\(\int x^4 \sqrt {\text {arctanh}(\tanh (a+b x))} \, dx\) [111]

Optimal result

Integrand size = 15, antiderivative size = 101 \[ \int x^4 \sqrt {\text {arctanh}(\tanh (a+b x))} \, dx=\frac {2 x^4 \text {arctanh}(\tanh (a+b x))^{3/2}}{3 b}-\frac {16 x^3 \text {arctanh}(\tanh (a+b x))^{5/2}}{15 b^2}+\frac {32 x^2 \text {arctanh}(\tanh (a+b x))^{7/2}}{35 b^3}-\frac {128 x \text {arctanh}(\tanh (a+b x))^{9/2}}{315 b^4}+\frac {256 \text {arctanh}(\tanh (a+b x))^{11/2}}{3465 b^5} \] Output:

2/3*x^4*arctanh(tanh(b*x+a))^(3/2)/b-16/15*x^3*arctanh(tanh(b*x+a))^(5/2)/ 
b^2+32/35*x^2*arctanh(tanh(b*x+a))^(7/2)/b^3-128/315*x*arctanh(tanh(b*x+a) 
)^(9/2)/b^4+256/3465*arctanh(tanh(b*x+a))^(11/2)/b^5
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.82 \[ \int x^4 \sqrt {\text {arctanh}(\tanh (a+b x))} \, dx=\frac {2 \text {arctanh}(\tanh (a+b x))^{3/2} \left (1155 b^4 x^4-1848 b^3 x^3 \text {arctanh}(\tanh (a+b x))+1584 b^2 x^2 \text {arctanh}(\tanh (a+b x))^2-704 b x \text {arctanh}(\tanh (a+b x))^3+128 \text {arctanh}(\tanh (a+b x))^4\right )}{3465 b^5} \] Input:

Integrate[x^4*Sqrt[ArcTanh[Tanh[a + b*x]]],x]
 

Output:

(2*ArcTanh[Tanh[a + b*x]]^(3/2)*(1155*b^4*x^4 - 1848*b^3*x^3*ArcTanh[Tanh[ 
a + b*x]] + 1584*b^2*x^2*ArcTanh[Tanh[a + b*x]]^2 - 704*b*x*ArcTanh[Tanh[a 
 + b*x]]^3 + 128*ArcTanh[Tanh[a + b*x]]^4))/(3465*b^5)
 

Rubi [A] (verified)

Time = 0.58 (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.

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

\(\Big \downarrow \) 2599

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

\(\Big \downarrow \) 2599

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

\(\Big \downarrow \) 2599

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

\(\Big \downarrow \) 2599

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

\(\Big \downarrow \) 2588

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

\(\Big \downarrow \) 15

\(\displaystyle \frac {2 x^4 \text {arctanh}(\tanh (a+b x))^{3/2}}{3 b}-\frac {8 \left (\frac {2 x^3 \text {arctanh}(\tanh (a+b x))^{5/2}}{5 b}-\frac {6 \left (\frac {2 x^2 \text {arctanh}(\tanh (a+b x))^{7/2}}{7 b}-\frac {4 \left (\frac {2 x \text {arctanh}(\tanh (a+b x))^{9/2}}{9 b}-\frac {4 \text {arctanh}(\tanh (a+b x))^{11/2}}{99 b^2}\right )}{7 b}\right )}{5 b}\right )}{3 b}\)

Input:

Int[x^4*Sqrt[ArcTanh[Tanh[a + b*x]]],x]
 

Output:

(2*x^4*ArcTanh[Tanh[a + b*x]]^(3/2))/(3*b) - (8*((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*ArcTanh[Tanh[a + b*x]]^(11/2) 
)/(99*b^2)))/(7*b)))/(5*b)))/(3*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 = 0.28 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.52

method	result	size
default	\(\frac {\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {11}{2}}}{11}+\frac {2 \left (-4 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )+4 b x \right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {9}{2}}}{9}+\frac {2 \left (2 \left (b x -\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )^{2}+\left (2 b x -2 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )^{2}\right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{7}+\frac {4 \left (b x -\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )^{2} \left (2 b x -2 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{5}+\frac {2 \left (b x -\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )^{4} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{3}}{b^{5}}\)	\(154\)

