\(\int \frac {1}{x^4 \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx\) [166]

Optimal result

Integrand size = 15, antiderivative size = 278 \[ \int \frac {1}{x^4 \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\frac {105 b^3 \arctan \left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right )}{8 (b x-\text {arctanh}(\tanh (a+b x)))^{11/2}}-\frac {35 b^2}{24 x \text {arctanh}(\tanh (a+b x))^{9/2}}+\frac {35 b^3}{24 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{9/2}}+\frac {5 b}{12 x^2 \text {arctanh}(\tanh (a+b x))^{7/2}}-\frac {15 b^3}{8 (b x-\text {arctanh}(\tanh (a+b x)))^2 \text {arctanh}(\tanh (a+b x))^{7/2}}-\frac {1}{3 x^3 \text {arctanh}(\tanh (a+b x))^{5/2}}+\frac {21 b^3}{8 (b x-\text {arctanh}(\tanh (a+b x)))^3 \text {arctanh}(\tanh (a+b x))^{5/2}}-\frac {35 b^3}{8 (b x-\text {arctanh}(\tanh (a+b x)))^4 \text {arctanh}(\tanh (a+b x))^{3/2}}+\frac {105 b^3}{8 (b x-\text {arctanh}(\tanh (a+b x)))^5 \sqrt {\text {arctanh}(\tanh (a+b x))}} \] Output:

105/8*b^3*arctan(arctanh(tanh(b*x+a))^(1/2)/(b*x-arctanh(tanh(b*x+a)))^(1/ 
2))/(b*x-arctanh(tanh(b*x+a)))^(11/2)-35/24*b^2/x/arctanh(tanh(b*x+a))^(9/ 
2)+35/24*b^3/(b*x-arctanh(tanh(b*x+a)))/arctanh(tanh(b*x+a))^(9/2)+5/12*b/ 
x^2/arctanh(tanh(b*x+a))^(7/2)-15/8*b^3/(b*x-arctanh(tanh(b*x+a)))^2/arcta 
nh(tanh(b*x+a))^(7/2)-1/3/x^3/arctanh(tanh(b*x+a))^(5/2)+21/8*b^3/(b*x-arc 
tanh(tanh(b*x+a)))^3/arctanh(tanh(b*x+a))^(5/2)-35/8*b^3/(b*x-arctanh(tanh 
(b*x+a)))^4/arctanh(tanh(b*x+a))^(3/2)+105/8*b^3/(b*x-arctanh(tanh(b*x+a)) 
)^5/arctanh(tanh(b*x+a))^(1/2)
 

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.54 \[ \int \frac {1}{x^4 \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\frac {1}{24} \left (\frac {315 b^3 \text {arctanh}\left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {-b x+\text {arctanh}(\tanh (a+b x))}}\right )}{(-b x+\text {arctanh}(\tanh (a+b x)))^{11/2}}+\frac {-16 b^4 x^4+208 b^3 x^3 \text {arctanh}(\tanh (a+b x))+165 b^2 x^2 \text {arctanh}(\tanh (a+b x))^2-50 b x \text {arctanh}(\tanh (a+b x))^3+8 \text {arctanh}(\tanh (a+b x))^4}{x^3 (b x-\text {arctanh}(\tanh (a+b x)))^5 \text {arctanh}(\tanh (a+b x))^{3/2}}\right ) \] Input:

Integrate[1/(x^4*ArcTanh[Tanh[a + b*x]]^(5/2)),x]
 

Output:

((315*b^3*ArcTanh[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[-(b*x) + ArcTanh[Tanh[ 
a + b*x]]]])/(-(b*x) + ArcTanh[Tanh[a + b*x]])^(11/2) + (-16*b^4*x^4 + 208 
*b^3*x^3*ArcTanh[Tanh[a + b*x]] + 165*b^2*x^2*ArcTanh[Tanh[a + b*x]]^2 - 5 
0*b*x*ArcTanh[Tanh[a + b*x]]^3 + 8*ArcTanh[Tanh[a + b*x]]^4)/(x^3*(b*x - A 
rcTanh[Tanh[a + b*x]])^5*ArcTanh[Tanh[a + b*x]]^(3/2)))/24
 

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 340, normalized size of antiderivative = 1.22, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {2599, 2599, 2599, 2594, 2594, 2594, 2594, 2594, 2592}

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.

