Integrand size = 15, antiderivative size = 145 \[ \int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))^2} \, dx=\frac {5 b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right )}{(b x-\text {arctanh}(\tanh (a+b x)))^{7/2}}-\frac {5 b}{\sqrt {x} (b x-\text {arctanh}(\tanh (a+b x)))^3}-\frac {5}{3 x^{3/2} (b x-\text {arctanh}(\tanh (a+b x)))^2}-\frac {1}{b x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))}-\frac {1}{b x^{5/2} \text {arctanh}(\tanh (a+b x))} \]
5*b^(3/2)*arctanh(b^(1/2)*x^(1/2)/(b*x-arctanh(tanh(b*x+a)))^(1/2))/(b*x-a rctanh(tanh(b*x+a)))^(7/2)-5/3/x^(3/2)/(b*x-arctanh(tanh(b*x+a)))^2-1/b/x^ (5/2)/(b*x-arctanh(tanh(b*x+a)))-1/b/x^(5/2)/arctanh(tanh(b*x+a))-5*b/(b*x -arctanh(tanh(b*x+a)))^3/x^(1/2)
Time = 0.16 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))^2} \, dx=\frac {2 (-7 b x+\text {arctanh}(\tanh (a+b x)))}{3 x^{3/2} (b x-\text {arctanh}(\tanh (a+b x)))^3}+\frac {5 b^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {-b x+\text {arctanh}(\tanh (a+b x))}}\right )}{(-b x+\text {arctanh}(\tanh (a+b x)))^{7/2}}+\frac {b^2 \sqrt {x}}{\text {arctanh}(\tanh (a+b x)) (-b x+\text {arctanh}(\tanh (a+b x)))^3} \]
(2*(-7*b*x + ArcTanh[Tanh[a + b*x]]))/(3*x^(3/2)*(b*x - ArcTanh[Tanh[a + b *x]])^3) + (5*b^(3/2)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[-(b*x) + ArcTanh[Tanh[ a + b*x]]]])/(-(b*x) + ArcTanh[Tanh[a + b*x]])^(7/2) + (b^2*Sqrt[x])/(ArcT anh[Tanh[a + b*x]]*(-(b*x) + ArcTanh[Tanh[a + b*x]])^3)
Time = 0.44 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.29, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2599, 2594, 2594, 2594, 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.
\(\displaystyle \int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))^2} \, dx\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle -\frac {5 \int \frac {1}{x^{7/2} \text {arctanh}(\tanh (a+b x))}dx}{2 b}-\frac {1}{b x^{5/2} \text {arctanh}(\tanh (a+b x))}\) |
\(\Big \downarrow \) 2594 |
\(\displaystyle -\frac {5 \left (\frac {b \int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))}dx}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{5 x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{2 b}-\frac {1}{b x^{5/2} \text {arctanh}(\tanh (a+b x))}\) |
\(\Big \downarrow \) 2594 |
\(\displaystyle -\frac {5 \left (\frac {b \left (\frac {b \int \frac {1}{x^{3/2} \text {arctanh}(\tanh (a+b x))}dx}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{3 x^{3/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{5 x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{2 b}-\frac {1}{b x^{5/2} \text {arctanh}(\tanh (a+b x))}\) |
\(\Big \downarrow \) 2594 |
\(\displaystyle -\frac {5 \left (\frac {b \left (\frac {b \left (\frac {b \int \frac {1}{\sqrt {x} \text {arctanh}(\tanh (a+b x))}dx}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{\sqrt {x} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{3 x^{3/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{5 x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{2 b}-\frac {1}{b x^{5/2} \text {arctanh}(\tanh (a+b x))}\) |
\(\Big \downarrow \) 2593 |
\(\displaystyle -\frac {5 \left (\frac {b \left (\frac {2}{3 x^{3/2} (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {b \left (\frac {2}{\sqrt {x} (b x-\text {arctanh}(\tanh (a+b x)))}-\frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right )}{(b x-\text {arctanh}(\tanh (a+b x)))^{3/2}}\right )}{b x-\text {arctanh}(\tanh (a+b x))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{5 x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{2 b}-\frac {1}{b x^{5/2} \text {arctanh}(\tanh (a+b x))}\) |
(-5*(2/(5*x^(5/2)*(b*x - ArcTanh[Tanh[a + b*x]])) + (b*(2/(3*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*x]])) + (b*((-2*Sqrt[b]*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqr t[b*x - ArcTanh[Tanh[a + b*x]]]])/(b*x - ArcTanh[Tanh[a + b*x]])^(3/2) + 2 /(Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]]))))/(b*x - ArcTanh[Tanh[a + b*x]]) ))/(b*x - ArcTanh[Tanh[a + b*x]])))/(2*b) - 1/(b*x^(5/2)*ArcTanh[Tanh[a + b*x]])
3.3.5.3.1 Defintions of rubi rules used
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[(v_)^(n_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[ D[v, x]]}, Simp[v^(n + 1)/((n + 1)*(b*u - a*v)), x] - Simp[a*((n + 1)/((n + 1)*(b*u - a*v))) Int[v^(n + 1)/u, x], x] /; NeQ[b*u - a*v, 0]] /; Piecew iseLinearQ[u, v, x] && LtQ[n, -1]
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 = 0.58 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.79
