\(\int \frac {x^3}{\sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx\) [141]

Optimal result

Integrand size = 15, antiderivative size = 76 \[ \int \frac {x^3}{\sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=\frac {2 x^3 \sqrt {\text {arctanh}(\tanh (a+b x))}}{b}-\frac {4 x^2 \text {arctanh}(\tanh (a+b x))^{3/2}}{b^2}+\frac {16 x \text {arctanh}(\tanh (a+b x))^{5/2}}{5 b^3}-\frac {32 \text {arctanh}(\tanh (a+b x))^{7/2}}{35 b^4} \] Output:

2*x^3*arctanh(tanh(b*x+a))^(1/2)/b-4*x^2*arctanh(tanh(b*x+a))^(3/2)/b^2+16 
/5*x*arctanh(tanh(b*x+a))^(5/2)/b^3-32/35*arctanh(tanh(b*x+a))^(7/2)/b^4
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.87 \[ \int \frac {x^3}{\sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=\frac {2 \sqrt {\text {arctanh}(\tanh (a+b x))} \left (35 b^3 x^3-70 b^2 x^2 \text {arctanh}(\tanh (a+b x))+56 b x \text {arctanh}(\tanh (a+b x))^2-16 \text {arctanh}(\tanh (a+b x))^3\right )}{35 b^4} \] Input:

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

Output:

(2*Sqrt[ArcTanh[Tanh[a + b*x]]]*(35*b^3*x^3 - 70*b^2*x^2*ArcTanh[Tanh[a + 
b*x]] + 56*b*x*ArcTanh[Tanh[a + b*x]]^2 - 16*ArcTanh[Tanh[a + b*x]]^3))/(3 
5*b^4)
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.21, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2599, 2599, 2599, 2588, 15}

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.

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

\(\Big \downarrow \) 2599

\(\displaystyle \frac {2 x^3 \sqrt {\text {arctanh}(\tanh (a+b x))}}{b}-\frac {6 \int x^2 \sqrt {\text {arctanh}(\tanh (a+b x))}dx}{b}\)

\(\Big \downarrow \) 2599

\(\displaystyle \frac {2 x^3 \sqrt {\text {arctanh}(\tanh (a+b x))}}{b}-\frac {6 \left (\frac {2 x^2 \text {arctanh}(\tanh (a+b x))^{3/2}}{3 b}-\frac {4 \int x \text {arctanh}(\tanh (a+b x))^{3/2}dx}{3 b}\right )}{b}\)

\(\Big \downarrow \) 2599

\(\displaystyle \frac {2 x^3 \sqrt {\text {arctanh}(\tanh (a+b x))}}{b}-\frac {6 \left (\frac {2 x^2 \text {arctanh}(\tanh (a+b x))^{3/2}}{3 b}-\frac {4 \left (\frac {2 x \text {arctanh}(\tanh (a+b x))^{5/2}}{5 b}-\frac {2 \int \text {arctanh}(\tanh (a+b x))^{5/2}dx}{5 b}\right )}{3 b}\right )}{b}\)

\(\Big \downarrow \) 2588

\(\displaystyle \frac {2 x^3 \sqrt {\text {arctanh}(\tanh (a+b x))}}{b}-\frac {6 \left (\frac {2 x^2 \text {arctanh}(\tanh (a+b x))^{3/2}}{3 b}-\frac {4 \left (\frac {2 x \text {arctanh}(\tanh (a+b x))^{5/2}}{5 b}-\frac {2 \int \text {arctanh}(\tanh (a+b x))^{5/2}d\text {arctanh}(\tanh (a+b x))}{5 b^2}\right )}{3 b}\right )}{b}\)

\(\Big \downarrow \) 15

\(\displaystyle \frac {2 x^3 \sqrt {\text {arctanh}(\tanh (a+b x))}}{b}-\frac {6 \left (\frac {2 x^2 \text {arctanh}(\tanh (a+b x))^{3/2}}{3 b}-\frac {4 \left (\frac {2 x \text {arctanh}(\tanh (a+b x))^{5/2}}{5 b}-\frac {4 \text {arctanh}(\tanh (a+b x))^{7/2}}{35 b^2}\right )}{3 b}\right )}{b}\)

Input:

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

Output:

(2*x^3*Sqrt[ArcTanh[Tanh[a + b*x]]])/b - (6*((2*x^2*ArcTanh[Tanh[a + b*x]] 
^(3/2))/(3*b) - (4*((2*x*ArcTanh[Tanh[a + b*x]]^(5/2))/(5*b) - (4*ArcTanh[ 
Tanh[a + b*x]]^(7/2))/(35*b^2)))/(3*b)))/b
 

