\(\int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx\) [263]

Optimal result

Integrand size = 17, antiderivative size = 146 \[ \int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=-\frac {16 b^2 \sqrt {x}}{3 (b x-\text {arctanh}(\tanh (a+b x)))^3 \text {arctanh}(\tanh (a+b x))^{3/2}}+\frac {4 b}{\sqrt {x} (b x-\text {arctanh}(\tanh (a+b x)))^2 \text {arctanh}(\tanh (a+b x))^{3/2}}+\frac {2}{3 x^{3/2} (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{3/2}}+\frac {32 b^2 \sqrt {x}}{3 (b x-\text {arctanh}(\tanh (a+b x)))^4 \sqrt {\text {arctanh}(\tanh (a+b x))}} \] Output:

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.54 \[ \int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=-\frac {2 \left (b^3 x^3-9 b^2 x^2 \text {arctanh}(\tanh (a+b x))-9 b x \text {arctanh}(\tanh (a+b x))^2+\text {arctanh}(\tanh (a+b x))^3\right )}{3 x^{3/2} \text {arctanh}(\tanh (a+b x))^{3/2} (-b x+\text {arctanh}(\tanh (a+b x)))^4} \] Input:

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

Output:

(-2*(b^3*x^3 - 9*b^2*x^2*ArcTanh[Tanh[a + b*x]] - 9*b*x*ArcTanh[Tanh[a + b 
*x]]^2 + ArcTanh[Tanh[a + b*x]]^3))/(3*x^(3/2)*ArcTanh[Tanh[a + b*x]]^(3/2 
)*(-(b*x) + ArcTanh[Tanh[a + b*x]])^4)
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.21, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {2602, 2602, 2602, 2598}

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.

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

\(\Big \downarrow \) 2602

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

\(\Big \downarrow \) 2602

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

\(\Big \downarrow \) 2602

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

\(\Big \downarrow \) 2598

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

Input:

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

Output:

(2*b*((4*b*((-2*Sqrt[x])/(3*(b*x - ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a 
+ b*x]]^(3/2)) + (4*Sqrt[x])/(3*(b*x - ArcTanh[Tanh[a + b*x]])^2*Sqrt[ArcT 
anh[Tanh[a + b*x]]])))/(b*x - ArcTanh[Tanh[a + b*x]]) + 2/(Sqrt[x]*(b*x - 
ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a + b*x]]^(3/2))))/(b*x - ArcTanh[Tan 
h[a + b*x]]) + 2/(3*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a 
+ b*x]]^(3/2))
 

Defintions of rubi rules used

rule 2598

Int[(u_)^(m_)*(v_)^(n_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simp 
lify[D[v, x]]}, Simp[(-u^(m + 1))*(v^(n + 1)/((m + 1)*(b*u - a*v))), x] /; 
NeQ[b*u - a*v, 0]] /; FreeQ[{m, n}, x] && PiecewiseLinearQ[u, v, x] && EqQ[ 
m + n + 2, 0] && NeQ[m, -1]
 

rule 2602

Int[(u_)^(m_)*(v_)^(n_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simp 
lify[D[v, x]]}, Simp[(-u^(m + 1))*(v^(n + 1)/((m + 1)*(b*u - a*v))), x] + S 
imp[b*((m + n + 2)/((m + 1)*(b*u - a*v)))   Int[u^(m + 1)*v^n, x], x] /; Ne 
Q[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && NeQ[m + n + 2, 0] && LtQ[m 
, -1]
 

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.03

method	result	size
derivativedivides	\(-\frac {2}{3 x^{\frac {3}{2}} \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}-\frac {4 b \left (-\frac {1}{\sqrt {x}\, \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}-\frac {4 b \left (\frac {\sqrt {x}}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}+\frac {2 \sqrt {x}}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}\right )}{\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}\right )}{\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}\)	\(150\)
default	\(-\frac {2}{3 x^{\frac {3}{2}} \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}-\frac {4 b \left (-\frac {1}{\sqrt {x}\, \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}-\frac {4 b \left (\frac {\sqrt {x}}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}+\frac {2 \sqrt {x}}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}\right )}{\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}\right )}{\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}\)	\(150\)

