Integrand size = 13, antiderivative size = 75 \[ \int \frac {x^3}{\text {arctanh}(\tanh (a+b x))^2} \, dx=\frac {3 x^2}{2 b^2}+\frac {3 x (b x-\text {arctanh}(\tanh (a+b x)))}{b^3}-\frac {x^3}{b \text {arctanh}(\tanh (a+b x))}+\frac {3 (b x-\text {arctanh}(\tanh (a+b x)))^2 \log (\text {arctanh}(\tanh (a+b x)))}{b^4} \]
3/2*x^2/b^2+3*x*(b*x-arctanh(tanh(b*x+a)))/b^3-x^3/b/arctanh(tanh(b*x+a))+ 3*(b*x-arctanh(tanh(b*x+a)))^2*ln(arctanh(tanh(b*x+a)))/b^4
Time = 0.04 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.11 \[ \int \frac {x^3}{\text {arctanh}(\tanh (a+b x))^2} \, dx=\frac {x^2}{2 b^2}-\frac {2 x (-b x+\text {arctanh}(\tanh (a+b x)))}{b^3}+\frac {(-b x+\text {arctanh}(\tanh (a+b x)))^3}{b^4 \text {arctanh}(\tanh (a+b x))}+\frac {3 (-b x+\text {arctanh}(\tanh (a+b x)))^2 \log (\text {arctanh}(\tanh (a+b x)))}{b^4} \]
x^2/(2*b^2) - (2*x*(-(b*x) + ArcTanh[Tanh[a + b*x]]))/b^3 + (-(b*x) + ArcT anh[Tanh[a + b*x]])^3/(b^4*ArcTanh[Tanh[a + b*x]]) + (3*(-(b*x) + ArcTanh[ Tanh[a + b*x]])^2*Log[ArcTanh[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.
\(\displaystyle \int \frac {x^3}{\text {arctanh}(\tanh (a+b x))^2} \, dx\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {3 \int \frac {x^2}{\text {arctanh}(\tanh (a+b x))}dx}{b}-\frac {x^3}{b \text {arctanh}(\tanh (a+b x))}\) |
\(\Big \downarrow \) 2590 |
\(\displaystyle \frac {3 \left (\frac {(b x-\text {arctanh}(\tanh (a+b x))) \int \frac {x}{\text {arctanh}(\tanh (a+b x))}dx}{b}+\frac {x^2}{2 b}\right )}{b}-\frac {x^3}{b \text {arctanh}(\tanh (a+b x))}\) |
\(\Big \downarrow \) 2589 |
\(\displaystyle \frac {3 \left (\frac {(b x-\text {arctanh}(\tanh (a+b x))) \left (\frac {(b x-\text {arctanh}(\tanh (a+b x))) \int \frac {1}{\text {arctanh}(\tanh (a+b x))}dx}{b}+\frac {x}{b}\right )}{b}+\frac {x^2}{2 b}\right )}{b}-\frac {x^3}{b \text {arctanh}(\tanh (a+b x))}\) |
\(\Big \downarrow \) 2588 |
\(\displaystyle \frac {3 \left (\frac {(b x-\text {arctanh}(\tanh (a+b x))) \left (\frac {(b x-\text {arctanh}(\tanh (a+b x))) \int \frac {1}{\text {arctanh}(\tanh (a+b x))}d\text {arctanh}(\tanh (a+b x))}{b^2}+\frac {x}{b}\right )}{b}+\frac {x^2}{2 b}\right )}{b}-\frac {x^3}{b \text {arctanh}(\tanh (a+b x))}\) |
\(\Big \downarrow \) 14 |
\(\displaystyle \frac {3 \left (\frac {(b x-\text {arctanh}(\tanh (a+b x))) \left (\frac {(b x-\text {arctanh}(\tanh (a+b x))) \log (\text {arctanh}(\tanh (a+b x)))}{b^2}+\frac {x}{b}\right )}{b}+\frac {x^2}{2 b}\right )}{b}-\frac {x^3}{b \text {arctanh}(\tanh (a+b x))}\) |
-(x^3/(b*ArcTanh[Tanh[a + b*x]])) + (3*(x^2/(2*b) + ((b*x - ArcTanh[Tanh[a + b*x]])*(x/b + ((b*x - ArcTanh[Tanh[a + b*x]])*Log[ArcTanh[Tanh[a + b*x] ]])/b^2))/b))/b
3.1.95.3.1 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]))
Leaf count of result is larger than twice the leaf count of optimal. \(168\) vs. \(2(73)=146\).
Time = 0.26 (sec) , antiderivative size = 169, normalized size of antiderivative = 2.25
method | result | size |
default | \(\frac {\frac {b \,x^{2}}{2}-2 a x -2 x \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )}{b^{3}}+\frac {\left (3 a^{2}+6 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )+3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}\right ) \ln \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )}{b^{4}}-\frac {-a^{3}-3 a^{2} \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )-3 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}-\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}}{b^{4} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}\) | \(169\) |
risch | \(\text {Expression too large to display}\) | \(14799\) |
1/b^3*(1/2*b*x^2-2*a*x-2*x*(arctanh(tanh(b*x+a))-b*x-a))+(3*a^2+6*a*(arcta nh(tanh(b*x+a))-b*x-a)+3*(arctanh(tanh(b*x+a))-b*x-a)^2)/b^4*ln(arctanh(ta nh(b*x+a)))-1/b^4*(-a^3-3*a^2*(arctanh(tanh(b*x+a))-b*x-a)-3*a*(arctanh(ta nh(b*x+a))-b*x-a)^2-(arctanh(tanh(b*x+a))-b*x-a)^3)/arctanh(tanh(b*x+a))
Time = 0.25 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.83 \[ \int \frac {x^3}{\text {arctanh}(\tanh (a+b x))^2} \, dx=\frac {b^{3} x^{3} - 3 \, a b^{2} x^{2} - 4 \, a^{2} b x + 2 \, a^{3} + 6 \, {\left (a^{2} b x + a^{3}\right )} \log \left (b x + a\right )}{2 \, {\left (b^{5} x + a b^{4}\right )}} \]
1/2*(b^3*x^3 - 3*a*b^2*x^2 - 4*a^2*b*x + 2*a^3 + 6*(a^2*b*x + a^3)*log(b*x + a))/(b^5*x + a*b^4)
\[ \int \frac {x^3}{\text {arctanh}(\tanh (a+b x))^2} \, dx=\int \frac {x^{3}}{\operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]
Time = 0.53 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.79 \[ \int \frac {x^3}{\text {arctanh}(\tanh (a+b x))^2} \, dx=\frac {b^{3} x^{3} - 3 \, a b^{2} x^{2} - 4 \, a^{2} b x + 2 \, a^{3}}{2 \, {\left (b^{5} x + a b^{4}\right )}} + \frac {3 \, a^{2} \log \left (b x + a\right )}{b^{4}} \]
Time = 0.27 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.64 \[ \int \frac {x^3}{\text {arctanh}(\tanh (a+b x))^2} \, dx=\frac {3 \, a^{2} \log \left ({\left | b x + a \right |}\right )}{b^{4}} + \frac {a^{3}}{{\left (b x + a\right )} b^{4}} + \frac {b^{2} x^{2} - 4 \, a b x}{2 \, b^{4}} \]
Time = 0.17 (sec) , antiderivative size = 490, normalized size of antiderivative = 6.53 \[ \int \frac {x^3}{\text {arctanh}(\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)) + log( 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)) + log(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 - l og((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + log(2/(exp(2*a)*ex p(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