\(\int \frac {x^{5/2}}{\text {arctanh}(\tanh (a+b x))^{3/2}} \, dx\) [250]

Optimal result

Integrand size = 17, antiderivative size = 128 \[ \int \frac {x^{5/2}}{\text {arctanh}(\tanh (a+b x))^{3/2}} \, dx=\frac {15 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\text {arctanh}(\tanh (a+b x))}}\right ) (b x-\text {arctanh}(\tanh (a+b x)))^2}{4 b^{7/2}}-\frac {2 x^{5/2}}{b \sqrt {\text {arctanh}(\tanh (a+b x))}}+\frac {5 x^{3/2} \sqrt {\text {arctanh}(\tanh (a+b x))}}{2 b^2}+\frac {15 \sqrt {x} (b x-\text {arctanh}(\tanh (a+b x))) \sqrt {\text {arctanh}(\tanh (a+b x))}}{4 b^3} \] Output:

15/4*arctanh(b^(1/2)*x^(1/2)/arctanh(tanh(b*x+a))^(1/2))*(b*x-arctanh(tanh 
(b*x+a)))^2/b^(7/2)-2*x^(5/2)/b/arctanh(tanh(b*x+a))^(1/2)+5/2*x^(3/2)*arc 
tanh(tanh(b*x+a))^(1/2)/b^2+15/4*x^(1/2)*(b*x-arctanh(tanh(b*x+a)))*arctan 
h(tanh(b*x+a))^(1/2)/b^3
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.81 \[ \int \frac {x^{5/2}}{\text {arctanh}(\tanh (a+b x))^{3/2}} \, dx=-\frac {\sqrt {x} \left (8 b^2 x^2-25 b x \text {arctanh}(\tanh (a+b x))+15 \text {arctanh}(\tanh (a+b x))^2\right )}{4 b^3 \sqrt {\text {arctanh}(\tanh (a+b x))}}+\frac {15 (-b x+\text {arctanh}(\tanh (a+b x)))^2 \log \left (b \sqrt {x}+\sqrt {b} \sqrt {\text {arctanh}(\tanh (a+b x))}\right )}{4 b^{7/2}} \] Input:

Integrate[x^(5/2)/ArcTanh[Tanh[a + b*x]]^(3/2),x]
 

Output:

-1/4*(Sqrt[x]*(8*b^2*x^2 - 25*b*x*ArcTanh[Tanh[a + b*x]] + 15*ArcTanh[Tanh 
[a + b*x]]^2))/(b^3*Sqrt[ArcTanh[Tanh[a + b*x]]]) + (15*(-(b*x) + ArcTanh[ 
Tanh[a + b*x]])^2*Log[b*Sqrt[x] + Sqrt[b]*Sqrt[ArcTanh[Tanh[a + b*x]]]])/( 
4*b^(7/2))
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.05, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {2599, 2600, 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.

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

\(\Big \downarrow \) 2599

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

\(\Big \downarrow \) 2600

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

\(\Big \downarrow \) 2600

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

\(\Big \downarrow \) 2596

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

Input:

Int[x^(5/2)/ArcTanh[Tanh[a + b*x]]^(3/2),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*x - ArcTanh[Tanh[a + b*x]]))/(4*b) + (x^(3/2)*Sqrt[ArcTanh[Tanh[a + b* 
x]]])/(2*b)))/b - (2*x^(5/2))/(b*Sqrt[ArcTanh[Tanh[a + b*x]]])
 

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 = 111, normalized size of antiderivative = 0.87

method	result	size
derivativedivides	\(\frac {x^{\frac {5}{2}}}{2 b \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}-\frac {5 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \left (\frac {x^{\frac {3}{2}}}{2 b \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}-\frac {3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \left (-\frac {\sqrt {x}}{b \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}+\frac {\ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}\right )}{b^{\frac {3}{2}}}\right )}{2 b}\right )}{2 b}\)	\(111\)
default	\(\frac {x^{\frac {5}{2}}}{2 b \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}-\frac {5 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \left (\frac {x^{\frac {3}{2}}}{2 b \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}-\frac {3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \left (-\frac {\sqrt {x}}{b \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}+\frac {\ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}\right )}{b^{\frac {3}{2}}}\right )}{2 b}\right )}{2 b}\)	\(111\)

