Integrand size = 15, antiderivative size = 176 \[ \int \frac {1}{x^{3/2} \text {arctanh}(\tanh (a+b x))^3} \, dx=-\frac {15 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right )}{4 (b x-\text {arctanh}(\tanh (a+b x)))^{7/2}}+\frac {15}{4 \sqrt {x} (b x-\text {arctanh}(\tanh (a+b x)))^3}+\frac {5}{4 b x^{3/2} (b x-\text {arctanh}(\tanh (a+b x)))^2}+\frac {3}{4 b^2 x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))}-\frac {1}{2 b x^{3/2} \text {arctanh}(\tanh (a+b x))^2}+\frac {3}{4 b^2 x^{5/2} \text {arctanh}(\tanh (a+b x))} \]
5/4/b/x^(3/2)/(b*x-arctanh(tanh(b*x+a)))^2+3/4/b^2/x^(5/2)/(b*x-arctanh(ta nh(b*x+a)))-1/2/b/x^(3/2)/arctanh(tanh(b*x+a))^2+3/4/b^2/x^(5/2)/arctanh(t anh(b*x+a))-15/4*arctanh(b^(1/2)*x^(1/2)/(b*x-arctanh(tanh(b*x+a)))^(1/2)) *b^(1/2)/(b*x-arctanh(tanh(b*x+a)))^(7/2)+15/4/(b*x-arctanh(tanh(b*x+a)))^ 3/x^(1/2)
Time = 0.12 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.80 \[ \int \frac {1}{x^{3/2} \text {arctanh}(\tanh (a+b x))^3} \, dx=-\frac {15 \sqrt {b} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {-b x+\text {arctanh}(\tanh (a+b x))}}\right )}{4 (-b x+\text {arctanh}(\tanh (a+b x)))^{7/2}}-\frac {2}{\sqrt {x} (-b x+\text {arctanh}(\tanh (a+b x)))^3}-\frac {7 b \sqrt {x}}{4 \text {arctanh}(\tanh (a+b x)) (-b x+\text {arctanh}(\tanh (a+b x)))^3}-\frac {b \sqrt {x}}{2 \text {arctanh}(\tanh (a+b x))^2 (-b x+\text {arctanh}(\tanh (a+b x)))^2} \]
(-15*Sqrt[b]*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]] ]])/(4*(-(b*x) + ArcTanh[Tanh[a + b*x]])^(7/2)) - 2/(Sqrt[x]*(-(b*x) + Arc Tanh[Tanh[a + b*x]])^3) - (7*b*Sqrt[x])/(4*ArcTanh[Tanh[a + b*x]]*(-(b*x) + ArcTanh[Tanh[a + b*x]])^3) - (b*Sqrt[x])/(2*ArcTanh[Tanh[a + b*x]]^2*(-( b*x) + ArcTanh[Tanh[a + b*x]])^2)
Time = 0.49 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.23, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2599, 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.
| \(\displaystyle \int \frac {1}{x^{3/2} \text {arctanh}(\tanh (a+b x))^3} \, dx\) |
| \(\Big \downarrow \) 2599 |
| \(\displaystyle -\frac {3 \int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))^2}dx}{4 b}-\frac {1}{2 b x^{3/2} \text {arctanh}(\tanh (a+b x))^2}\) |
| \(\Big \downarrow \) 2599 |
| \(\displaystyle -\frac {3 \left (-\frac {5 \int \frac {1}{x^{7/2} \text {arctanh}(\tanh (a+b x))}dx}{2 b}-\frac {1}{b x^{5/2} \text {arctanh}(\tanh (a+b x))}\right )}{4 b}-\frac {1}{2 b x^{3/2} \text {arctanh}(\tanh (a+b x))^2}\) |
| \(\Big \downarrow \) 2594 |
| \(\displaystyle -\frac {3 \left (-\frac {5 \left (\frac {b \int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))}dx}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{5 x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{2 b}-\frac {1}{b x^{5/2} \text {arctanh}(\tanh (a+b x))}\right )}{4 b}-\frac {1}{2 b x^{3/2} \text {arctanh}(\tanh (a+b x))^2}\) |
| \(\Big \downarrow \) 2594 |
| \(\displaystyle -\frac {3 \left (-\frac {5 \left (\frac {b \left (\frac {b \int \frac {1}{x^{3/2} \text {arctanh}(\tanh (a+b x))}dx}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{3 x^{3/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{5 x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{2 b}-\frac {1}{b x^{5/2} \text {arctanh}(\tanh (a+b x))}\right )}{4 b}-\frac {1}{2 b x^{3/2} \text {arctanh}(\tanh (a+b x))^2}\) |
| \(\Big \downarrow \) 2594 |
| \(\displaystyle -\frac {3 \left (-\frac {5 \left (\frac {b \left (\frac {b \left (\frac {b \int \frac {1}{\sqrt {x} \text {arctanh}(\tanh (a+b x))}dx}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{\sqrt {x} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{3 x^{3/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{5 x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{2 b}-\frac {1}{b x^{5/2} \text {arctanh}(\tanh (a+b x))}\right )}{4 b}-\frac {1}{2 b x^{3/2} \text {arctanh}(\tanh (a+b x))^2}\) |
| \(\Big \downarrow \) 2593 |
| \(\displaystyle -\frac {3 \left (-\frac {5 \left (\frac {b \left (\frac {2}{3 x^{3/2} (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {b \left (\frac {2}{\sqrt {x} (b x-\text {arctanh}(\tanh (a+b x)))}-\frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right )}{(b x-\text {arctanh}(\tanh (a+b x)))^{3/2}}\right )}{b x-\text {arctanh}(\tanh (a+b x))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{5 x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{2 b}-\frac {1}{b x^{5/2} \text {arctanh}(\tanh (a+b x))}\right )}{4 b}-\frac {1}{2 b x^{3/2} \text {arctanh}(\tanh (a+b x))^2}\) |
(-3*((-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] )/Sqrt[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]])))/(4*b) - 1/(2*b*x^(3/2)*ArcTanh[Tanh[a + b*x]]^2)