method | result | size |
derivativedivides | \(-\frac {2}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2} x^{\frac {3}{2}}}+\frac {4 b}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \sqrt {x}}+\frac {2 b^{2} \left (\frac {\sqrt {x}}{2 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}+\frac {5 \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{2 \sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{3}}\) | \(115\) |
default | \(-\frac {2}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2} x^{\frac {3}{2}}}+\frac {4 b}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \sqrt {x}}+\frac {2 b^{2} \left (\frac {\sqrt {x}}{2 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}+\frac {5 \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{2 \sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{3}}\) | \(115\) |
risch | \(\text {Expression too large to display}\) | \(4100\) |
-2/3/(arctanh(tanh(b*x+a))-b*x)^2/x^(3/2)+4/(arctanh(tanh(b*x+a))-b*x)^3*b /x^(1/2)+2/(arctanh(tanh(b*x+a))-b*x)^3*b^2*(1/2*x^(1/2)/arctanh(tanh(b*x+ a))+5/2/((arctanh(tanh(b*x+a))-b*x)*b)^(1/2)*arctan(b*x^(1/2)/((arctanh(ta nh(b*x+a))-b*x)*b)^(1/2)))
Time = 0.27 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.27 \[ \int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))^2} \, dx=\left [\frac {15 \, {\left (b^{2} x^{3} + a b x^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x + 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) + 2 \, {\left (15 \, b^{2} x^{2} + 10 \, a b x - 2 \, a^{2}\right )} \sqrt {x}}{6 \, {\left (a^{3} b x^{3} + a^{4} x^{2}\right )}}, -\frac {15 \, {\left (b^{2} x^{3} + a b x^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b \sqrt {x}}\right ) - {\left (15 \, b^{2} x^{2} + 10 \, a b x - 2 \, a^{2}\right )} \sqrt {x}}{3 \, {\left (a^{3} b x^{3} + a^{4} x^{2}\right )}}\right ] \]
[1/6*(15*(b^2*x^3 + a*b*x^2)*sqrt(-b/a)*log((b*x + 2*a*sqrt(x)*sqrt(-b/a) - a)/(b*x + a)) + 2*(15*b^2*x^2 + 10*a*b*x - 2*a^2)*sqrt(x))/(a^3*b*x^3 + a^4*x^2), -1/3*(15*(b^2*x^3 + a*b*x^2)*sqrt(b/a)*arctan(a*sqrt(b/a)/(b*sqr t(x))) - (15*b^2*x^2 + 10*a*b*x - 2*a^2)*sqrt(x))/(a^3*b*x^3 + a^4*x^2)]
\[ \int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))^2} \, dx=\int \frac {1}{x^{\frac {5}{2}} \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]
Time = 0.30 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.44 \[ \int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))^2} \, dx=\frac {15 \, b^{2} x^{2} + 10 \, a b x - 2 \, a^{2}}{3 \, {\left (a^{3} b x^{\frac {5}{2}} + a^{4} x^{\frac {3}{2}}\right )}} + \frac {5 \, b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} \]
1/3*(15*b^2*x^2 + 10*a*b*x - 2*a^2)/(a^3*b*x^(5/2) + a^4*x^(3/2)) + 5*b^2* arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^3)
Time = 0.27 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.40 \[ \int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))^2} \, dx=\frac {5 \, b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} + \frac {b^{2} \sqrt {x}}{{\left (b x + a\right )} a^{3}} + \frac {2 \, {\left (6 \, b x - a\right )}}{3 \, a^{3} x^{\frac {3}{2}}} \]
5*b^2*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^3) + b^2*sqrt(x)/((b*x + a) *a^3) + 2/3*(6*b*x - a)/(a^3*x^(3/2))
Time = 5.39 (sec) , antiderivative size = 871, normalized size of antiderivative = 6.01 \[ \int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))^2} \, dx=\text {Too large to display} \]
((32*b)/(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(e xp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^2 - (80*b^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) ^3)/(x^(1/2)*(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)))) - 8/(3*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)^2 ) + (20*2^(1/2)*b^(3/2)*log(-(b^(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^( 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) + 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)*((2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*ex p(2*b*x) + 1)) + log(2/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^6 + 60*a^2*(2*a - 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)) + 2*b*x)^4 - 160*a^3*(2*a - 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)) + 2*b*x)^3 + 240*a^4*(2*a - 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)) + 2*b*x)^2 + 64*a^6 - 12*a*(2*a - 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)) + 2*b*x)^5 - 192*a^5*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2...