\(\displaystyle \int \frac {1}{x^4 \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx\)

\(\Big \downarrow \) 2599

\(\displaystyle -\frac {5}{6} b \int \frac {1}{x^3 \text {arctanh}(\tanh (a+b x))^{7/2}}dx-\frac {1}{3 x^3 \text {arctanh}(\tanh (a+b x))^{5/2}}\)

\(\Big \downarrow \) 2599

\(\displaystyle -\frac {5}{6} b \left (-\frac {7}{4} b \int \frac {1}{x^2 \text {arctanh}(\tanh (a+b x))^{9/2}}dx-\frac {1}{2 x^2 \text {arctanh}(\tanh (a+b x))^{7/2}}\right )-\frac {1}{3 x^3 \text {arctanh}(\tanh (a+b x))^{5/2}}\)

\(\Big \downarrow \) 2599

\(\displaystyle -\frac {5}{6} b \left (-\frac {7}{4} b \left (-\frac {9}{2} b \int \frac {1}{x \text {arctanh}(\tanh (a+b x))^{11/2}}dx-\frac {1}{x \text {arctanh}(\tanh (a+b x))^{9/2}}\right )-\frac {1}{2 x^2 \text {arctanh}(\tanh (a+b x))^{7/2}}\right )-\frac {1}{3 x^3 \text {arctanh}(\tanh (a+b x))^{5/2}}\)

\(\Big \downarrow \) 2594

\(\displaystyle -\frac {5}{6} b \left (-\frac {7}{4} b \left (-\frac {9}{2} b \left (-\frac {\int \frac {1}{x \text {arctanh}(\tanh (a+b x))^{9/2}}dx}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{9 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{9/2}}\right )-\frac {1}{x \text {arctanh}(\tanh (a+b x))^{9/2}}\right )-\frac {1}{2 x^2 \text {arctanh}(\tanh (a+b x))^{7/2}}\right )-\frac {1}{3 x^3 \text {arctanh}(\tanh (a+b x))^{5/2}}\)

\(\Big \downarrow \) 2594

\(\displaystyle -\frac {5}{6} b \left (-\frac {7}{4} b \left (-\frac {9}{2} b \left (-\frac {-\frac {\int \frac {1}{x \text {arctanh}(\tanh (a+b x))^{7/2}}dx}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{7 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{7/2}}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{9 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{9/2}}\right )-\frac {1}{x \text {arctanh}(\tanh (a+b x))^{9/2}}\right )-\frac {1}{2 x^2 \text {arctanh}(\tanh (a+b x))^{7/2}}\right )-\frac {1}{3 x^3 \text {arctanh}(\tanh (a+b x))^{5/2}}\)

\(\Big \downarrow \) 2594

\(\displaystyle -\frac {5}{6} b \left (-\frac {7}{4} b \left (-\frac {9}{2} b \left (-\frac {-\frac {-\frac {\int \frac {1}{x \text {arctanh}(\tanh (a+b x))^{5/2}}dx}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{5 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{5/2}}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{7 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{7/2}}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{9 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{9/2}}\right )-\frac {1}{x \text {arctanh}(\tanh (a+b x))^{9/2}}\right )-\frac {1}{2 x^2 \text {arctanh}(\tanh (a+b x))^{7/2}}\right )-\frac {1}{3 x^3 \text {arctanh}(\tanh (a+b x))^{5/2}}\)

