Integrand size = 15, antiderivative size = 108 \[ \int \frac {x^{5/2}}{\text {arctanh}(\tanh (a+b x))^2} \, dx=\frac {5 x^{3/2}}{3 b^2}+\frac {5 \sqrt {x} (b x-\text {arctanh}(\tanh (a+b x)))}{b^3}-\frac {5 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right ) (b x-\text {arctanh}(\tanh (a+b x)))^{3/2}}{b^{7/2}}-\frac {x^{5/2}}{b \text {arctanh}(\tanh (a+b x))} \]
5/3*x^(3/2)/b^2-5*arctanh(b^(1/2)*x^(1/2)/(b*x-arctanh(tanh(b*x+a)))^(1/2) )*(b*x-arctanh(tanh(b*x+a)))^(3/2)/b^(7/2)-x^(5/2)/b/arctanh(tanh(b*x+a))+ 5*(b*x-arctanh(tanh(b*x+a)))*x^(1/2)/b^3
Time = 0.11 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.10 \[ \int \frac {x^{5/2}}{\text {arctanh}(\tanh (a+b x))^2} \, dx=\frac {2 x^{3/2}}{3 b^2}-\frac {4 \sqrt {x} (-b x+\text {arctanh}(\tanh (a+b x)))}{b^3}+\frac {5 \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {-b x+\text {arctanh}(\tanh (a+b x))}}\right ) (-b x+\text {arctanh}(\tanh (a+b x)))^{3/2}}{b^{7/2}}-\frac {\sqrt {x} (-b x+\text {arctanh}(\tanh (a+b x)))^2}{b^3 \text {arctanh}(\tanh (a+b x))} \]
(2*x^(3/2))/(3*b^2) - (4*Sqrt[x]*(-(b*x) + ArcTanh[Tanh[a + b*x]]))/b^3 + (5*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]]*(-(b*x) + ArcTanh[Tanh[a + b*x]])^(3/2))/b^(7/2) - (Sqrt[x]*(-(b*x) + ArcTanh[Tan h[a + b*x]])^2)/(b^3*ArcTanh[Tanh[a + b*x]])
Time = 0.32 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.12, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {2599, 2590, 2590, 2593}
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.
| \(\displaystyle \int \frac {x^{5/2}}{\text {arctanh}(\tanh (a+b x))^2} \, dx\) |
| \(\Big \downarrow \) 2599 |
| \(\displaystyle \frac {5 \int \frac {x^{3/2}}{\text {arctanh}(\tanh (a+b x))}dx}{2 b}-\frac {x^{5/2}}{b \text {arctanh}(\tanh (a+b x))}\) |
| \(\Big \downarrow \) 2590 |
| \(\displaystyle \frac {5 \left (\frac {(b x-\text {arctanh}(\tanh (a+b x))) \int \frac {\sqrt {x}}{\text {arctanh}(\tanh (a+b x))}dx}{b}+\frac {2 x^{3/2}}{3 b}\right )}{2 b}-\frac {x^{5/2}}{b \text {arctanh}(\tanh (a+b x))}\) |
| \(\Big \downarrow \) 2590 |
| \(\displaystyle \frac {5 \left (\frac {(b x-\text {arctanh}(\tanh (a+b x))) \left (\frac {(b x-\text {arctanh}(\tanh (a+b x))) \int \frac {1}{\sqrt {x} \text {arctanh}(\tanh (a+b x))}dx}{b}+\frac {2 \sqrt {x}}{b}\right )}{b}+\frac {2 x^{3/2}}{3 b}\right )}{2 b}-\frac {x^{5/2}}{b \text {arctanh}(\tanh (a+b x))}\) |
| \(\Big \downarrow \) 2593 |
| \(\displaystyle \frac {5 \left (\frac {\left (\frac {2 \sqrt {x}}{b}-\frac {2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right ) \sqrt {b x-\text {arctanh}(\tanh (a+b x))}}{b^{3/2}}\right ) (b x-\text {arctanh}(\tanh (a+b x)))}{b}+\frac {2 x^{3/2}}{3 b}\right )}{2 b}-\frac {x^{5/2}}{b \text {arctanh}(\tanh (a+b x))}\) |
(5*((2*x^(3/2))/(3*b) + (((2*Sqrt[x])/b - (2*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqr t[b*x - ArcTanh[Tanh[a + b*x]]]]*Sqrt[b*x - ArcTanh[Tanh[a + b*x]]])/b^(3/ 2))*(b*x - ArcTanh[Tanh[a + b*x]]))/b))/(2*b) - x^(5/2)/(b*ArcTanh[Tanh[a + b*x]])
3.2.100.3.1 Defintions of rubi rules used
Int[(v_)^(n_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[ D[v, x]]}, Simp[v^n/(a*n), x] - Simp[(b*u - a*v)/a Int[v^(n - 1)/u, x], x ] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && GtQ[n, 0] && NeQ[n, 1]
