\(\int \frac {1}{x^3 \text {arctanh}(\tanh (a+b x))^{3/2}} \, dx\) [156]

Optimal result

Integrand size = 15, antiderivative size = 191 \[ \int \frac {1}{x^3 \text {arctanh}(\tanh (a+b x))^{3/2}} \, dx=-\frac {15 b^2 \arctan \left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right )}{4 (b x-\text {arctanh}(\tanh (a+b x)))^{7/2}}+\frac {3 b}{4 x \text {arctanh}(\tanh (a+b x))^{5/2}}-\frac {3 b^2}{4 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{5/2}}-\frac {1}{2 x^2 \text {arctanh}(\tanh (a+b x))^{3/2}}+\frac {5 b^2}{4 (b x-\text {arctanh}(\tanh (a+b x)))^2 \text {arctanh}(\tanh (a+b x))^{3/2}}-\frac {15 b^2}{4 (b x-\text {arctanh}(\tanh (a+b x)))^3 \sqrt {\text {arctanh}(\tanh (a+b x))}} \] Output:

-15/4*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)))^(7/2)+3/4*b/x/arctanh(tanh(b*x+a))^(5/2)-3/ 
4*b^2/(b*x-arctanh(tanh(b*x+a)))/arctanh(tanh(b*x+a))^(5/2)-1/2/x^2/arctan 
h(tanh(b*x+a))^(3/2)+5/4*b^2/(b*x-arctanh(tanh(b*x+a)))^2/arctanh(tanh(b*x 
+a))^(3/2)-15/4*b^2/(b*x-arctanh(tanh(b*x+a)))^3/arctanh(tanh(b*x+a))^(1/2 
)
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.60 \[ \int \frac {1}{x^3 \text {arctanh}(\tanh (a+b x))^{3/2}} \, dx=\frac {1}{4} \left (-\frac {15 b^2 \text {arctanh}\left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {-b x+\text {arctanh}(\tanh (a+b x))}}\right )}{(-b x+\text {arctanh}(\tanh (a+b x)))^{7/2}}+\frac {8 b^2 x^2+9 b x \text {arctanh}(\tanh (a+b x))-2 \text {arctanh}(\tanh (a+b x))^2}{x^2 \sqrt {\text {arctanh}(\tanh (a+b x))} (-b x+\text {arctanh}(\tanh (a+b x)))^3}\right ) \] Input:

Integrate[1/(x^3*ArcTanh[Tanh[a + b*x]]^(3/2)),x]
 

Output:

((-15*b^2*ArcTanh[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[-(b*x) + ArcTanh[Tanh[ 
a + b*x]]]])/(-(b*x) + ArcTanh[Tanh[a + b*x]])^(7/2) + (8*b^2*x^2 + 9*b*x* 
ArcTanh[Tanh[a + b*x]] - 2*ArcTanh[Tanh[a + b*x]]^2)/(x^2*Sqrt[ArcTanh[Tan 
h[a + b*x]]]*(-(b*x) + ArcTanh[Tanh[a + b*x]])^3))/4
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.15, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2599, 2599, 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.

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

\(\Big \downarrow \) 2599

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

\(\Big \downarrow \) 2599

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

\(\Big \downarrow \) 2594

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

\(\Big \downarrow \) 2594

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

\(\Big \downarrow \) 2594

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

\(\Big \downarrow \) 2592

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

Input:

Int[1/(x^3*ArcTanh[Tanh[a + b*x]]^(3/2)),x]
 

Output:

(-3*b*((-5*b*(-((-(((-2*ArcTan[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[b*x - Arc 
Tanh[Tanh[a + b*x]]]])/(b*x - ArcTanh[Tanh[a + b*x]])^(3/2) - 2/((b*x - Ar 
cTanh[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)))/(b*x - ArcTanh[Tanh[a + b*x]])) - 2/(5*(b*x - ArcTanh[Tanh[a + b*x]] 
)*ArcTanh[Tanh[a + b*x]]^(5/2))))/2 - 1/(x*ArcTanh[Tanh[a + b*x]]^(5/2)))) 
/4 - 1/(2*x^2*ArcTanh[Tanh[a + b*x]]^(3/2))
 

