\(\int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^{5/2}} \, dx\) [235]

Optimal result

Integrand size = 17, antiderivative size = 106 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^{5/2}} \, dx=-5 b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\text {arctanh}(\tanh (a+b x))}}\right ) (b x-\text {arctanh}(\tanh (a+b x)))+5 b^2 \sqrt {x} \sqrt {\text {arctanh}(\tanh (a+b x))}-\frac {10 b \text {arctanh}(\tanh (a+b x))^{3/2}}{3 \sqrt {x}}-\frac {2 \text {arctanh}(\tanh (a+b x))^{5/2}}{3 x^{3/2}} \] Output:

-5*b^(3/2)*arctanh(b^(1/2)*x^(1/2)/arctanh(tanh(b*x+a))^(1/2))*(b*x-arctan 
h(tanh(b*x+a)))+5*b^2*x^(1/2)*arctanh(tanh(b*x+a))^(1/2)-10/3*b*arctanh(ta 
nh(b*x+a))^(3/2)/x^(1/2)-2/3*arctanh(tanh(b*x+a))^(5/2)/x^(3/2)
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.92 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^{5/2}} \, dx=\frac {\sqrt {\text {arctanh}(\tanh (a+b x))} \left (15 b^2 x^2-10 b x \text {arctanh}(\tanh (a+b x))-2 \text {arctanh}(\tanh (a+b x))^2\right )}{3 x^{3/2}}+5 b^{3/2} (-b x+\text {arctanh}(\tanh (a+b x))) \log \left (b \sqrt {x}+\sqrt {b} \sqrt {\text {arctanh}(\tanh (a+b x))}\right ) \] Input:

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

Output:

(Sqrt[ArcTanh[Tanh[a + b*x]]]*(15*b^2*x^2 - 10*b*x*ArcTanh[Tanh[a + b*x]] 
- 2*ArcTanh[Tanh[a + b*x]]^2))/(3*x^(3/2)) + 5*b^(3/2)*(-(b*x) + ArcTanh[T 
anh[a + b*x]])*Log[b*Sqrt[x] + Sqrt[b]*Sqrt[ArcTanh[Tanh[a + b*x]]]]
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {2599, 2599, 2600, 2596}

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.

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

\(\Big \downarrow \) 2599

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

\(\Big \downarrow \) 2599

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

\(\Big \downarrow \) 2600

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

\(\Big \downarrow \) 2596

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 2596

Int[1/(Sqrt[u_]*Sqrt[v_]), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Si 
mplify[D[v, x]]}, Simp[(2/Rt[a*b, 2])*ArcTanh[Rt[a*b, 2]*(Sqrt[u]/(a*Sqrt[v 
]))], x] /; NeQ[b*u - a*v, 0] && PosQ[a*b]] /; PiecewiseLinearQ[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]))
 

rule 2600

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 + n + 1))), x] - Simp[n*((b*u - 
a*v)/(a*(m + n + 1)))   Int[u^m*v^(n - 1), x], x] /; NeQ[b*u - a*v, 0]] /; 
PiecewiseLinearQ[u, v, x] && NeQ[m + n + 2, 0] && GtQ[n, 0] && NeQ[m + n + 
1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || LtQ[0, m, n])) &&  !ILtQ[m + n, 
 -2]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(200\) vs. \(2(82)=164\).

Time = 0.38 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.90

method	result	size
derivativedivides	\(-\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {3}{2}}}+\frac {8 b \left (-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \sqrt {x}}+\frac {6 b \left (\frac {\sqrt {x}\, \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{6}+\frac {5 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \left (\frac {\sqrt {x}\, \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{4}+\frac {3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \left (\frac {\sqrt {x}\, \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{2}+\frac {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6}\right )}{\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}\right )}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )}\)	\(201\)
default	\(-\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {3}{2}}}+\frac {8 b \left (-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \sqrt {x}}+\frac {6 b \left (\frac {\sqrt {x}\, \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{6}+\frac {5 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \left (\frac {\sqrt {x}\, \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{4}+\frac {3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \left (\frac {\sqrt {x}\, \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{2}+\frac {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6}\right )}{\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}\right )}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )}\)	\(201\)