Defintions of rubi rules used

rule 15

Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ 
{a, m}, x] && NeQ[m, -1]
 

rule 2588

Int[(u_)^(m_.), x_Symbol] :> With[{c = Simplify[D[u, x]]}, Simp[1/c   Subst 
[Int[x^m, x], x, u], x]] /; FreeQ[m, x] && PiecewiseLinearQ[u, 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 = 0.21 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.62

method	result	size
default	\(\frac {\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{7}+\frac {2 \left (-3 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )+3 b x \right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{5}+\frac {2 \left (\left (b x -\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right ) \left (2 b x -2 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )+\left (b x -\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )^{2}\right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{3}+2 \left (b x -\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )^{3} \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{b^{4}}\)	\(123\)

Input:

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

Output:

2/b^4*(1/7*arctanh(tanh(b*x+a))^(7/2)+1/5*(-3*arctanh(tanh(b*x+a))+3*b*x)* 
arctanh(tanh(b*x+a))^(5/2)+1/3*((b*x-arctanh(tanh(b*x+a)))*(2*b*x-2*arctan 
h(tanh(b*x+a)))+(b*x-arctanh(tanh(b*x+a)))^2)*arctanh(tanh(b*x+a))^(3/2)+( 
b*x-arctanh(tanh(b*x+a)))^3*arctanh(tanh(b*x+a))^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.55 \[ \int \frac {x^3}{\sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=\frac {2 \, {\left (5 \, b^{3} x^{3} - 6 \, a b^{2} x^{2} + 8 \, a^{2} b x - 16 \, a^{3}\right )} \sqrt {b x + a}}{35 \, b^{4}} \] Input:

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

Output:

2/35*(5*b^3*x^3 - 6*a*b^2*x^2 + 8*a^2*b*x - 16*a^3)*sqrt(b*x + a)/b^4
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.70 \[ \int \frac {x^3}{\sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=\frac {2 \, {\left (5 \, b^{4} x^{4} - a b^{3} x^{3} + 2 \, a^{2} b^{2} x^{2} - 8 \, a^{3} b x - 16 \, a^{4}\right )}}{35 \, \sqrt {b x + a} b^{4}} \] Input:

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

Output:

2/35*(5*b^4*x^4 - a*b^3*x^3 + 2*a^2*b^2*x^2 - 8*a^3*b*x - 16*a^4)/(sqrt(b* 
x + a)*b^4)
 

Giac [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.64 \[ \int \frac {x^3}{\sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=\frac {2 \, {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )}}{35 \, b^{4}} \] Input:

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

Output:

2/35*(5*(b*x + a)^(7/2) - 21*(b*x + a)^(5/2)*a + 35*(b*x + a)^(3/2)*a^2 - 
35*sqrt(b*x + a)*a^3)/b^4
 

Mupad [B] (verification not implemented)

Time = 3.01 (sec) , antiderivative size = 385, normalized size of antiderivative = 5.07 \[ \int \frac {x^3}{\sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=\frac {2\,x^3\,\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}}{7\,b}+\frac {32\,\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,{\left (\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}+b\,x\right )}^3}{35\,b^4}+\frac {12\,x^2\,\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\left (\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}+b\,x\right )}{35\,b^2}+\frac {16\,x\,\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,{\left (\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}+b\,x\right )}^2}{35\,b^3} \] Input:

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

Output:

(2*x^3*(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/( 
exp(2*a)*exp(2*b*x) + 1))/2)^(1/2))/(7*b) + (32*(log((2*exp(2*a)*exp(2*b*x 
))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(exp(2*a)*exp(2*b*x) + 1))/2)^(1/2 
)*(log(2/(exp(2*a)*exp(2*b*x) + 1))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2 
*a)*exp(2*b*x) + 1))/2 + b*x)^3)/(35*b^4) + (12*x^2*(log((2*exp(2*a)*exp(2 
*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(exp(2*a)*exp(2*b*x) + 1))/2)^ 
(1/2)*(log(2/(exp(2*a)*exp(2*b*x) + 1))/2 - log((2*exp(2*a)*exp(2*b*x))/(e 
xp(2*a)*exp(2*b*x) + 1))/2 + b*x))/(35*b^2) + (16*x*(log((2*exp(2*a)*exp(2 
*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(exp(2*a)*exp(2*b*x) + 1))/2)^ 
(1/2)*(log(2/(exp(2*a)*exp(2*b*x) + 1))/2 - log((2*exp(2*a)*exp(2*b*x))/(e 
xp(2*a)*exp(2*b*x) + 1))/2 + b*x)^2)/(35*b^3)
 

Reduce [F]

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

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

Output:

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