\(\int \frac {x^4}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx\) [158]

Optimal result

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

-2/3*x^4/b/arctanh(tanh(b*x+a))^(3/2)-16/3*x^3/b^2/arctanh(tanh(b*x+a))^(1 
/2)+32*x^2*arctanh(tanh(b*x+a))^(1/2)/b^3-128/3*x*arctanh(tanh(b*x+a))^(3/ 
2)/b^4+256/15*arctanh(tanh(b*x+a))^(5/2)/b^5
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.84 \[ \int \frac {x^4}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=-\frac {2 \left (5 b^4 x^4+40 b^3 x^3 \text {arctanh}(\tanh (a+b x))-240 b^2 x^2 \text {arctanh}(\tanh (a+b x))^2+320 b x \text {arctanh}(\tanh (a+b x))^3-128 \text {arctanh}(\tanh (a+b x))^4\right )}{15 b^5 \text {arctanh}(\tanh (a+b x))^{3/2}} \] Input:

Integrate[x^4/ArcTanh[Tanh[a + b*x]]^(5/2),x]
 

Output:

(-2*(5*b^4*x^4 + 40*b^3*x^3*ArcTanh[Tanh[a + b*x]] - 240*b^2*x^2*ArcTanh[T 
anh[a + b*x]]^2 + 320*b*x*ArcTanh[Tanh[a + b*x]]^3 - 128*ArcTanh[Tanh[a + 
b*x]]^4))/(15*b^5*ArcTanh[Tanh[a + b*x]]^(3/2))
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.18, 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.

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

\(\Big \downarrow \) 2599

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

\(\Big \downarrow \) 2599

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

\(\Big \downarrow \) 2599

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

\(\Big \downarrow \) 2599

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

\(\Big \downarrow \) 2588

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

\(\Big \downarrow \) 15

\(\displaystyle \frac {8 \left (\frac {6 \left (\frac {2 x^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}{b}-\frac {4 \left (\frac {2 x \text {arctanh}(\tanh (a+b x))^{3/2}}{3 b}-\frac {4 \text {arctanh}(\tanh (a+b x))^{5/2}}{15 b^2}\right )}{b}\right )}{b}-\frac {2 x^3}{b \sqrt {\text {arctanh}(\tanh (a+b x))}}\right )}{3 b}-\frac {2 x^4}{3 b \text {arctanh}(\tanh (a+b x))^{3/2}}\)

Input:

Int[x^4/ArcTanh[Tanh[a + b*x]]^(5/2),x]
 

Output:

(-2*x^4)/(3*b*ArcTanh[Tanh[a + b*x]]^(3/2)) + (8*((-2*x^3)/(b*Sqrt[ArcTanh 
[Tanh[a + b*x]]]) + (6*((2*x^2*Sqrt[ArcTanh[Tanh[a + b*x]]])/b - (4*((2*x* 
ArcTanh[Tanh[a + b*x]]^(3/2))/(3*b) - (4*ArcTanh[Tanh[a + b*x]]^(5/2))/(15 
*b^2)))/b))/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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(294\) vs. \(2(81)=162\).

Time = 0.21 (sec) , antiderivative size = 295, normalized size of antiderivative = 2.98

method	result	size
default	\(\frac {\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{5}-\frac {8 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}} a}{3}-\frac {8 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}} \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )}{3}+12 \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}\, a^{2}+24 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right ) \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}+12 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2} \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}-\frac {2 \left (-4 a^{3}-12 a^{2} \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )-12 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}-4 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}\right )}{\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}-\frac {2 \left (a^{4}+4 a^{3} \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )+6 a^{2} \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}+4 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}+\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{4}\right )}{3 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}}{b^{5}}\)	\(295\)

Input:

int(x^4/arctanh(tanh(b*x+a))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

