\(\int x^4 \text {arctanh}(\tanh (a+b x))^{3/2} \, dx\) [120]

Optimal result

Integrand size = 15, antiderivative size = 101 \[ \int x^4 \text {arctanh}(\tanh (a+b x))^{3/2} \, dx=\frac {2 x^4 \text {arctanh}(\tanh (a+b x))^{5/2}}{5 b}-\frac {16 x^3 \text {arctanh}(\tanh (a+b x))^{7/2}}{35 b^2}+\frac {32 x^2 \text {arctanh}(\tanh (a+b x))^{9/2}}{105 b^3}-\frac {128 x \text {arctanh}(\tanh (a+b x))^{11/2}}{1155 b^4}+\frac {256 \text {arctanh}(\tanh (a+b x))^{13/2}}{15015 b^5} \] Output:

2/5*x^4*arctanh(tanh(b*x+a))^(5/2)/b-16/35*x^3*arctanh(tanh(b*x+a))^(7/2)/ 
b^2+32/105*x^2*arctanh(tanh(b*x+a))^(9/2)/b^3-128/1155*x*arctanh(tanh(b*x+ 
a))^(11/2)/b^4+256/15015*arctanh(tanh(b*x+a))^(13/2)/b^5
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.82 \[ \int x^4 \text {arctanh}(\tanh (a+b x))^{3/2} \, dx=\frac {2 \text {arctanh}(\tanh (a+b x))^{5/2} \left (3003 b^4 x^4-3432 b^3 x^3 \text {arctanh}(\tanh (a+b x))+2288 b^2 x^2 \text {arctanh}(\tanh (a+b x))^2-832 b x \text {arctanh}(\tanh (a+b x))^3+128 \text {arctanh}(\tanh (a+b x))^4\right )}{15015 b^5} \] Input:

Integrate[x^4*ArcTanh[Tanh[a + b*x]]^(3/2),x]
 

Output:

(2*ArcTanh[Tanh[a + b*x]]^(5/2)*(3003*b^4*x^4 - 3432*b^3*x^3*ArcTanh[Tanh[ 
a + b*x]] + 2288*b^2*x^2*ArcTanh[Tanh[a + b*x]]^2 - 832*b*x*ArcTanh[Tanh[a 
 + b*x]]^3 + 128*ArcTanh[Tanh[a + b*x]]^4))/(15015*b^5)
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.24, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2599, 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.

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

\(\Big \downarrow \) 2599

\(\displaystyle \frac {2 x^4 \text {arctanh}(\tanh (a+b x))^{5/2}}{5 b}-\frac {8 \int x^3 \text {arctanh}(\tanh (a+b x))^{5/2}dx}{5 b}\)

\(\Big \downarrow \) 2599

\(\displaystyle \frac {2 x^4 \text {arctanh}(\tanh (a+b x))^{5/2}}{5 b}-\frac {8 \left (\frac {2 x^3 \text {arctanh}(\tanh (a+b x))^{7/2}}{7 b}-\frac {6 \int x^2 \text {arctanh}(\tanh (a+b x))^{7/2}dx}{7 b}\right )}{5 b}\)

\(\Big \downarrow \) 2599

\(\displaystyle \frac {2 x^4 \text {arctanh}(\tanh (a+b x))^{5/2}}{5 b}-\frac {8 \left (\frac {2 x^3 \text {arctanh}(\tanh (a+b x))^{7/2}}{7 b}-\frac {6 \left (\frac {2 x^2 \text {arctanh}(\tanh (a+b x))^{9/2}}{9 b}-\frac {4 \int x \text {arctanh}(\tanh (a+b x))^{9/2}dx}{9 b}\right )}{7 b}\right )}{5 b}\)

\(\Big \downarrow \) 2599

\(\displaystyle \frac {2 x^4 \text {arctanh}(\tanh (a+b x))^{5/2}}{5 b}-\frac {8 \left (\frac {2 x^3 \text {arctanh}(\tanh (a+b x))^{7/2}}{7 b}-\frac {6 \left (\frac {2 x^2 \text {arctanh}(\tanh (a+b x))^{9/2}}{9 b}-\frac {4 \left (\frac {2 x \text {arctanh}(\tanh (a+b x))^{11/2}}{11 b}-\frac {2 \int \text {arctanh}(\tanh (a+b x))^{11/2}dx}{11 b}\right )}{9 b}\right )}{7 b}\right )}{5 b}\)

