Integrand size = 15, antiderivative size = 221 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^6} \, dx=\frac {3 b^5 \arctan \left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right )}{128 (b x-\text {arctanh}(\tanh (a+b x)))^{5/2}}+\frac {b^4}{128 x \text {arctanh}(\tanh (a+b x))^{3/2}}-\frac {b^5}{128 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{3/2}}-\frac {b^3}{64 x^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}+\frac {3 b^5}{128 (b x-\text {arctanh}(\tanh (a+b x)))^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}-\frac {b^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}{16 x^3}-\frac {b \text {arctanh}(\tanh (a+b x))^{3/2}}{8 x^4}-\frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{5 x^5} \]
3/128*b^5*arctan(arctanh(tanh(b*x+a))^(1/2)/(b*x-arctanh(tanh(b*x+a)))^(1/ 2))/(b*x-arctanh(tanh(b*x+a)))^(5/2)+1/128*b^4/x/arctanh(tanh(b*x+a))^(3/2 )-1/128*b^5/(b*x-arctanh(tanh(b*x+a)))/arctanh(tanh(b*x+a))^(3/2)-1/8*b*ar ctanh(tanh(b*x+a))^(3/2)/x^4-1/5*arctanh(tanh(b*x+a))^(5/2)/x^5-1/64*b^3/x ^2/arctanh(tanh(b*x+a))^(1/2)+3/128*b^5/(b*x-arctanh(tanh(b*x+a)))^2/arcta nh(tanh(b*x+a))^(1/2)-1/16*b^2*arctanh(tanh(b*x+a))^(1/2)/x^3
Time = 0.09 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.68 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^6} \, dx=\frac {1}{640} \left (-\frac {15 b^5 \text {arctanh}\left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {-b x+\text {arctanh}(\tanh (a+b x))}}\right )}{(-b x+\text {arctanh}(\tanh (a+b x)))^{5/2}}-\frac {\sqrt {\text {arctanh}(\tanh (a+b x))} \left (15 b^4 x^4+10 b^3 x^3 \text {arctanh}(\tanh (a+b x))+8 b^2 x^2 \text {arctanh}(\tanh (a+b x))^2-176 b x \text {arctanh}(\tanh (a+b x))^3+128 \text {arctanh}(\tanh (a+b x))^4\right )}{x^5 (-b x+\text {arctanh}(\tanh (a+b x)))^2}\right ) \]
((-15*b^5*ArcTanh[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[-(b*x) + ArcTanh[Tanh[ a + b*x]]]])/(-(b*x) + ArcTanh[Tanh[a + b*x]])^(5/2) - (Sqrt[ArcTanh[Tanh[ a + b*x]]]*(15*b^4*x^4 + 10*b^3*x^3*ArcTanh[Tanh[a + b*x]] + 8*b^2*x^2*Arc Tanh[Tanh[a + b*x]]^2 - 176*b*x*ArcTanh[Tanh[a + b*x]]^3 + 128*ArcTanh[Tan h[a + b*x]]^4))/(x^5*(-(b*x) + ArcTanh[Tanh[a + b*x]])^2))/640
Time = 0.57 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.10, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {2599, 2599, 2599, 2599, 2599, 2594, 2594, 2592}
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.