Input:

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

Output:

-2/3/(arctanh(tanh(b*x+a))-b*x)/x^(3/2)*arctanh(tanh(b*x+a))^(7/2)+8/3*b/( 
arctanh(tanh(b*x+a))-b*x)*(-1/(arctanh(tanh(b*x+a))-b*x)/x^(1/2)*arctanh(t 
anh(b*x+a))^(7/2)+6*b/(arctanh(tanh(b*x+a))-b*x)*(1/6*x^(1/2)*arctanh(tanh 
(b*x+a))^(5/2)+5/6*(arctanh(tanh(b*x+a))-b*x)*(1/4*x^(1/2)*arctanh(tanh(b* 
x+a))^(3/2)+3/4*(arctanh(tanh(b*x+a))-b*x)*(1/2*x^(1/2)*arctanh(tanh(b*x+a 
))^(1/2)+1/2/b^(1/2)*(arctanh(tanh(b*x+a))-b*x)*ln(b^(1/2)*x^(1/2)+arctanh 
(tanh(b*x+a))^(1/2))))))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.27 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^{5/2}} \, dx=\left [\frac {15 \, a b^{\frac {3}{2}} x^{2} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (3 \, b^{2} x^{2} - 14 \, a b x - 2 \, a^{2}\right )} \sqrt {b x + a} \sqrt {x}}{6 \, x^{2}}, -\frac {15 \, a \sqrt {-b} b x^{2} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {b x + a}}\right ) - {\left (3 \, b^{2} x^{2} - 14 \, a b x - 2 \, a^{2}\right )} \sqrt {b x + a} \sqrt {x}}{3 \, x^{2}}\right ] \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 73.12 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.03 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^{5/2}} \, dx=-\frac {\sqrt {2} {\left (\frac {15 \, \sqrt {2} a \log \left ({\left | -\sqrt {b x + a} \sqrt {b} + \sqrt {{\left (b x + a\right )} b - a b} \right |}\right )}{\sqrt {b}} - \frac {{\left (15 \, \sqrt {2} a^{2} b + {\left (3 \, \sqrt {2} {\left (b x + a\right )} b - 20 \, \sqrt {2} a b\right )} {\left (b x + a\right )}\right )} \sqrt {b x + a}}{{\left ({\left (b x + a\right )} b - a b\right )}^{\frac {3}{2}}}\right )} b^{3}}{6 \, {\left | b \right |}} \] Input:

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

Output:

-1/6*sqrt(2)*(15*sqrt(2)*a*log(abs(-sqrt(b*x + a)*sqrt(b) + sqrt((b*x + a) 
*b - a*b)))/sqrt(b) - (15*sqrt(2)*a^2*b + (3*sqrt(2)*(b*x + a)*b - 20*sqrt 
(2)*a*b)*(b*x + a))*sqrt(b*x + a)/((b*x + a)*b - a*b)^(3/2))*b^3/abs(b)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^{5/2}} \, dx=\frac {-2 \sqrt {x}\, \sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}\, \mathit {atanh} \left (\tanh \left (b x +a \right )\right )^{2}-10 \sqrt {x}\, \sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}\, \mathit {atanh} \left (\tanh \left (b x +a \right )\right ) b x +15 \left (\int \frac {\sqrt {x}\, \sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}}{x}d x \right ) b^{2} x^{2}}{3 x^{2}} \] Input:

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

Output:

( - 2*sqrt(x)*sqrt(atanh(tanh(a + b*x)))*atanh(tanh(a + b*x))**2 - 10*sqrt 
(x)*sqrt(atanh(tanh(a + b*x)))*atanh(tanh(a + b*x))*b*x + 15*int((sqrt(x)* 
sqrt(atanh(tanh(a + b*x))))/x,x)*b**2*x**2)/(3*x**2)