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 = 131, normalized size of antiderivative = 0.69

method	result	size
default	\(2 b^{2} \left (\frac {1}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}+\frac {\frac {\frac {7 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{8}+\left (-\frac {9 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}{8}+\frac {9 b x}{8}\right ) \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{b^{2} x^{2}}-\frac {15 \,\operatorname {arctanh}\left (\frac {\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}}\right )}{8 \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}}}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{3}}\right )\)	\(131\)

Input:

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

Output:

2*b^2*(1/(arctanh(tanh(b*x+a))-b*x)^3/arctanh(tanh(b*x+a))^(1/2)+1/(arctan 
h(tanh(b*x+a))-b*x)^3*((7/8*arctanh(tanh(b*x+a))^(3/2)+(-9/8*arctanh(tanh( 
b*x+a))+9/8*b*x)*arctanh(tanh(b*x+a))^(1/2))/b^2/x^2-15/8/(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 = 0.10 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.97 \[ \int \frac {1}{x^3 \text {arctanh}(\tanh (a+b x))^{3/2}} \, dx=\left [\frac {15 \, {\left (b^{3} x^{3} + a b^{2} x^{2}\right )} \sqrt {a} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (15 \, a b^{2} x^{2} + 5 \, a^{2} b x - 2 \, a^{3}\right )} \sqrt {b x + a}}{8 \, {\left (a^{4} b x^{3} + a^{5} x^{2}\right )}}, \frac {15 \, {\left (b^{3} x^{3} + a b^{2} x^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x + a}}\right ) + {\left (15 \, a b^{2} x^{2} + 5 \, a^{2} b x - 2 \, a^{3}\right )} \sqrt {b x + a}}{4 \, {\left (a^{4} b x^{3} + a^{5} x^{2}\right )}}\right ] \] Input:

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

Output:

[1/8*(15*(b^3*x^3 + a*b^2*x^2)*sqrt(a)*log((b*x - 2*sqrt(b*x + a)*sqrt(a) 
+ 2*a)/x) + 2*(15*a*b^2*x^2 + 5*a^2*b*x - 2*a^3)*sqrt(b*x + a))/(a^4*b*x^3 
 + a^5*x^2), 1/4*(15*(b^3*x^3 + a*b^2*x^2)*sqrt(-a)*arctan(sqrt(-a)/sqrt(b 
*x + a)) + (15*a*b^2*x^2 + 5*a^2*b*x - 2*a^3)*sqrt(b*x + a))/(a^4*b*x^3 + 
a^5*x^2)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.42 \[ \int \frac {1}{x^3 \text {arctanh}(\tanh (a+b x))^{3/2}} \, dx=\frac {15 \, b^{2} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{4 \, \sqrt {-a} a^{3}} + \frac {2 \, b^{2}}{\sqrt {b x + a} a^{3}} + \frac {7 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{2} - 9 \, \sqrt {b x + a} a b^{2}}{4 \, a^{3} b^{2} x^{2}} \] Input:

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

Output:

15/4*b^2*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^3) + 2*b^2/(sqrt(b*x + 
 a)*a^3) + 1/4*(7*(b*x + a)^(3/2)*b^2 - 9*sqrt(b*x + a)*a*b^2)/(a^3*b^2*x^ 
2)
 

Mupad [B] (verification not implemented)

Time = 8.62 (sec) , antiderivative size = 1028, normalized size of antiderivative = 5.38 \[ \int \frac {1}{x^3 \text {arctanh}(\tanh (a+b x))^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

(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)) - log((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))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x) - 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)^7 + 84*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)^5 - 280*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)^4 + 560*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) 
^3 - 672*a^5*(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 - 128*a^7 - 14*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)^6 + 448*a^6*(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)/(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)^(1/2)))*15i)/(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)^(7/2) - 
 (2*(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(...
 

Reduce [F]

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

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

Output:

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