Integrand size = 15, antiderivative size = 212 \[ \int \frac {1}{x^4 \sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=\frac {5 b^3 \arctan \left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right )}{8 (b x-\text {arctanh}(\tanh (a+b x)))^{7/2}}-\frac {b^2}{8 x \text {arctanh}(\tanh (a+b x))^{5/2}}+\frac {b^3}{8 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{5/2}}+\frac {b}{12 x^2 \text {arctanh}(\tanh (a+b x))^{3/2}}-\frac {5 b^3}{24 (b x-\text {arctanh}(\tanh (a+b x)))^2 \text {arctanh}(\tanh (a+b x))^{3/2}}-\frac {1}{3 x^3 \sqrt {\text {arctanh}(\tanh (a+b x))}}+\frac {5 b^3}{8 (b x-\text {arctanh}(\tanh (a+b x)))^3 \sqrt {\text {arctanh}(\tanh (a+b x))}} \]
5/8*b^3*arctan(arctanh(tanh(b*x+a))^(1/2)/(b*x-arctanh(tanh(b*x+a)))^(1/2) )/(b*x-arctanh(tanh(b*x+a)))^(7/2)-1/8*b^2/x/arctanh(tanh(b*x+a))^(5/2)+1/ 8*b^3/(b*x-arctanh(tanh(b*x+a)))/arctanh(tanh(b*x+a))^(5/2)+1/12*b/x^2/arc tanh(tanh(b*x+a))^(3/2)-5/24*b^3/(b*x-arctanh(tanh(b*x+a)))^2/arctanh(tanh (b*x+a))^(3/2)-1/3/x^3/arctanh(tanh(b*x+a))^(1/2)+5/8*b^3/(b*x-arctanh(tan h(b*x+a)))^3/arctanh(tanh(b*x+a))^(1/2)
Time = 0.08 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.55 \[ \int \frac {1}{x^4 \sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=\frac {5 b^3 \text {arctanh}\left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {-b x+\text {arctanh}(\tanh (a+b x))}}\right )}{8 (-b x+\text {arctanh}(\tanh (a+b x)))^{7/2}}+\frac {\sqrt {\text {arctanh}(\tanh (a+b x))} \left (33 b^2 x^2-26 b x \text {arctanh}(\tanh (a+b x))+8 \text {arctanh}(\tanh (a+b x))^2\right )}{24 x^3 (b x-\text {arctanh}(\tanh (a+b x)))^3} \]
(5*b^3*ArcTanh[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]])/(8*(-(b*x) + ArcTanh[Tanh[a + b*x]])^(7/2)) + (Sqrt[ArcTanh[Tanh [a + b*x]]]*(33*b^2*x^2 - 26*b*x*ArcTanh[Tanh[a + b*x]] + 8*ArcTanh[Tanh[a + b*x]]^2))/(24*x^3*(b*x - ArcTanh[Tanh[a + b*x]])^3)
Time = 0.56 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.15, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {2599, 2599, 2599, 2594, 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.
