Integrand size = 11, antiderivative size = 28 \[ \int \frac {x}{\text {arctanh}(\tanh (a+b x))^2} \, dx=-\frac {x}{b \text {arctanh}(\tanh (a+b x))}+\frac {\log (\text {arctanh}(\tanh (a+b x)))}{b^2} \] Output:
-x/b/arctanh(tanh(b*x+a))+ln(arctanh(tanh(b*x+a)))/b^2
Time = 0.10 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {x}{\text {arctanh}(\tanh (a+b x))^2} \, dx=\frac {1-\frac {b x}{\text {arctanh}(\tanh (a+b x))}+\log (\text {arctanh}(\tanh (a+b x)))}{b^2} \] Input:
Integrate[x/ArcTanh[Tanh[a + b*x]]^2,x]
Output:
(1 - (b*x)/ArcTanh[Tanh[a + b*x]] + Log[ArcTanh[Tanh[a + b*x]]])/b^2
Time = 0.32 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {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.
\(\displaystyle \int \frac {x}{\text {arctanh}(\tanh (a+b x))^2} \, dx\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {\int \frac {1}{\text {arctanh}(\tanh (a+b x))}dx}{b}-\frac {x}{b \text {arctanh}(\tanh (a+b x))}\) |
\(\Big \downarrow \) 2588 |
\(\displaystyle \frac {\int \frac {1}{\text {arctanh}(\tanh (a+b x))}d\text {arctanh}(\tanh (a+b x))}{b^2}-\frac {x}{b \text {arctanh}(\tanh (a+b x))}\) |
\(\Big \downarrow \) 14 |
\(\displaystyle \frac {\log (\text {arctanh}(\tanh (a+b x)))}{b^2}-\frac {x}{b \text {arctanh}(\tanh (a+b x))}\) |
Input:
Int[x/ArcTanh[Tanh[a + b*x]]^2,x]
Output:
-(x/(b*ArcTanh[Tanh[a + b*x]])) + Log[ArcTanh[Tanh[a + b*x]]]/b^2
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 = 0.14 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.25
method | result | size |
parallelrisch | \(\frac {\ln \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}{b^{2} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}\) | \(35\) |
default | \(\frac {\ln \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )}{b^{2}}-\frac {b x -\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}{b^{2} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}\) | \(41\) |
risch | \(\frac {4 i x}{b \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 )}+\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^{2}}\) | \(541\) |
Input:
int(x/arctanh(tanh(b*x+a))^2,x,method=_RETURNVERBOSE)
Output:
(ln(arctanh(tanh(b*x+a)))*arctanh(tanh(b*x+a))-b*x)/b^2/arctanh(tanh(b*x+a ))
Time = 0.08 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {x}{\text {arctanh}(\tanh (a+b x))^2} \, dx=\frac {{\left (b x + a\right )} \log \left (b x + a\right ) + a}{b^{3} x + a b^{2}} \] Input:
integrate(x/arctanh(tanh(b*x+a))^2,x, algorithm="fricas")
Output:
((b*x + a)*log(b*x + a) + a)/(b^3*x + a*b^2)
Leaf count of result is larger than twice the leaf count of optimal. 94 vs. \(2 (24) = 48\).
Time = 13.32 (sec) , antiderivative size = 94, normalized size of antiderivative = 3.36 \[ \int \frac {x}{\text {arctanh}(\tanh (a+b x))^2} \, dx=\begin {cases} \frac {x^{2}}{2 \operatorname {atanh}^{2}{\left (\tanh {\left (a \right )} \right )}} & \text {for}\: b = 0 \\\frac {x^{2}}{2 \operatorname {atanh}^{2}{\left (\tanh {\left (b x + \log {\left (- e^{- b x} \right )} \right )} \right )}} & \text {for}\: a = \log {\left (- e^{- b x} \right )} \\\frac {x^{2}}{2 \operatorname {atanh}^{2}{\left (\tanh {\left (b x + \log {\left (e^{- b x} \right )} \right )} \right )}} & \text {for}\: a = \log {\left (e^{- b x} \right )} \\- \frac {x}{b \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}} + \frac {\log {\left (\operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )} \right )}}{b^{2}} & \text {otherwise} \end {cases} \] Input:
integrate(x/atanh(tanh(b*x+a))**2,x)
Output:
Piecewise((x**2/(2*atanh(tanh(a))**2), Eq(b, 0)), (x**2/(2*atanh(tanh(b*x + log(-exp(-b*x))))**2), Eq(a, log(-exp(-b*x)))), (x**2/(2*atanh(tanh(b*x + log(exp(-b*x))))**2), Eq(a, log(exp(-b*x)))), (-x/(b*atanh(tanh(a + b*x) )) + log(atanh(tanh(a + b*x)))/b**2, True))
Time = 0.33 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {x}{\text {arctanh}(\tanh (a+b x))^2} \, dx=\frac {a}{b^{3} x + a b^{2}} + \frac {\log \left (b x + a\right )}{b^{2}} \] Input:
integrate(x/arctanh(tanh(b*x+a))^2,x, algorithm="maxima")
Output:
a/(b^3*x + a*b^2) + log(b*x + a)/b^2
Time = 0.11 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {x}{\text {arctanh}(\tanh (a+b x))^2} \, dx=\frac {\log \left ({\left | b x + a \right |}\right )}{b^{2}} + \frac {a}{{\left (b x + a\right )} b^{2}} \] Input:
integrate(x/arctanh(tanh(b*x+a))^2,x, algorithm="giac")
Output:
log(abs(b*x + a))/b^2 + a/((b*x + a)*b^2)
Time = 0.09 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {x}{\text {arctanh}(\tanh (a+b x))^2} \, dx=\frac {\ln \left (\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )\right )}{b^2}-\frac {x}{b\,\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )} \] Input:
int(x/atanh(tanh(a + b*x))^2,x)
Output:
log(atanh(tanh(a + b*x)))/b^2 - x/(b*atanh(tanh(a + b*x)))
Time = 0.17 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \[ \int \frac {x}{\text {arctanh}(\tanh (a+b x))^2} \, dx=\frac {\mathit {atanh} \left (\tanh \left (b x +a \right )\right ) \mathrm {log}\left (\mathit {atanh} \left (\tanh \left (b x +a \right )\right )\right )-b x}{\mathit {atanh} \left (\tanh \left (b x +a \right )\right ) b^{2}} \] Input:
int(x/atanh(tanh(b*x+a))^2,x)
Output:
(atanh(tanh(a + b*x))*log(atanh(tanh(a + b*x))) - b*x)/(atanh(tanh(a + b*x ))*b**2)