3.2.97 \(\int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))} \, dx\) [197]

3.2.97.1 Optimal result

Integrand size = 15, antiderivative size = 101 \[ \int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))} \, dx=-\frac {2 b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right )}{(b x-\text {arctanh}(\tanh (a+b x)))^{5/2}}+\frac {2 b}{\sqrt {x} (b x-\text {arctanh}(\tanh (a+b x)))^2}+\frac {2}{3 x^{3/2} (b x-\text {arctanh}(\tanh (a+b x)))} \]

output

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

3.2.97.2 Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.88 \[ \int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))} \, dx=\frac {2 b^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {-b x+\text {arctanh}(\tanh (a+b x))}}\right )}{(-b x+\text {arctanh}(\tanh (a+b x)))^{5/2}}+\frac {2 (4 b x-\text {arctanh}(\tanh (a+b x)))}{3 x^{3/2} (-b x+\text {arctanh}(\tanh (a+b x)))^2} \]

input

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

output

(2*b^(3/2)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]] 
)/(-(b*x) + ArcTanh[Tanh[a + b*x]])^(5/2) + (2*(4*b*x - ArcTanh[Tanh[a + b 
*x]]))/(3*x^(3/2)*(-(b*x) + ArcTanh[Tanh[a + b*x]])^2)
 

3.2.97.3 Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.17, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2594, 2594, 2593}

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.

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

\(\Big \downarrow \) 2594

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

\(\Big \downarrow \) 2594

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

\(\Big \downarrow \) 2593

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

input

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

output

2/(3*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*x]])) + (b*((-2*Sqrt[b]*ArcTanh[(Sq 
rt[b]*Sqrt[x])/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]])/(b*x - ArcTanh[Tanh[a 
+ b*x]])^(3/2) + 2/(Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]]))))/(b*x - ArcTa 
nh[Tanh[a + b*x]])
 

3.2.97.3.1 Defintions of rubi rules used

rule 2593

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

3.2.97.4 Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.97

method	result	size
derivativedivides	\(-\frac {2}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {3}{2}}}+\frac {2 b}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \sqrt {x}}+\frac {2 b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\)	\(98\)
default	\(-\frac {2}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {3}{2}}}+\frac {2 b}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \sqrt {x}}+\frac {2 b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\)	\(98\)

input

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

output

-2/3/(arctanh(tanh(b*x+a))-b*x)/x^(3/2)+2/(arctanh(tanh(b*x+a))-b*x)^2*b/x 
^(1/2)+2*b^2/(arctanh(tanh(b*x+a))-b*x)^2/((arctanh(tanh(b*x+a))-b*x)*b)^( 
1/2)*arctan(b*x^(1/2)/((arctanh(tanh(b*x+a))-b*x)*b)^(1/2))
 

3.2.97.5 Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.17 \[ \int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))} \, dx=\left [\frac {3 \, b x^{2} \sqrt {-\frac {b}{a}} \log \left (\frac {b x + 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) + 2 \, {\left (3 \, b x - a\right )} \sqrt {x}}{3 \, a^{2} x^{2}}, -\frac {2 \, {\left (3 \, b x^{2} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b \sqrt {x}}\right ) - {\left (3 \, b x - a\right )} \sqrt {x}\right )}}{3 \, a^{2} x^{2}}\right ] \]

input

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

output

[1/3*(3*b*x^2*sqrt(-b/a)*log((b*x + 2*a*sqrt(x)*sqrt(-b/a) - a)/(b*x + a)) 
 + 2*(3*b*x - a)*sqrt(x))/(a^2*x^2), -2/3*(3*b*x^2*sqrt(b/a)*arctan(a*sqrt 
(b/a)/(b*sqrt(x))) - (3*b*x - a)*sqrt(x))/(a^2*x^2)]
 

3.2.97.6 Sympy [F]

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

input

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

output

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

3.2.97.7 Maxima [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.41 \[ \int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))} \, dx=\frac {2 \, b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{2}} + \frac {2 \, {\left (3 \, b x - a\right )}}{3 \, a^{2} x^{\frac {3}{2}}} \]

input

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

output

2*b^2*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^2) + 2/3*(3*b*x - a)/(a^2*x 
^(3/2))
 

3.2.97.8 Giac [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.41 \[ \int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))} \, dx=\frac {2 \, b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{2}} + \frac {2 \, {\left (3 \, b x - a\right )}}{3 \, a^{2} x^{\frac {3}{2}}} \]

input

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

output

2*b^2*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^2) + 2/3*(3*b*x - a)/(a^2*x 
^(3/2))
 

3.2.97.9 Mupad [B] (verification not implemented)

Time = 5.10 (sec) , antiderivative size = 642, normalized size of antiderivative = 6.36 \[ \int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))} \, dx=\frac {4}{3\,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 )}+\frac {8\,b}{\sqrt {x}\,{\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 {4\,\sqrt {2}\,b^{3/2}\,\ln \left (\frac {\sqrt {b}\,\sqrt {\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}\,\left (\sqrt {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 )-4\,\sqrt {b}\,\sqrt {x}\,\sqrt {\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}+2\,\sqrt {2}\,b\,x\right )\,\left ({\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^4+24\,a^2\,{\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2+16\,a^4-8\,a\,{\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3-32\,a^3\,\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )\right )}{2\,\left (\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\right )}\right )}{{\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 )}^{5/2}} \]

input

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

output

4/(3*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)) + (8*b)/(x^(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)^2) + (4*2^(1/2)*b^(3/2)*log((b^(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)^(1 
/2)*(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) - 4*b^(1/2)*x^(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)^(1/2) + 2*2^(1/2)*b*x)*((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)^4 + 24* 
a^2*(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)^2 + 16*a^4 - 8*a*(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)^3 - 32*a^3*(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)))/(2*(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) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp 
(2*a)*exp(2*b*x) + 1)) + 2*b*x)^(5/2)