\(\int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^{3/2}} \, dx\) [234]

Optimal result

Integrand size = 17, antiderivative size = 121 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^{3/2}} \, dx=\frac {15}{4} \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\text {arctanh}(\tanh (a+b x))}}\right ) (b x-\text {arctanh}(\tanh (a+b x)))^2-\frac {15}{4} b \sqrt {x} (b x-\text {arctanh}(\tanh (a+b x))) \sqrt {\text {arctanh}(\tanh (a+b x))}+\frac {5}{2} b \sqrt {x} \text {arctanh}(\tanh (a+b x))^{3/2}-\frac {2 \text {arctanh}(\tanh (a+b x))^{5/2}}{\sqrt {x}} \] Output:

15/4*b^(1/2)*arctanh(b^(1/2)*x^(1/2)/arctanh(tanh(b*x+a))^(1/2))*(b*x-arct 
anh(tanh(b*x+a)))^2-15/4*b*x^(1/2)*(b*x-arctanh(tanh(b*x+a)))*arctanh(tanh 
(b*x+a))^(1/2)+5/2*b*x^(1/2)*arctanh(tanh(b*x+a))^(3/2)-2*arctanh(tanh(b*x 
+a))^(5/2)/x^(1/2)
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.83 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^{3/2}} \, dx=-\frac {\sqrt {\text {arctanh}(\tanh (a+b x))} \left (15 b^2 x^2-25 b x \text {arctanh}(\tanh (a+b x))+8 \text {arctanh}(\tanh (a+b x))^2\right )}{4 \sqrt {x}}+\frac {15}{4} \sqrt {b} (-b x+\text {arctanh}(\tanh (a+b x)))^2 \log \left (b \sqrt {x}+\sqrt {b} \sqrt {\text {arctanh}(\tanh (a+b x))}\right ) \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00, 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.

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

\(\Big \downarrow \) 2599

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

\(\Big \downarrow \) 2600

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

\(\Big \downarrow \) 2600

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

\(\Big \downarrow \) 2596

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

Input:

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

Output:

(-2*ArcTanh[Tanh[a + b*x]]^(5/2))/Sqrt[x] + 5*b*((-3*(-((ArcTanh[(Sqrt[b]* 
Sqrt[x])/Sqrt[ArcTanh[Tanh[a + b*x]]]]*(b*x - ArcTanh[Tanh[a + b*x]]))/Sqr 
t[b]) + Sqrt[x]*Sqrt[ArcTanh[Tanh[a + b*x]]])*(b*x - ArcTanh[Tanh[a + b*x] 
]))/4 + (Sqrt[x]*ArcTanh[Tanh[a + b*x]]^(3/2))/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 = 0.45 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.28

method	result	size
derivativedivides	\(-\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \sqrt {x}}+\frac {12 b \left (\frac {\sqrt {x}\, \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{6}+\frac {5 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \left (\frac {\sqrt {x}\, \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{4}+\frac {3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \left (\frac {\sqrt {x}\, \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{2}+\frac {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6}\right )}{\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}\)	\(155\)
default	\(-\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \sqrt {x}}+\frac {12 b \left (\frac {\sqrt {x}\, \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{6}+\frac {5 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \left (\frac {\sqrt {x}\, \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{4}+\frac {3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \left (\frac {\sqrt {x}\, \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{2}+\frac {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6}\right )}{\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}\)	\(155\)

Input:

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

Output:

-2/(arctanh(tanh(b*x+a))-b*x)/x^(1/2)*arctanh(tanh(b*x+a))^(7/2)+12*b/(arc 
tanh(tanh(b*x+a))-b*x)*(1/6*x^(1/2)*arctanh(tanh(b*x+a))^(5/2)+5/6*(arctan 
h(tanh(b*x+a))-b*x)*(1/4*x^(1/2)*arctanh(tanh(b*x+a))^(3/2)+3/4*(arctanh(t 
anh(b*x+a))-b*x)*(1/2*x^(1/2)*arctanh(tanh(b*x+a))^(1/2)+1/2/b^(1/2)*(arct 
anh(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 = 0.09 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.11 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^{3/2}} \, dx=\left [\frac {15 \, a^{2} \sqrt {b} x \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (2 \, b^{2} x^{2} + 9 \, a b x - 8 \, a^{2}\right )} \sqrt {b x + a} \sqrt {x}}{8 \, x}, -\frac {15 \, a^{2} \sqrt {-b} x \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {b x + a}}\right ) - {\left (2 \, b^{2} x^{2} + 9 \, a b x - 8 \, a^{2}\right )} \sqrt {b x + a} \sqrt {x}}{4 \, x}\right ] \] Input:

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

Output:

[1/8*(15*a^2*sqrt(b)*x*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 
2*(2*b^2*x^2 + 9*a*b*x - 8*a^2)*sqrt(b*x + a)*sqrt(x))/x, -1/4*(15*a^2*sqr 
t(-b)*x*arctan(sqrt(-b)*sqrt(x)/sqrt(b*x + a)) - (2*b^2*x^2 + 9*a*b*x - 8* 
a^2)*sqrt(b*x + a)*sqrt(x))/x]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 73.41 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.89 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^{3/2}} \, dx=-\frac {\sqrt {2} {\left (\frac {15 \, \sqrt {2} a^{2} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} + \sqrt {{\left (b x + a\right )} b - a b} \right |}\right )}{\sqrt {b}} + \frac {{\left (15 \, \sqrt {2} a^{2} - {\left (b x + a\right )} {\left (2 \, \sqrt {2} {\left (b x + a\right )} + 5 \, \sqrt {2} a\right )}\right )} \sqrt {b x + a}}{\sqrt {{\left (b x + a\right )} b - a b}}\right )} b^{2}}{8 \, {\left | b \right |}} \] Input:

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

Output:

-1/8*sqrt(2)*(15*sqrt(2)*a^2*log(abs(-sqrt(b*x + a)*sqrt(b) + sqrt((b*x + 
a)*b - a*b)))/sqrt(b) + (15*sqrt(2)*a^2 - (b*x + a)*(2*sqrt(2)*(b*x + a) + 
 5*sqrt(2)*a))*sqrt(b*x + a)/sqrt((b*x + a)*b - a*b))*b^2/abs(b)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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