Integrand size = 15, antiderivative size = 74 \[ \int \frac {x^3}{\text {arctanh}(\tanh (a+b x))^{3/2}} \, dx=-\frac {2 x^3}{b \sqrt {\text {arctanh}(\tanh (a+b x))}}+\frac {12 x^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}{b^2}-\frac {16 x \text {arctanh}(\tanh (a+b x))^{3/2}}{b^3}+\frac {32 \text {arctanh}(\tanh (a+b x))^{5/2}}{5 b^4} \]
-16*x*arctanh(tanh(b*x+a))^(3/2)/b^3+32/5*arctanh(tanh(b*x+a))^(5/2)/b^4-2 *x^3/b/arctanh(tanh(b*x+a))^(1/2)+12*x^2*arctanh(tanh(b*x+a))^(1/2)/b^2
Time = 0.03 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.89 \[ \int \frac {x^3}{\text {arctanh}(\tanh (a+b x))^{3/2}} \, dx=\frac {2 \left (-5 b^3 x^3+30 b^2 x^2 \text {arctanh}(\tanh (a+b x))-40 b x \text {arctanh}(\tanh (a+b x))^2+16 \text {arctanh}(\tanh (a+b x))^3\right )}{5 b^4 \sqrt {\text {arctanh}(\tanh (a+b x))}} \]
(2*(-5*b^3*x^3 + 30*b^2*x^2*ArcTanh[Tanh[a + b*x]] - 40*b*x*ArcTanh[Tanh[a + b*x]]^2 + 16*ArcTanh[Tanh[a + b*x]]^3))/(5*b^4*Sqrt[ArcTanh[Tanh[a + b* x]]])
Time = 0.29 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.19, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2599, 2599, 2599, 2588, 15}
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.
| \(\displaystyle \int \frac {x^3}{\text {arctanh}(\tanh (a+b x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 2599 |
| \(\displaystyle \frac {6 \int \frac {x^2}{\sqrt {\text {arctanh}(\tanh (a+b x))}}dx}{b}-\frac {2 x^3}{b \sqrt {\text {arctanh}(\tanh (a+b x))}}\) |
| \(\Big \downarrow \) 2599 |
| \(\displaystyle \frac {6 \left (\frac {2 x^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}{b}-\frac {4 \int x \sqrt {\text {arctanh}(\tanh (a+b x))}dx}{b}\right )}{b}-\frac {2 x^3}{b \sqrt {\text {arctanh}(\tanh (a+b x))}}\) |
| \(\Big \downarrow \) 2599 |
| \(\displaystyle \frac {6 \left (\frac {2 x^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}{b}-\frac {4 \left (\frac {2 x \text {arctanh}(\tanh (a+b x))^{3/2}}{3 b}-\frac {2 \int \text {arctanh}(\tanh (a+b x))^{3/2}dx}{3 b}\right )}{b}\right )}{b}-\frac {2 x^3}{b \sqrt {\text {arctanh}(\tanh (a+b x))}}\) |
| \(\Big \downarrow \) 2588 |
| \(\displaystyle \frac {6 \left (\frac {2 x^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}{b}-\frac {4 \left (\frac {2 x \text {arctanh}(\tanh (a+b x))^{3/2}}{3 b}-\frac {2 \int \text {arctanh}(\tanh (a+b x))^{3/2}d\text {arctanh}(\tanh (a+b x))}{3 b^2}\right )}{b}\right )}{b}-\frac {2 x^3}{b \sqrt {\text {arctanh}(\tanh (a+b x))}}\) |
| \(\Big \downarrow \) 15 |
| \(\displaystyle \frac {6 \left (\frac {2 x^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}{b}-\frac {4 \left (\frac {2 x \text {arctanh}(\tanh (a+b x))^{3/2}}{3 b}-\frac {4 \text {arctanh}(\tanh (a+b x))^{5/2}}{15 b^2}\right )}{b}\right )}{b}-\frac {2 x^3}{b \sqrt {\text {arctanh}(\tanh (a+b x))}}\) |
(-2*x^3)/(b*Sqrt[ArcTanh[Tanh[a + b*x]]]) + (6*((2*x^2*Sqrt[ArcTanh[Tanh[a + b*x]]])/b - (4*((2*x*ArcTanh[Tanh[a + b*x]]^(3/2))/(3*b) - (4*ArcTanh[T anh[a + b*x]]^(5/2))/(15*b^2)))/b))/b
3.2.50.3.1 Defintions of rubi rules used
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[(u_)^(m_.), x_Symbol] :> With[{c = Simplify[D[u, x]]}, Simp[1/c Subst [Int[x^m, x], x, u], x]] /; FreeQ[m, x] && PiecewiseLinearQ[u, 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]))
Leaf count of result is larger than twice the leaf count of optimal. \(200\) vs. \(2(64)=128\).