Input:

int(x^(5/2)/arctanh(tanh(b*x+a))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.34 \[ \int \frac {x^{5/2}}{\text {arctanh}(\tanh (a+b x))^{3/2}} \, dx=\left [\frac {15 \, {\left (a^{2} b x + a^{3}\right )} \sqrt {b} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (2 \, b^{3} x^{2} - 5 \, a b^{2} x - 15 \, a^{2} b\right )} \sqrt {b x + a} \sqrt {x}}{8 \, {\left (b^{5} x + a b^{4}\right )}}, -\frac {15 \, {\left (a^{2} b x + a^{3}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {b x + a}}\right ) - {\left (2 \, b^{3} x^{2} - 5 \, a b^{2} x - 15 \, a^{2} b\right )} \sqrt {b x + a} \sqrt {x}}{4 \, {\left (b^{5} x + a b^{4}\right )}}\right ] \] Input:

integrate(x^(5/2)/arctanh(tanh(b*x+a))^(3/2),x, algorithm="fricas")
 

Output:

[1/8*(15*(a^2*b*x + a^3)*sqrt(b)*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt( 
x) + a) + 2*(2*b^3*x^2 - 5*a*b^2*x - 15*a^2*b)*sqrt(b*x + a)*sqrt(x))/(b^5 
*x + a*b^4), -1/4*(15*(a^2*b*x + a^3)*sqrt(-b)*arctan(sqrt(-b)*sqrt(x)/sqr 
t(b*x + a)) - (2*b^3*x^2 - 5*a*b^2*x - 15*a^2*b)*sqrt(b*x + a)*sqrt(x))/(b 
^5*x + a*b^4)]
 

Sympy [F]

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

integrate(x**(5/2)/atanh(tanh(b*x+a))**(3/2),x)
 

Output:

Integral(x**(5/2)/atanh(tanh(a + b*x))**(3/2), x)
 

Maxima [F]

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

integrate(x^(5/2)/arctanh(tanh(b*x+a))^(3/2),x, algorithm="maxima")
 

Output:

integrate(x^(5/2)/arctanh(tanh(b*x + a))^(3/2), x)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.49 \[ \int \frac {x^{5/2}}{\text {arctanh}(\tanh (a+b x))^{3/2}} \, dx=\frac {{\left (x {\left (\frac {2 \, x}{b} - \frac {5 \, a}{b^{2}}\right )} - \frac {15 \, a^{2}}{b^{3}}\right )} \sqrt {x}}{4 \, \sqrt {b x + a}} - \frac {15 \, a^{2} \log \left ({\left | -\sqrt {b} \sqrt {x} + \sqrt {b x + a} \right |}\right )}{4 \, b^{\frac {7}{2}}} \] Input:

integrate(x^(5/2)/arctanh(tanh(b*x+a))^(3/2),x, algorithm="giac")
 

Output:

1/4*(x*(2*x/b - 5*a/b^2) - 15*a^2/b^3)*sqrt(x)/sqrt(b*x + a) - 15/4*a^2*lo 
g(abs(-sqrt(b)*sqrt(x) + sqrt(b*x + a)))/b^(7/2)
 

Mupad [F(-1)]

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

int(x^(5/2)/atanh(tanh(a + b*x))^(3/2),x)
 

Output:

int(x^(5/2)/atanh(tanh(a + b*x))^(3/2), x)
 

Reduce [F]

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

int(x^(5/2)/atanh(tanh(b*x+a))^(3/2),x)
 

Output:

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