3.1.90 \(\int \frac {1}{x \tanh ^{-1}(\tanh (a+b x))} \, dx\) [90]

Optimal. Leaf size=44 \[ -\frac {\log (x)}{b x-\tanh ^{-1}(\tanh (a+b x))}+\frac {\log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b x-\tanh ^{-1}(\tanh (a+b x))} \]

[Out]

-ln(x)/(b*x-arctanh(tanh(b*x+a)))+ln(arctanh(tanh(b*x+a)))/(b*x-arctanh(tanh(b*x+a)))

________________________________________________________________________________________

Rubi [A]
time = 0.02, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2191, 2188, 29} \begin {gather*} \frac {\log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b x-\tanh ^{-1}(\tanh (a+b x))}-\frac {\log (x)}{b x-\tanh ^{-1}(\tanh (a+b x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x*ArcTanh[Tanh[a + b*x]]),x]

[Out]

-(Log[x]/(b*x - ArcTanh[Tanh[a + b*x]])) + Log[ArcTanh[Tanh[a + b*x]]]/(b*x - ArcTanh[Tanh[a + b*x]])

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2188

Int[(u_)^(m_.), x_Symbol] :> With[{c = Simplify[D[u, x]]}, Dist[1/c, Subst[Int[x^m, x], x, u], x]] /; FreeQ[m,
x] && PiecewiseLinearQ[u, x]

Rule 2191

Int[1/((u_)*(v_)), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Dist[b/(b*u - a*v), Int[1
/v, x], x] - Dist[a/(b*u - a*v), Int[1/u, x], x] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x]

Rubi steps

\begin {align*} \int \frac {1}{x \tanh ^{-1}(\tanh (a+b x))} \, dx &=-\frac {\int \frac {1}{x} \, dx}{b x-\tanh ^{-1}(\tanh (a+b x))}+\frac {b \int \frac {1}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{b x-\tanh ^{-1}(\tanh (a+b x))}\\ &=-\frac {\log (x)}{b x-\tanh ^{-1}(\tanh (a+b x))}+\frac {\text {Subst}\left (\int \frac {1}{x} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{b x-\tanh ^{-1}(\tanh (a+b x))}\\ &=-\frac {\log (x)}{b x-\tanh ^{-1}(\tanh (a+b x))}+\frac {\log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b x-\tanh ^{-1}(\tanh (a+b x))}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.02, size = 29, normalized size = 0.66 \begin {gather*} \frac {-\log (x)+\log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b x-\tanh ^{-1}(\tanh (a+b x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x*ArcTanh[Tanh[a + b*x]]),x]

[Out]

(-Log[x] + Log[ArcTanh[Tanh[a + b*x]]])/(b*x - ArcTanh[Tanh[a + b*x]])

________________________________________________________________________________________

Maple [A]
time = 7.72, size = 43, normalized size = 0.98

method	result	size
default	\(-\frac {\ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right )}{\arctanh \left (\tanh \left (b x +a \right )\right )-b x}+\frac {\ln \left (x \right )}{\arctanh \left (\tanh \left (b x +a \right )\right )-b x}\)	\(43\)
risch	\(\frac {4 i \ln \left (-\pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )+\pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}-\pi \mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )+2 \pi \,\mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2}-\pi \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3}+\pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}-\pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}-4 i b x -4 i \left (\ln \left ({\mathrm e}^{b x +a}\right )-b x -a \right )-4 i a \right )}{-\pi \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3}+2 \pi \,\mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2}+\pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}-\pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )-\pi \mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )-\pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}+\pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+4 i b x -4 i \ln \left ({\mathrm e}^{b x +a}\right )}+\frac {4 i \ln \left (x \right )}{\pi \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3}-2 \pi \,\mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2}-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+\pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )+\pi \mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )+\pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}-\pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}-4 i b x +4 i \ln \left ({\mathrm e}^{b x +a}\right )}\)	\(835\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/arctanh(tanh(b*x+a)),x,method=_RETURNVERBOSE)

[Out]

-1/(arctanh(tanh(b*x+a))-b*x)*ln(arctanh(tanh(b*x+a)))+1/(arctanh(tanh(b*x+a))-b*x)*ln(x)

________________________________________________________________________________________

Maxima [A]
time = 0.52, size = 18, normalized size = 0.41 \begin {gather*} -\frac {\log \left (b x + a\right )}{a} + \frac {\log \left (x\right )}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/arctanh(tanh(b*x+a)),x, algorithm="maxima")

[Out]

-log(b*x + a)/a + log(x)/a

________________________________________________________________________________________

Fricas [A]
time = 0.34, size = 16, normalized size = 0.36 \begin {gather*} -\frac {\log \left (b x + a\right ) - \log \left (x\right )}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/arctanh(tanh(b*x+a)),x, algorithm="fricas")

[Out]

-(log(b*x + a) - log(x))/a

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/atanh(tanh(b*x+a)),x)

[Out]

Integral(1/(x*atanh(tanh(a + b*x))), x)

________________________________________________________________________________________

Giac [A]
time = 0.39, size = 20, normalized size = 0.45 \begin {gather*} -\frac {\log \left ({\left | b x + a \right |}\right )}{a} + \frac {\log \left ({\left | x \right |}\right )}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/arctanh(tanh(b*x+a)),x, algorithm="giac")

[Out]

-log(abs(b*x + a))/a + log(abs(x))/a

________________________________________________________________________________________

Mupad [B]
time = 2.87, size = 107, normalized size = 2.43 \begin {gather*} -\frac {4\,\mathrm {atanh}\left (\frac {4\,b\,x}{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x}-1\right )}{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*atanh(tanh(a + b*x))),x)

[Out]

-(4*atanh((4*b*x)/(log(1/(exp(2*a)*exp(2*b*x) + 1)) - log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2
*b*x) - 1))/(log(1/(exp(2*a)*exp(2*b*x) + 1)) - log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)

________________________________________________________________________________________