\(\Big \downarrow \) 2594

\(\displaystyle -\frac {5}{6} b \left (-\frac {7}{4} b \left (-\frac {9}{2} b \left (-\frac {-\frac {-\frac {-\frac {\int \frac {1}{x \text {arctanh}(\tanh (a+b x))^{3/2}}dx}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{3 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{3/2}}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{5 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{5/2}}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{7 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{7/2}}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{9 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{9/2}}\right )-\frac {1}{x \text {arctanh}(\tanh (a+b x))^{9/2}}\right )-\frac {1}{2 x^2 \text {arctanh}(\tanh (a+b x))^{7/2}}\right )-\frac {1}{3 x^3 \text {arctanh}(\tanh (a+b x))^{5/2}}\)

\(\Big \downarrow \) 2594

\(\displaystyle -\frac {5}{6} b \left (-\frac {7}{4} b \left (-\frac {9}{2} b \left (-\frac {-\frac {-\frac {-\frac {-\frac {\int \frac {1}{x \sqrt {\text {arctanh}(\tanh (a+b x))}}dx}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{(b x-\text {arctanh}(\tanh (a+b x))) \sqrt {\text {arctanh}(\tanh (a+b x))}}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{3 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{3/2}}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{5 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{5/2}}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{7 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{7/2}}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{9 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{9/2}}\right )-\frac {1}{x \text {arctanh}(\tanh (a+b x))^{9/2}}\right )-\frac {1}{2 x^2 \text {arctanh}(\tanh (a+b x))^{7/2}}\right )-\frac {1}{3 x^3 \text {arctanh}(\tanh (a+b x))^{5/2}}\)

\(\Big \downarrow \) 2592

\(\displaystyle -\frac {5}{6} b \left (-\frac {7}{4} b \left (-\frac {9}{2} b \left (-\frac {-\frac {-\frac {-\frac {-\frac {2 \arctan \left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right )}{(b x-\text {arctanh}(\tanh (a+b x)))^{3/2}}-\frac {2}{(b x-\text {arctanh}(\tanh (a+b x))) \sqrt {\text {arctanh}(\tanh (a+b x))}}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{3 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{3/2}}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{5 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{5/2}}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{7 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{7/2}}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{9 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{9/2}}\right )-\frac {1}{x \text {arctanh}(\tanh (a+b x))^{9/2}}\right )-\frac {1}{2 x^2 \text {arctanh}(\tanh (a+b x))^{7/2}}\right )-\frac {1}{3 x^3 \text {arctanh}(\tanh (a+b x))^{5/2}}\)

Input:

Int[1/(x^4*ArcTanh[Tanh[a + b*x]]^(5/2)),x]
 

Output:

(-5*b*((-7*b*((-9*b*(-((-((-((-(((-2*ArcTan[Sqrt[ArcTanh[Tanh[a + b*x]]]/S 
qrt[b*x - ArcTanh[Tanh[a + b*x]]]])/(b*x - ArcTanh[Tanh[a + b*x]])^(3/2) - 
 2/((b*x - ArcTanh[Tanh[a + b*x]])*Sqrt[ArcTanh[Tanh[a + b*x]]]))/(b*x - A 
rcTanh[Tanh[a + b*x]])) - 2/(3*(b*x - ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh 
[a + b*x]]^(3/2)))/(b*x - ArcTanh[Tanh[a + b*x]])) - 2/(5*(b*x - ArcTanh[T 
anh[a + b*x]])*ArcTanh[Tanh[a + b*x]]^(5/2)))/(b*x - ArcTanh[Tanh[a + b*x] 
])) - 2/(7*(b*x - ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a + b*x]]^(7/2)))/( 
b*x - ArcTanh[Tanh[a + b*x]])) - 2/(9*(b*x - ArcTanh[Tanh[a + b*x]])*ArcTa 
nh[Tanh[a + b*x]]^(9/2))))/2 - 1/(x*ArcTanh[Tanh[a + b*x]]^(9/2))))/4 - 1/ 
(2*x^2*ArcTanh[Tanh[a + b*x]]^(7/2))))/6 - 1/(3*x^3*ArcTanh[Tanh[a + b*x]] 
^(5/2))
 

