\(\int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^4} \, dx\) [137]

Optimal result

Integrand size = 15, antiderivative size = 113 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^4} \, dx=\frac {5 b^3 \arctan \left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right )}{8 \sqrt {b x-\text {arctanh}(\tanh (a+b x))}}-\frac {5 b^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}{8 x}-\frac {5 b \text {arctanh}(\tanh (a+b x))^{3/2}}{12 x^2}-\frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{3 x^3} \] Output:

5/8*b^3*arctan(arctanh(tanh(b*x+a))^(1/2)/(b*x-arctanh(tanh(b*x+a)))^(1/2) 
)/(b*x-arctanh(tanh(b*x+a)))^(1/2)-5/8*b^2*arctanh(tanh(b*x+a))^(1/2)/x-5/ 
12*b*arctanh(tanh(b*x+a))^(3/2)/x^2-1/3*arctanh(tanh(b*x+a))^(5/2)/x^3
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.95 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^4} \, dx=\frac {1}{24} \left (-\frac {15 b^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}{x}-\frac {10 b \text {arctanh}(\tanh (a+b x))^{3/2}}{x^2}-\frac {8 \text {arctanh}(\tanh (a+b x))^{5/2}}{x^3}-\frac {15 b^3 \text {arctanh}\left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {-b x+\text {arctanh}(\tanh (a+b x))}}\right )}{\sqrt {-b x+\text {arctanh}(\tanh (a+b x))}}\right ) \] Input:

Integrate[ArcTanh[Tanh[a + b*x]]^(5/2)/x^4,x]
 

Output:

((-15*b^2*Sqrt[ArcTanh[Tanh[a + b*x]]])/x - (10*b*ArcTanh[Tanh[a + b*x]]^( 
3/2))/x^2 - (8*ArcTanh[Tanh[a + b*x]]^(5/2))/x^3 - (15*b^3*ArcTanh[Sqrt[Ar 
cTanh[Tanh[a + b*x]]]/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]])/Sqrt[-(b*x) 
+ ArcTanh[Tanh[a + b*x]]])/24
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.01, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {2599, 2599, 2599, 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.

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

\(\Big \downarrow \) 2599

\(\displaystyle \frac {5}{6} b \int \frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{x^3}dx-\frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{3 x^3}\)

\(\Big \downarrow \) 2599

\(\displaystyle \frac {5}{6} b \left (\frac {3}{4} b \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^2}dx-\frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{2 x^2}\right )-\frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{3 x^3}\)

\(\Big \downarrow \) 2599

\(\displaystyle \frac {5}{6} b \left (\frac {3}{4} b \left (\frac {1}{2} b \int \frac {1}{x \sqrt {\text {arctanh}(\tanh (a+b x))}}dx-\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x}\right )-\frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{2 x^2}\right )-\frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{3 x^3}\)

\(\Big \downarrow \) 2592

\(\displaystyle \frac {5}{6} b \left (\frac {3}{4} b \left (\frac {b \arctan \left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right )}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}-\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x}\right )-\frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{2 x^2}\right )-\frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{3 x^3}\)

Input:

Int[ArcTanh[Tanh[a + b*x]]^(5/2)/x^4,x]
 

Output:

-1/3*ArcTanh[Tanh[a + b*x]]^(5/2)/x^3 + (5*b*((3*b*((b*ArcTan[Sqrt[ArcTanh 
[Tanh[a + b*x]]]/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]])/Sqrt[b*x - ArcTanh[T 
anh[a + b*x]]] - Sqrt[ArcTanh[Tanh[a + b*x]]]/x))/4 - ArcTanh[Tanh[a + b*x 
]]^(3/2)/(2*x^2)))/6
 

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 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 = 0.34 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.27

method	result	size
default	\(2 b^{3} \left (\frac {-\frac {11 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{16}+\left (\frac {5 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}{6}-\frac {5 b x}{6}\right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}+\left (-\frac {5 a^{2}}{16}-\frac {5 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )}{8}-\frac {5 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{16}\right ) \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{b^{3} x^{3}}-\frac {5 \,\operatorname {arctanh}\left (\frac {\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}}\right )}{16 \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}}\right )\)	\(144\)

Input:

int(arctanh(tanh(b*x+a))^(5/2)/x^4,x,method=_RETURNVERBOSE)
 

Output:

2*b^3*((-11/16*arctanh(tanh(b*x+a))^(5/2)+(5/6*arctanh(tanh(b*x+a))-5/6*b* 
x)*arctanh(tanh(b*x+a))^(3/2)+(-5/16*a^2-5/8*a*(arctanh(tanh(b*x+a))-b*x-a 
)-5/16*(arctanh(tanh(b*x+a))-b*x-a)^2)*arctanh(tanh(b*x+a))^(1/2))/b^3/x^3 
-5/16/(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 = 143, normalized size of antiderivative = 1.27 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^4} \, dx=\left [\frac {15 \, \sqrt {a} b^{3} x^{3} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) - 2 \, {\left (33 \, a b^{2} x^{2} + 26 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt {b x + a}}{48 \, a x^{3}}, \frac {15 \, \sqrt {-a} b^{3} x^{3} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x + a}}\right ) - {\left (33 \, a b^{2} x^{2} + 26 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt {b x + a}}{24 \, a x^{3}}\right ] \] Input:

integrate(arctanh(tanh(b*x+a))^(5/2)/x^4,x, algorithm="fricas")
 

Output:

[1/48*(15*sqrt(a)*b^3*x^3*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) - 2 
*(33*a*b^2*x^2 + 26*a^2*b*x + 8*a^3)*sqrt(b*x + a))/(a*x^3), 1/24*(15*sqrt 
(-a)*b^3*x^3*arctan(sqrt(-a)/sqrt(b*x + a)) - (33*a*b^2*x^2 + 26*a^2*b*x + 
 8*a^3)*sqrt(b*x + a))/(a*x^3)]
 

Sympy [F]

\[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^4} \, dx=\int \frac {\operatorname {atanh}^{\frac {5}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}{x^{4}}\, dx \] Input:

integrate(atanh(tanh(b*x+a))**(5/2)/x**4,x)
 

Output:

Integral(atanh(tanh(a + b*x))**(5/2)/x**4, x)
 

Maxima [F]

\[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^4} \, dx=\int { \frac {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{\frac {5}{2}}}{x^{4}} \,d x } \] Input:

integrate(arctanh(tanh(b*x+a))^(5/2)/x^4,x, algorithm="maxima")
 

Output:

integrate(arctanh(tanh(b*x + a))^(5/2)/x^4, x)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.67 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^4} \, dx=\frac {1}{48} \, \sqrt {2} b^{3} {\left (\frac {15 \, \sqrt {2} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {\sqrt {2} {\left (33 \, {\left (b x + a\right )}^{\frac {5}{2}} - 40 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )}}{b^{3} x^{3}}\right )} \] Input:

integrate(arctanh(tanh(b*x+a))^(5/2)/x^4,x, algorithm="giac")
 

Output:

1/48*sqrt(2)*b^3*(15*sqrt(2)*arctan(sqrt(b*x + a)/sqrt(-a))/sqrt(-a) - sqr 
t(2)*(33*(b*x + a)^(5/2) - 40*(b*x + a)^(3/2)*a + 15*sqrt(b*x + a)*a^2)/(b 
^3*x^3))
 

Mupad [B] (verification not implemented)

Time = 7.74 (sec) , antiderivative size = 669, normalized size of antiderivative = 5.92 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^4} \, dx =\text {Too large to display} \] Input:

int(atanh(tanh(a + b*x))^(5/2)/x^4,x)
 

Output:

(2^(1/2)*b^3*log(((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))/(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)*64i)/x)*5i)/(16*(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))/(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^3*(3*log(2/(exp(2 
*a)*exp(2*b*x) + 1)) - 3*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) 
+ 1)) + 6*b*x)) - (11*b^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))/(8*x) + (13*b*(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)^2)/(12*x^2*(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))
 

Reduce [F]

\[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^4} \, dx=\frac {-16 \sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}\, \mathit {atanh} \left (\tanh \left (b x +a \right )\right )^{2}-20 \sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}\, \mathit {atanh} \left (\tanh \left (b x +a \right )\right ) b x -30 \sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}\, b^{2} x^{2}+15 \left (\int \frac {\sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}}{\mathit {atanh} \left (\tanh \left (b x +a \right )\right ) x}d x \right ) b^{3} x^{3}}{48 x^{3}} \] Input:

int(atanh(tanh(b*x+a))^(5/2)/x^4,x)
 

Output:

( - 16*sqrt(atanh(tanh(a + b*x)))*atanh(tanh(a + b*x))**2 - 20*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),x)*b 
**3*x**3)/(48*x**3)