Integrand size = 15, antiderivative size = 76 \[ \int \frac {x^3}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=-\frac {2 x^3}{3 b \text {arctanh}(\tanh (a+b x))^{3/2}}-\frac {4 x^2}{b^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}+\frac {16 x \sqrt {\text {arctanh}(\tanh (a+b x))}}{b^3}-\frac {32 \text {arctanh}(\tanh (a+b x))^{3/2}}{3 b^4} \]
-2/3*x^3/b/arctanh(tanh(b*x+a))^(3/2)-32/3*arctanh(tanh(b*x+a))^(3/2)/b^4- 4*x^2/b^2/arctanh(tanh(b*x+a))^(1/2)+16*x*arctanh(tanh(b*x+a))^(1/2)/b^3
Time = 0.03 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.86 \[ \int \frac {x^3}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=-\frac {2 \left (b^3 x^3+6 b^2 x^2 \text {arctanh}(\tanh (a+b x))-24 b x \text {arctanh}(\tanh (a+b x))^2+16 \text {arctanh}(\tanh (a+b x))^3\right )}{3 b^4 \text {arctanh}(\tanh (a+b x))^{3/2}} \]
(-2*(b^3*x^3 + 6*b^2*x^2*ArcTanh[Tanh[a + b*x]] - 24*b*x*ArcTanh[Tanh[a + b*x]]^2 + 16*ArcTanh[Tanh[a + b*x]]^3))/(3*b^4*ArcTanh[Tanh[a + b*x]]^(3/2 ))
Time = 0.29 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.16, 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.
\(\displaystyle \int \frac {x^3}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {2 \int \frac {x^2}{\text {arctanh}(\tanh (a+b x))^{3/2}}dx}{b}-\frac {2 x^3}{3 b \text {arctanh}(\tanh (a+b x))^{3/2}}\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {2 \left (\frac {4 \int \frac {x}{\sqrt {\text {arctanh}(\tanh (a+b x))}}dx}{b}-\frac {2 x^2}{b \sqrt {\text {arctanh}(\tanh (a+b x))}}\right )}{b}-\frac {2 x^3}{3 b \text {arctanh}(\tanh (a+b x))^{3/2}}\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {2 \left (\frac {4 \left (\frac {2 x \sqrt {\text {arctanh}(\tanh (a+b x))}}{b}-\frac {2 \int \sqrt {\text {arctanh}(\tanh (a+b x))}dx}{b}\right )}{b}-\frac {2 x^2}{b \sqrt {\text {arctanh}(\tanh (a+b x))}}\right )}{b}-\frac {2 x^3}{3 b \text {arctanh}(\tanh (a+b x))^{3/2}}\) |
\(\Big \downarrow \) 2588 |
\(\displaystyle \frac {2 \left (\frac {4 \left (\frac {2 x \sqrt {\text {arctanh}(\tanh (a+b x))}}{b}-\frac {2 \int \sqrt {\text {arctanh}(\tanh (a+b x))}d\text {arctanh}(\tanh (a+b x))}{b^2}\right )}{b}-\frac {2 x^2}{b \sqrt {\text {arctanh}(\tanh (a+b x))}}\right )}{b}-\frac {2 x^3}{3 b \text {arctanh}(\tanh (a+b x))^{3/2}}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {2 \left (\frac {4 \left (\frac {2 x \sqrt {\text {arctanh}(\tanh (a+b x))}}{b}-\frac {4 \text {arctanh}(\tanh (a+b x))^{3/2}}{3 b^2}\right )}{b}-\frac {2 x^2}{b \sqrt {\text {arctanh}(\tanh (a+b x))}}\right )}{b}-\frac {2 x^3}{3 b \text {arctanh}(\tanh (a+b x))^{3/2}}\) |
(-2*x^3)/(3*b*ArcTanh[Tanh[a + b*x]]^(3/2)) + (2*((-2*x^2)/(b*Sqrt[ArcTanh [Tanh[a + b*x]]]) + (4*((2*x*Sqrt[ArcTanh[Tanh[a + b*x]]])/b - (4*ArcTanh[ Tanh[a + b*x]]^(3/2))/(3*b^2)))/b))/b
3.2.59.3.1 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. \(185\) vs. \(2(64)=128\).
