\(\int x^3 \text {arctanh}(\tanh (a+b x))^{5/2} \, dx\) [130]

Optimal result

Integrand size = 15, antiderivative size = 80 \[ \int x^3 \text {arctanh}(\tanh (a+b x))^{5/2} \, dx=\frac {2 x^3 \text {arctanh}(\tanh (a+b x))^{7/2}}{7 b}-\frac {4 x^2 \text {arctanh}(\tanh (a+b x))^{9/2}}{21 b^2}+\frac {16 x \text {arctanh}(\tanh (a+b x))^{11/2}}{231 b^3}-\frac {32 \text {arctanh}(\tanh (a+b x))^{13/2}}{3003 b^4} \] Output:

2/7*x^3*arctanh(tanh(b*x+a))^(7/2)/b-4/21*x^2*arctanh(tanh(b*x+a))^(9/2)/b 
^2+16/231*x*arctanh(tanh(b*x+a))^(11/2)/b^3-32/3003*arctanh(tanh(b*x+a))^( 
13/2)/b^4
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.82 \[ \int x^3 \text {arctanh}(\tanh (a+b x))^{5/2} \, dx=\frac {2 \text {arctanh}(\tanh (a+b x))^{7/2} \left (429 b^3 x^3-286 b^2 x^2 \text {arctanh}(\tanh (a+b x))+104 b x \text {arctanh}(\tanh (a+b x))^2-16 \text {arctanh}(\tanh (a+b x))^3\right )}{3003 b^4} \] Input:

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

Output:

(2*ArcTanh[Tanh[a + b*x]]^(7/2)*(429*b^3*x^3 - 286*b^2*x^2*ArcTanh[Tanh[a 
+ b*x]] + 104*b*x*ArcTanh[Tanh[a + b*x]]^2 - 16*ArcTanh[Tanh[a + b*x]]^3)) 
/(3003*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.

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

\(\Big \downarrow \) 2599

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

\(\Big \downarrow \) 2599

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

\(\Big \downarrow \) 2599

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

\(\Big \downarrow \) 2588

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

\(\Big \downarrow \) 15

\(\displaystyle \frac {2 x^3 \text {arctanh}(\tanh (a+b x))^{7/2}}{7 b}-\frac {6 \left (\frac {2 x^2 \text {arctanh}(\tanh (a+b x))^{9/2}}{9 b}-\frac {4 \left (\frac {2 x \text {arctanh}(\tanh (a+b x))^{11/2}}{11 b}-\frac {4 \text {arctanh}(\tanh (a+b x))^{13/2}}{143 b^2}\right )}{9 b}\right )}{7 b}\)

Input:

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

Output:

(2*x^3*ArcTanh[Tanh[a + b*x]]^(7/2))/(7*b) - (6*((2*x^2*ArcTanh[Tanh[a + b 
*x]]^(9/2))/(9*b) - (4*((2*x*ArcTanh[Tanh[a + b*x]]^(11/2))/(11*b) - (4*Ar 
cTanh[Tanh[a + b*x]]^(13/2))/(143*b^2)))/(9*b)))/(7*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 = 0.34 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.55

method	result	size
default	\(\frac {\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {13}{2}}}{13}+\frac {2 \left (-3 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )+3 b x \right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {11}{2}}}{11}+\frac {2 \left (\left (b x -\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right ) \left (2 b x -2 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )+\left (b x -\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )^{2}\right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {9}{2}}}{9}+\frac {2 \left (b x -\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )^{3} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{7}}{b^{4}}\)	\(124\)

Input:

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

Output:

2/b^4*(1/13*arctanh(tanh(b*x+a))^(13/2)+1/11*(-3*arctanh(tanh(b*x+a))+3*b* 
x)*arctanh(tanh(b*x+a))^(11/2)+1/9*((b*x-arctanh(tanh(b*x+a)))*(2*b*x-2*ar 
ctanh(tanh(b*x+a)))+(b*x-arctanh(tanh(b*x+a)))^2)*arctanh(tanh(b*x+a))^(9/ 
2)+1/7*(b*x-arctanh(tanh(b*x+a)))^3*arctanh(tanh(b*x+a))^(7/2))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.94 \[ \int x^3 \text {arctanh}(\tanh (a+b x))^{5/2} \, dx=\frac {2 \, {\left (231 \, b^{6} x^{6} + 567 \, a b^{5} x^{5} + 371 \, a^{2} b^{4} x^{4} + 5 \, a^{3} b^{3} x^{3} - 6 \, a^{4} b^{2} x^{2} + 8 \, a^{5} b x - 16 \, a^{6}\right )} \sqrt {b x + a}}{3003 \, b^{4}} \] Input:

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

Output:

2/3003*(231*b^6*x^6 + 567*a*b^5*x^5 + 371*a^2*b^4*x^4 + 5*a^3*b^3*x^3 - 6* 
a^4*b^2*x^2 + 8*a^5*b*x - 16*a^6)*sqrt(b*x + a)/b^4
 

