\(\int \frac {1}{x^3 \sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx\) [147]

Optimal result

Integrand size = 15, antiderivative size = 158 \[ \int \frac {1}{x^3 \sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=\frac {3 b^2 \arctan \left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right )}{4 (b x-\text {arctanh}(\tanh (a+b x)))^{5/2}}+\frac {b}{4 x \text {arctanh}(\tanh (a+b x))^{3/2}}-\frac {b^2}{4 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{3/2}}-\frac {1}{2 x^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}+\frac {3 b^2}{4 (b x-\text {arctanh}(\tanh (a+b x)))^2 \sqrt {\text {arctanh}(\tanh (a+b x))}} \] Output:

3/4*b^2*arctan(arctanh(tanh(b*x+a))^(1/2)/(b*x-arctanh(tanh(b*x+a)))^(1/2) 
)/(b*x-arctanh(tanh(b*x+a)))^(5/2)+1/4*b/x/arctanh(tanh(b*x+a))^(3/2)-1/4* 
b^2/(b*x-arctanh(tanh(b*x+a)))/arctanh(tanh(b*x+a))^(3/2)-1/2/x^2/arctanh( 
tanh(b*x+a))^(1/2)+3/4*b^2/(b*x-arctanh(tanh(b*x+a)))^2/arctanh(tanh(b*x+a 
))^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.62 \[ \int \frac {1}{x^3 \sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=\frac {1}{4} \left (-\frac {3 b^2 \text {arctanh}\left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {-b x+\text {arctanh}(\tanh (a+b x))}}\right )}{(-b x+\text {arctanh}(\tanh (a+b x)))^{5/2}}+\frac {(5 b x-2 \text {arctanh}(\tanh (a+b x))) \sqrt {\text {arctanh}(\tanh (a+b x))}}{x^2 (-b x+\text {arctanh}(\tanh (a+b x)))^2}\right ) \] Input:

Integrate[1/(x^3*Sqrt[ArcTanh[Tanh[a + b*x]]]),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.09, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2599, 2599, 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.

\(\displaystyle \int \frac {1}{x^3 \sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx\)

\(\Big \downarrow \) 2599

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

\(\Big \downarrow \) 2599

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

\(\Big \downarrow \) 2594

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

\(\Big \downarrow \) 2594

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

\(\Big \downarrow \) 2592

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

Input:

Int[1/(x^3*Sqrt[ArcTanh[Tanh[a + b*x]]]),x]
 

Output:

-1/4*(b*((-3*b*(-(((-2*ArcTan[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[b*x - ArcT 
anh[Tanh[a + b*x]]]])/(b*x - ArcTanh[Tanh[a + b*x]])^(3/2) - 2/((b*x - Arc 
Tanh[Tanh[a + b*x]])*Sqrt[ArcTanh[Tanh[a + b*x]]]))/(b*x - ArcTanh[Tanh[a 
+ b*x]])) - 2/(3*(b*x - ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a + b*x]]^(3/ 
2))))/2 - 1/(x*ArcTanh[Tanh[a + b*x]]^(3/2)))) - 1/(2*x^2*Sqrt[ArcTanh[Tan 
h[a + b*x]]])
 

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 = 0.22 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.94

method	result	size
default	\(2 b^{2} \left (\frac {\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{\left (-4 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )+4 b x \right ) b^{2} x^{2}}+\frac {\frac {6 \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{\left (-4 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )+4 b x \right ) b x}-\frac {6 \,\operatorname {arctanh}\left (\frac {\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}}\right )}{\left (-4 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )+4 b x \right ) \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}}}{-4 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )+4 b x}\right )\)	\(148\)

Input:

int(1/x^3/arctanh(tanh(b*x+a))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.76 \[ \int \frac {1}{x^3 \sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=\left [\frac {3 \, \sqrt {a} b^{2} x^{2} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (3 \, a b x - 2 \, a^{2}\right )} \sqrt {b x + a}}{8 \, a^{3} x^{2}}, \frac {3 \, \sqrt {-a} b^{2} x^{2} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x + a}}\right ) + {\left (3 \, a b x - 2 \, a^{2}\right )} \sqrt {b x + a}}{4 \, a^{3} x^{2}}\right ] \] Input:

integrate(1/x^3/arctanh(tanh(b*x+a))^(1/2),x, algorithm="fricas")
 

Output:

[1/8*(3*sqrt(a)*b^2*x^2*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*( 
3*a*b*x - 2*a^2)*sqrt(b*x + a))/(a^3*x^2), 1/4*(3*sqrt(-a)*b^2*x^2*arctan( 
sqrt(-a)/sqrt(b*x + a)) + (3*a*b*x - 2*a^2)*sqrt(b*x + a))/(a^3*x^2)]
 

Sympy [F]

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

integrate(1/x**3/atanh(tanh(b*x+a))**(1/2),x)
 

Output:

Integral(1/(x**3*sqrt(atanh(tanh(a + b*x)))), x)
 

Maxima [F]

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

integrate(1/x^3/arctanh(tanh(b*x+a))^(1/2),x, algorithm="maxima")
 

Output:

integrate(1/(x^3*sqrt(arctanh(tanh(b*x + a)))), x)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.44 \[ \int \frac {1}{x^3 \sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=\frac {\frac {3 \, b^{3} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {3 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{3} - 5 \, \sqrt {b x + a} a b^{3}}{a^{2} b^{2} x^{2}}}{4 \, b} \] Input:

integrate(1/x^3/arctanh(tanh(b*x+a))^(1/2),x, algorithm="giac")
 

Output:

1/4*(3*b^3*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^2) + (3*(b*x + a)^(3 
/2)*b^3 - 5*sqrt(b*x + a)*a*b^3)/(a^2*b^2*x^2))/b
 

Mupad [B] (verification not implemented)

Time = 7.99 (sec) , antiderivative size = 802, normalized size of antiderivative = 5.08 \[ \int \frac {1}{x^3 \sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=\text {Too large to display} \] Input:

int(1/(x^3*atanh(tanh(a + b*x))^(1/2)),x)
 

Output:

(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))/(x^2*(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)) + (3*b 
*(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*(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^(1/2)*b 
^2*log((((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)) - l 
og((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^(1/2)*2i - 
2^(1/2)*(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) + 2^(1/2)*b*x)*((2*a - log((2*exp(2*a)*e 
xp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + log(2/(exp(2*a)*exp(2*b*x) + 1)) + 
 2*b*x)^5 + 40*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)^3 - 80*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 - 32*a^5 - 10*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)^4 + 
80*a^4*(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))*1i)/(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...
 

Reduce [F]

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

int(1/x^3/atanh(tanh(b*x+a))^(1/2),x)
 

Output:

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