\(\int \frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{\sqrt {x}} \, dx\) [225]

Optimal result

Integrand size = 17, antiderivative size = 101 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{\sqrt {x}} \, dx=\frac {3 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\text {arctanh}(\tanh (a+b x))}}\right ) (b x-\text {arctanh}(\tanh (a+b x)))^2}{4 \sqrt {b}}-\frac {3}{4} \sqrt {x} (b x-\text {arctanh}(\tanh (a+b x))) \sqrt {\text {arctanh}(\tanh (a+b x))}+\frac {1}{2} \sqrt {x} \text {arctanh}(\tanh (a+b x))^{3/2} \] Output:

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.82 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{\sqrt {x}} \, dx=\frac {1}{4} \left (\sqrt {x} \sqrt {\text {arctanh}(\tanh (a+b x))} (-3 b x+5 \text {arctanh}(\tanh (a+b x)))+\frac {3 (-b x+\text {arctanh}(\tanh (a+b x)))^2 \log \left (b \sqrt {x}+\sqrt {b} \sqrt {\text {arctanh}(\tanh (a+b x))}\right )}{\sqrt {b}}\right ) \] Input:

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

Output:

(Sqrt[x]*Sqrt[ArcTanh[Tanh[a + b*x]]]*(-3*b*x + 5*ArcTanh[Tanh[a + b*x]]) 
+ (3*(-(b*x) + ArcTanh[Tanh[a + b*x]])^2*Log[b*Sqrt[x] + Sqrt[b]*Sqrt[ArcT 
anh[Tanh[a + b*x]]]])/Sqrt[b])/4
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.98, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {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.

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

\(\Big \downarrow \) 2600

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

\(\Big \downarrow \) 2600

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

\(\Big \downarrow \) 2596

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

Input:

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

Output:

(-3*(-((ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[ArcTanh[Tanh[a + b*x]]]]*(b*x - Arc 
Tanh[Tanh[a + b*x]]))/Sqrt[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 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 = 0.40 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.79

method	result	size
derivativedivides	\(\frac {\sqrt {x}\, \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{2}+\frac {3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \left (\frac {\sqrt {x}\, \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{2}+\frac {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}\right )}{2 \sqrt {b}}\right )}{2}\)	\(80\)
default	\(\frac {\sqrt {x}\, \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{2}+\frac {3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \left (\frac {\sqrt {x}\, \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{2}+\frac {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}\right )}{2 \sqrt {b}}\right )}{2}\)	\(80\)

Input:

int(arctanh(tanh(b*x+a))^(3/2)/x^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/2*x^(1/2)*arctanh(tanh(b*x+a))^(3/2)+3/2*(arctanh(tanh(b*x+a))-b*x)*(1/2 
*x^(1/2)*arctanh(tanh(b*x+a))^(1/2)+1/2/b^(1/2)*(arctanh(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 = 0.10 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.15 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{\sqrt {x}} \, dx=\left [\frac {3 \, a^{2} \sqrt {b} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (2 \, b^{2} x + 5 \, a b\right )} \sqrt {b x + a} \sqrt {x}}{8 \, b}, -\frac {3 \, a^{2} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {b x + a}}\right ) - {\left (2 \, b^{2} x + 5 \, a b\right )} \sqrt {b x + a} \sqrt {x}}{4 \, b}\right ] \] Input:

integrate(arctanh(tanh(b*x+a))^(3/2)/x^(1/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

integrate(atanh(tanh(b*x+a))**(3/2)/x**(1/2),x)
 

Output:

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

Maxima [F]

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

integrate(arctanh(tanh(b*x+a))^(3/2)/x^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 72.36 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.96 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{\sqrt {x}} \, dx=-\frac {\sqrt {2} {\left (\frac {3 \, \sqrt {2} a^{2} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} + \sqrt {{\left (b x + a\right )} b - a b} \right |}\right )}{\sqrt {b}} - \sqrt {{\left (b x + a\right )} b - a b} \sqrt {b x + a} {\left (\frac {2 \, \sqrt {2} {\left (b x + a\right )}}{b} + \frac {3 \, \sqrt {2} a}{b}\right )}\right )} b}{8 \, {\left | b \right |}} \] Input:

integrate(arctanh(tanh(b*x+a))^(3/2)/x^(1/2),x, algorithm="giac")
 

Output:

-1/8*sqrt(2)*(3*sqrt(2)*a^2*log(abs(-sqrt(b*x + a)*sqrt(b) + sqrt((b*x + a 
)*b - a*b)))/sqrt(b) - sqrt((b*x + a)*b - a*b)*sqrt(b*x + a)*(2*sqrt(2)*(b 
*x + a)/b + 3*sqrt(2)*a/b))*b/abs(b)
 

Mupad [F(-1)]

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

int(atanh(tanh(a + b*x))^(3/2)/x^(1/2),x)
 

Output:

int(atanh(tanh(a + b*x))^(3/2)/x^(1/2), x)
 

Reduce [F]

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

int(atanh(tanh(b*x+a))^(3/2)/x^(1/2),x)
 

Output:

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