Sympy [A] (verification not implemented)

Time = 76.16 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.18 \[ \int x^3 \text {arctanh}(\tanh (a+b x))^{5/2} \, dx=\begin {cases} \frac {2 x^{3} \operatorname {atanh}^{\frac {7}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}{7 b} - \frac {4 x^{2} \operatorname {atanh}^{\frac {9}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}{21 b^{2}} + \frac {16 x \operatorname {atanh}^{\frac {11}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}{231 b^{3}} - \frac {32 \operatorname {atanh}^{\frac {13}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}{3003 b^{4}} & \text {for}\: b \neq 0 \\\frac {x^{4} \operatorname {atanh}^{\frac {5}{2}}{\left (\tanh {\left (a \right )} \right )}}{4} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((2*x**3*atanh(tanh(a + b*x))**(7/2)/(7*b) - 4*x**2*atanh(tanh(a 
+ b*x))**(9/2)/(21*b**2) + 16*x*atanh(tanh(a + b*x))**(11/2)/(231*b**3) - 
32*atanh(tanh(a + b*x))**(13/2)/(3003*b**4), Ne(b, 0)), (x**4*atanh(tanh(a 
))**(5/2)/4, True))
 

Maxima [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.66 \[ \int x^3 \text {arctanh}(\tanh (a+b x))^{5/2} \, dx=\frac {2 \, {\left (231 \, b^{4} x^{4} + 105 \, a b^{3} x^{3} - 70 \, a^{2} b^{2} x^{2} + 40 \, a^{3} b x - 16 \, a^{4}\right )} {\left (b x + a\right )}^{\frac {5}{2}}}{3003 \, b^{4}} \] Input:

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

Output:

2/3003*(231*b^4*x^4 + 105*a*b^3*x^3 - 70*a^2*b^2*x^2 + 40*a^3*b*x - 16*a^4 
)*(b*x + a)^(5/2)/b^4
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 296 vs. \(2 (64) = 128\).

Time = 0.12 (sec) , antiderivative size = 296, normalized size of antiderivative = 3.70 \[ \int x^3 \text {arctanh}(\tanh (a+b x))^{5/2} \, dx=\frac {\sqrt {2} {\left (\frac {429 \, \sqrt {2} {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )} a^{3}}{b^{3}} + \frac {143 \, \sqrt {2} {\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} - 180 \, {\left (b x + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b x + a} a^{4}\right )} a^{2}}{b^{3}} + \frac {65 \, \sqrt {2} {\left (63 \, {\left (b x + a\right )}^{\frac {11}{2}} - 385 \, {\left (b x + a\right )}^{\frac {9}{2}} a + 990 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{2} - 1386 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{3} + 1155 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{4} - 693 \, \sqrt {b x + a} a^{5}\right )} a}{b^{3}} + \frac {5 \, \sqrt {2} {\left (231 \, {\left (b x + a\right )}^{\frac {13}{2}} - 1638 \, {\left (b x + a\right )}^{\frac {11}{2}} a + 5005 \, {\left (b x + a\right )}^{\frac {9}{2}} a^{2} - 8580 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{3} + 9009 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{4} - 6006 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{5} + 3003 \, \sqrt {b x + a} a^{6}\right )}}{b^{3}}\right )}}{15015 \, b} \] Input:

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

Output:

1/15015*sqrt(2)*(429*sqrt(2)*(5*(b*x + a)^(7/2) - 21*(b*x + a)^(5/2)*a + 3 
5*(b*x + a)^(3/2)*a^2 - 35*sqrt(b*x + a)*a^3)*a^3/b^3 + 143*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^2/b^3 + 65*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)*a/b^3 + 5* 
sqrt(2)*(231*(b*x + a)^(13/2) - 1638*(b*x + a)^(11/2)*a + 5005*(b*x + a)^( 
9/2)*a^2 - 8580*(b*x + a)^(7/2)*a^3 + 9009*(b*x + a)^(5/2)*a^4 - 6006*(b*x 
 + a)^(3/2)*a^5 + 3003*sqrt(b*x + a)*a^6)/b^3)/b
 

Mupad [B] (verification not implemented)

Time = 3.35 (sec) , antiderivative size = 2235, normalized size of antiderivative = 27.94 \[ \int x^3 \text {arctanh}(\tanh (a+b x))^{5/2} \, dx=\text {Too large to display} \] Input:

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

Output:

(2*b^2*x^6*(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))/13 + (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)*((3*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)^2)/2 - (10*(3*b^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) - (24*b^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))/13)*(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*b)))/(9*b) - (x^5*(3*b^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) - (24*b^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))/13)*(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*b) - (x^3*(lo 
g((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(exp(2*a)*e 
xp(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)^3/4 - (8*((3*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)^2)/2 - (10*(3*b^2*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - l 
og((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x) - (24*b^...
 

Reduce [F]

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

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

Output:

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