\(\int \frac {x^{5/2}}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx\) [258]

Optimal result

Integrand size = 17, antiderivative size = 111 \[ \int \frac {x^{5/2}}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\frac {5 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\text {arctanh}(\tanh (a+b x))}}\right ) (b x-\text {arctanh}(\tanh (a+b x)))}{b^{7/2}}-\frac {2 x^{5/2}}{3 b \text {arctanh}(\tanh (a+b x))^{3/2}}-\frac {10 x^{3/2}}{3 b^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}+\frac {5 \sqrt {x} \sqrt {\text {arctanh}(\tanh (a+b x))}}{b^3} \] Output:

5*arctanh(b^(1/2)*x^(1/2)/arctanh(tanh(b*x+a))^(1/2))*(b*x-arctanh(tanh(b* 
x+a)))/b^(7/2)-2/3*x^(5/2)/b/arctanh(tanh(b*x+a))^(3/2)-10/3*x^(3/2)/b^2/a 
rctanh(tanh(b*x+a))^(1/2)+5*x^(1/2)*arctanh(tanh(b*x+a))^(1/2)/b^3
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.91 \[ \int \frac {x^{5/2}}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=-\frac {\sqrt {x} \left (2 b^2 x^2+10 b x \text {arctanh}(\tanh (a+b x))-15 \text {arctanh}(\tanh (a+b x))^2\right )}{3 b^3 \text {arctanh}(\tanh (a+b x))^{3/2}}+\frac {5 (b x-\text {arctanh}(\tanh (a+b x))) \log \left (b \sqrt {x}+\sqrt {b} \sqrt {\text {arctanh}(\tanh (a+b x))}\right )}{b^{7/2}} \] Input:

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

Output:

-1/3*(Sqrt[x]*(2*b^2*x^2 + 10*b*x*ArcTanh[Tanh[a + b*x]] - 15*ArcTanh[Tanh 
[a + b*x]]^2))/(b^3*ArcTanh[Tanh[a + b*x]]^(3/2)) + (5*(b*x - ArcTanh[Tanh 
[a + b*x]])*Log[b*Sqrt[x] + Sqrt[b]*Sqrt[ArcTanh[Tanh[a + b*x]]]])/b^(7/2)
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.09, 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.

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

\(\Big \downarrow \) 2599

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

\(\Big \downarrow \) 2599

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

\(\Big \downarrow \) 2600

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

\(\Big \downarrow \) 2596

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

Input:

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

Output:

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

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 [A] (verified)

Time = 0.39 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.87

method	result	size
derivativedivides	\(\frac {x^{\frac {5}{2}}}{b \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}-\frac {5 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \left (-\frac {x^{\frac {3}{2}}}{3 b \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}+\frac {-\frac {\sqrt {x}}{b \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}+\frac {\ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}\right )}{b^{\frac {3}{2}}}}{b}\right )}{b}\)	\(97\)
default	\(\frac {x^{\frac {5}{2}}}{b \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}-\frac {5 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \left (-\frac {x^{\frac {3}{2}}}{3 b \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}+\frac {-\frac {\sqrt {x}}{b \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}+\frac {\ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}\right )}{b^{\frac {3}{2}}}}{b}\right )}{b}\)	\(97\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.90 \[ \int \frac {x^{5/2}}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\left [\frac {15 \, {\left (a b^{2} x^{2} + 2 \, a^{2} b x + a^{3}\right )} \sqrt {b} \log \left (2 \, b x - 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (3 \, b^{3} x^{2} + 20 \, a b^{2} x + 15 \, a^{2} b\right )} \sqrt {b x + a} \sqrt {x}}{6 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}}, \frac {15 \, {\left (a b^{2} x^{2} + 2 \, a^{2} b x + a^{3}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {b x + a}}\right ) + {\left (3 \, b^{3} x^{2} + 20 \, a b^{2} x + 15 \, a^{2} b\right )} \sqrt {b x + a} \sqrt {x}}{3 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}}\right ] \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.55 \[ \int \frac {x^{5/2}}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\frac {{\left (x {\left (\frac {3 \, x}{b} + \frac {20 \, a}{b^{2}}\right )} + \frac {15 \, a^{2}}{b^{3}}\right )} \sqrt {x}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}}} + \frac {5 \, a \log \left ({\left | -\sqrt {b} \sqrt {x} + \sqrt {b x + a} \right |}\right )}{b^{\frac {7}{2}}} \] Input:

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

Output:

1/3*(x*(3*x/b + 20*a/b^2) + 15*a^2/b^3)*sqrt(x)/(b*x + a)^(3/2) + 5*a*log( 
abs(-sqrt(b)*sqrt(x) + sqrt(b*x + a)))/b^(7/2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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