| \(\displaystyle \int \frac {1}{x^4 \sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx\) |
| \(\Big \downarrow \) 2599 |
| \(\displaystyle -\frac {1}{6} b \int \frac {1}{x^3 \text {arctanh}(\tanh (a+b x))^{3/2}}dx-\frac {1}{3 x^3 \sqrt {\text {arctanh}(\tanh (a+b x))}}\) |
| \(\Big \downarrow \) 2599 |
| \(\displaystyle -\frac {1}{6} b \left (-\frac {3}{4} b \int \frac {1}{x^2 \text {arctanh}(\tanh (a+b x))^{5/2}}dx-\frac {1}{2 x^2 \text {arctanh}(\tanh (a+b x))^{3/2}}\right )-\frac {1}{3 x^3 \sqrt {\text {arctanh}(\tanh (a+b x))}}\) |
| \(\Big \downarrow \) 2599 |
| \(\displaystyle -\frac {1}{6} b \left (-\frac {3}{4} b \left (-\frac {5}{2} b \int \frac {1}{x \text {arctanh}(\tanh (a+b x))^{7/2}}dx-\frac {1}{x \text {arctanh}(\tanh (a+b x))^{5/2}}\right )-\frac {1}{2 x^2 \text {arctanh}(\tanh (a+b x))^{3/2}}\right )-\frac {1}{3 x^3 \sqrt {\text {arctanh}(\tanh (a+b x))}}\) |
| \(\Big \downarrow \) 2594 |
| \(\displaystyle -\frac {1}{6} b \left (-\frac {3}{4} b \left (-\frac {5}{2} b \left (-\frac {\int \frac {1}{x \text {arctanh}(\tanh (a+b x))^{5/2}}dx}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{5 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{5/2}}\right )-\frac {1}{x \text {arctanh}(\tanh (a+b x))^{5/2}}\right )-\frac {1}{2 x^2 \text {arctanh}(\tanh (a+b x))^{3/2}}\right )-\frac {1}{3 x^3 \sqrt {\text {arctanh}(\tanh (a+b x))}}\) |
| \(\Big \downarrow \) 2594 |
| \(\displaystyle -\frac {1}{6} b \left (-\frac {3}{4} b \left (-\frac {5}{2} b \left (-\frac {-\frac {\int \frac {1}{x \text {arctanh}(\tanh (a+b x))^{3/2}}dx}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{3 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{3/2}}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{5 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{5/2}}\right )-\frac {1}{x \text {arctanh}(\tanh (a+b x))^{5/2}}\right )-\frac {1}{2 x^2 \text {arctanh}(\tanh (a+b x))^{3/2}}\right )-\frac {1}{3 x^3 \sqrt {\text {arctanh}(\tanh (a+b x))}}\) |
| \(\Big \downarrow \) 2594 |
| \(\displaystyle -\frac {1}{6} b \left (-\frac {3}{4} b \left (-\frac {5}{2} b \left (-\frac {-\frac {-\frac {\int \frac {1}{x \sqrt {\text {arctanh}(\tanh (a+b x))}}dx}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{(b x-\text {arctanh}(\tanh (a+b x))) \sqrt {\text {arctanh}(\tanh (a+b x))}}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{3 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{3/2}}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{5 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{5/2}}\right )-\frac {1}{x \text {arctanh}(\tanh (a+b x))^{5/2}}\right )-\frac {1}{2 x^2 \text {arctanh}(\tanh (a+b x))^{3/2}}\right )-\frac {1}{3 x^3 \sqrt {\text {arctanh}(\tanh (a+b x))}}\) |
| \(\Big \downarrow \) 2592 |
| \(\displaystyle -\frac {1}{6} b \left (-\frac {3}{4} b \left (-\frac {5}{2} b \left (-\frac {-\frac {-\frac {2 \arctan \left (\frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right )}{(b x-\text {arctanh}(\tanh (a+b x)))^{3/2}}-\frac {2}{(b x-\text {arctanh}(\tanh (a+b x))) \sqrt {\text {arctanh}(\tanh (a+b x))}}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{3 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{3/2}}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {2}{5 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^{5/2}}\right )-\frac {1}{x \text {arctanh}(\tanh (a+b x))^{5/2}}\right )-\frac {1}{2 x^2 \text {arctanh}(\tanh (a+b x))^{3/2}}\right )-\frac {1}{3 x^3 \sqrt {\text {arctanh}(\tanh (a+b x))}}\) |
-1/6*(b*((-3*b*((-5*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 - ArcTa nh[Tanh[a + b*x]])) - 2/(3*(b*x - ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a + b*x]]^(3/2)))/(b*x - ArcTanh[Tanh[a + b*x]])) - 2/(5*(b*x - ArcTanh[Tanh[ a + b*x]])*ArcTanh[Tanh[a + b*x]]^(5/2))))/2 - 1/(x*ArcTanh[Tanh[a + b*x]] ^(5/2))))/4 - 1/(2*x^2*ArcTanh[Tanh[a + b*x]]^(3/2)))) - 1/(3*x^3*Sqrt[Arc Tanh[Tanh[a + b*x]]])
3.2.48.3.1 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 = 0.10 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.94
| method | result | size |