Int[1/((u_)*Sqrt[v_]), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simpli fy[D[v, x]]}, Simp[-2*(ArcTanh[Sqrt[v]/Rt[-(b*u - a*v)/a, 2]]/(a*Rt[-(b*u - a*v)/a, 2])), x] /; NeQ[b*u - a*v, 0] && NegQ[(b*u - a*v)/a]] /; Piecewise LinearQ[u, v, x]
Int[(u_)^(m_)*(v_)^(n_.), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Sim plify[D[v, x]]}, Simp[u^(m + 1)*(v^n/(a*(m + 1))), x] - Simp[b*(n/(a*(m + 1 ))) Int[u^(m + 1)*v^(n - 1), x], x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m, n} , x] && PiecewiseLinearQ[u, v, x] && NeQ[m, -1] && ((LtQ[m, -1] && GtQ[n, 0 ] && !(ILtQ[m + n, -2] && (FractionQ[m] || GeQ[2*n + m + 1, 0]))) || (IGtQ [n, 0] && IGtQ[m, 0] && LeQ[n, m]) || (IGtQ[n, 0] && !IntegerQ[m]) || (ILt Q[m, 0] && !IntegerQ[n]))
Time = 0.55 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.68
| method | result | size |
| derivativedivides | \(-\frac {2 \left (-\frac {b \,x^{\frac {3}{2}}}{3}+2 a \sqrt {x}+2 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right ) \sqrt {x}\right )}{b^{3}}+\frac {\frac {2 \left (-\frac {a^{2}}{2}-a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )-\frac {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{2}\right ) \sqrt {x}}{\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}+\frac {5 \left (a^{2}+2 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )+\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}\right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{\sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}}{b^{3}}\) | \(181\) |
| default | \(-\frac {2 \left (-\frac {b \,x^{\frac {3}{2}}}{3}+2 a \sqrt {x}+2 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right ) \sqrt {x}\right )}{b^{3}}+\frac {\frac {2 \left (-\frac {a^{2}}{2}-a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )-\frac {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{2}\right ) \sqrt {x}}{\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}+\frac {5 \left (a^{2}+2 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )+\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}\right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{\sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}}{b^{3}}\) | \(181\) |
| risch | \(\text {Expression too large to display}\) | \(3073\) |
-2/b^3*(-1/3*b*x^(3/2)+2*a*x^(1/2)+2*(arctanh(tanh(b*x+a))-b*x-a)*x^(1/2)) +2/b^3*((-1/2*a^2-a*(arctanh(tanh(b*x+a))-b*x-a)-1/2*(arctanh(tanh(b*x+a)) -b*x-a)^2)*x^(1/2)/arctanh(tanh(b*x+a))+5/2*(a^2+2*a*(arctanh(tanh(b*x+a)) -b*x-a)+(arctanh(tanh(b*x+a))-b*x-a)^2)/((arctanh(tanh(b*x+a))-b*x)*b)^(1/ 2)*arctan(b*x^(1/2)/((arctanh(tanh(b*x+a))-b*x)*b)^(1/2)))
Time = 0.25 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.49 \[ \int \frac {x^{5/2}}{\text {arctanh}(\tanh (a+b x))^2} \, dx=\left [\frac {15 \, {\left (a b x + a^{2}\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x + 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - a}{b x + a}\right ) + 2 \, {\left (2 \, b^{2} x^{2} - 10 \, a b x - 15 \, a^{2}\right )} \sqrt {x}}{6 \, {\left (b^{4} x + a b^{3}\right )}}, \frac {15 \, {\left (a b x + a^{2}\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) + {\left (2 \, b^{2} x^{2} - 10 \, a b x - 15 \, a^{2}\right )} \sqrt {x}}{3 \, {\left (b^{4} x + a b^{3}\right )}}\right ] \]
