Integrand size = 13, antiderivative size = 75 \[ \int \frac {x^3}{\coth ^{-1}(\tanh (a+b x))^2} \, dx=\frac {3 x^2}{2 b^2}+\frac {3 x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}{b^3}-\frac {x^3}{b \coth ^{-1}(\tanh (a+b x))}+\frac {3 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b^4} \]
3/2*x^2/b^2+3*x*(b*x-arccoth(tanh(b*x+a)))/b^3-x^3/b/arccoth(tanh(b*x+a))+ 3*(b*x-arccoth(tanh(b*x+a)))^2*ln(arccoth(tanh(b*x+a)))/b^4
Time = 0.04 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.11 \[ \int \frac {x^3}{\coth ^{-1}(\tanh (a+b x))^2} \, dx=\frac {x^2}{2 b^2}-\frac {2 x \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )}{b^3}+\frac {\left (-b x+\coth ^{-1}(\tanh (a+b x))\right )^3}{b^4 \coth ^{-1}(\tanh (a+b x))}+\frac {3 \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )^2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b^4} \]
x^2/(2*b^2) - (2*x*(-(b*x) + ArcCoth[Tanh[a + b*x]]))/b^3 + (-(b*x) + ArcC oth[Tanh[a + b*x]])^3/(b^4*ArcCoth[Tanh[a + b*x]]) + (3*(-(b*x) + ArcCoth[ Tanh[a + b*x]])^2*Log[ArcCoth[Tanh[a + b*x]]])/b^4
Time = 0.29 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {2599, 2590, 2589, 2588, 14}
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.
\(\displaystyle \int \frac {x^3}{\coth ^{-1}(\tanh (a+b x))^2} \, dx\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {3 \int \frac {x^2}{\coth ^{-1}(\tanh (a+b x))}dx}{b}-\frac {x^3}{b \coth ^{-1}(\tanh (a+b x))}\) |
\(\Big \downarrow \) 2590 |
\(\displaystyle \frac {3 \left (\frac {\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \int \frac {x}{\coth ^{-1}(\tanh (a+b x))}dx}{b}+\frac {x^2}{2 b}\right )}{b}-\frac {x^3}{b \coth ^{-1}(\tanh (a+b x))}\) |
\(\Big \downarrow \) 2589 |
\(\displaystyle \frac {3 \left (\frac {\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \left (\frac {\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \int \frac {1}{\coth ^{-1}(\tanh (a+b x))}dx}{b}+\frac {x}{b}\right )}{b}+\frac {x^2}{2 b}\right )}{b}-\frac {x^3}{b \coth ^{-1}(\tanh (a+b x))}\) |
\(\Big \downarrow \) 2588 |
\(\displaystyle \frac {3 \left (\frac {\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \left (\frac {\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \int \frac {1}{\coth ^{-1}(\tanh (a+b x))}d\coth ^{-1}(\tanh (a+b x))}{b^2}+\frac {x}{b}\right )}{b}+\frac {x^2}{2 b}\right )}{b}-\frac {x^3}{b \coth ^{-1}(\tanh (a+b x))}\) |
\(\Big \downarrow \) 14 |
\(\displaystyle \frac {3 \left (\frac {\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \left (\frac {\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b^2}+\frac {x}{b}\right )}{b}+\frac {x^2}{2 b}\right )}{b}-\frac {x^3}{b \coth ^{-1}(\tanh (a+b x))}\) |
-(x^3/(b*ArcCoth[Tanh[a + b*x]])) + (3*(x^2/(2*b) + ((b*x - ArcCoth[Tanh[a + b*x]])*(x/b + ((b*x - ArcCoth[Tanh[a + b*x]])*Log[ArcCoth[Tanh[a + b*x] ]])/b^2))/b))/b
3.2.68.3.1 Defintions of rubi rules used
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[(v_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[b*(x/a), x] - Simp[(b*u - a*v)/a Int[1/u, x], x] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x]
Int[(v_)^(n_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[ D[v, x]]}, Simp[v^n/(a*n), x] - Simp[(b*u - a*v)/a Int[v^(n - 1)/u, x], x ] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && GtQ[n, 0] && NeQ[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]))
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.38 (sec) , antiderivative size = 29109, normalized size of antiderivative = 388.12
Result contains complex when optimal does not.
