Integrand size = 17, antiderivative size = 148 \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{11/2}} \, dx=\frac {32 b^3 \text {arctanh}(\tanh (a+b x))^{3/2}}{315 x^{3/2} (b x-\text {arctanh}(\tanh (a+b x)))^4}+\frac {16 b^2 \text {arctanh}(\tanh (a+b x))^{3/2}}{105 x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))^3}+\frac {4 b \text {arctanh}(\tanh (a+b x))^{3/2}}{21 x^{7/2} (b x-\text {arctanh}(\tanh (a+b x)))^2}+\frac {2 \text {arctanh}(\tanh (a+b x))^{3/2}}{9 x^{9/2} (b x-\text {arctanh}(\tanh (a+b x)))} \]
32/315*b^3*arctanh(tanh(b*x+a))^(3/2)/x^(3/2)/(b*x-arctanh(tanh(b*x+a)))^4 +16/105*b^2*arctanh(tanh(b*x+a))^(3/2)/x^(5/2)/(b*x-arctanh(tanh(b*x+a)))^ 3+4/21*b*arctanh(tanh(b*x+a))^(3/2)/x^(7/2)/(b*x-arctanh(tanh(b*x+a)))^2+2 /9*arctanh(tanh(b*x+a))^(3/2)/x^(9/2)/(b*x-arctanh(tanh(b*x+a)))
Time = 0.06 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.55 \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{11/2}} \, dx=\frac {2 \text {arctanh}(\tanh (a+b x))^{3/2} \left (105 b^3 x^3-189 b^2 x^2 \text {arctanh}(\tanh (a+b x))+135 b x \text {arctanh}(\tanh (a+b x))^2-35 \text {arctanh}(\tanh (a+b x))^3\right )}{315 x^{9/2} (-b x+\text {arctanh}(\tanh (a+b x)))^4} \]
(2*ArcTanh[Tanh[a + b*x]]^(3/2)*(105*b^3*x^3 - 189*b^2*x^2*ArcTanh[Tanh[a + b*x]] + 135*b*x*ArcTanh[Tanh[a + b*x]]^2 - 35*ArcTanh[Tanh[a + b*x]]^3)) /(315*x^(9/2)*(-(b*x) + ArcTanh[Tanh[a + b*x]])^4)
Time = 0.39 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.24, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {2602, 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.
\(\displaystyle \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{11/2}} \, dx\) |
\(\Big \downarrow \) 2602 |
\(\displaystyle \frac {2 b \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{9/2}}dx}{3 (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {2 \text {arctanh}(\tanh (a+b x))^{3/2}}{9 x^{9/2} (b x-\text {arctanh}(\tanh (a+b x)))}\) |
\(\Big \downarrow \) 2602 |
\(\displaystyle \frac {2 b \left (\frac {4 b \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{7/2}}dx}{7 (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {2 \text {arctanh}(\tanh (a+b x))^{3/2}}{7 x^{7/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{3 (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {2 \text {arctanh}(\tanh (a+b x))^{3/2}}{9 x^{9/2} (b x-\text {arctanh}(\tanh (a+b x)))}\) |
\(\Big \downarrow \) 2602 |
\(\displaystyle \frac {2 b \left (\frac {4 b \left (\frac {2 b \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{5/2}}dx}{5 (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {2 \text {arctanh}(\tanh (a+b x))^{3/2}}{5 x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{7 (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {2 \text {arctanh}(\tanh (a+b x))^{3/2}}{7 x^{7/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{3 (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {2 \text {arctanh}(\tanh (a+b x))^{3/2}}{9 x^{9/2} (b x-\text {arctanh}(\tanh (a+b x)))}\) |
\(\Big \downarrow \) 2598 |
\(\displaystyle \frac {2 \text {arctanh}(\tanh (a+b x))^{3/2}}{9 x^{9/2} (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {2 b \left (\frac {2 \text {arctanh}(\tanh (a+b x))^{3/2}}{7 x^{7/2} (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {4 b \left (\frac {2 \text {arctanh}(\tanh (a+b x))^{3/2}}{5 x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {4 b \text {arctanh}(\tanh (a+b x))^{3/2}}{15 x^{3/2} (b x-\text {arctanh}(\tanh (a+b x)))^2}\right )}{7 (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{3 (b x-\text {arctanh}(\tanh (a+b x)))}\) |
(2*ArcTanh[Tanh[a + b*x]]^(3/2))/(9*x^(9/2)*(b*x - ArcTanh[Tanh[a + b*x]]) ) + (2*b*((2*ArcTanh[Tanh[a + b*x]]^(3/2))/(7*x^(7/2)*(b*x - ArcTanh[Tanh[ a + b*x]])) + (4*b*((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]]))))/(7*(b*x - ArcTanh[Tanh[a + b*x]]))))/(3*( b*x - ArcTanh[Tanh[a + b*x]]))
3.3.22.3.1 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 = 0.19 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.02
method | result | size |