2/b^5*(1/5*arctanh(tanh(b*x+a))^(5/2)-4/3*arctanh(tanh(b*x+a))^(3/2)*a-4/3 
*arctanh(tanh(b*x+a))^(3/2)*(arctanh(tanh(b*x+a))-b*x-a)+6*arctanh(tanh(b* 
x+a))^(1/2)*a^2+12*a*(arctanh(tanh(b*x+a))-b*x-a)*arctanh(tanh(b*x+a))^(1/ 
2)+6*(arctanh(tanh(b*x+a))-b*x-a)^2*arctanh(tanh(b*x+a))^(1/2)-(-4*a^3-12* 
a^2*(arctanh(tanh(b*x+a))-b*x-a)-12*a*(arctanh(tanh(b*x+a))-b*x-a)^2-4*(ar 
ctanh(tanh(b*x+a))-b*x-a)^3)/arctanh(tanh(b*x+a))^(1/2)-1/3*(a^4+4*a^3*(ar 
ctanh(tanh(b*x+a))-b*x-a)+6*a^2*(arctanh(tanh(b*x+a))-b*x-a)^2+4*a*(arctan 
h(tanh(b*x+a))-b*x-a)^3+(arctanh(tanh(b*x+a))-b*x-a)^4)/arctanh(tanh(b*x+a 
))^(3/2))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.75 \[ \int \frac {x^4}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\frac {2 \, {\left (3 \, b^{4} x^{4} - 8 \, a b^{3} x^{3} + 48 \, a^{2} b^{2} x^{2} + 192 \, a^{3} b x + 128 \, a^{4}\right )} \sqrt {b x + a}}{15 \, {\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}} \] Input:

integrate(x^4/arctanh(tanh(b*x+a))^(5/2),x, algorithm="fricas")
 

Output:

2/15*(3*b^4*x^4 - 8*a*b^3*x^3 + 48*a^2*b^2*x^2 + 192*a^3*b*x + 128*a^4)*sq 
rt(b*x + a)/(b^7*x^2 + 2*a*b^6*x + a^2*b^5)
 

Sympy [F]

\[ \int \frac {x^4}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\int \frac {x^{4}}{\operatorname {atanh}^{\frac {5}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \] Input:

integrate(x**4/atanh(tanh(b*x+a))**(5/2),x)
 

Output:

Integral(x**4/atanh(tanh(a + b*x))**(5/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.65 \[ \int \frac {x^4}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\frac {2 \, {\left (3 \, b^{5} x^{5} - 5 \, a b^{4} x^{4} + 40 \, a^{2} b^{3} x^{3} + 240 \, a^{3} b^{2} x^{2} + 320 \, a^{4} b x + 128 \, a^{5}\right )}}{15 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{5}} \] Input:

integrate(x^4/arctanh(tanh(b*x+a))^(5/2),x, algorithm="maxima")
 

Output:

2/15*(3*b^5*x^5 - 5*a*b^4*x^4 + 40*a^2*b^3*x^3 + 240*a^3*b^2*x^2 + 320*a^4 
*b*x + 128*a^5)/((b*x + a)^(5/2)*b^5)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.80 \[ \int \frac {x^4}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\frac {2 \, {\left (\frac {5 \, {\left (12 \, {\left (b x + a\right )} a^{3} - a^{4}\right )}}{{\left (b x + a\right )}^{\frac {3}{2}} b} + \frac {3 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{4} - 20 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{4} + 90 \, \sqrt {b x + a} a^{2} b^{4}}{b^{5}}\right )}}{15 \, b^{4}} \] Input:

integrate(x^4/arctanh(tanh(b*x+a))^(5/2),x, algorithm="giac")
 

Output:

2/15*(5*(12*(b*x + a)*a^3 - a^4)/((b*x + a)^(3/2)*b) + (3*(b*x + a)^(5/2)* 
b^4 - 20*(b*x + a)^(3/2)*a*b^4 + 90*sqrt(b*x + a)*a^2*b^4)/b^5)/b^4
 

Mupad [B] (verification not implemented)

Time = 3.51 (sec) , antiderivative size = 817, normalized size of antiderivative = 8.25 \[ \int \frac {x^4}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\text {Too large to display} \] Input:

int(x^4/atanh(tanh(a + b*x))^(5/2),x)
 

Output:

((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*(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*b^4) + (2 
*((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))/b^3 + (8*(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))/(5*b 
^3))*(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))/(3*b)))/b + (2*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))/(5*b^3) + (x*((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))/b^3 + (8*(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))/(5*b^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))/(3*b) - (2*(log((2*e 
xp(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)^3)/(b^5*(log((2*exp(2*a)*exp( 
2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) - 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 - 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...
 

Reduce [F]

\[ \int \frac {x^4}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\int \frac {\sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}\, x^{4}}{\mathit {atanh} \left (\tanh \left (b x +a \right )\right )^{3}}d x \] Input:

int(x^4/atanh(tanh(b*x+a))^(5/2),x)
 

Output:

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