Integrand size = 15, antiderivative size = 65 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^{5/2}} \, dx=-\frac {32}{3} b^3 x^{3/2}+16 b^2 \sqrt {x} \text {arctanh}(\tanh (a+b x))-\frac {4 b \text {arctanh}(\tanh (a+b x))^2}{\sqrt {x}}-\frac {2 \text {arctanh}(\tanh (a+b x))^3}{3 x^{3/2}} \] Output:
-32/3*b^3*x^(3/2)+16*b^2*x^(1/2)*arctanh(tanh(b*x+a))-4*b*arctanh(tanh(b*x +a))^2/x^(1/2)-2/3*arctanh(tanh(b*x+a))^3/x^(3/2)
Time = 0.02 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.85 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^{5/2}} \, dx=-\frac {2 \left (16 b^3 x^3-24 b^2 x^2 \text {arctanh}(\tanh (a+b x))+6 b x \text {arctanh}(\tanh (a+b x))^2+\text {arctanh}(\tanh (a+b x))^3\right )}{3 x^{3/2}} \] Input:
Integrate[ArcTanh[Tanh[a + b*x]]^3/x^(5/2),x]
Output:
(-2*(16*b^3*x^3 - 24*b^2*x^2*ArcTanh[Tanh[a + b*x]] + 6*b*x*ArcTanh[Tanh[a + b*x]]^2 + ArcTanh[Tanh[a + b*x]]^3))/(3*x^(3/2))
Time = 0.24 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {2599, 2599, 2599, 15}
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.
\(\displaystyle \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^{5/2}} \, dx\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle 2 b \int \frac {\text {arctanh}(\tanh (a+b x))^2}{x^{3/2}}dx-\frac {2 \text {arctanh}(\tanh (a+b x))^3}{3 x^{3/2}}\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle 2 b \left (4 b \int \frac {\text {arctanh}(\tanh (a+b x))}{\sqrt {x}}dx-\frac {2 \text {arctanh}(\tanh (a+b x))^2}{\sqrt {x}}\right )-\frac {2 \text {arctanh}(\tanh (a+b x))^3}{3 x^{3/2}}\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle 2 b \left (4 b \left (2 \sqrt {x} \text {arctanh}(\tanh (a+b x))-2 b \int \sqrt {x}dx\right )-\frac {2 \text {arctanh}(\tanh (a+b x))^2}{\sqrt {x}}\right )-\frac {2 \text {arctanh}(\tanh (a+b x))^3}{3 x^{3/2}}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle 2 b \left (4 b \left (2 \sqrt {x} \text {arctanh}(\tanh (a+b x))-\frac {4}{3} b x^{3/2}\right )-\frac {2 \text {arctanh}(\tanh (a+b x))^2}{\sqrt {x}}\right )-\frac {2 \text {arctanh}(\tanh (a+b x))^3}{3 x^{3/2}}\) |
Input:
Int[ArcTanh[Tanh[a + b*x]]^3/x^(5/2),x]
Output:
(-2*ArcTanh[Tanh[a + b*x]]^3)/(3*x^(3/2)) + 2*b*((-2*ArcTanh[Tanh[a + b*x] ]^2)/Sqrt[x] + 4*b*((-4*b*x^(3/2))/3 + 2*Sqrt[x]*ArcTanh[Tanh[a + b*x]]))
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -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 = 0.99 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.85
method | result | size |
derivativedivides | \(-\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{3}}{3 x^{\frac {3}{2}}}+4 b \left (-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2}}{\sqrt {x}}+4 b \left (\sqrt {x}\, \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-\frac {2 b \,x^{\frac {3}{2}}}{3}\right )\right )\) | \(55\) |
default | \(-\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{3}}{3 x^{\frac {3}{2}}}+4 b \left (-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2}}{\sqrt {x}}+4 b \left (\sqrt {x}\, \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-\frac {2 b \,x^{\frac {3}{2}}}{3}\right )\right )\) | \(55\) |
risch | \(\text {Expression too large to display}\) | \(7808\) |
Input:
int(arctanh(tanh(b*x+a))^3/x^(5/2),x,method=_RETURNVERBOSE)
Output:
-2/3*arctanh(tanh(b*x+a))^3/x^(3/2)+4*b*(-arctanh(tanh(b*x+a))^2/x^(1/2)+4 *b*(x^(1/2)*arctanh(tanh(b*x+a))-2/3*b*x^(3/2)))
Time = 0.08 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.52 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^{5/2}} \, dx=\frac {2 \, {\left (b^{3} x^{3} + 9 \, a b^{2} x^{2} - 9 \, a^{2} b x - a^{3}\right )}}{3 \, x^{\frac {3}{2}}} \] Input:
integrate(arctanh(tanh(b*x+a))^3/x^(5/2),x, algorithm="fricas")
Output:
2/3*(b^3*x^3 + 9*a*b^2*x^2 - 9*a^2*b*x - a^3)/x^(3/2)
Time = 1.28 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.02 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^{5/2}} \, dx=- \frac {32 b^{3} x^{\frac {3}{2}}}{3} + 16 b^{2} \sqrt {x} \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )} - \frac {4 b \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{\sqrt {x}} - \frac {2 \operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{3 x^{\frac {3}{2}}} \] Input:
integrate(atanh(tanh(b*x+a))**3/x**(5/2),x)
Output:
-32*b**3*x**(3/2)/3 + 16*b**2*sqrt(x)*atanh(tanh(a + b*x)) - 4*b*atanh(tan h(a + b*x))**2/sqrt(x) - 2*atanh(tanh(a + b*x))**3/(3*x**(3/2))
Time = 0.06 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.85 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^{5/2}} \, dx=-\frac {4 \, b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2}}{\sqrt {x}} - \frac {16}{3} \, {\left (2 \, b^{2} x^{\frac {3}{2}} - 3 \, b \sqrt {x} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )\right )} b - \frac {2 \, \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3}}{3 \, x^{\frac {3}{2}}} \] Input:
integrate(arctanh(tanh(b*x+a))^3/x^(5/2),x, algorithm="maxima")
Output:
-4*b*arctanh(tanh(b*x + a))^2/sqrt(x) - 16/3*(2*b^2*x^(3/2) - 3*b*sqrt(x)* arctanh(tanh(b*x + a)))*b - 2/3*arctanh(tanh(b*x + a))^3/x^(3/2)
Time = 0.11 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.52 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^{5/2}} \, dx=\frac {2}{3} \, b^{3} x^{\frac {3}{2}} + 6 \, a b^{2} \sqrt {x} - \frac {2 \, {\left (9 \, a^{2} b x + a^{3}\right )}}{3 \, x^{\frac {3}{2}}} \] Input:
integrate(arctanh(tanh(b*x+a))^3/x^(5/2),x, algorithm="giac")
Output:
2/3*b^3*x^(3/2) + 6*a*b^2*sqrt(x) - 2/3*(9*a^2*b*x + a^3)/x^(3/2)
Time = 3.16 (sec) , antiderivative size = 182, normalized size of antiderivative = 2.80 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^{5/2}} \, dx=\frac {{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}{12\,x^{3/2}}+\frac {2\,b^3\,x^{3/2}}{3}-\frac {3\,b\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{2\,\sqrt {x}}-3\,b^2\,\sqrt {x}\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right ) \] Input:
int(atanh(tanh(a + b*x))^3/x^(5/2),x)
Output:
(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)^3/(12*x^(3/2)) + (2*b^3*x^(3/2))/3 - (3*b*(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*x^(1/2)) - 3*b^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)
\[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^{5/2}} \, dx=\int \frac {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )^{3}}{\sqrt {x}\, x^{2}}d x \] Input:
int(atanh(tanh(b*x+a))^3/x^(5/2),x)
Output:
int(atanh(tanh(a + b*x))**3/(sqrt(x)*x**2),x)