Time = 0.10 (sec) , antiderivative size = 186, normalized size of antiderivative = 2.45
method | result | size |
default | \(\frac {\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{3}-6 \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}\, a -6 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right ) \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}-\frac {2 \left (3 a^{2}+6 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )+3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}\right )}{\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}-\frac {2 \left (-a^{3}-3 a^{2} \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )-3 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}-\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}\right )}{3 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}}{b^{4}}\) | \(186\) |
2/b^4*(1/3*arctanh(tanh(b*x+a))^(3/2)-3*arctanh(tanh(b*x+a))^(1/2)*a-3*(ar ctanh(tanh(b*x+a))-b*x-a)*arctanh(tanh(b*x+a))^(1/2)-(3*a^2+6*a*(arctanh(t anh(b*x+a))-b*x-a)+3*(arctanh(tanh(b*x+a))-b*x-a)^2)/arctanh(tanh(b*x+a))^ (1/2)-1/3*(-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))^(3/2))
Time = 0.25 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.82 \[ \int \frac {x^3}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\frac {2 \, {\left (b^{3} x^{3} - 6 \, a b^{2} x^{2} - 24 \, a^{2} b x - 16 \, a^{3}\right )} \sqrt {b x + a}}{3 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} \]
2/3*(b^3*x^3 - 6*a*b^2*x^2 - 24*a^2*b*x - 16*a^3)*sqrt(b*x + a)/(b^6*x^2 + 2*a*b^5*x + a^2*b^4)
Time = 2.23 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.18 \[ \int \frac {x^3}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\begin {cases} - \frac {2 x^{3}}{3 b \operatorname {atanh}^{\frac {3}{2}}{\left (\tanh {\left (a + b x \right )} \right )}} - \frac {4 x^{2}}{b^{2} \sqrt {\operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}} + \frac {16 x \sqrt {\operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}}{b^{3}} - \frac {32 \operatorname {atanh}^{\frac {3}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}{3 b^{4}} & \text {for}\: b \neq 0 \\\frac {x^{4}}{4 \operatorname {atanh}^{\frac {5}{2}}{\left (\tanh {\left (a \right )} \right )}} & \text {otherwise} \end {cases} \]
Piecewise((-2*x**3/(3*b*atanh(tanh(a + b*x))**(3/2)) - 4*x**2/(b**2*sqrt(a tanh(tanh(a + b*x)))) + 16*x*sqrt(atanh(tanh(a + b*x)))/b**3 - 32*atanh(ta nh(a + b*x))**(3/2)/(3*b**4), Ne(b, 0)), (x**4/(4*atanh(tanh(a))**(5/2)), True))
Time = 0.36 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.68 \[ \int \frac {x^3}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\frac {2 \, {\left (b^{4} x^{4} - 5 \, a b^{3} x^{3} - 30 \, a^{2} b^{2} x^{2} - 40 \, a^{3} b x - 16 \, a^{4}\right )}}{3 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{4}} \]
Time = 0.27 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.78 \[ \int \frac {x^3}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=-\frac {2 \, {\left (9 \, {\left (b x + a\right )} a^{2} - a^{3}\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{4}} + \frac {2 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} b^{8} - 9 \, \sqrt {b x + a} a b^{8}\right )}}{3 \, b^{12}} \]
-2/3*(9*(b*x + a)*a^2 - a^3)/((b*x + a)^(3/2)*b^4) + 2/3*((b*x + a)^(3/2)* b^8 - 9*sqrt(b*x + a)*a*b^8)/b^12
Time = 4.53 (sec) , antiderivative size = 533, normalized size of antiderivative = 7.01 \[ \int \frac {x^3}{\text {arctanh}(\tanh (a+b x))^{5/2}} \, dx=\frac {2\,x\,\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^3}+\frac {\left (\frac {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 )}{b^3}+\frac {4\,\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^3}\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}}}{b}-\frac {3\,\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 )}^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 )}-\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}{3\,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 )}^2} \]
(2*x*(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(ex p(2*a)*exp(2*b*x) + 1))/2)^(1/2))/(3*b^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) )/b^3 + (4*(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))/(3*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))/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)*(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)/(b^4*( log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) - log(2/(exp(2*a)*e xp(2*b*x) + 1)))) - ((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)/(3*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)))^2)