10*b^2*x^2*ArcTanh[Tanh[a + b*x]] + 8*b*x*ArcTa 
nh[Tanh[a + b*x]]^2 - 48*ArcTanh[Tanh[a + b*x]]^3))/(192*x^4*(b*x - ArcTan 
h[Tanh[a + b*x]]))
 

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2599, 2599, 2599, 2599, 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.

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

\(\Big \downarrow \) 2599

\(\displaystyle \frac {5}{8} b \int \frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{x^4}dx-\frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{4 x^4}\)

\(\Big \downarrow \) 2599

\(\displaystyle \frac {5}{8} b \left (\frac {1}{2} b \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^3}dx-\frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{3 x^3}\right )-\frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{4 x^4}\)

\(\Big \downarrow \) 2599

\(\displaystyle \frac {5}{8} b \left (\frac {1}{2} b \left (\frac {1}{4} b \int \frac {1}{x^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}dx-\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{2 x^2}\right )-\frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{3 x^3}\right )-\frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{4 x^4}\)

\(\Big \downarrow \) 2599

\(\displaystyle \frac {5}{8} b \left (\frac {1}{2} b \left (\frac {1}{4} b \left (-\frac {1}{2} b \int \frac {1}{x \text {arctanh}(\tanh (a+b x))^{3/2}}dx-\frac {1}{x \sqrt {\text {arctanh}(\tanh (a+b x))}}\right )-\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{2 x^2}\right )-\frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{3 x^3}\right )-\frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{4 x^4}\)

\(\Big \downarrow \) 2594

\(\displaystyle \frac {5}{8} b \left (\frac {1}{2} b \left (\frac {1}{4} b \left (-\frac {1}{2} b \left (-\frac {\int \frac {1}{x \sqrt {\text {arctanh}(\tanh (a+b x))}}dx}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{(b x-\text {arctanh}(\tanh (a+b x))) \sqrt {\text {arctanh}(\tanh (a+b x))}}\right )-\frac {1}{x \sqrt {\text {arctanh}(\tanh (a+b x))}}\right )-\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{2 x^2}\right )-\frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{3 x^3}\right )-\frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{4 x^4}\)

\(\Big \downarrow \) 2592

\(\displaystyle \frac {5}{8} b \left (\frac {1}{2} b \left (\frac {1}{4} b \left (-\frac {1}{2} b \left (-\frac {2 \arctan \left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right )}{(b x-\text {arctanh}(\tanh (a+b x)))^{3/2}}-\frac {2}{(b x-\text {arctanh}(\tanh (a+b x))) \sqrt {\text {arctanh}(\tanh (a+b x))}}\right )-\frac {1}{x \sqrt {\text {arctanh}(\tanh (a+b x))}}\right )-\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{2 x^2}\right )-\frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{3 x^3}\right )-\frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{4 x^4}\)

Input:

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

Output:

-1/4*ArcTanh[Tanh[a + b*x]]^(5/2)/x^4 + (5*b*((b*((b*(-1/2*(b*((-2*ArcTan[ 
Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]])/(b*x - A 
rcTanh[Tanh[a + b*x]])^(3/2) - 2/((b*x - ArcTanh[Tanh[a + b*x]])*Sqrt[ArcT 
anh[Tanh[a + b*x]]]))) - 1/(x*Sqrt[ArcTanh[Tanh[a + b*x]]])))/4 - Sqrt[Arc 
Tanh[Tanh[a + b*x]]]/(2*x^2)))/2 - ArcTanh[Tanh[a + b*x]]^(3/2)/(3*x^3)))/ 
8
 

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 = 0.63 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.01

method	result	size
default	\(2 b^{4} \left (\frac {-\frac {5 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{128 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )}-\frac {73 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{384}+\left (\frac {55 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}{384}-\frac {55 b x}{384}\right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}+\left (-\frac {5 a^{2}}{128}-\frac {5 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )}{64}-\frac {5 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{128}\right ) \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{b^{4} x^{4}}+\frac {5 \,\operatorname {arctanh}\left (\frac {\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}}\right )}{128 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{\frac {3}{2}}}\right )\)	\(169\)

Input:

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

Output:

2*b^4*((-5/128/(arctanh(tanh(b*x+a))-b*x)*arctanh(tanh(b*x+a))^(7/2)-73/38 
4*arctanh(tanh(b*x+a))^(5/2)+(55/384*arctanh(tanh(b*x+a))-55/384*b*x)*arct 
anh(tanh(b*x+a))^(3/2)+(-5/128*a^2-5/64*a*(arctanh(tanh(b*x+a))-b*x-a)-5/1 
28*(arctanh(tanh(b*x+a))-b*x-a)^2)*arctanh(tanh(b*x+a))^(1/2))/b^4/x^4+5/1 
28/(arctanh(tanh(b*x+a))-b*x)^(3/2)*arctanh(arctanh(tanh(b*x+a))^(1/2)/(ar 
ctanh(tanh(b*x+a))-b*x)^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.98 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^5} \, dx=\left [\frac {15 \, \sqrt {a} b^{4} x^{4} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) - 2 \, {\left (15 \, a b^{3} x^{3} + 118 \, a^{2} b^{2} x^{2} + 136 \, a^{3} b x + 48 \, a^{4}\right )} \sqrt {b x + a}}{384 \, a^{2} x^{4}}, -\frac {15 \, \sqrt {-a} b^{4} x^{4} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x + a}}\right ) + {\left (15 \, a b^{3} x^{3} + 118 \, a^{2} b^{2} x^{2} + 136 \, a^{3} b x + 48 \, a^{4}\right )} \sqrt {b x + a}}{192 \, a^{2} x^{4}}\right ] \] Input:

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

Output:

[1/384*(15*sqrt(a)*b^4*x^4*log((b*x + 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) - 
2*(15*a*b^3*x^3 + 118*a^2*b^2*x^2 + 136*a^3*b*x + 48*a^4)*sqrt(b*x + a))/( 
a^2*x^4), -1/192*(15*sqrt(-a)*b^4*x^4*arctan(sqrt(-a)/sqrt(b*x + a)) + (15 
*a*b^3*x^3 + 118*a^2*b^2*x^2 + 136*a^3*b*x + 48*a^4)*sqrt(b*x + a))/(a^2*x 
^4)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.65 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^5} \, dx=-\frac {\sqrt {2} {\left (\frac {15 \, \sqrt {2} b^{5} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} + \frac {\sqrt {2} {\left (15 \, {\left (b x + a\right )}^{\frac {7}{2}} b^{5} + 73 \, {\left (b x + a\right )}^{\frac {5}{2}} a b^{5} - 55 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} b^{5} + 15 \, \sqrt {b x + a} a^{3} b^{5}\right )}}{a b^{4} x^{4}}\right )}}{384 \, b} \] Input:

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

Output:

-1/384*sqrt(2)*(15*sqrt(2)*b^5*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a) 
 + sqrt(2)*(15*(b*x + a)^(7/2)*b^5 + 73*(b*x + a)^(5/2)*a*b^5 - 55*(b*x + 
a)^(3/2)*a^2*b^5 + 15*sqrt(b*x + a)*a^3*b^5)/(a*b^4*x^4))/b
 

Mupad [B] (verification not implemented)

Time = 7.84 (sec) , antiderivative size = 1069, normalized size of antiderivative = 6.40 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^5} \, dx=\text {Too large to display} \] Input:

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

Output:

(5*b^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))/(32*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)) - ((l 
og((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)^3)/(4*x^4*(4*log(2/(ex 
p(2*a)*exp(2*b*x) + 1)) - 4*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b* 
x) + 1)) + 8*b*x)) + (2^(1/2)*b^4*log(((2*2^(1/2)*a + (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))/(e 
xp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^(1/2)*2i - 2^(1/2)*(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^(1/2)*b*x)*((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)^3 - 8*a^3 - 6*a 
*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + log(2/(ex 
p(2*a)*exp(2*b*x) + 1)) + 2*b*x)^2 + 12*a^2*(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 
))*1024i)/(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)))*5i)/(64*(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))...
 

Reduce [F]

\[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^5} \, dx=\frac {-48 \sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}\, \mathit {atanh} \left (\tanh \left (b x +a \right )\right )^{2}-40 \sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}\, \mathit {atanh} \left (\tanh \left (b x +a \right )\right ) b x -30 \sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}\, b^{2} x^{2}+15 \left (\int \frac {\sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}}{\mathit {atanh} \left (\tanh \left (b x +a \right )\right ) x^{2}}d x \right ) b^{3} x^{4}}{192 x^{4}} \] Input:

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

Output:

( - 48*sqrt(atanh(tanh(a + b*x)))*atanh(tanh(a + b*x))**2 - 40*sqrt(atanh( 
tanh(a + b*x)))*atanh(tanh(a + b*x))*b*x - 30*sqrt(atanh(tanh(a + b*x)))*b 
**2*x**2 + 15*int(sqrt(atanh(tanh(a + b*x)))/(atanh(tanh(a + b*x))*x**2),x 
)*b**3*x**4)/(192*x**4)