\(\int \frac {x^{3/2}}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx\) [259]

Optimal result

Integrand size = 17, antiderivative size = 75 \[ \int \frac {x^{3/2}}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\text {arctanh}(\tanh (a+b x))}}\right )}{b^{5/2}}-\frac {2 x^{3/2}}{3 b \text {arctanh}(\tanh (a+b x))^{3/2}}-\frac {2 \sqrt {x}}{b^2 \sqrt {\text {arctanh}(\tanh (a+b x))}} \] Output:

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.04 \[ \int \frac {x^{3/2}}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=-\frac {2 x^{3/2}}{3 b \text {arctanh}(\tanh (a+b x))^{3/2}}-\frac {2 \sqrt {x}}{b^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}+\frac {2 \log \left (b \sqrt {x}+\sqrt {b} \sqrt {\text {arctanh}(\tanh (a+b x))}\right )}{b^{5/2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.07, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2599, 2599, 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.

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

\(\Big \downarrow \) 2599

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

\(\Big \downarrow \) 2599

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

\(\Big \downarrow \) 2596

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

Input:

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

Output:

((2*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[ArcTanh[Tanh[a + b*x]]]])/b^(3/2) - (2* 
Sqrt[x])/(b*Sqrt[ArcTanh[Tanh[a + b*x]]]))/b - (2*x^(3/2))/(3*b*ArcTanh[Ta 
nh[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]))
 

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.79

method	result	size
derivativedivides	\(-\frac {2 x^{\frac {3}{2}}}{3 b \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}-\frac {2 \sqrt {x}}{b^{2} \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}+\frac {2 \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}\right )}{b^{\frac {5}{2}}}\)	\(59\)
default	\(-\frac {2 x^{\frac {3}{2}}}{3 b \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}-\frac {2 \sqrt {x}}{b^{2} \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}+\frac {2 \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}\right )}{b^{\frac {5}{2}}}\)	\(59\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 183, normalized size of antiderivative = 2.44 \[ \int \frac {x^{3/2}}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\left [\frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {b} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, {\left (4 \, b^{2} x + 3 \, a b\right )} \sqrt {b x + a} \sqrt {x}}{3 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}, -\frac {2 \, {\left (3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {b x + a}}\right ) + {\left (4 \, b^{2} x + 3 \, a b\right )} \sqrt {b x + a} \sqrt {x}\right )}}{3 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}\right ] \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.65 \[ \int \frac {x^{3/2}}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=-\frac {2 \, \sqrt {x} {\left (\frac {4 \, x}{b} + \frac {3 \, a}{b^{2}}\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}}} - \frac {2 \, \log \left ({\left | -\sqrt {b} \sqrt {x} + \sqrt {b x + a} \right |}\right )}{b^{\frac {5}{2}}} \] Input:

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

Output:

-2/3*sqrt(x)*(4*x/b + 3*a/b^2)/(b*x + a)^(3/2) - 2*log(abs(-sqrt(b)*sqrt(x 
) + sqrt(b*x + a)))/b^(5/2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {x^{3/2}}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\frac {3 \mathit {atanh} \left (\tanh \left (b x +a \right )\right )^{2} \left (\int \frac {\sqrt {x}\, \sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}}{\mathit {atanh} \left (\tanh \left (b x +a \right )\right ) x}d x \right )-6 \sqrt {x}\, \sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}\, \mathit {atanh} \left (\tanh \left (b x +a \right )\right )-2 \sqrt {x}\, \sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}\, b x}{3 \mathit {atanh} \left (\tanh \left (b x +a \right )\right )^{2} b^{2}} \] Input:

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

Output:

(3*atanh(tanh(a + b*x))**2*int((sqrt(x)*sqrt(atanh(tanh(a + b*x))))/(atanh 
(tanh(a + b*x))*x),x) - 6*sqrt(x)*sqrt(atanh(tanh(a + b*x)))*atanh(tanh(a 
+ b*x)) - 2*sqrt(x)*sqrt(atanh(tanh(a + b*x)))*b*x)/(3*atanh(tanh(a + b*x) 
)**2*b**2)