Time = 0.09 (sec) , antiderivative size = 201, normalized size of antiderivative = 2.72
| method | result | size |
| default | \(\frac {\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{5}-2 a \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}-2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}} \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )+6 a^{2} \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}+12 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right ) \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}+6 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2} \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}-\frac {2 \left (-a^{3}-3 a^{2} \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )-3 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}-\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}\right )}{\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}}{b^{4}}\) | \(201\) |
2/b^4*(1/5*arctanh(tanh(b*x+a))^(5/2)-a*arctanh(tanh(b*x+a))^(3/2)-arctanh (tanh(b*x+a))^(3/2)*(arctanh(tanh(b*x+a))-b*x-a)+3*a^2*arctanh(tanh(b*x+a) )^(1/2)+6*a*(arctanh(tanh(b*x+a))-b*x-a)*arctanh(tanh(b*x+a))^(1/2)+3*(arc tanh(tanh(b*x+a))-b*x-a)^2*arctanh(tanh(b*x+a))^(1/2)-(-a^3-3*a^2*(arctanh (tanh(b*x+a))-b*x-a)-3*a*(arctanh(tanh(b*x+a))-b*x-a)^2-(arctanh(tanh(b*x+ a))-b*x-a)^3)/arctanh(tanh(b*x+a))^(1/2))
Time = 0.25 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.69 \[ \int \frac {x^3}{\text {arctanh}(\tanh (a+b x))^{3/2}} \, dx=\frac {2 \, {\left (b^{3} x^{3} - 2 \, a b^{2} x^{2} + 8 \, a^{2} b x + 16 \, a^{3}\right )} \sqrt {b x + a}}{5 \, {\left (b^{5} x + a b^{4}\right )}} \]
\[ \int \frac {x^3}{\text {arctanh}(\tanh (a+b x))^{3/2}} \, dx=\int \frac {x^{3}}{\operatorname {atanh}^{\frac {3}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]
Time = 0.36 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.70 \[ \int \frac {x^3}{\text {arctanh}(\tanh (a+b x))^{3/2}} \, dx=\frac {2 \, {\left (b^{4} x^{4} - a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} + 24 \, a^{3} b x + 16 \, a^{4}\right )}}{5 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{4}} \]
Time = 0.27 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.82 \[ \int \frac {x^3}{\text {arctanh}(\tanh (a+b x))^{3/2}} \, dx=\frac {2 \, a^{3}}{\sqrt {b x + a} b^{4}} + \frac {2 \, {\left ({\left (b x + a\right )}^{\frac {5}{2}} b^{16} - 5 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{16} + 15 \, \sqrt {b x + a} a^{2} b^{16}\right )}}{5 \, b^{20}} \]
2*a^3/(sqrt(b*x + a)*b^4) + 2/5*((b*x + a)^(5/2)*b^16 - 5*(b*x + a)^(3/2)* a*b^16 + 15*sqrt(b*x + a)*a^2*b^16)/b^20
Time = 3.83 (sec) , antiderivative size = 660, normalized size of antiderivative = 8.92 \[ \int \frac {x^3}{\text {arctanh}(\tanh (a+b x))^{3/2}} \, dx=\frac {\sqrt {\frac {\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}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\left (\frac {{\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}+\frac {2\,\left (\frac {\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}{b^2}+\frac {8\,\left (\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\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 )}{5\,b^2}\right )\,\left (\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\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\,b}\right )}{b}+\frac {2\,x^2\,\sqrt {\frac {\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}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}}{5\,b^2}+\frac {x\,\left (\frac {\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}{b^2}+\frac {8\,\left (\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\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 )}{5\,b^2}\right )\,\sqrt {\frac {\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}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}}{3\,b}-\frac {\sqrt {\frac {\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}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{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 )}^3}{2\,b^4\,\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 )} \]
((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*e xp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^2/(2*b^3) + (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)/b^2 + (8*(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))/2 + b*x))/(5*b^2))*(lo g(2/(exp(2*a)*exp(2*b*x) + 1))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*e xp(2*b*x) + 1))/2 + b*x))/(3*b)))/b + (2*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))/ (5*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)/b^2 + (8*(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))/2 + b* x))/(5*b^2))*(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - l og(2/(exp(2*a)*exp(2*b*x) + 1))/2)^(1/2))/(3*b) - ((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)/(2*b^4*(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))))