\(\int x^3 \text {arctanh}(\tanh (a+b x))^{3/2} \, dx\) [121]

Optimal result

Integrand size = 15, antiderivative size = 80 \[ \int x^3 \text {arctanh}(\tanh (a+b x))^{3/2} \, dx=\frac {2 x^3 \text {arctanh}(\tanh (a+b x))^{5/2}}{5 b}-\frac {12 x^2 \text {arctanh}(\tanh (a+b x))^{7/2}}{35 b^2}+\frac {16 x \text {arctanh}(\tanh (a+b x))^{9/2}}{105 b^3}-\frac {32 \text {arctanh}(\tanh (a+b x))^{11/2}}{1155 b^4} \] Output:

2/5*x^3*arctanh(tanh(b*x+a))^(5/2)/b-12/35*x^2*arctanh(tanh(b*x+a))^(7/2)/ 
b^2+16/105*x*arctanh(tanh(b*x+a))^(9/2)/b^3-32/1155*arctanh(tanh(b*x+a))^( 
11/2)/b^4
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.82 \[ \int x^3 \text {arctanh}(\tanh (a+b x))^{3/2} \, dx=\frac {2 \text {arctanh}(\tanh (a+b x))^{5/2} \left (231 b^3 x^3-198 b^2 x^2 \text {arctanh}(\tanh (a+b x))+88 b x \text {arctanh}(\tanh (a+b x))^2-16 \text {arctanh}(\tanh (a+b x))^3\right )}{1155 b^4} \] Input:

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

Output:

(2*ArcTanh[Tanh[a + b*x]]^(5/2)*(231*b^3*x^3 - 198*b^2*x^2*ArcTanh[Tanh[a 
+ b*x]] + 88*b*x*ArcTanh[Tanh[a + b*x]]^2 - 16*ArcTanh[Tanh[a + b*x]]^3))/ 
(1155*b^4)
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.20, 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.

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

\(\Big \downarrow \) 2599

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

\(\Big \downarrow \) 2599

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

\(\Big \downarrow \) 2599

\(\displaystyle \frac {2 x^3 \text {arctanh}(\tanh (a+b x))^{5/2}}{5 b}-\frac {6 \left (\frac {2 x^2 \text {arctanh}(\tanh (a+b x))^{7/2}}{7 b}-\frac {4 \left (\frac {2 x \text {arctanh}(\tanh (a+b x))^{9/2}}{9 b}-\frac {2 \int \text {arctanh}(\tanh (a+b x))^{9/2}dx}{9 b}\right )}{7 b}\right )}{5 b}\)

\(\Big \downarrow \) 2588

\(\displaystyle \frac {2 x^3 \text {arctanh}(\tanh (a+b x))^{5/2}}{5 b}-\frac {6 \left (\frac {2 x^2 \text {arctanh}(\tanh (a+b x))^{7/2}}{7 b}-\frac {4 \left (\frac {2 x \text {arctanh}(\tanh (a+b x))^{9/2}}{9 b}-\frac {2 \int \text {arctanh}(\tanh (a+b x))^{9/2}d\text {arctanh}(\tanh (a+b x))}{9 b^2}\right )}{7 b}\right )}{5 b}\)

\(\Big \downarrow \) 15

\(\displaystyle \frac {2 x^3 \text {arctanh}(\tanh (a+b x))^{5/2}}{5 b}-\frac {6 \left (\frac {2 x^2 \text {arctanh}(\tanh (a+b x))^{7/2}}{7 b}-\frac {4 \left (\frac {2 x \text {arctanh}(\tanh (a+b x))^{9/2}}{9 b}-\frac {4 \text {arctanh}(\tanh (a+b x))^{11/2}}{99 b^2}\right )}{7 b}\right )}{5 b}\)

Input:

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

Output:

(2*x^3*ArcTanh[Tanh[a + b*x]]^(5/2))/(5*b) - (6*((2*x^2*ArcTanh[Tanh[a + b 
*x]]^(7/2))/(7*b) - (4*((2*x*ArcTanh[Tanh[a + b*x]]^(9/2))/(9*b) - (4*ArcT 
anh[Tanh[a + b*x]]^(11/2))/(99*b^2)))/(7*b)))/(5*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 = 0.24 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.55

method	result	size
default	\(\frac {\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {11}{2}}}{11}+\frac {2 \left (-3 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )+3 b x \right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {9}{2}}}{9}+\frac {2 \left (\left (b x -\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right ) \left (2 b x -2 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )+\left (b x -\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )^{2}\right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{7}+\frac {2 \left (b x -\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )^{3} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{5}}{b^{4}}\)	\(124\)

Input:

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

Output:

2/b^4*(1/11*arctanh(tanh(b*x+a))^(11/2)+1/9*(-3*arctanh(tanh(b*x+a))+3*b*x 
)*arctanh(tanh(b*x+a))^(9/2)+1/7*((b*x-arctanh(tanh(b*x+a)))*(2*b*x-2*arct 
anh(tanh(b*x+a)))+(b*x-arctanh(tanh(b*x+a)))^2)*arctanh(tanh(b*x+a))^(7/2) 
+1/5*(b*x-arctanh(tanh(b*x+a)))^3*arctanh(tanh(b*x+a))^(5/2))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.80 \[ \int x^3 \text {arctanh}(\tanh (a+b x))^{3/2} \, dx=\frac {2 \, {\left (105 \, b^{5} x^{5} + 140 \, a b^{4} x^{4} + 5 \, a^{2} b^{3} x^{3} - 6 \, a^{3} b^{2} x^{2} + 8 \, a^{4} b x - 16 \, a^{5}\right )} \sqrt {b x + a}}{1155 \, b^{4}} \] Input:

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

Output:

2/1155*(105*b^5*x^5 + 140*a*b^4*x^4 + 5*a^2*b^3*x^3 - 6*a^3*b^2*x^2 + 8*a^ 
4*b*x - 16*a^5)*sqrt(b*x + a)/b^4
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.66 \[ \int x^3 \text {arctanh}(\tanh (a+b x))^{3/2} \, dx=\frac {2 \, {\left (105 \, b^{4} x^{4} + 35 \, a b^{3} x^{3} - 30 \, a^{2} b^{2} x^{2} + 24 \, a^{3} b x - 16 \, a^{4}\right )} {\left (b x + a\right )}^{\frac {3}{2}}}{1155 \, b^{4}} \] Input:

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

Output:

2/1155*(105*b^4*x^4 + 35*a*b^3*x^3 - 30*a^2*b^2*x^2 + 24*a^3*b*x - 16*a^4) 
*(b*x + a)^(3/2)/b^4
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 205 vs. \(2 (64) = 128\).

Time = 0.12 (sec) , antiderivative size = 205, normalized size of antiderivative = 2.56 \[ \int x^3 \text {arctanh}(\tanh (a+b x))^{3/2} \, dx=\frac {\sqrt {2} {\left (\frac {99 \, \sqrt {2} {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )} a^{2}}{b^{3}} + \frac {22 \, \sqrt {2} {\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} - 180 \, {\left (b x + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b x + a} a^{4}\right )} a}{b^{3}} + \frac {5 \, \sqrt {2} {\left (63 \, {\left (b x + a\right )}^{\frac {11}{2}} - 385 \, {\left (b x + a\right )}^{\frac {9}{2}} a + 990 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{2} - 1386 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{3} + 1155 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{4} - 693 \, \sqrt {b x + a} a^{5}\right )}}{b^{3}}\right )}}{3465 \, b} \] Input:

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

Output:

1/3465*sqrt(2)*(99*sqrt(2)*(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)*a^2/b^3 + 22*sqrt(2)*(35*(b*x 
+ a)^(9/2) - 180*(b*x + a)^(7/2)*a + 378*(b*x + a)^(5/2)*a^2 - 420*(b*x + 
a)^(3/2)*a^3 + 315*sqrt(b*x + a)*a^4)*a/b^3 + 5*sqrt(2)*(63*(b*x + a)^(11/ 
2) - 385*(b*x + a)^(9/2)*a + 990*(b*x + a)^(7/2)*a^2 - 1386*(b*x + a)^(5/2 
)*a^3 + 1155*(b*x + a)^(3/2)*a^4 - 693*sqrt(b*x + a)*a^5)/b^3)/b
 

Mupad [B] (verification not implemented)

Time = 3.24 (sec) , antiderivative size = 1483, normalized size of antiderivative = 18.54 \[ \int x^3 \text {arctanh}(\tanh (a+b x))^{3/2} \, dx=\text {Too large to display} \] Input:

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

Output:

(2*b*x^5*(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))/11 + (x^4*(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 
)*((20*b*(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))/11 - 2*b*(log(2/(exp(2*a)*exp(2*b*x) 
 + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x))) 
/(9*b) + (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)*((log(2/(exp(2*a)*exp(2*b*x) + 1 
)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^2/2 + 
 (8*((20*b*(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))/11 - 2*b*(log(2/(exp(2*a)*exp(2*b* 
x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x) 
)*(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))/(9*b)))/(7*b) + (16*(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)) - log((2*exp(2*a)*exp(2*b*x))/(exp 
(2*a)*exp(2*b*x) + 1)) + 2*b*x)^2/2 + (8*((20*b*(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 
))/11 - 2*b*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x) 
)/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x))*(log(2/(exp(2*a)*exp(2*b*x) + 1)...
 

Reduce [F]

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

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

Output:

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