\(\Big \downarrow \) 2588

\(\displaystyle \frac {2 x^4 \text {arctanh}(\tanh (a+b x))^{5/2}}{5 b}-\frac {8 \left (\frac {2 x^3 \text {arctanh}(\tanh (a+b x))^{7/2}}{7 b}-\frac {6 \left (\frac {2 x^2 \text {arctanh}(\tanh (a+b x))^{9/2}}{9 b}-\frac {4 \left (\frac {2 x \text {arctanh}(\tanh (a+b x))^{11/2}}{11 b}-\frac {2 \int \text {arctanh}(\tanh (a+b x))^{11/2}d\text {arctanh}(\tanh (a+b x))}{11 b^2}\right )}{9 b}\right )}{7 b}\right )}{5 b}\)

\(\Big \downarrow \) 15

\(\displaystyle \frac {2 x^4 \text {arctanh}(\tanh (a+b x))^{5/2}}{5 b}-\frac {8 \left (\frac {2 x^3 \text {arctanh}(\tanh (a+b x))^{7/2}}{7 b}-\frac {6 \left (\frac {2 x^2 \text {arctanh}(\tanh (a+b x))^{9/2}}{9 b}-\frac {4 \left (\frac {2 x \text {arctanh}(\tanh (a+b x))^{11/2}}{11 b}-\frac {4 \text {arctanh}(\tanh (a+b x))^{13/2}}{143 b^2}\right )}{9 b}\right )}{7 b}\right )}{5 b}\)

Input:

Int[x^4*ArcTanh[Tanh[a + b*x]]^(3/2),x]
 

Output:

(2*x^4*ArcTanh[Tanh[a + b*x]]^(5/2))/(5*b) - (8*((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*ArcTanh[Tanh[a + b*x]]^(13/ 
2))/(143*b^2)))/(9*b)))/(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 = 0.25 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.52

method	result	size
default	\(\frac {\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {13}{2}}}{13}+\frac {2 \left (-4 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )+4 b x \right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {11}{2}}}{11}+\frac {2 \left (2 \left (b x -\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )^{2}+\left (2 b x -2 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )^{2}\right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {9}{2}}}{9}+\frac {4 \left (b x -\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )^{2} \left (2 b x -2 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{7}+\frac {2 \left (b x -\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )^{4} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{5}}{b^{5}}\)	\(154\)

Input:

int(x^4*arctanh(tanh(b*x+a))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

2/b^5*(1/13*arctanh(tanh(b*x+a))^(13/2)+1/11*(-4*arctanh(tanh(b*x+a))+4*b* 
x)*arctanh(tanh(b*x+a))^(11/2)+1/9*(2*(b*x-arctanh(tanh(b*x+a)))^2+(2*b*x- 
2*arctanh(tanh(b*x+a)))^2)*arctanh(tanh(b*x+a))^(9/2)+2/7*(b*x-arctanh(tan 
h(b*x+a)))^2*(2*b*x-2*arctanh(tanh(b*x+a)))*arctanh(tanh(b*x+a))^(7/2)+1/5 
*(b*x-arctanh(tanh(b*x+a)))^4*arctanh(tanh(b*x+a))^(5/2))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.74 \[ \int x^4 \text {arctanh}(\tanh (a+b x))^{3/2} \, dx=\frac {2 \, {\left (1155 \, b^{6} x^{6} + 1470 \, a b^{5} x^{5} + 35 \, a^{2} b^{4} x^{4} - 40 \, a^{3} b^{3} x^{3} + 48 \, a^{4} b^{2} x^{2} - 64 \, a^{5} b x + 128 \, a^{6}\right )} \sqrt {b x + a}}{15015 \, b^{5}} \] Input:

integrate(x^4*arctanh(tanh(b*x+a))^(3/2),x, algorithm="fricas")
 

Output:

2/15015*(1155*b^6*x^6 + 1470*a*b^5*x^5 + 35*a^2*b^4*x^4 - 40*a^3*b^3*x^3 + 
 48*a^4*b^2*x^2 - 64*a^5*b*x + 128*a^6)*sqrt(b*x + a)/b^5
 

Sympy [F]

\[ \int x^4 \text {arctanh}(\tanh (a+b x))^{3/2} \, dx=\int x^{4} \operatorname {atanh}^{\frac {3}{2}}{\left (\tanh {\left (a + b x \right )} \right )}\, dx \] Input:

integrate(x**4*atanh(tanh(b*x+a))**(3/2),x)
 

Output:

Integral(x**4*atanh(tanh(a + b*x))**(3/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.63 \[ \int x^4 \text {arctanh}(\tanh (a+b x))^{3/2} \, dx=\frac {2 \, {\left (1155 \, b^{5} x^{5} + 315 \, a b^{4} x^{4} - 280 \, a^{2} b^{3} x^{3} + 240 \, a^{3} b^{2} x^{2} - 192 \, a^{4} b x + 128 \, a^{5}\right )} {\left (b x + a\right )}^{\frac {3}{2}}}{15015 \, b^{5}} \] Input:

integrate(x^4*arctanh(tanh(b*x+a))^(3/2),x, algorithm="maxima")
 

Output:

2/15015*(1155*b^5*x^5 + 315*a*b^4*x^4 - 280*a^2*b^3*x^3 + 240*a^3*b^2*x^2 
- 192*a^4*b*x + 128*a^5)*(b*x + a)^(3/2)/b^5
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 241 vs. \(2 (81) = 162\).

Time = 0.11 (sec) , antiderivative size = 241, normalized size of antiderivative = 2.39 \[ \int x^4 \text {arctanh}(\tanh (a+b x))^{3/2} \, dx=\frac {\sqrt {2} {\left (\frac {143 \, \sqrt {2} {\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} - 180 \, {\left (b x + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b x + a} a^{4}\right )} a^{2}}{b^{4}} + \frac {130 \, \sqrt {2} {\left (63 \, {\left (b x + a\right )}^{\frac {11}{2}} - 385 \, {\left (b x + a\right )}^{\frac {9}{2}} a + 990 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{2} - 1386 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{3} + 1155 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{4} - 693 \, \sqrt {b x + a} a^{5}\right )} a}{b^{4}} + \frac {15 \, \sqrt {2} {\left (231 \, {\left (b x + a\right )}^{\frac {13}{2}} - 1638 \, {\left (b x + a\right )}^{\frac {11}{2}} a + 5005 \, {\left (b x + a\right )}^{\frac {9}{2}} a^{2} - 8580 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{3} + 9009 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{4} - 6006 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{5} + 3003 \, \sqrt {b x + a} a^{6}\right )}}{b^{4}}\right )}}{45045 \, b} \] Input:

integrate(x^4*arctanh(tanh(b*x+a))^(3/2),x, algorithm="giac")
 

Output:

1/45045*sqrt(2)*(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^4 + 130*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^4 + 15*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^4)/b
 

Mupad [B] (verification not implemented)

Time = 3.23 (sec) , antiderivative size = 1813, normalized size of antiderivative = 17.95 \[ \int x^4 \text {arctanh}(\tanh (a+b x))^{3/2} \, dx=\text {Too large to display} \] Input:

int(x^4*atanh(tanh(a + b*x))^(3/2),x)
 

Output:

(2*b*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^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 
)*((24*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))/13 - 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))) 
/(11*b) + (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)*((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*((24*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))/13 - 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))/(11*b)))/(9*b) + (128*(log((2*exp(2*a)*ex 
p(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 + (10*((24*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))/13 - 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)...
 

Reduce [F]

\[ \int x^4 \text {arctanh}(\tanh (a+b x))^{3/2} \, dx=\int \sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}\, \mathit {atanh} \left (\tanh \left (b x +a \right )\right ) x^{4}d x \] Input:

int(x^4*atanh(tanh(b*x+a))^(3/2),x)
 

Output:

int(sqrt(atanh(tanh(a + b*x)))*atanh(tanh(a + b*x))*x**4,x)