Input:

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

Output:

-2/3/x^(3/2)/(arctanh(tanh(b*x+a))-b*x)/arctanh(tanh(b*x+a))^(3/2)-4*b/(ar 
ctanh(tanh(b*x+a))-b*x)*(-1/x^(1/2)/(arctanh(tanh(b*x+a))-b*x)/arctanh(tan 
h(b*x+a))^(3/2)-4*b/(arctanh(tanh(b*x+a))-b*x)*(1/3*x^(1/2)/(arctanh(tanh( 
b*x+a))-b*x)/arctanh(tanh(b*x+a))^(3/2)+2/3/(arctanh(tanh(b*x+a))-b*x)^2*x 
^(1/2)/arctanh(tanh(b*x+a))^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.49 \[ \int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\frac {2 \, {\left (16 \, b^{3} x^{3} + 24 \, a b^{2} x^{2} + 6 \, a^{2} b x - a^{3}\right )} \sqrt {b x + a} \sqrt {x}}{3 \, {\left (a^{4} b^{2} x^{4} + 2 \, a^{5} b x^{3} + a^{6} x^{2}\right )}} \] Input:

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

Output:

2/3*(16*b^3*x^3 + 24*a*b^2*x^2 + 6*a^2*b*x - a^3)*sqrt(b*x + a)*sqrt(x)/(a 
^4*b^2*x^4 + 2*a^5*b*x^3 + a^6*x^2)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.38 \[ \int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\frac {2 \, {\left (16 \, b^{4} x^{4} + 40 \, a b^{3} x^{3} + 30 \, a^{2} b^{2} x^{2} + 5 \, a^{3} b x - a^{4}\right )}}{3 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{4} x^{\frac {3}{2}}} \] Input:

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

Output:

2/3*(16*b^4*x^4 + 40*a*b^3*x^3 + 30*a^2*b^2*x^2 + 5*a^3*b*x - a^4)/((b*x + 
 a)^(5/2)*a^4*x^(3/2))
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.82 \[ \int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\frac {2 \, \sqrt {x} {\left (\frac {8 \, b^{3} x}{a^{4}} + \frac {9 \, b^{2}}{a^{3}}\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}}} - \frac {8 \, {\left (3 \, b^{\frac {3}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{4} - 9 \, a b^{\frac {3}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} + 4 \, a^{2} b^{\frac {3}{2}}\right )}}{3 \, {\left ({\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} - a\right )}^{3} a^{3}} \] Input:

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

Output:

2/3*sqrt(x)*(8*b^3*x/a^4 + 9*b^2/a^3)/(b*x + a)^(3/2) - 8/3*(3*b^(3/2)*(sq 
rt(b)*sqrt(x) - sqrt(b*x + a))^4 - 9*a*b^(3/2)*(sqrt(b)*sqrt(x) - sqrt(b*x 
 + a))^2 + 4*a^2*b^(3/2))/(((sqrt(b)*sqrt(x) - sqrt(b*x + a))^2 - a)^3*a^3 
)
 

Mupad [B] (verification not implemented)

Time = 3.93 (sec) , antiderivative size = 406, normalized size of antiderivative = 2.78 \[ \int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\frac {\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\left (\frac {4}{3\,b^2\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}-\frac {128\,x^2}{{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}+\frac {16\,x}{b\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}+\frac {512\,b\,x^3}{3\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^4}\right )}{x^{7/2}-\frac {x^{5/2}\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{b}+\frac {x^{3/2}\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{4\,b^2}} \] Input:

int(1/(x^(5/2)*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)*(4/(3*b^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)) - (128*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)^3 + (16*x)/(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)^2) + 
 (512*b*x^3)/(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)^4)))/(x^(7/2) - (x^(5/2)*(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 + (x^(3/2)*(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)/(4*b^2))
 

Reduce [F]

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

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

Output:

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