Integrand size = 13, antiderivative size = 47 \[ \int \frac {x^2}{\text {arctanh}(\tanh (a+b x))^3} \, dx=-\frac {x^2}{2 b \text {arctanh}(\tanh (a+b x))^2}-\frac {x}{b^2 \text {arctanh}(\tanh (a+b x))}+\frac {\log (\text {arctanh}(\tanh (a+b x)))}{b^3} \]
Time = 0.04 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.04 \[ \int \frac {x^2}{\text {arctanh}(\tanh (a+b x))^3} \, dx=\frac {3-\frac {b^2 x^2}{\text {arctanh}(\tanh (a+b x))^2}-\frac {2 b x}{\text {arctanh}(\tanh (a+b x))}+2 \log (\text {arctanh}(\tanh (a+b x)))}{2 b^3} \]
(3 - (b^2*x^2)/ArcTanh[Tanh[a + b*x]]^2 - (2*b*x)/ArcTanh[Tanh[a + b*x]] + 2*Log[ArcTanh[Tanh[a + b*x]]])/(2*b^3)
Time = 0.24 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.11, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2599, 2599, 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.
\(\displaystyle \int \frac {x^2}{\text {arctanh}(\tanh (a+b x))^3} \, dx\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {\int \frac {x}{\text {arctanh}(\tanh (a+b x))^2}dx}{b}-\frac {x^2}{2 b \text {arctanh}(\tanh (a+b x))^2}\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {\frac {\int \frac {1}{\text {arctanh}(\tanh (a+b x))}dx}{b}-\frac {x}{b \text {arctanh}(\tanh (a+b x))}}{b}-\frac {x^2}{2 b \text {arctanh}(\tanh (a+b x))^2}\) |
\(\Big \downarrow \) 2588 |
\(\displaystyle \frac {\frac {\int \frac {1}{\text {arctanh}(\tanh (a+b x))}d\text {arctanh}(\tanh (a+b x))}{b^2}-\frac {x}{b \text {arctanh}(\tanh (a+b x))}}{b}-\frac {x^2}{2 b \text {arctanh}(\tanh (a+b x))^2}\) |
\(\Big \downarrow \) 14 |
\(\displaystyle \frac {\frac {\log (\text {arctanh}(\tanh (a+b x)))}{b^2}-\frac {x}{b \text {arctanh}(\tanh (a+b x))}}{b}-\frac {x^2}{2 b \text {arctanh}(\tanh (a+b x))^2}\) |
-1/2*x^2/(b*ArcTanh[Tanh[a + b*x]]^2) + (-(x/(b*ArcTanh[Tanh[a + b*x]])) + Log[ArcTanh[Tanh[a + b*x]]]/b^2)/b
3.2.5.3.1 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[(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 = 0.10 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.15
method | result | size |
parallelrisch | \(\frac {-b^{2} x^{2}+2 \ln \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2}-2 b x \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}{2 b^{3} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2}}\) | \(54\) |
default | \(-\frac {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2}}{2 b^{3} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2}}-\frac {-2 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )+2 b x}{b^{3} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}+\frac {\ln \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )}{b^{3}}\) | \(70\) |
risch | \(-\frac {4 i \left (\pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right ) x -\pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2} x +\pi \operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) x -2 \pi \,\operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2} x +\pi \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3} x -\pi \,\operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2} x +\pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3} x +4 i x \ln \left ({\mathrm e}^{b x +a}\right )+2 i b \,x^{2}\right )}{b^{2} {\left (\pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )-\pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+\pi \operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )-2 \pi \,\operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2}+\pi \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3}-\pi \,\operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+\pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}+4 i \ln \left ({\mathrm e}^{b x +a}\right )\right )}^{2}}+\frac {\ln \left (\ln \left ({\mathrm e}^{b x +a}\right )-\frac {i \pi \left (\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )-\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+\operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )-2 \,\operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2}+\operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3}-\operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+\operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}\right )}{4}\right )}{b^{3}}\) | \(812\) |
1/2*(-b^2*x^2+2*ln(arctanh(tanh(b*x+a)))*arctanh(tanh(b*x+a))^2-2*b*x*arct anh(tanh(b*x+a)))/b^3/arctanh(tanh(b*x+a))^2
Time = 0.25 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.30 \[ \int \frac {x^2}{\text {arctanh}(\tanh (a+b x))^3} \, dx=\frac {4 \, a b x + 3 \, a^{2} + 2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \log \left (b x + a\right )}{2 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} \]
1/2*(4*a*b*x + 3*a^2 + 2*(b^2*x^2 + 2*a*b*x + a^2)*log(b*x + a))/(b^5*x^2 + 2*a*b^4*x + a^2*b^3)
Leaf count of result is larger than twice the leaf count of optimal. 112 vs. \(2 (42) = 84\).
Time = 28.82 (sec) , antiderivative size = 112, normalized size of antiderivative = 2.38 \[ \int \frac {x^2}{\text {arctanh}(\tanh (a+b x))^3} \, dx=\begin {cases} \frac {x^{3}}{3 \operatorname {atanh}^{3}{\left (\tanh {\left (a \right )} \right )}} & \text {for}\: b = 0 \\\frac {x^{3}}{3 \operatorname {atanh}^{3}{\left (\tanh {\left (b x + \log {\left (- e^{- b x} \right )} \right )} \right )}} & \text {for}\: a = \log {\left (- e^{- b x} \right )} \\\frac {x^{3}}{3 \operatorname {atanh}^{3}{\left (\tanh {\left (b x + \log {\left (e^{- b x} \right )} \right )} \right )}} & \text {for}\: a = \log {\left (e^{- b x} \right )} \\- \frac {x^{2}}{2 b \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}} - \frac {x}{b^{2} \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}} + \frac {\log {\left (\operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )} \right )}}{b^{3}} & \text {otherwise} \end {cases} \]
Piecewise((x**3/(3*atanh(tanh(a))**3), Eq(b, 0)), (x**3/(3*atanh(tanh(b*x + log(-exp(-b*x))))**3), Eq(a, log(-exp(-b*x)))), (x**3/(3*atanh(tanh(b*x + log(exp(-b*x))))**3), Eq(a, log(exp(-b*x)))), (-x**2/(2*b*atanh(tanh(a + b*x))**2) - x/(b**2*atanh(tanh(a + b*x))) + log(atanh(tanh(a + b*x)))/b** 3, True))
Time = 0.76 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.02 \[ \int \frac {x^2}{\text {arctanh}(\tanh (a+b x))^3} \, dx=\frac {4 \, a b x + 3 \, a^{2}}{2 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} + \frac {\log \left (b x + a\right )}{b^{3}} \]
Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.79 \[ \int \frac {x^2}{\text {arctanh}(\tanh (a+b x))^3} \, dx=\frac {\log \left ({\left | b x + a \right |}\right )}{b^{3}} + \frac {4 \, a x + \frac {3 \, a^{2}}{b}}{2 \, {\left (b x + a\right )}^{2} b^{2}} \]
Time = 3.54 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.98 \[ \int \frac {x^2}{\text {arctanh}(\tanh (a+b x))^3} \, dx=\frac {\ln \left (\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )\right )}{b^3}-\frac {\frac {b^2\,x^2}{2}+b\,x\,\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}{b^3\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2} \]