Integrand size = 17, antiderivative size = 35 \[ \int \frac {\sqrt {x}}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=-\frac {2 x^{3/2}}{3 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{3/2}} \] Output:
-2/3*x^(3/2)/(b*x-arctanh(tanh(b*x+a)))/arctanh(tanh(b*x+a))^(3/2)
Time = 0.04 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.97 \[ \int \frac {\sqrt {x}}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\frac {2 x^{3/2}}{3 \text {arctanh}(\tanh (a+b x))^{3/2} (-b x+\text {arctanh}(\tanh (a+b x)))} \] Input:
Integrate[Sqrt[x]/ArcTanh[Tanh[a + b*x]]^(5/2),x]
Output:
(2*x^(3/2))/(3*ArcTanh[Tanh[a + b*x]]^(3/2)*(-(b*x) + ArcTanh[Tanh[a + b*x ]]))
Time = 0.16 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {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.
\(\displaystyle \int \frac {\sqrt {x}}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 2598 |
\(\displaystyle -\frac {2 x^{3/2}}{3 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{3/2}}\) |
Input:
Int[Sqrt[x]/ArcTanh[Tanh[a + b*x]]^(5/2),x]
Output:
(-2*x^(3/2))/(3*(b*x - ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a + b*x]]^(3/2 ))
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]
Leaf count of result is larger than twice the leaf count of optimal. \(91\) vs. \(2(29)=58\).
Time = 0.40 (sec) , antiderivative size = 92, normalized size of antiderivative = 2.63
method | result | size |
derivativedivides | \(-\frac {\sqrt {x}}{b \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}+\frac {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \left (\frac {\sqrt {x}}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}+\frac {2 \sqrt {x}}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}\right )}{b}\) | \(92\) |
default | \(-\frac {\sqrt {x}}{b \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}+\frac {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \left (\frac {\sqrt {x}}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}+\frac {2 \sqrt {x}}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}\right )}{b}\) | \(92\) |
Input:
int(x^(1/2)/arctanh(tanh(b*x+a))^(5/2),x,method=_RETURNVERBOSE)
Output:
-x^(1/2)/b/arctanh(tanh(b*x+a))^(3/2)+(arctanh(tanh(b*x+a))-b*x)/b*(1/3*x^ (1/2)/(arctanh(tanh(b*x+a))-b*x)/arctanh(tanh(b*x+a))^(3/2)+2/3/(arctanh(t anh(b*x+a))-b*x)^2*x^(1/2)/arctanh(tanh(b*x+a))^(1/2))
Time = 0.10 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.94 \[ \int \frac {\sqrt {x}}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\frac {2 \, \sqrt {b x + a} x^{\frac {3}{2}}}{3 \, {\left (a b^{2} x^{2} + 2 \, a^{2} b x + a^{3}\right )}} \] Input:
integrate(x^(1/2)/arctanh(tanh(b*x+a))^(5/2),x, algorithm="fricas")
Output:
2/3*sqrt(b*x + a)*x^(3/2)/(a*b^2*x^2 + 2*a^2*b*x + a^3)
\[ \int \frac {\sqrt {x}}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\int \frac {\sqrt {x}}{\operatorname {atanh}^{\frac {5}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \] Input:
integrate(x**(1/2)/atanh(tanh(b*x+a))**(5/2),x)
Output:
Integral(sqrt(x)/atanh(tanh(a + b*x))**(5/2), x)
\[ \int \frac {\sqrt {x}}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\int { \frac {\sqrt {x}}{\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate(x^(1/2)/arctanh(tanh(b*x+a))^(5/2),x, algorithm="maxima")
Output:
integrate(sqrt(x)/arctanh(tanh(b*x + a))^(5/2), x)
Time = 0.12 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.43 \[ \int \frac {\sqrt {x}}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\frac {2 \, x^{\frac {3}{2}}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} a} \] Input:
integrate(x^(1/2)/arctanh(tanh(b*x+a))^(5/2),x, algorithm="giac")
Output:
2/3*x^(3/2)/((b*x + a)^(3/2)*a)
Time = 3.83 (sec) , antiderivative size = 229, normalized size of antiderivative = 6.54 \[ \int \frac {\sqrt {x}}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=-\frac {4\,x^{3/2}\,\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}}{3\,b^2\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )\,\left (\frac {{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{4\,b^2}+x^2-\frac {x\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{b}\right )} \] Input:
int(x^(1/2)/atanh(tanh(a + b*x))^(5/2),x)
Output:
-(4*x^(3/2)*(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - lo g(2/(exp(2*a)*exp(2*b*x) + 1))/2)^(1/2))/(3*b^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)* ((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^2) + x^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)) /b))
\[ \int \frac {\sqrt {x}}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\frac {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )^{2} \left (\int \frac {\sqrt {x}\, \sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}}{\mathit {atanh} \left (\tanh \left (b x +a \right )\right )^{2} x}d x \right )-2 \sqrt {x}\, \sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}}{3 \mathit {atanh} \left (\tanh \left (b x +a \right )\right )^{2} b} \] Input:
int(x^(1/2)/atanh(tanh(b*x+a))^(5/2),x)
Output:
(atanh(tanh(a + b*x))**2*int((sqrt(x)*sqrt(atanh(tanh(a + b*x))))/(atanh(t anh(a + b*x))**2*x),x) - 2*sqrt(x)*sqrt(atanh(tanh(a + b*x))))/(3*atanh(ta nh(a + b*x))**2*b)