Integrand size = 17, antiderivative size = 71 \[ \int \frac {1}{\sqrt {x} \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=-\frac {2 \sqrt {x}}{3 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{3/2}}+\frac {4 \sqrt {x}}{3 (b x-\text {arctanh}(\tanh (a+b x)))^2 \sqrt {\text {arctanh}(\tanh (a+b x))}} \]
-2/3*x^(1/2)/(b*x-arctanh(tanh(b*x+a)))/arctanh(tanh(b*x+a))^(3/2)+4/3*x^( 1/2)/(b*x-arctanh(tanh(b*x+a)))^2/arctanh(tanh(b*x+a))^(1/2)
Time = 0.04 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.66 \[ \int \frac {1}{\sqrt {x} \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=-\frac {2 \sqrt {x} (b x-3 \text {arctanh}(\tanh (a+b x)))}{3 \text {arctanh}(\tanh (a+b x))^{3/2} (-b x+\text {arctanh}(\tanh (a+b x)))^2} \]
(-2*Sqrt[x]*(b*x - 3*ArcTanh[Tanh[a + b*x]]))/(3*ArcTanh[Tanh[a + b*x]]^(3 /2)*(-(b*x) + ArcTanh[Tanh[a + b*x]])^2)
Time = 0.23 (sec) , antiderivative size = 71, 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.
\(\displaystyle \int \frac {1}{\sqrt {x} \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 2602 |
\(\displaystyle -\frac {2 \int \frac {1}{\sqrt {x} \text {arctanh}(\tanh (a+b x))^{3/2}}dx}{3 (b x-\text {arctanh}(\tanh (a+b x)))}-\frac {2 \sqrt {x}}{3 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{3/2}}\) |
\(\Big \downarrow \) 2598 |
\(\displaystyle \frac {4 \sqrt {x}}{3 (b x-\text {arctanh}(\tanh (a+b x)))^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}-\frac {2 \sqrt {x}}{3 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{3/2}}\) |
(-2*Sqrt[x])/(3*(b*x - ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a + b*x]]^(3/2 )) + (4*Sqrt[x])/(3*(b*x - ArcTanh[Tanh[a + b*x]])^2*Sqrt[ArcTanh[Tanh[a + b*x]]])
3.3.61.3.1 Defintions of rubi rules used
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 = 0.18 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.82
method | result | size |
derivativedivides | \(\frac {2 \sqrt {x}}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}+\frac {4 \sqrt {x}}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}\) | \(58\) |
default | \(\frac {2 \sqrt {x}}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}+\frac {4 \sqrt {x}}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}\) | \(58\) |
2/3*x^(1/2)/(arctanh(tanh(b*x+a))-b*x)/arctanh(tanh(b*x+a))^(3/2)+4/3/(arc tanh(tanh(b*x+a))-b*x)^2*x^(1/2)/arctanh(tanh(b*x+a))^(1/2)
Time = 0.26 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.61 \[ \int \frac {1}{\sqrt {x} \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\frac {2 \, {\left (2 \, b x + 3 \, a\right )} \sqrt {b x + a} \sqrt {x}}{3 \, {\left (a^{2} b^{2} x^{2} + 2 \, a^{3} b x + a^{4}\right )}} \]
\[ \int \frac {1}{\sqrt {x} \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\int \frac {1}{\sqrt {x} \operatorname {atanh}^{\frac {5}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]
\[ \int \frac {1}{\sqrt {x} \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\int { \frac {1}{\sqrt {x} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{\frac {5}{2}}} \,d x } \]
Time = 0.28 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.35 \[ \int \frac {1}{\sqrt {x} \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\frac {2 \, \sqrt {x} {\left (\frac {2 \, b x}{a^{2}} + \frac {3}{a}\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}}} \]
Time = 4.61 (sec) , antiderivative size = 346, normalized size of antiderivative = 4.87 \[ \int \frac {1}{\sqrt {x} \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\frac {\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\left (\frac {16\,x^2}{3\,b\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}-\frac {x\,\left (48\,\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-48\,\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+96\,b\,x\right )}{12\,b^2\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}\right )}{x^{5/2}-\frac {x^{3/2}\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{b}+\frac {\sqrt {x}\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{4\,b^2}} \]
((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)*((16*x^2)/(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) - (x*(48*log(2/(exp(2*a)*exp(2*b*x) + 1)) - 48*log((2*exp(2*a)*exp(2*b*x))/ (exp(2*a)*exp(2*b*x) + 1)) + 96*b*x))/(12*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)^2) ))/(x^(5/2) - (x^(3/2)*(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))/b + (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)^2)/(4*b^2))