\(\displaystyle \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^6} \, dx\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {1}{2} b \int \frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{x^5}dx-\frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{5 x^5}\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {1}{2} b \left (\frac {3}{8} b \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^4}dx-\frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{4 x^4}\right )-\frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{5 x^5}\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {1}{2} b \left (\frac {3}{8} b \left (\frac {1}{6} b \int \frac {1}{x^3 \sqrt {\text {arctanh}(\tanh (a+b x))}}dx-\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{3 x^3}\right )-\frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{4 x^4}\right )-\frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{5 x^5}\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {1}{2} b \left (\frac {3}{8} b \left (\frac {1}{6} b \left (-\frac {1}{4} b \int \frac {1}{x^2 \text {arctanh}(\tanh (a+b x))^{3/2}}dx-\frac {1}{2 x^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}\right )-\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{3 x^3}\right )-\frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{4 x^4}\right )-\frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{5 x^5}\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {1}{2} b \left (\frac {3}{8} b \left (\frac {1}{6} b \left (-\frac {1}{4} b \left (-\frac {3}{2} b \int \frac {1}{x \text {arctanh}(\tanh (a+b x))^{5/2}}dx-\frac {1}{x \text {arctanh}(\tanh (a+b x))^{3/2}}\right )-\frac {1}{2 x^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}\right )-\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{3 x^3}\right )-\frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{4 x^4}\right )-\frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{5 x^5}\) |
\(\Big \downarrow \) 2594 |
\(\displaystyle \frac {1}{2} b \left (\frac {3}{8} b \left (\frac {1}{6} b \left (-\frac {1}{4} b \left (-\frac {3}{2} b \left (-\frac {\int \frac {1}{x \text {arctanh}(\tanh (a+b x))^{3/2}}dx}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{3 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{3/2}}\right )-\frac {1}{x \text {arctanh}(\tanh (a+b x))^{3/2}}\right )-\frac {1}{2 x^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}\right )-\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{3 x^3}\right )-\frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{4 x^4}\right )-\frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{5 x^5}\) |
\(\Big \downarrow \) 2594 |
\(\displaystyle \frac {1}{2} b \left (\frac {3}{8} b \left (\frac {1}{6} b \left (-\frac {1}{4} b \left (-\frac {3}{2} b \left (-\frac {-\frac {\int \frac {1}{x \sqrt {\text {arctanh}(\tanh (a+b x))}}dx}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{(b x-\text {arctanh}(\tanh (a+b x))) \sqrt {\text {arctanh}(\tanh (a+b x))}}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{3 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{3/2}}\right )-\frac {1}{x \text {arctanh}(\tanh (a+b x))^{3/2}}\right )-\frac {1}{2 x^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}\right )-\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{3 x^3}\right )-\frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{4 x^4}\right )-\frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{5 x^5}\) |
\(\Big \downarrow \) 2592 |
\(\displaystyle \frac {1}{2} b \left (\frac {3}{8} b \left (\frac {1}{6} b \left (-\frac {1}{4} b \left (-\frac {3}{2} b \left (-\frac {-\frac {2 \arctan \left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right )}{(b x-\text {arctanh}(\tanh (a+b x)))^{3/2}}-\frac {2}{(b x-\text {arctanh}(\tanh (a+b x))) \sqrt {\text {arctanh}(\tanh (a+b x))}}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{3 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{3/2}}\right )-\frac {1}{x \text {arctanh}(\tanh (a+b x))^{3/2}}\right )-\frac {1}{2 x^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}\right )-\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{3 x^3}\right )-\frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{4 x^4}\right )-\frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{5 x^5}\) |
-1/5*ArcTanh[Tanh[a + b*x]]^(5/2)/x^5 + (b*((3*b*((b*(-1/4*(b*((-3*b*(-((( -2*ArcTan[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]] )/(b*x - ArcTanh[Tanh[a + b*x]])^(3/2) - 2/((b*x - ArcTanh[Tanh[a + b*x]]) *Sqrt[ArcTanh[Tanh[a + b*x]]]))/(b*x - ArcTanh[Tanh[a + b*x]])) - 2/(3*(b* x - ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a + b*x]]^(3/2))))/2 - 1/(x*ArcTa nh[Tanh[a + b*x]]^(3/2)))) - 1/(2*x^2*Sqrt[ArcTanh[Tanh[a + b*x]]])))/6 - Sqrt[ArcTanh[Tanh[a + b*x]]]/(3*x^3)))/8 - ArcTanh[Tanh[a + b*x]]^(3/2)/(4 *x^4)))/2
3.2.39.3.1 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*(ArcTan[Sqrt[v]/Rt[(b*u - a*v)/a, 2]]/(a*Rt[(b*u - a*v )/a, 2])), x] /; NeQ[b*u - a*v, 0] && PosQ[(b*u - a*v)/a]] /; PiecewiseLine arQ[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 = 1.23 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.19
