Integrand size = 17, antiderivative size = 106 \[ \int \frac {1}{x^{3/2} \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=-\frac {8 b \sqrt {x}}{3 (b x-\text {arctanh}(\tanh (a+b x)))^2 \text {arctanh}(\tanh (a+b x))^{3/2}}+\frac {2}{\sqrt {x} (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{3/2}}+\frac {16 b \sqrt {x}}{3 (b x-\text {arctanh}(\tanh (a+b x)))^3 \sqrt {\text {arctanh}(\tanh (a+b x))}} \]
2/(b*x-arctanh(tanh(b*x+a)))/arctanh(tanh(b*x+a))^(3/2)/x^(1/2)-8/3*b*x^(1 /2)/(b*x-arctanh(tanh(b*x+a)))^2/arctanh(tanh(b*x+a))^(3/2)+16/3*b*x^(1/2) /(b*x-arctanh(tanh(b*x+a)))^3/arctanh(tanh(b*x+a))^(1/2)
Time = 0.06 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.62 \[ \int \frac {1}{x^{3/2} \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\frac {2 \left (-b^2 x^2+6 b x \text {arctanh}(\tanh (a+b x))+3 \text {arctanh}(\tanh (a+b x))^2\right )}{3 \sqrt {x} (b x-\text {arctanh}(\tanh (a+b x)))^3 \text {arctanh}(\tanh (a+b x))^{3/2}} \]
(2*(-(b^2*x^2) + 6*b*x*ArcTanh[Tanh[a + b*x]] + 3*ArcTanh[Tanh[a + b*x]]^2 ))/(3*Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]])^3*ArcTanh[Tanh[a + b*x]]^(3/2 ))
Time = 0.30 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.16, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2602, 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.
\(\displaystyle \int \frac {1}{x^{3/2} \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 2602 |
\(\displaystyle \frac {4 b \int \frac {1}{\sqrt {x} \text {arctanh}(\tanh (a+b x))^{5/2}}dx}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{\sqrt {x} (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{3/2}}\) |
\(\Big \downarrow \) 2602 |
\(\displaystyle \frac {4 b \left (-\frac {2 \int \frac {1}{\sqrt {x} \text {arctanh}(\tanh (a+b x))^{3/2}}dx}{3 (b x-\text {arctanh}(\tanh (a+b x)))}-\frac {2 \sqrt {x}}{3 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{3/2}}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{\sqrt {x} (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{3/2}}\) |
\(\Big \downarrow \) 2598 |
\(\displaystyle \frac {4 b \left (\frac {4 \sqrt {x}}{3 (b x-\text {arctanh}(\tanh (a+b x)))^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}-\frac {2 \sqrt {x}}{3 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{3/2}}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{\sqrt {x} (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{3/2}}\) |
(4*b*((-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[Ta nh[a + b*x]]])))/(b*x - ArcTanh[Tanh[a + b*x]]) + 2/(Sqrt[x]*(b*x - ArcTan h[Tanh[a + b*x]])*ArcTanh[Tanh[a + b*x]]^(3/2))
3.3.62.3.1 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 = 104, normalized size of antiderivative = 0.98
method | result | size |
derivativedivides | \(-\frac {2}{\sqrt {x}\, \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}-\frac {8 b \left (\frac {\sqrt {x}}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}+\frac {2 \sqrt {x}}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}\right )}{\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}\) | \(104\) |
default | \(-\frac {2}{\sqrt {x}\, \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}-\frac {8 b \left (\frac {\sqrt {x}}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}+\frac {2 \sqrt {x}}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}\right )}{\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}\) | \(104\) |
-2/x^(1/2)/(arctanh(tanh(b*x+a))-b*x)/arctanh(tanh(b*x+a))^(3/2)-8*b/(arct anh(tanh(b*x+a))-b*x)*(1/3*x^(1/2)/(arctanh(tanh(b*x+a))-b*x)/arctanh(tanh (b*x+a))^(3/2)+2/3/(arctanh(tanh(b*x+a))-b*x)^2*x^(1/2)/arctanh(tanh(b*x+a ))^(1/2))
Time = 0.26 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.55 \[ \int \frac {1}{x^{3/2} \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=-\frac {2 \, {\left (8 \, b^{2} x^{2} + 12 \, a b x + 3 \, a^{2}\right )} \sqrt {b x + a} \sqrt {x}}{3 \, {\left (a^{3} b^{2} x^{3} + 2 \, a^{4} b x^{2} + a^{5} x\right )}} \]
-2/3*(8*b^2*x^2 + 12*a*b*x + 3*a^2)*sqrt(b*x + a)*sqrt(x)/(a^3*b^2*x^3 + 2 *a^4*b*x^2 + a^5*x)
\[ \int \frac {1}{x^{3/2} \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\int \frac {1}{x^{\frac {3}{2}} \operatorname {atanh}^{\frac {5}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]
Time = 0.30 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.42 \[ \int \frac {1}{x^{3/2} \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=-\frac {2 \, {\left (8 \, b^{3} x^{3} + 20 \, a b^{2} x^{2} + 15 \, a^{2} b x + 3 \, a^{3}\right )}}{3 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{3} \sqrt {x}} \]
Time = 0.29 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.58 \[ \int \frac {1}{x^{3/2} \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=-\frac {2 \, \sqrt {x} {\left (\frac {5 \, b^{2} x}{a^{3}} + \frac {6 \, b}{a^{2}}\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}}} + \frac {4 \, \sqrt {b}}{{\left ({\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} - a\right )} a^{2}} \]
-2/3*sqrt(x)*(5*b^2*x/a^3 + 6*b/a^2)/(b*x + a)^(3/2) + 4*sqrt(b)/(((sqrt(b )*sqrt(x) - sqrt(b*x + a))^2 - a)*a^2)
Time = 4.58 (sec) , antiderivative size = 348, normalized size of antiderivative = 3.28 \[ \int \frac {1}{x^{3/2} \text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\frac {\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\left (\frac {4}{b^2\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}+\frac {128\,x^2}{3\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}-\frac {32\,x}{b\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}\right )}{x^{5/2}-\frac {x^{3/2}\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{b}+\frac {\sqrt {x}\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{4\,b^2}} \]
((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)*(4/(b^2*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - l og((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)) + (128*x^2 )/(3*(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) - (32*x)/(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^(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))