Integrand size = 17, antiderivative size = 72 \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{7/2}} \, dx=\frac {4 b \text {arctanh}(\tanh (a+b x))^{3/2}}{15 x^{3/2} (b x-\text {arctanh}(\tanh (a+b x)))^2}+\frac {2 \text {arctanh}(\tanh (a+b x))^{3/2}}{5 x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))} \] Output:
4/15*b*arctanh(tanh(b*x+a))^(3/2)/x^(3/2)/(b*x-arctanh(tanh(b*x+a)))^2+2/5 *arctanh(tanh(b*x+a))^(3/2)/x^(5/2)/(b*x-arctanh(tanh(b*x+a)))
Time = 0.03 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.67 \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{7/2}} \, dx=\frac {2 (5 b x-3 \text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{3/2}}{15 x^{5/2} (-b x+\text {arctanh}(\tanh (a+b x)))^2} \] Input:
Integrate[Sqrt[ArcTanh[Tanh[a + b*x]]]/x^(7/2),x]
Output:
(2*(5*b*x - 3*ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a + b*x]]^(3/2))/(15*x^ (5/2)*(-(b*x) + ArcTanh[Tanh[a + b*x]])^2)
Time = 0.31 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2602, 2598}
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.
\(\displaystyle \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{7/2}} \, dx\) |
\(\Big \downarrow \) 2602 |
\(\displaystyle \frac {2 b \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{5/2}}dx}{5 (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {2 \text {arctanh}(\tanh (a+b x))^{3/2}}{5 x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))}\) |
\(\Big \downarrow \) 2598 |
\(\displaystyle \frac {2 \text {arctanh}(\tanh (a+b x))^{3/2}}{5 x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {4 b \text {arctanh}(\tanh (a+b x))^{3/2}}{15 x^{3/2} (b x-\text {arctanh}(\tanh (a+b x)))^2}\) |
Input:
Int[Sqrt[ArcTanh[Tanh[a + b*x]]]/x^(7/2),x]
Output:
(4*b*ArcTanh[Tanh[a + b*x]]^(3/2))/(15*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*x ]])^2) + (2*ArcTanh[Tanh[a + b*x]]^(3/2))/(5*x^(5/2)*(b*x - ArcTanh[Tanh[a + b*x]]))
Int[(u_)^(m_)*(v_)^(n_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simp lify[D[v, x]]}, Simp[(-u^(m + 1))*(v^(n + 1)/((m + 1)*(b*u - a*v))), x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m, n}, x] && PiecewiseLinearQ[u, v, x] && EqQ[ m + n + 2, 0] && NeQ[m, -1]
Int[(u_)^(m_)*(v_)^(n_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simp lify[D[v, x]]}, Simp[(-u^(m + 1))*(v^(n + 1)/((m + 1)*(b*u - a*v))), x] + S imp[b*((m + n + 2)/((m + 1)*(b*u - a*v))) Int[u^(m + 1)*v^n, x], x] /; Ne Q[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && NeQ[m + n + 2, 0] && LtQ[m , -1]
Time = 0.42 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.82
method | result | size |
derivativedivides | \(-\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{5 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {5}{2}}}+\frac {4 b \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{15 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2} x^{\frac {3}{2}}}\) | \(59\) |
default | \(-\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{5 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {5}{2}}}+\frac {4 b \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{15 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2} x^{\frac {3}{2}}}\) | \(59\) |
Input:
int(arctanh(tanh(b*x+a))^(1/2)/x^(7/2),x,method=_RETURNVERBOSE)
Output:
-2/5/(arctanh(tanh(b*x+a))-b*x)/x^(5/2)*arctanh(tanh(b*x+a))^(3/2)+4/15*b/ (arctanh(tanh(b*x+a))-b*x)^2/x^(3/2)*arctanh(tanh(b*x+a))^(3/2)
Time = 0.08 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.47 \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{7/2}} \, dx=\frac {2 \, {\left (2 \, b^{2} x^{2} - a b x - 3 \, a^{2}\right )} \sqrt {b x + a}}{15 \, a^{2} x^{\frac {5}{2}}} \] Input:
integrate(arctanh(tanh(b*x+a))^(1/2)/x^(7/2),x, algorithm="fricas")
Output:
2/15*(2*b^2*x^2 - a*b*x - 3*a^2)*sqrt(b*x + a)/(a^2*x^(5/2))
Timed out. \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{7/2}} \, dx=\text {Timed out} \] Input:
integrate(atanh(tanh(b*x+a))**(1/2)/x**(7/2),x)
Output:
Timed out
Time = 0.12 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.47 \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{7/2}} \, dx=\frac {2 \, {\left (2 \, b^{2} x^{2} - a b x - 3 \, a^{2}\right )} \sqrt {b x + a}}{15 \, a^{2} x^{\frac {5}{2}}} \] Input:
integrate(arctanh(tanh(b*x+a))^(1/2)/x^(7/2),x, algorithm="maxima")
Output:
2/15*(2*b^2*x^2 - a*b*x - 3*a^2)*sqrt(b*x + a)/(a^2*x^(5/2))
Time = 0.12 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.56 \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{7/2}} \, dx=\frac {8 \, {\left (15 \, b^{\frac {5}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{6} + 5 \, a b^{\frac {5}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{4} + 5 \, a^{2} b^{\frac {5}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} - a^{3} b^{\frac {5}{2}}\right )}}{15 \, {\left ({\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} - a\right )}^{5}} \] Input:
integrate(arctanh(tanh(b*x+a))^(1/2)/x^(7/2),x, algorithm="giac")
Output:
8/15*(15*b^(5/2)*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^6 + 5*a*b^(5/2)*(sqrt(b )*sqrt(x) - sqrt(b*x + a))^4 + 5*a^2*b^(5/2)*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^2 - a^3*b^(5/2))/((sqrt(b)*sqrt(x) - sqrt(b*x + a))^2 - a)^5
Time = 3.59 (sec) , antiderivative size = 174, normalized size of antiderivative = 2.42 \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{7/2}} \, dx=\frac {\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\left (\frac {16\,b^2\,x^2}{15\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}+\frac {4\,b\,x}{15\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}-\frac {2}{5}\right )}{x^{5/2}} \] Input:
int(atanh(tanh(a + b*x))^(1/2)/x^(7/2),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)*((16*b^2*x^2)/(15*(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 ) + (4*b*x)/(15*(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)) - 2/5))/x^(5/2)
\[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{7/2}} \, dx=\frac {-2 \sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}+\sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}}{\mathit {atanh} \left (\tanh \left (b x +a \right )\right ) x^{3}}d x \right ) b \,x^{2}}{5 \sqrt {x}\, x^{2}} \] Input:
int(atanh(tanh(b*x+a))^(1/2)/x^(7/2),x)
Output:
( - 2*sqrt(atanh(tanh(a + b*x))) + sqrt(x)*int((sqrt(x)*sqrt(atanh(tanh(a + b*x))))/(atanh(tanh(a + b*x))*x**3),x)*b*x**2)/(5*sqrt(x)*x**2)