3.3.12.3.1 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 = 0.62 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.65
| method | result | size |
| derivativedivides | \(-\frac {2}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \sqrt {x}}-\frac {2 b \left (\frac {\frac {7 b \,x^{\frac {3}{2}}}{8}+\left (\frac {9 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}{8}-\frac {9 b x}{8}\right ) \sqrt {x}}{\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2}}+\frac {15 \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{8 \sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{3}}\) | \(114\) |
| default | \(-\frac {2}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \sqrt {x}}-\frac {2 b \left (\frac {\frac {7 b \,x^{\frac {3}{2}}}{8}+\left (\frac {9 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}{8}-\frac {9 b x}{8}\right ) \sqrt {x}}{\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2}}+\frac {15 \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{8 \sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{3}}\) | \(114\) |
| risch | \(\text {Expression too large to display}\) | \(4370\) |
-2/(arctanh(tanh(b*x+a))-b*x)^3/x^(1/2)-2/(arctanh(tanh(b*x+a))-b*x)^3*b*( (7/8*b*x^(3/2)+(9/8*arctanh(tanh(b*x+a))-9/8*b*x)*x^(1/2))/arctanh(tanh(b* x+a))^2+15/8/((arctanh(tanh(b*x+a))-b*x)*b)^(1/2)*arctan(b*x^(1/2)/((arcta nh(tanh(b*x+a))-b*x)*b)^(1/2)))
Time = 0.26 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.22 \[ \int \frac {1}{x^{3/2} \text {arctanh}(\tanh (a+b x))^3} \, dx=\left [\frac {15 \, {\left (b^{2} x^{3} + 2 \, a b x^{2} + a^{2} x\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x - 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) - 2 \, {\left (15 \, b^{2} x^{2} + 25 \, a b x + 8 \, a^{2}\right )} \sqrt {x}}{8 \, {\left (a^{3} b^{2} x^{3} + 2 \, a^{4} b x^{2} + a^{5} x\right )}}, \frac {15 \, {\left (b^{2} x^{3} + 2 \, a b x^{2} + a^{2} x\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b \sqrt {x}}\right ) - {\left (15 \, b^{2} x^{2} + 25 \, a b x + 8 \, a^{2}\right )} \sqrt {x}}{4 \, {\left (a^{3} b^{2} x^{3} + 2 \, a^{4} b x^{2} + a^{5} x\right )}}\right ] \]
[1/8*(15*(b^2*x^3 + 2*a*b*x^2 + a^2*x)*sqrt(-b/a)*log((b*x - 2*a*sqrt(x)*s qrt(-b/a) - a)/(b*x + a)) - 2*(15*b^2*x^2 + 25*a*b*x + 8*a^2)*sqrt(x))/(a^ 3*b^2*x^3 + 2*a^4*b*x^2 + a^5*x), 1/4*(15*(b^2*x^3 + 2*a*b*x^2 + a^2*x)*sq rt(b/a)*arctan(a*sqrt(b/a)/(b*sqrt(x))) - (15*b^2*x^2 + 25*a*b*x + 8*a^2)* sqrt(x))/(a^3*b^2*x^3 + 2*a^4*b*x^2 + a^5*x)]
\[ \int \frac {1}{x^{3/2} \text {arctanh}(\tanh (a+b x))^3} \, dx=\int \frac {1}{x^{\frac {3}{2}} \operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]
Time = 0.31 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.41 \[ \int \frac {1}{x^{3/2} \text {arctanh}(\tanh (a+b x))^3} \, dx=-\frac {15 \, b^{2} x^{2} + 25 \, a b x + 8 \, a^{2}}{4 \, {\left (a^{3} b^{2} x^{\frac {5}{2}} + 2 \, a^{4} b x^{\frac {3}{2}} + a^{5} \sqrt {x}\right )}} - \frac {15 \, b \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{3}} \]
-1/4*(15*b^2*x^2 + 25*a*b*x + 8*a^2)/(a^3*b^2*x^(5/2) + 2*a^4*b*x^(3/2) + a^5*sqrt(x)) - 15/4*b*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^3)
Time = 0.29 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.34 \[ \int \frac {1}{x^{3/2} \text {arctanh}(\tanh (a+b x))^3} \, dx=-\frac {15 \, b \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{3}} - \frac {2}{a^{3} \sqrt {x}} - \frac {7 \, b^{2} x^{\frac {3}{2}} + 9 \, a b \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} a^{3}} \]
-15/4*b*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^3) - 2/(a^3*sqrt(x)) - 1/ 4*(7*b^2*x^(3/2) + 9*a*b*sqrt(x))/((b*x + a)^2*a^3)
Time = 5.02 (sec) , antiderivative size = 1077, normalized size of antiderivative = 6.12 \[ \int \frac {1}{x^{3/2} \text {arctanh}(\tanh (a+b x))^3} \, dx=\text {Too large to display} \]
(x*((12*b)/(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 + (3*b*(16*log(2/(exp(2*a)*exp(2*b* x) + 1)) - 16*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 32* b*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)^4) - (16*log(2/(exp(2*a)*exp(2*b*x) + 1)) - 16*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 32*b*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)*e xp(2*b*x) + 1)) - log(2/(exp(2*a)*exp(2*b*x) + 1)))) + (15*2^(1/2)*b^(1/2) *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)*ex p(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)*exp(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...