\(\int \frac {x^4}{\sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx\) [140]

Optimal result

Integrand size = 15, antiderivative size = 99 \[ \int \frac {x^4}{\sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=\frac {2 x^4 \sqrt {\text {arctanh}(\tanh (a+b x))}}{b}-\frac {16 x^3 \text {arctanh}(\tanh (a+b x))^{3/2}}{3 b^2}+\frac {32 x^2 \text {arctanh}(\tanh (a+b x))^{5/2}}{5 b^3}-\frac {128 x \text {arctanh}(\tanh (a+b x))^{7/2}}{35 b^4}+\frac {256 \text {arctanh}(\tanh (a+b x))^{9/2}}{315 b^5} \] Output:

2*x^4*arctanh(tanh(b*x+a))^(1/2)/b-16/3*x^3*arctanh(tanh(b*x+a))^(3/2)/b^2 
+32/5*x^2*arctanh(tanh(b*x+a))^(5/2)/b^3-128/35*x*arctanh(tanh(b*x+a))^(7/ 
2)/b^4+256/315*arctanh(tanh(b*x+a))^(9/2)/b^5
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.84 \[ \int \frac {x^4}{\sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=\frac {2 \sqrt {\text {arctanh}(\tanh (a+b x))} \left (315 b^4 x^4-840 b^3 x^3 \text {arctanh}(\tanh (a+b x))+1008 b^2 x^2 \text {arctanh}(\tanh (a+b x))^2-576 b x \text {arctanh}(\tanh (a+b x))^3+128 \text {arctanh}(\tanh (a+b x))^4\right )}{315 b^5} \] Input:

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

Output:

(2*Sqrt[ArcTanh[Tanh[a + b*x]]]*(315*b^4*x^4 - 840*b^3*x^3*ArcTanh[Tanh[a 
+ b*x]] + 1008*b^2*x^2*ArcTanh[Tanh[a + b*x]]^2 - 576*b*x*ArcTanh[Tanh[a + 
 b*x]]^3 + 128*ArcTanh[Tanh[a + b*x]]^4))/(315*b^5)
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.20, 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.

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

\(\Big \downarrow \) 2599

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

\(\Big \downarrow \) 2599

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

\(\Big \downarrow \) 2599

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

\(\Big \downarrow \) 2599

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

\(\Big \downarrow \) 2588

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

\(\Big \downarrow \) 15

\(\displaystyle \frac {2 x^4 \sqrt {\text {arctanh}(\tanh (a+b x))}}{b}-\frac {8 \left (\frac {2 x^3 \text {arctanh}(\tanh (a+b x))^{3/2}}{3 b}-\frac {2 \left (\frac {2 x^2 \text {arctanh}(\tanh (a+b x))^{5/2}}{5 b}-\frac {4 \left (\frac {2 x \text {arctanh}(\tanh (a+b x))^{7/2}}{7 b}-\frac {4 \text {arctanh}(\tanh (a+b x))^{9/2}}{63 b^2}\right )}{5 b}\right )}{b}\right )}{b}\)

Input:

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

Output:

(2*x^4*Sqrt[ArcTanh[Tanh[a + b*x]]])/b - (8*((2*x^3*ArcTanh[Tanh[a + b*x]] 
^(3/2))/(3*b) - (2*((2*x^2*ArcTanh[Tanh[a + b*x]]^(5/2))/(5*b) - (4*((2*x* 
ArcTanh[Tanh[a + b*x]]^(7/2))/(7*b) - (4*ArcTanh[Tanh[a + b*x]]^(9/2))/(63 
*b^2)))/(5*b)))/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 = 0.20 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.55

method	result	size
default	\(\frac {\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {9}{2}}}{9}+\frac {2 \left (-4 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )+4 b x \right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{7}+\frac {2 \left (2 \left (b x -\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )^{2}+\left (2 b x -2 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )^{2}\right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{5}+\frac {4 \left (b x -\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )^{2} \left (2 b x -2 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{3}+2 \left (b x -\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )^{4} \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{b^{5}}\)	\(153\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.54 \[ \int \frac {x^4}{\sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=\frac {2 \, {\left (35 \, b^{4} x^{4} - 40 \, a b^{3} x^{3} + 48 \, a^{2} b^{2} x^{2} - 64 \, a^{3} b x + 128 \, a^{4}\right )} \sqrt {b x + a}}{315 \, b^{5}} \] Input:

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

Output:

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

Sympy [F]

\[ \int \frac {x^4}{\sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=\int \frac {x^{4}}{\sqrt {\operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}}\, dx \] 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 = 0.65 \[ \int \frac {x^4}{\sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=\frac {2 \, {\left (35 \, b^{5} x^{5} - 5 \, a b^{4} x^{4} + 8 \, a^{2} b^{3} x^{3} - 16 \, a^{3} b^{2} x^{2} + 64 \, a^{4} b x + 128 \, a^{5}\right )}}{315 \, \sqrt {b x + a} b^{5}} \] Input:

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

Output:

2/315*(35*b^5*x^5 - 5*a*b^4*x^4 + 8*a^2*b^3*x^3 - 16*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 = 0.11 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.62 \[ \int \frac {x^4}{\sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=\frac {2 \, {\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} - 180 \, {\left (b x + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b x + a} a^{4}\right )}}{315 \, b^{5}} \] Input:

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

Output:

2/315*(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)/b^5
 

Mupad [B] (verification not implemented)

Time = 3.12 (sec) , antiderivative size = 496, normalized size of antiderivative = 5.01 \[ \int \frac {x^4}{\sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx =\text {Too large to display} \] Input:

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

Output:

(2*x^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))/(9*b) + (256*(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)^4)/(315*b^5) + (16*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))/(63*b^2) + (128*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)/(315*b^4) + (32*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))/(exp(2*a)*exp(2*b*x) + 1))/2 + b*x)^2)/(105*b^3)
 

Reduce [F]

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

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

Output:

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