method | result | size |
default | \(2 b^{5} \left (\frac {\frac {3 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {9}{2}}}{256 \left (a^{2}+2 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )+\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}\right )}-\frac {7 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{128 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )}-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{10}+\left (\frac {7 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}{128}-\frac {7 b x}{128}\right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}+\left (-\frac {3 a^{2}}{256}-\frac {3 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )}{128}-\frac {3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{256}\right ) \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{x^{5} b^{5}}-\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}}\right )}{256 \left (a^{2}+2 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )+\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}\right ) \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}}\right )\) | \(262\) |
2*b^5*((3/256/(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))^(9/2)-7/128/(arctanh(tanh(b*x+a))-b*x)*arct anh(tanh(b*x+a))^(7/2)-1/10*arctanh(tanh(b*x+a))^(5/2)+(7/128*arctanh(tanh (b*x+a))-7/128*b*x)*arctanh(tanh(b*x+a))^(3/2)+(-3/256*a^2-3/128*a*(arctan h(tanh(b*x+a))-b*x-a)-3/256*(arctanh(tanh(b*x+a))-b*x-a)^2)*arctanh(tanh(b *x+a))^(1/2))/x^5/b^5-3/256/(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)^(1/2)*arctanh(arctanh(t anh(b*x+a))^(1/2)/(arctanh(tanh(b*x+a))-b*x)^(1/2)))
Time = 0.26 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.86 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^6} \, dx=\left [\frac {15 \, \sqrt {a} b^{5} x^{5} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (15 \, a b^{4} x^{4} - 10 \, a^{2} b^{3} x^{3} - 248 \, a^{3} b^{2} x^{2} - 336 \, a^{4} b x - 128 \, a^{5}\right )} \sqrt {b x + a}}{1280 \, a^{3} x^{5}}, \frac {15 \, \sqrt {-a} b^{5} x^{5} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (15 \, a b^{4} x^{4} - 10 \, a^{2} b^{3} x^{3} - 248 \, a^{3} b^{2} x^{2} - 336 \, a^{4} b x - 128 \, a^{5}\right )} \sqrt {b x + a}}{640 \, a^{3} x^{5}}\right ] \]
[1/1280*(15*sqrt(a)*b^5*x^5*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(15*a*b^4*x^4 - 10*a^2*b^3*x^3 - 248*a^3*b^2*x^2 - 336*a^4*b*x - 128*a^ 5)*sqrt(b*x + a))/(a^3*x^5), 1/640*(15*sqrt(-a)*b^5*x^5*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (15*a*b^4*x^4 - 10*a^2*b^3*x^3 - 248*a^3*b^2*x^2 - 336*a^ 4*b*x - 128*a^5)*sqrt(b*x + a))/(a^3*x^5)]
\[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^6} \, dx=\int \frac {\operatorname {atanh}^{\frac {5}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}{x^{6}}\, dx \]
\[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^6} \, dx=\int { \frac {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{\frac {5}{2}}}{x^{6}} \,d x } \]
Time = 0.27 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.56 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^6} \, dx=\frac {\sqrt {2} {\left (\frac {15 \, \sqrt {2} b^{6} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {\sqrt {2} {\left (15 \, {\left (b x + a\right )}^{\frac {9}{2}} b^{6} - 70 \, {\left (b x + a\right )}^{\frac {7}{2}} a b^{6} - 128 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} b^{6} + 70 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} b^{6} - 15 \, \sqrt {b x + a} a^{4} b^{6}\right )}}{a^{2} b^{5} x^{5}}\right )}}{1280 \, b} \]
1/1280*sqrt(2)*(15*sqrt(2)*b^6*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^ 2) + sqrt(2)*(15*(b*x + a)^(9/2)*b^6 - 70*(b*x + a)^(7/2)*a*b^6 - 128*(b*x + a)^(5/2)*a^2*b^6 + 70*(b*x + a)^(3/2)*a^3*b^6 - 15*sqrt(b*x + a)*a^4*b^ 6)/(a^2*b^5*x^5))/b
Time = 9.72 (sec) , antiderivative size = 1292, normalized size of antiderivative = 5.85 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^6} \, dx=\text {Too large to display} \]
(3*b^4*(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))/(32*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)^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)*(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)/(4*x^5*(5*log(2/( exp(2*a)*exp(2*b*x) + 1)) - 5*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2* b*x) + 1)) + 10*b*x)) + (b^3*(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*(2*log(2 /(exp(2*a)*exp(2*b*x) + 1)) - 2*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp( 2*b*x) + 1)) + 4*b*x)) + (2^(1/2)*b^5*log((((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)*(l og(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*ex p(2*b*x) + 1)) + 2*b*x)^(1/2)*2i - 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) + 2^( 1/2)*b*x)*((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 + 40*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)^3 - 80*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)^2 - 32*a^5 - 10*...