Integrand size = 15, antiderivative size = 92 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{x^3} \, dx=\frac {3 b^2 \arctan \left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right )}{4 \sqrt {b x-\text {arctanh}(\tanh (a+b x))}}-\frac {3 b \sqrt {\text {arctanh}(\tanh (a+b x))}}{4 x}-\frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{2 x^2} \] 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)))^(1/2)-3/4*b*arctanh(tanh(b*x+a))^(1/2)/x-1/2* arctanh(tanh(b*x+a))^(3/2)/x^2
Time = 0.04 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.96 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{x^3} \, dx=\frac {1}{4} \left (-\frac {3 b \sqrt {\text {arctanh}(\tanh (a+b x))}}{x}-\frac {2 \text {arctanh}(\tanh (a+b x))^{3/2}}{x^2}-\frac {3 b^2 \text {arctanh}\left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {-b x+\text {arctanh}(\tanh (a+b x))}}\right )}{\sqrt {-b x+\text {arctanh}(\tanh (a+b x))}}\right ) \] Input:
Integrate[ArcTanh[Tanh[a + b*x]]^(3/2)/x^3,x]
Output:
((-3*b*Sqrt[ArcTanh[Tanh[a + b*x]]])/x - (2*ArcTanh[Tanh[a + b*x]]^(3/2))/ x^2 - (3*b^2*ArcTanh[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[-(b*x) + ArcTanh[Ta nh[a + b*x]]]])/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]])/4
Time = 0.37 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.98, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2599, 2599, 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.
\(\displaystyle \int \frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{x^3} \, dx\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {3}{4} b \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^2}dx-\frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{2 x^2}\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {3}{4} b \left (\frac {1}{2} b \int \frac {1}{x \sqrt {\text {arctanh}(\tanh (a+b x))}}dx-\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x}\right )-\frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{2 x^2}\) |
\(\Big \downarrow \) 2592 |
\(\displaystyle \frac {3}{4} b \left (\frac {b \arctan \left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right )}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}-\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x}\right )-\frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{2 x^2}\) |
Input:
Int[ArcTanh[Tanh[a + b*x]]^(3/2)/x^3,x]
Output:
(3*b*((b*ArcTan[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[b*x - ArcTanh[Tanh[a + b *x]]]])/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]] - Sqrt[ArcTanh[Tanh[a + b*x]]]/ x))/4 - ArcTanh[Tanh[a + b*x]]^(3/2)/(2*x^2)
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[(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 = 91, normalized size of antiderivative = 0.99
method | result | size |
default | \(2 b^{2} \left (\frac {-\frac {5 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{8}+\left (\frac {3 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}{8}-\frac {3 b x}{8}\right ) \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{b^{2} x^{2}}-\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}}\right )}{8 \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}}\right )\) | \(91\) |
Input:
int(arctanh(tanh(b*x+a))^(3/2)/x^3,x,method=_RETURNVERBOSE)
Output:
2*b^2*((-5/8*arctanh(tanh(b*x+a))^(3/2)+(3/8*arctanh(tanh(b*x+a))-3/8*b*x) *arctanh(tanh(b*x+a))^(1/2))/b^2/x^2-3/8/(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)))
Time = 0.09 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.32 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{x^3} \, dx=\left [\frac {3 \, \sqrt {a} b^{2} x^{2} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) - 2 \, {\left (5 \, a b x + 2 \, a^{2}\right )} \sqrt {b x + a}}{8 \, a x^{2}}, \frac {3 \, \sqrt {-a} b^{2} x^{2} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x + a}}\right ) - {\left (5 \, a b x + 2 \, a^{2}\right )} \sqrt {b x + a}}{4 \, a x^{2}}\right ] \] Input:
integrate(arctanh(tanh(b*x+a))^(3/2)/x^3,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*( 5*a*b*x + 2*a^2)*sqrt(b*x + a))/(a*x^2), 1/4*(3*sqrt(-a)*b^2*x^2*arctan(sq rt(-a)/sqrt(b*x + a)) - (5*a*b*x + 2*a^2)*sqrt(b*x + a))/(a*x^2)]
\[ \int \frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{x^3} \, dx=\int \frac {\operatorname {atanh}^{\frac {3}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}{x^{3}}\, dx \] Input:
integrate(atanh(tanh(b*x+a))**(3/2)/x**3,x)
Output:
Integral(atanh(tanh(a + b*x))**(3/2)/x**3, x)
\[ \int \frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{x^3} \, dx=\int { \frac {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{\frac {3}{2}}}{x^{3}} \,d x } \] Input:
integrate(arctanh(tanh(b*x+a))^(3/2)/x^3,x, algorithm="maxima")
Output:
integrate(arctanh(tanh(b*x + a))^(3/2)/x^3, x)
Time = 0.12 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.79 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{x^3} \, dx=\frac {\sqrt {2} {\left (\frac {3 \, \sqrt {2} b^{3} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {\sqrt {2} {\left (5 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{3} - 3 \, \sqrt {b x + a} a b^{3}\right )}}{b^{2} x^{2}}\right )}}{8 \, b} \] Input:
integrate(arctanh(tanh(b*x+a))^(3/2)/x^3,x, algorithm="giac")
Output:
1/8*sqrt(2)*(3*sqrt(2)*b^3*arctan(sqrt(b*x + a)/sqrt(-a))/sqrt(-a) - sqrt( 2)*(5*(b*x + a)^(3/2)*b^3 - 3*sqrt(b*x + a)*a*b^3)/(b^2*x^2))/b
Time = 8.04 (sec) , antiderivative size = 609, normalized size of antiderivative = 6.62 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{x^3} \, dx =\text {Too large to display} \] Input:
int(atanh(tanh(a + b*x))^(3/2)/x^3,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)*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*ex p(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^2)/(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)) + (2^(1/2)*b^2*log(((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)* ((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*ex p(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)*16i)/x)*3i)/(8*(log(2/(exp(2*a)*e xp(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)) - (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)*((5*log(2/(exp(2*a)*exp(2* b*x) + 1)))/4 - (5*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))) /4 + (5*b*x)/2))/(x*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*ex p(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x))
\[ \int \frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{x^3} \, dx=\frac {-4 \sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}\, \mathit {atanh} \left (\tanh \left (b x +a \right )\right )-6 \sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}\, b x +3 \left (\int \frac {\sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}}{\mathit {atanh} \left (\tanh \left (b x +a \right )\right ) x}d x \right ) b^{2} x^{2}}{8 x^{2}} \] Input:
int(atanh(tanh(b*x+a))^(3/2)/x^3,x)
Output:
( - 4*sqrt(atanh(tanh(a + b*x)))*atanh(tanh(a + b*x)) - 6*sqrt(atanh(tanh( a + b*x)))*b*x + 3*int(sqrt(atanh(tanh(a + b*x)))/(atanh(tanh(a + b*x))*x) ,x)*b**2*x**2)/(8*x**2)