3.3.24 \(\int \sqrt {x} \text {arctanh}(\tanh (a+b x))^{3/2} \, dx\) [224]

3.3.24.1 Optimal result

Integrand size = 17, antiderivative size = 139 \[ \int \sqrt {x} \text {arctanh}(\tanh (a+b x))^{3/2} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\text {arctanh}(\tanh (a+b x))}}\right ) (b x-\text {arctanh}(\tanh (a+b x)))^3}{8 b^{3/2}}-\frac {1}{4} x^{3/2} (b x-\text {arctanh}(\tanh (a+b x))) \sqrt {\text {arctanh}(\tanh (a+b x))}+\frac {\sqrt {x} (b x-\text {arctanh}(\tanh (a+b x)))^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}{8 b}+\frac {1}{3} x^{3/2} \text {arctanh}(\tanh (a+b x))^{3/2} \]

output

1/8*arctanh(b^(1/2)*x^(1/2)/arctanh(tanh(b*x+a))^(1/2))*(b*x-arctanh(tanh( 
b*x+a)))^3/b^(3/2)+1/3*x^(3/2)*arctanh(tanh(b*x+a))^(3/2)-1/4*x^(3/2)*(b*x 
-arctanh(tanh(b*x+a)))*arctanh(tanh(b*x+a))^(1/2)+1/8*(b*x-arctanh(tanh(b* 
x+a)))^2*x^(1/2)*arctanh(tanh(b*x+a))^(1/2)/b
 

3.3.24.2 Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.76 \[ \int \sqrt {x} \text {arctanh}(\tanh (a+b x))^{3/2} \, dx=\frac {\sqrt {x} \sqrt {\text {arctanh}(\tanh (a+b x))} \left (-3 b^2 x^2+8 b x \text {arctanh}(\tanh (a+b x))+3 \text {arctanh}(\tanh (a+b x))^2\right )}{24 b}+\frac {(b x-\text {arctanh}(\tanh (a+b x)))^3 \log \left (b \sqrt {x}+\sqrt {b} \sqrt {\text {arctanh}(\tanh (a+b x))}\right )}{8 b^{3/2}} \]

input

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

output

(Sqrt[x]*Sqrt[ArcTanh[Tanh[a + b*x]]]*(-3*b^2*x^2 + 8*b*x*ArcTanh[Tanh[a + 
 b*x]] + 3*ArcTanh[Tanh[a + b*x]]^2))/(24*b) + ((b*x - ArcTanh[Tanh[a + b* 
x]])^3*Log[b*Sqrt[x] + Sqrt[b]*Sqrt[ArcTanh[Tanh[a + b*x]]]])/(8*b^(3/2))
 

3.3.24.3 Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 139, 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 = {2600, 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.

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

\(\Big \downarrow \) 2600

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

\(\Big \downarrow \) 2600

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

\(\Big \downarrow \) 2600

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

\(\Big \downarrow \) 2596

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

input

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

output

-1/2*((-1/4*(((ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[ArcTanh[Tanh[a + b*x]]]]*(b* 
x - ArcTanh[Tanh[a + b*x]]))/b^(3/2) + (Sqrt[x]*Sqrt[ArcTanh[Tanh[a + b*x] 
]])/b)*(b*x - ArcTanh[Tanh[a + b*x]])) + (x^(3/2)*Sqrt[ArcTanh[Tanh[a + b* 
x]]])/2)*(b*x - ArcTanh[Tanh[a + b*x]])) + (x^(3/2)*ArcTanh[Tanh[a + b*x]] 
^(3/2))/3
 

3.3.24.3.1 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]
 

3.3.24.4 Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.83

method	result	size
derivativedivides	\(\frac {\sqrt {x}\, \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{3 b}-\frac {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \left (\frac {\sqrt {x}\, \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{4}+\frac {3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \left (\frac {\sqrt {x}\, \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{2}+\frac {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}\right )}{2 \sqrt {b}}\right )}{4}\right )}{3 b}\)	\(115\)
default	\(\frac {\sqrt {x}\, \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{3 b}-\frac {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \left (\frac {\sqrt {x}\, \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{4}+\frac {3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \left (\frac {\sqrt {x}\, \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{2}+\frac {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}\right )}{2 \sqrt {b}}\right )}{4}\right )}{3 b}\)	\(115\)

input

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

output

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

3.3.24.5 Fricas [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.01 \[ \int \sqrt {x} \text {arctanh}(\tanh (a+b x))^{3/2} \, dx=\left [\frac {3 \, a^{3} \sqrt {b} \log \left (2 \, b x - 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (8 \, b^{3} x^{2} + 14 \, a b^{2} x + 3 \, a^{2} b\right )} \sqrt {b x + a} \sqrt {x}}{48 \, b^{2}}, \frac {3 \, a^{3} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (8 \, b^{3} x^{2} + 14 \, a b^{2} x + 3 \, a^{2} b\right )} \sqrt {b x + a} \sqrt {x}}{24 \, b^{2}}\right ] \]

input

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

output

[1/48*(3*a^3*sqrt(b)*log(2*b*x - 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2* 
(8*b^3*x^2 + 14*a*b^2*x + 3*a^2*b)*sqrt(b*x + a)*sqrt(x))/b^2, 1/24*(3*a^3 
*sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) + (8*b^3*x^2 + 14*a*b 
^2*x + 3*a^2*b)*sqrt(b*x + a)*sqrt(x))/b^2]
 

3.3.24.6 Sympy [F]

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

input

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

output

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

3.3.24.7 Maxima [F]

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

input

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

output

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

3.3.24.8 Giac [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.88 \[ \int \sqrt {x} \text {arctanh}(\tanh (a+b x))^{3/2} \, dx=\frac {1}{48} \, \sqrt {2} {\left (6 \, \sqrt {2} {\left (\sqrt {b x + a} {\left (2 \, x + \frac {a}{b}\right )} \sqrt {x} + \frac {a^{2} \log \left ({\left | -\sqrt {b} \sqrt {x} + \sqrt {b x + a} \right |}\right )}{b^{\frac {3}{2}}}\right )} a + \sqrt {2} {\left (\sqrt {b x + a} {\left (2 \, {\left (4 \, x + \frac {a}{b}\right )} x - \frac {3 \, a^{2}}{b^{2}}\right )} \sqrt {x} - \frac {3 \, a^{3} \log \left ({\left | -\sqrt {b} \sqrt {x} + \sqrt {b x + a} \right |}\right )}{b^{\frac {5}{2}}}\right )} b\right )} \]

input

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

output

1/48*sqrt(2)*(6*sqrt(2)*(sqrt(b*x + a)*(2*x + a/b)*sqrt(x) + a^2*log(abs(- 
sqrt(b)*sqrt(x) + sqrt(b*x + a)))/b^(3/2))*a + sqrt(2)*(sqrt(b*x + a)*(2*( 
4*x + a/b)*x - 3*a^2/b^2)*sqrt(x) - 3*a^3*log(abs(-sqrt(b)*sqrt(x) + sqrt( 
b*x + a)))/b^(5/2))*b)
 

3.3.24.9 Mupad [F(-1)]

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

input

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

output

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