Time = 0.25 (sec) , antiderivative size = 151, normalized size of antiderivative = 2.01 \[ \int \frac {x^3}{\coth ^{-1}(\tanh (a+b x))^2} \, dx=\frac {4 \, b^{3} x^{3} - 12 \, a b^{2} x^{2} - 16 \, a^{2} b x - i \, \pi ^{3} + 2 \, \pi ^{2} {\left (2 \, b x - 3 \, a\right )} + 8 \, a^{3} - 2 i \, \pi {\left (3 \, b^{2} x^{2} + 8 \, a b x - 6 \, a^{2}\right )} + 3 \, {\left (8 \, a^{2} b x - i \, \pi ^{3} - 2 \, \pi ^{2} {\left (b x + 3 \, a\right )} + 8 \, a^{3} + 4 i \, \pi {\left (2 \, a b x + 3 \, a^{2}\right )}\right )} \log \left (i \, \pi + 2 \, b x + 2 \, a\right )}{4 \, {\left (2 \, b^{5} x + i \, \pi b^{4} + 2 \, a b^{4}\right )}} \]
1/4*(4*b^3*x^3 - 12*a*b^2*x^2 - 16*a^2*b*x - I*pi^3 + 2*pi^2*(2*b*x - 3*a) + 8*a^3 - 2*I*pi*(3*b^2*x^2 + 8*a*b*x - 6*a^2) + 3*(8*a^2*b*x - I*pi^3 - 2*pi^2*(b*x + 3*a) + 8*a^3 + 4*I*pi*(2*a*b*x + 3*a^2))*log(I*pi + 2*b*x + 2*a))/(2*b^5*x + I*pi*b^4 + 2*a*b^4)
\[ \int \frac {x^3}{\coth ^{-1}(\tanh (a+b x))^2} \, dx=\int \frac {x^{3}}{\operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]
Result contains complex when optimal does not.
Time = 0.52 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.64 \[ \int \frac {x^3}{\coth ^{-1}(\tanh (a+b x))^2} \, dx=\frac {4 \, b^{3} x^{3} + i \, \pi ^{3} - 6 \, \pi ^{2} a - 12 i \, \pi a^{2} + 8 \, a^{3} - 6 \, {\left (-i \, \pi b^{2} + 2 \, a b^{2}\right )} x^{2} + 4 \, {\left (\pi ^{2} b + 4 i \, \pi a b - 4 \, a^{2} b\right )} x}{4 \, {\left (2 \, b^{5} x - i \, \pi b^{4} + 2 \, a b^{4}\right )}} - \frac {3 \, {\left (\pi ^{2} + 4 i \, \pi a - 4 \, a^{2}\right )} \log \left (-i \, \pi + 2 \, b x + 2 \, a\right )}{4 \, b^{4}} \]
1/4*(4*b^3*x^3 + I*pi^3 - 6*pi^2*a - 12*I*pi*a^2 + 8*a^3 - 6*(-I*pi*b^2 + 2*a*b^2)*x^2 + 4*(pi^2*b + 4*I*pi*a*b - 4*a^2*b)*x)/(2*b^5*x - I*pi*b^4 + 2*a*b^4) - 3/4*(pi^2 + 4*I*pi*a - 4*a^2)*log(-I*pi + 2*b*x + 2*a)/b^4
Result contains complex when optimal does not.
Time = 0.29 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.29 \[ \int \frac {x^3}{\coth ^{-1}(\tanh (a+b x))^2} \, dx=-\frac {i \, \pi ^{3} + 6 \, \pi ^{2} a - 12 i \, \pi a^{2} - 8 \, a^{3}}{4 \, {\left (2 \, b^{5} x + i \, \pi b^{4} + 2 \, a b^{4}\right )}} + \frac {x^{2}}{2 \, b^{2}} + \frac {{\left (-i \, \pi - 2 \, a\right )} x}{b^{3}} - \frac {3 \, {\left (\pi ^{2} - 4 i \, \pi a - 4 \, a^{2}\right )} \log \left (i \, \pi + 2 \, b x + 2 \, a\right )}{4 \, b^{4}} \]
-1/4*(I*pi^3 + 6*pi^2*a - 12*I*pi*a^2 - 8*a^3)/(2*b^5*x + I*pi*b^4 + 2*a*b ^4) + 1/2*x^2/b^2 + (-I*pi - 2*a)*x/b^3 - 3/4*(pi^2 - 4*I*pi*a - 4*a^2)*lo g(I*pi + 2*b*x + 2*a)/b^4
Time = 0.18 (sec) , antiderivative size = 490, normalized size of antiderivative = 6.53 \[ \int \frac {x^3}{\coth ^{-1}(\tanh (a+b x))^2} \, dx=\frac {x^2}{2\,b^2}+\frac {\ln \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 )\,\left (3\,{\left (2\,a-\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 )+2\,b\,x\right )}^2-12\,a\,\left (2\,a-\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 )+2\,b\,x\right )+12\,a^2\right )}{4\,b^4}-\frac {{\left (2\,a-\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 )+2\,b\,x\right )}^3-8\,a^3-6\,a\,{\left (2\,a-\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 )+2\,b\,x\right )}^2+12\,a^2\,\left (2\,a-\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 )+2\,b\,x\right )}{4\,b\,\left (2\,a\,b^3+2\,b^4\,x-b^3\,\left (2\,a-\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 )+2\,b\,x\right )\right )}+\frac {x\,\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} \]
x^2/(2*b^2) + (log(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)))*(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 - 12*a*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + lo g(-2/(exp(2*a)*exp(2*b*x) - 1)) + 2*b*x) + 12*a^2))/(4*b^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 - 8*a^3 - 6*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) - 1)) + 2*b*x)^2 + 12*a^2*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + lo g(-2/(exp(2*a)*exp(2*b*x) - 1)) + 2*b*x))/(4*b*(2*a*b^3 + 2*b^4*x - b^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))) + (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^3