Input:

int(x^4*arctanh(tanh(b*x+a))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.63 \[ \int x^4 \sqrt {\text {arctanh}(\tanh (a+b x))} \, dx=\frac {2 \, {\left (315 \, b^{5} x^{5} + 35 \, a b^{4} x^{4} - 40 \, a^{2} b^{3} x^{3} + 48 \, a^{3} b^{2} x^{2} - 64 \, a^{4} b x + 128 \, a^{5}\right )} \sqrt {b x + a}}{3465 \, b^{5}} \] Input:

integrate(x^4*arctanh(tanh(b*x+a))^(1/2),x, algorithm="fricas")
 

Output:

2/3465*(315*b^5*x^5 + 35*a*b^4*x^4 - 40*a^2*b^3*x^3 + 48*a^3*b^2*x^2 - 64* 
a^4*b*x + 128*a^5)*sqrt(b*x + a)/b^5
 

Sympy [F]

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

integrate(x**4*atanh(tanh(b*x+a))**(1/2),x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.63 \[ \int x^4 \sqrt {\text {arctanh}(\tanh (a+b x))} \, dx=\frac {2 \, {\left (315 \, b^{5} x^{5} + 35 \, a b^{4} x^{4} - 40 \, a^{2} b^{3} x^{3} + 48 \, a^{3} b^{2} x^{2} - 64 \, a^{4} b x + 128 \, a^{5}\right )} \sqrt {b x + a}}{3465 \, b^{5}} \] Input:

integrate(x^4*arctanh(tanh(b*x+a))^(1/2),x, algorithm="maxima")
 

Output:

2/3465*(315*b^5*x^5 + 35*a*b^4*x^4 - 40*a^2*b^3*x^3 + 48*a^3*b^2*x^2 - 64* 
a^4*b*x + 128*a^5)*sqrt(b*x + a)/b^5
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.49 \[ \int x^4 \sqrt {\text {arctanh}(\tanh (a+b x))} \, dx=\frac {\sqrt {2} {\left (\frac {11 \, \sqrt {2} {\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} - 180 \, {\left (b x + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b x + a} a^{4}\right )} a}{b^{4}} + \frac {5 \, \sqrt {2} {\left (63 \, {\left (b x + a\right )}^{\frac {11}{2}} - 385 \, {\left (b x + a\right )}^{\frac {9}{2}} a + 990 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{2} - 1386 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{3} + 1155 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{4} - 693 \, \sqrt {b x + a} a^{5}\right )}}{b^{4}}\right )}}{3465 \, b} \] Input:

integrate(x^4*arctanh(tanh(b*x+a))^(1/2),x, algorithm="giac")
 

Output:

1/3465*sqrt(2)*(11*sqrt(2)*(35*(b*x + a)^(9/2) - 180*(b*x + a)^(7/2)*a + 3 
78*(b*x + a)^(5/2)*a^2 - 420*(b*x + a)^(3/2)*a^3 + 315*sqrt(b*x + a)*a^4)* 
a/b^4 + 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^4)/b
 

Mupad [B] (verification not implemented)

Time = 3.03 (sec) , antiderivative size = 811, normalized size of antiderivative = 8.03 \[ \int x^4 \sqrt {\text {arctanh}(\tanh (a+b x))} \, dx=\text {Too large to display} \] Input:

int(x^4*atanh(tanh(a + b*x))^(1/2),x)
 

Output:

(2*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)* 
(log(2/(exp(2*a)*exp(2*b*x) + 1))/11 - log((2*exp(2*a)*exp(2*b*x))/(exp(2* 
a)*exp(2*b*x) + 1))/11 + (2*b*x)/11))/(9*b) - (128*(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))/(ex 
p(2*a)*exp(2*b*x) + 1))/2 + b*x)^4*(log(2/(exp(2*a)*exp(2*b*x) + 1))/11 - 
log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/11 + (2*b*x)/11))/( 
315*b^5) - (8*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))/2 - 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))/11 - log((2*exp(2*a)*exp(2*b*x))/(exp(2* 
a)*exp(2*b*x) + 1))/11 + (2*b*x)/11))/(63*b^2) - (64*x*(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))/2 - log((2*exp(2*a)*exp(2*b*x)) 
/(exp(2*a)*exp(2*b*x) + 1))/2 + b*x)^3*(log(2/(exp(2*a)*exp(2*b*x) + 1))/1 
1 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/11 + (2*b*x)/11 
))/(315*b^4) - (16*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...
 

Reduce [F]

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

int(x^4*atanh(tanh(b*x+a))^(1/2),x)
 

Output:

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