Integrand size = 15, antiderivative size = 94 \[ \int \frac {1}{x^2 \sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=\frac {b \arctan \left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right )}{(b x-\text {arctanh}(\tanh (a+b x)))^{3/2}}-\frac {1}{x \sqrt {\text {arctanh}(\tanh (a+b x))}}+\frac {b}{(b x-\text {arctanh}(\tanh (a+b x))) \sqrt {\text {arctanh}(\tanh (a+b x))}} \] Output:
b*arctan(arctanh(tanh(b*x+a))^(1/2)/(b*x-arctanh(tanh(b*x+a)))^(1/2))/(b*x -arctanh(tanh(b*x+a)))^(3/2)-1/x/arctanh(tanh(b*x+a))^(1/2)+b/(b*x-arctanh (tanh(b*x+a)))/arctanh(tanh(b*x+a))^(1/2)
Time = 0.04 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x^2 \sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=\frac {b \text {arctanh}\left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {-b x+\text {arctanh}(\tanh (a+b x))}}\right )}{(-b x+\text {arctanh}(\tanh (a+b x)))^{3/2}}-\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x (-b x+\text {arctanh}(\tanh (a+b x)))} \] Input:
Integrate[1/(x^2*Sqrt[ArcTanh[Tanh[a + b*x]]]),x]
Output:
(b*ArcTanh[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x ]]]])/(-(b*x) + ArcTanh[Tanh[a + b*x]])^(3/2) - Sqrt[ArcTanh[Tanh[a + b*x] ]]/(x*(-(b*x) + ArcTanh[Tanh[a + b*x]]))
Time = 0.27 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.06, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2599, 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.
\(\displaystyle \int \frac {1}{x^2 \sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle -\frac {1}{2} b \int \frac {1}{x \text {arctanh}(\tanh (a+b x))^{3/2}}dx-\frac {1}{x \sqrt {\text {arctanh}(\tanh (a+b x))}}\) |
\(\Big \downarrow \) 2594 |
\(\displaystyle -\frac {1}{2} b \left (-\frac {\int \frac {1}{x \sqrt {\text {arctanh}(\tanh (a+b x))}}dx}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{(b x-\text {arctanh}(\tanh (a+b x))) \sqrt {\text {arctanh}(\tanh (a+b x))}}\right )-\frac {1}{x \sqrt {\text {arctanh}(\tanh (a+b x))}}\) |
\(\Big \downarrow \) 2592 |
\(\displaystyle -\frac {1}{2} b \left (-\frac {2 \arctan \left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right )}{(b x-\text {arctanh}(\tanh (a+b x)))^{3/2}}-\frac {2}{(b x-\text {arctanh}(\tanh (a+b x))) \sqrt {\text {arctanh}(\tanh (a+b x))}}\right )-\frac {1}{x \sqrt {\text {arctanh}(\tanh (a+b x))}}\) |
Input:
Int[1/(x^2*Sqrt[ArcTanh[Tanh[a + b*x]]]),x]
Output:
-1/2*(b*((-2*ArcTan[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[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]]]))) - 1/(x*Sqrt[ArcTanh[Tanh[a + b* x]]])
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]
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]
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]))
Time = 0.20 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.01
method | result | size |
default | \(2 b \left (\frac {2 \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{\left (-4 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )+4 b x \right ) b x}-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}}\right )}{\left (-4 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )+4 b x \right ) \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}}\right )\) | \(95\) |
Input:
int(1/x^2/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/x-2/(- 4*arctanh(tanh(b*x+a))+4*b*x)/(arctanh(tanh(b*x+a))-b*x)^(1/2)*arctanh(arc tanh(tanh(b*x+a))^(1/2)/(arctanh(tanh(b*x+a))-b*x)^(1/2)))
Time = 0.10 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.96 \[ \int \frac {1}{x^2 \sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=\left [\frac {\sqrt {a} b x \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) - 2 \, \sqrt {b x + a} a}{2 \, a^{2} x}, -\frac {\sqrt {-a} b x \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x + a}}\right ) + \sqrt {b x + a} a}{a^{2} x}\right ] \] Input:
integrate(1/x^2/arctanh(tanh(b*x+a))^(1/2),x, algorithm="fricas")
Output:
[1/2*(sqrt(a)*b*x*log((b*x + 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) - 2*sqrt(b* x + a)*a)/(a^2*x), -(sqrt(-a)*b*x*arctan(sqrt(-a)/sqrt(b*x + a)) + sqrt(b* x + a)*a)/(a^2*x)]
\[ \int \frac {1}{x^2 \sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=\int \frac {1}{x^{2} \sqrt {\operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}}\, dx \] Input:
integrate(1/x**2/atanh(tanh(b*x+a))**(1/2),x)
Output:
Integral(1/(x**2*sqrt(atanh(tanh(a + b*x)))), x)
\[ \int \frac {1}{x^2 \sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=\int { \frac {1}{x^{2} \sqrt {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )}} \,d x } \] Input:
integrate(1/x^2/arctanh(tanh(b*x+a))^(1/2),x, algorithm="maxima")
Output:
integrate(1/(x^2*sqrt(arctanh(tanh(b*x + a)))), x)
Time = 0.12 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.47 \[ \int \frac {1}{x^2 \sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=-b {\left (\frac {\arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} + \frac {\sqrt {b x + a}}{a b x}\right )} \] Input:
integrate(1/x^2/arctanh(tanh(b*x+a))^(1/2),x, algorithm="giac")
Output:
-b*(arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a) + sqrt(b*x + a)/(a*b*x))
Time = 8.71 (sec) , antiderivative size = 570, normalized size of antiderivative = 6.06 \[ \int \frac {1}{x^2 \sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx =\text {Too large to display} \] Input:
int(1/(x^2*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*(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 *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)^3 - 8*a^3 - 6*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)^2 + 12*a^2*(2*a - lo g((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)))*1i)/(lo g(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)^(3/2)
\[ \int \frac {1}{x^2 \sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=\int \frac {\sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}}{\mathit {atanh} \left (\tanh \left (b x +a \right )\right ) x^{2}}d x \] Input:
int(1/x^2/atanh(tanh(b*x+a))^(1/2),x)
Output:
int(sqrt(atanh(tanh(a + b*x)))/(atanh(tanh(a + b*x))*x**2),x)