Defintions of rubi rules used

rule 2592

Int[1/((u_)*Sqrt[v_]), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simpli 
fy[D[v, x]]}, Simp[2*(ArcTan[Sqrt[v]/Rt[(b*u - a*v)/a, 2]]/(a*Rt[(b*u - a*v 
)/a, 2])), x] /; NeQ[b*u - a*v, 0] && PosQ[(b*u - a*v)/a]] /; PiecewiseLine 
arQ[u, v, x]
 

rule 2594

Int[(v_)^(n_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[ 
D[v, x]]}, Simp[v^(n + 1)/((n + 1)*(b*u - a*v)), x] - Simp[a*((n + 1)/((n + 
 1)*(b*u - a*v)))   Int[v^(n + 1)/u, x], x] /; NeQ[b*u - a*v, 0]] /; Piecew 
iseLinearQ[u, v, x] && LtQ[n, -1]
 

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 = 0.21 (sec) , antiderivative size = 211, normalized size of antiderivative = 0.76

method	result	size
default	\(2 b^{3} \left (-\frac {\frac {\frac {41 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{16}+\left (-\frac {35 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}{6}+\frac {35 b x}{6}\right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}+\left (\frac {55 a^{2}}{16}+\frac {55 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )}{8}+\frac {55 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{16}\right ) \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{b^{3} x^{3}}-\frac {105 \,\operatorname {arctanh}\left (\frac {\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}}\right )}{16 \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}}}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{5}}-\frac {1}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{4} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}-\frac {4}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{5} \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}\right )\)	\(211\)

Input:

int(1/x^4/arctanh(tanh(b*x+a))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

2*b^3*(-1/(arctanh(tanh(b*x+a))-b*x)^5*((41/16*arctanh(tanh(b*x+a))^(5/2)+ 
(-35/6*arctanh(tanh(b*x+a))+35/6*b*x)*arctanh(tanh(b*x+a))^(3/2)+(55/16*a^ 
2+55/8*a*(arctanh(tanh(b*x+a))-b*x-a)+55/16*(arctanh(tanh(b*x+a))-b*x-a)^2 
)*arctanh(tanh(b*x+a))^(1/2))/b^3/x^3-105/16/(arctanh(tanh(b*x+a))-b*x)^(1 
/2)*arctanh(arctanh(tanh(b*x+a))^(1/2)/(arctanh(tanh(b*x+a))-b*x)^(1/2)))- 
1/3/(arctanh(tanh(b*x+a))-b*x)^4/arctanh(tanh(b*x+a))^(3/2)-4/(arctanh(tan 
h(b*x+a))-b*x)^5/arctanh(tanh(b*x+a))^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 274, normalized size of antiderivative = 0.99 \[ \int \frac {1}{x^4 \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\left [\frac {315 \, {\left (b^{5} x^{5} + 2 \, a b^{4} x^{4} + a^{2} b^{3} x^{3}\right )} \sqrt {a} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) - 2 \, {\left (315 \, a b^{4} x^{4} + 420 \, a^{2} b^{3} x^{3} + 63 \, a^{3} b^{2} x^{2} - 18 \, a^{4} b x + 8 \, a^{5}\right )} \sqrt {b x + a}}{48 \, {\left (a^{6} b^{2} x^{5} + 2 \, a^{7} b x^{4} + a^{8} x^{3}\right )}}, -\frac {315 \, {\left (b^{5} x^{5} + 2 \, a b^{4} x^{4} + a^{2} b^{3} x^{3}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x + a}}\right ) + {\left (315 \, a b^{4} x^{4} + 420 \, a^{2} b^{3} x^{3} + 63 \, a^{3} b^{2} x^{2} - 18 \, a^{4} b x + 8 \, a^{5}\right )} \sqrt {b x + a}}{24 \, {\left (a^{6} b^{2} x^{5} + 2 \, a^{7} b x^{4} + a^{8} x^{3}\right )}}\right ] \] Input:

integrate(1/x^4/arctanh(tanh(b*x+a))^(5/2),x, algorithm="fricas")
 

Output:

[1/48*(315*(b^5*x^5 + 2*a*b^4*x^4 + a^2*b^3*x^3)*sqrt(a)*log((b*x + 2*sqrt 
(b*x + a)*sqrt(a) + 2*a)/x) - 2*(315*a*b^4*x^4 + 420*a^2*b^3*x^3 + 63*a^3* 
b^2*x^2 - 18*a^4*b*x + 8*a^5)*sqrt(b*x + a))/(a^6*b^2*x^5 + 2*a^7*b*x^4 + 
a^8*x^3), -1/24*(315*(b^5*x^5 + 2*a*b^4*x^4 + a^2*b^3*x^3)*sqrt(-a)*arctan 
(sqrt(-a)/sqrt(b*x + a)) + (315*a*b^4*x^4 + 420*a^2*b^3*x^3 + 63*a^3*b^2*x 
^2 - 18*a^4*b*x + 8*a^5)*sqrt(b*x + a))/(a^6*b^2*x^5 + 2*a^7*b*x^4 + a^8*x 
^3)]
 

Sympy [F]

\[ \int \frac {1}{x^4 \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\int \frac {1}{x^{4} \operatorname {atanh}^{\frac {5}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \] Input:

integrate(1/x**4/atanh(tanh(b*x+a))**(5/2),x)
 

Output:

Integral(1/(x**4*atanh(tanh(a + b*x))**(5/2)), x)
 

Maxima [F]

\[ \int \frac {1}{x^4 \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\int { \frac {1}{x^{4} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate(1/x^4/arctanh(tanh(b*x+a))^(5/2),x, algorithm="maxima")
 

Output:

integrate(1/(x^4*arctanh(tanh(b*x + a))^(5/2)), x)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.41 \[ \int \frac {1}{x^4 \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=-\frac {105 \, b^{3} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{8 \, \sqrt {-a} a^{5}} - \frac {315 \, {\left (b x + a\right )}^{4} b^{3} - 840 \, {\left (b x + a\right )}^{3} a b^{3} + 693 \, {\left (b x + a\right )}^{2} a^{2} b^{3} - 144 \, {\left (b x + a\right )} a^{3} b^{3} - 16 \, a^{4} b^{3}}{24 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} - \sqrt {b x + a} a\right )}^{3} a^{5}} \] Input:

integrate(1/x^4/arctanh(tanh(b*x+a))^(5/2),x, algorithm="giac")
 

Output:

-105/8*b^3*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^5) - 1/24*(315*(b*x 
+ a)^4*b^3 - 840*(b*x + a)^3*a*b^3 + 693*(b*x + a)^2*a^2*b^3 - 144*(b*x + 
a)*a^3*b^3 - 16*a^4*b^3)/(((b*x + a)^(3/2) - sqrt(b*x + a)*a)^3*a^5)
 

Mupad [B] (verification not implemented)

Time = 7.92 (sec) , antiderivative size = 2359, normalized size of antiderivative = 8.49 \[ \int \frac {1}{x^4 \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\text {Too large to display} \] Input:

int(1/(x^4*atanh(tanh(a + b*x))^(5/2)),x)
 

Output:

(8*(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*x^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)^3) - (((1 
40*b^2)/(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(e 
xp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^4 - (840*b^3*x)/(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 
)^5)*(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(ex 
p(2*a)*exp(2*b*x) + 1))/2)^(1/2))/(x*(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)*(x*((((1232*b^4)/(9*(2*a*b - b*(2*a - 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)) + 2*b*x 
))*(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) - (80*b^3*(2*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)) + 4*b* 
x) - 7*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)))/(9*(2*a*b - b*(2*a - 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)) 
+ 2*b*x))*(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)^4))*(log(2/(exp(2*a)*exp(2*b*x) + 1...
 

Reduce [F]

\[ \int \frac {1}{x^4 \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\int \frac {\sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}}{\mathit {atanh} \left (\tanh \left (b x +a \right )\right )^{3} x^{4}}d x \] Input:

int(1/x^4/atanh(tanh(b*x+a))^(5/2),x)
 

Output:

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