\(\int \frac {x^{3/2}}{\text {arctanh}(\tanh (a+b x))^{3/2}} \, dx\) [251]

Optimal result

Integrand size = 17, antiderivative size = 86 \[ \int \frac {x^{3/2}}{\text {arctanh}(\tanh (a+b x))^{3/2}} \, dx=\frac {3 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\text {arctanh}(\tanh (a+b x))}}\right ) (b x-\text {arctanh}(\tanh (a+b x)))}{b^{5/2}}-\frac {2 x^{3/2}}{b \sqrt {\text {arctanh}(\tanh (a+b x))}}+\frac {3 \sqrt {x} \sqrt {\text {arctanh}(\tanh (a+b x))}}{b^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.94 \[ \int \frac {x^{3/2}}{\text {arctanh}(\tanh (a+b x))^{3/2}} \, dx=\frac {\sqrt {x} (-2 b x+3 \text {arctanh}(\tanh (a+b x)))}{b^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}+\frac {3 (b x-\text {arctanh}(\tanh (a+b x))) \log \left (b \sqrt {x}+\sqrt {b} \sqrt {\text {arctanh}(\tanh (a+b x))}\right )}{b^{5/2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.05, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {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.

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

\(\Big \downarrow \) 2599

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

\(\Big \downarrow \) 2600

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

\(\Big \downarrow \) 2596

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

Input:

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

Output:

(3*((ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[ArcTanh[Tanh[a + b*x]]]]*(b*x - ArcTan 
h[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]]])
 

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 = 75, normalized size of antiderivative = 0.87

method	result	size
derivativedivides	\(\frac {x^{\frac {3}{2}}}{b \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}-\frac {3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \left (-\frac {\sqrt {x}}{b \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}+\frac {\ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}\right )}{b^{\frac {3}{2}}}\right )}{b}\)	\(75\)
default	\(\frac {x^{\frac {3}{2}}}{b \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}-\frac {3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \left (-\frac {\sqrt {x}}{b \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}+\frac {\ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}\right )}{b^{\frac {3}{2}}}\right )}{b}\)	\(75\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.65 \[ \int \frac {x^{3/2}}{\text {arctanh}(\tanh (a+b x))^{3/2}} \, dx=\left [\frac {3 \, {\left (a b x + a^{2}\right )} \sqrt {b} \log \left (2 \, b x - 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (b^{2} x + 3 \, a b\right )} \sqrt {b x + a} \sqrt {x}}{2 \, {\left (b^{4} x + a b^{3}\right )}}, \frac {3 \, {\left (a b x + a^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {b x + a}}\right ) + {\left (b^{2} x + 3 \, a b\right )} \sqrt {b x + a} \sqrt {x}}{b^{4} x + a b^{3}}\right ] \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.56 \[ \int \frac {x^{3/2}}{\text {arctanh}(\tanh (a+b x))^{3/2}} \, dx=\frac {\sqrt {x} {\left (\frac {x}{b} + \frac {3 \, a}{b^{2}}\right )}}{\sqrt {b x + a}} + \frac {3 \, a \log \left ({\left | -\sqrt {b} \sqrt {x} + \sqrt {b x + a} \right |}\right )}{b^{\frac {5}{2}}} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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