derivativedivides | \(-\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{9 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {9}{2}}}-\frac {4 b \left (-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{7 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {7}{2}}}-\frac {4 b \left (-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{5 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {5}{2}}}+\frac {2 b \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{15 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2} x^{\frac {3}{2}}}\right )}{7 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )}\right )}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )}\) | \(151\) |
default | \(-\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{9 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {9}{2}}}-\frac {4 b \left (-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{7 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {7}{2}}}-\frac {4 b \left (-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{5 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {5}{2}}}+\frac {2 b \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{15 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2} x^{\frac {3}{2}}}\right )}{7 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )}\right )}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )}\) | \(151\) |
-2/9/(arctanh(tanh(b*x+a))-b*x)/x^(9/2)*arctanh(tanh(b*x+a))^(3/2)-4/3*b/( arctanh(tanh(b*x+a))-b*x)*(-1/7/(arctanh(tanh(b*x+a))-b*x)/x^(7/2)*arctanh (tanh(b*x+a))^(3/2)-4/7*b/(arctanh(tanh(b*x+a))-b*x)*(-1/5/(arctanh(tanh(b *x+a))-b*x)/x^(5/2)*arctanh(tanh(b*x+a))^(3/2)+2/15*b/(arctanh(tanh(b*x+a) )-b*x)^2/x^(3/2)*arctanh(tanh(b*x+a))^(3/2)))
Time = 0.26 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.38 \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{11/2}} \, dx=\frac {2 \, {\left (16 \, b^{4} x^{4} - 8 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} - 5 \, a^{3} b x - 35 \, a^{4}\right )} \sqrt {b x + a}}{315 \, a^{4} x^{\frac {9}{2}}} \]
2/315*(16*b^4*x^4 - 8*a*b^3*x^3 + 6*a^2*b^2*x^2 - 5*a^3*b*x - 35*a^4)*sqrt (b*x + a)/(a^4*x^(9/2))
Timed out. \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{11/2}} \, dx=\text {Timed out} \]
Time = 0.29 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.38 \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{11/2}} \, dx=\frac {2 \, {\left (16 \, b^{4} x^{4} - 8 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} - 5 \, a^{3} b x - 35 \, a^{4}\right )} \sqrt {b x + a}}{315 \, a^{4} x^{\frac {9}{2}}} \]
2/315*(16*b^4*x^4 - 8*a*b^3*x^3 + 6*a^2*b^2*x^2 - 5*a^3*b*x - 35*a^4)*sqrt (b*x + a)/(a^4*x^(9/2))
Time = 0.28 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.12 \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{11/2}} \, dx=\frac {64 \, {\left (315 \, b^{\frac {9}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{10} + 189 \, a b^{\frac {9}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{8} + 84 \, a^{2} b^{\frac {9}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{6} - 36 \, a^{3} b^{\frac {9}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{4} + 9 \, a^{4} b^{\frac {9}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} - a^{5} b^{\frac {9}{2}}\right )}}{315 \, {\left ({\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} - a\right )}^{9}} \]
64/315*(315*b^(9/2)*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^10 + 189*a*b^(9/2)*( sqrt(b)*sqrt(x) - sqrt(b*x + a))^8 + 84*a^2*b^(9/2)*(sqrt(b)*sqrt(x) - sqr t(b*x + a))^6 - 36*a^3*b^(9/2)*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^4 + 9*a^4 *b^(9/2)*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^2 - a^5*b^(9/2))/((sqrt(b)*sqrt (x) - sqrt(b*x + a))^2 - a)^9
Time = 4.40 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.99 \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{11/2}} \, dx=\frac {\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\left (\frac {16\,b^2\,x^2}{105\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}+\frac {128\,b^3\,x^3}{315\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}+\frac {512\,b^4\,x^4}{315\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^4}+\frac {4\,b\,x}{63\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}-\frac {2}{9}\right )}{x^{9/2}} \]
((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*b^2*x^2)/(105*(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) + (128*b^3*x^3)/(315*(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) + (512*b^4*x^4)/(315* (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)^4) + (4*b*x)/(63*(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/ 9))/x^(9/2)