[1/6*(15*(a*b*x + a^2)*sqrt(-a/b)*log((b*x + 2*b*sqrt(x)*sqrt(-a/b) - a)/( b*x + a)) + 2*(2*b^2*x^2 - 10*a*b*x - 15*a^2)*sqrt(x))/(b^4*x + a*b^3), 1/ 3*(15*(a*b*x + a^2)*sqrt(a/b)*arctan(b*sqrt(x)*sqrt(a/b)/a) + (2*b^2*x^2 - 10*a*b*x - 15*a^2)*sqrt(x))/(b^4*x + a*b^3)]
\[ \int \frac {x^{5/2}}{\text {arctanh}(\tanh (a+b x))^2} \, dx=\int \frac {x^{\frac {5}{2}}}{\operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]
Time = 0.32 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.59 \[ \int \frac {x^{5/2}}{\text {arctanh}(\tanh (a+b x))^2} \, dx=\frac {2 \, b^{2} x^{\frac {5}{2}} - 10 \, a b x^{\frac {3}{2}} - 15 \, a^{2} \sqrt {x}}{3 \, {\left (b^{4} x + a b^{3}\right )}} + \frac {5 \, a^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{3}} \]
1/3*(2*b^2*x^(5/2) - 10*a*b*x^(3/2) - 15*a^2*sqrt(x))/(b^4*x + a*b^3) + 5* a^2*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^3)
Time = 0.27 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.60 \[ \int \frac {x^{5/2}}{\text {arctanh}(\tanh (a+b x))^2} \, dx=\frac {5 \, a^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{3}} - \frac {a^{2} \sqrt {x}}{{\left (b x + a\right )} b^{3}} + \frac {2 \, {\left (b^{4} x^{\frac {3}{2}} - 6 \, a b^{3} \sqrt {x}\right )}}{3 \, b^{6}} \]
5*a^2*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^3) - a^2*sqrt(x)/((b*x + a) *b^3) + 2/3*(b^4*x^(3/2) - 6*a*b^3*sqrt(x))/b^6
Time = 4.38 (sec) , antiderivative size = 463, normalized size of antiderivative = 4.29 \[ \int \frac {x^{5/2}}{\text {arctanh}(\tanh (a+b x))^2} \, dx=\frac {2\,x^{3/2}}{3\,b^2}+\frac {2\,\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 )}{b^3}+\frac {5\,\sqrt {2}\,\ln \left (\frac {16\,b^{15/2}\,\left (\sqrt {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 )-4\,\sqrt {b}\,\sqrt {x}\,\sqrt {\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}+2\,\sqrt {2}\,b\,x\right )}{\left (\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\right )\,\sqrt {\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 (\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 )}^{3/2}}{8\,b^{7/2}}-\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}{2\,b^3\,\left (\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\right )} \]
(2*x^(3/2))/(3*b^2) + (2*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))/b^3 + (5*2^(1/ 2)*log((16*b^(15/2)*(2^(1/2)*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*ex p(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x) - 4*b^(1/2)*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)^(1/2) + 2*2^(1/2)*b*x))/((log((2*exp(2*a)*exp(2 *b*x))/(exp(2*a)*exp(2*b*x) + 1)) - log(2/(exp(2*a)*exp(2*b*x) + 1)))*(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)^(1/2)))*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*e xp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^(3/2))/(8*b^(7/2)) - (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)/(2*b^3*(log((2*exp(2*a)*exp(2*b*x ))/(exp(2*a)*exp(2*b*x) + 1)) - log(2/(exp(2*a)*exp(2*b*x) + 1))))