| default | \(2 b^{3} \left (\frac {2 \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{3 \left (-4 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )+4 b x \right ) x^{3} b^{3}}+\frac {\frac {10 \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{3 \left (-4 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )+4 b x \right ) x^{2} b^{2}}+\frac {10 \left (\frac {6 \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{\left (-4 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )+4 b x \right ) x b}-\frac {6 \,\operatorname {arctanh}\left (\frac {\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}}\right )}{\left (-4 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )+4 b x \right ) \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}}\right )}{3 \left (-4 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )+4 b x \right )}}{-4 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )+4 b x}\right )\) | \(200\) |
2*b^3*(2/3*arctanh(tanh(b*x+a))^(1/2)/(-4*arctanh(tanh(b*x+a))+4*b*x)/x^3/ b^3+10/3/(-4*arctanh(tanh(b*x+a))+4*b*x)*(arctanh(tanh(b*x+a))^(1/2)/(-4*a rctanh(tanh(b*x+a))+4*b*x)/x^2/b^2+3/(-4*arctanh(tanh(b*x+a))+4*b*x)*(2*ar ctanh(tanh(b*x+a))^(1/2)/(-4*arctanh(tanh(b*x+a))+4*b*x)/x/b-2/(-4*arctanh (tanh(b*x+a))+4*b*x)/(arctanh(tanh(b*x+a))-b*x)^(1/2)*arctanh(arctanh(tanh (b*x+a))^(1/2)/(arctanh(tanh(b*x+a))-b*x)^(1/2)))))
Time = 0.25 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.68 \[ \int \frac {1}{x^4 \sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=\left [\frac {15 \, \sqrt {a} b^{3} x^{3} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) - 2 \, {\left (15 \, a b^{2} x^{2} - 10 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt {b x + a}}{48 \, a^{4} x^{3}}, -\frac {15 \, \sqrt {-a} b^{3} x^{3} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (15 \, a b^{2} x^{2} - 10 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt {b x + a}}{24 \, a^{4} x^{3}}\right ] \]
[1/48*(15*sqrt(a)*b^3*x^3*log((b*x + 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) - 2 *(15*a*b^2*x^2 - 10*a^2*b*x + 8*a^3)*sqrt(b*x + a))/(a^4*x^3), -1/24*(15*s qrt(-a)*b^3*x^3*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (15*a*b^2*x^2 - 10*a^2* b*x + 8*a^3)*sqrt(b*x + a))/(a^4*x^3)]
\[ \int \frac {1}{x^4 \sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=\int \frac {1}{x^{4} \sqrt {\operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}}\, dx \]
\[ \int \frac {1}{x^4 \sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=\int { \frac {1}{x^{4} \sqrt {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )}} \,d x } \]
Time = 0.26 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.40 \[ \int \frac {1}{x^4 \sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=-\frac {\frac {15 \, b^{4} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{3}} + \frac {15 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{4} - 40 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{4} + 33 \, \sqrt {b x + a} a^{2} b^{4}}{a^{3} b^{3} x^{3}}}{24 \, b} \]
-1/24*(15*b^4*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^3) + (15*(b*x + a )^(5/2)*b^4 - 40*(b*x + a)^(3/2)*a*b^4 + 33*sqrt(b*x + a)*a^2*b^4)/(a^3*b^ 3*x^3))/b
Time = 8.10 (sec) , antiderivative size = 1086, normalized size of antiderivative = 5.12 \[ \int \frac {1}{x^4 \sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=\text {Too large to display} \]
(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))/(x^3*(3*log(2/(exp(2*a)*exp(2*b*x) + 1)) - 3*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 6*b*x)) + (5*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))/(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) + (2^(1/2) *b^3*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)*(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)*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)^7 + 84*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)^5 - 280*a^3*(2*a - lo g((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 + 560*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)^3 - 672 *a^5*(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 - 128*a^7 - 14*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...