3.1.88 \(\int \frac {x}{\tanh ^{-1}(\tanh (a+b x))} \, dx\) [88]

Optimal. Leaf size=31 \[ \frac {x}{b}+\frac {\left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^2} \]

[Out]

x/b+(b*x-arctanh(tanh(b*x+a)))*ln(arctanh(tanh(b*x+a)))/b^2

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Rubi [A]
time = 0.01, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2189, 2188, 29} \begin {gather*} \frac {\left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^2}+\frac {x}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/ArcTanh[Tanh[a + b*x]],x]

[Out]

x/b + ((b*x - ArcTanh[Tanh[a + b*x]])*Log[ArcTanh[Tanh[a + b*x]]])/b^2

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 2189

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

Rubi steps

\begin {align*} \int \frac {x}{\tanh ^{-1}(\tanh (a+b x))} \, dx &=\frac {x}{b}-\frac {\left (-b x+\tanh ^{-1}(\tanh (a+b x))\right ) \int \frac {1}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{b}\\ &=\frac {x}{b}-\frac {\left (-b x+\tanh ^{-1}(\tanh (a+b x))\right ) \text {Subst}\left (\int \frac {1}{x} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{b^2}\\ &=\frac {x}{b}+\frac {\left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^2}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 31, normalized size = 1.00 \begin {gather*} \frac {x}{b}-\frac {\left (-b x+\tanh ^{-1}(\tanh (a+b x))\right ) \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/ArcTanh[Tanh[a + b*x]],x]

[Out]

x/b - ((-(b*x) + ArcTanh[Tanh[a + b*x]])*Log[ArcTanh[Tanh[a + b*x]]])/b^2

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Maple [A]
time = 0.19, size = 49, normalized size = 1.58

method	result	size
default	\(\frac {x}{b}-\frac {\ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right ) a}{b^{2}}-\frac {\ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right ) \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{b^{2}}\)	\(49\)
risch	\(\text {Expression too large to display}\)	\(2837\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/b-1/b^2*ln(arctanh(tanh(b*x+a)))*a-1/b^2*ln(arctanh(tanh(b*x+a)))*(arctanh(tanh(b*x+a))-b*x-a)

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Maxima [A]
time = 0.53, size = 18, normalized size = 0.58 \begin {gather*} \frac {x}{b} - \frac {a \log \left (b x + a\right )}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/b - a*log(b*x + a)/b^2

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Fricas [A]
time = 0.34, size = 17, normalized size = 0.55 \begin {gather*} \frac {b x - a \log \left (b x + a\right )}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*x - a*log(b*x + a))/b^2

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {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(x/atanh(tanh(b*x+a)),x)

[Out]

Integral(x/atanh(tanh(a + b*x)), x)

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Giac [A]
time = 0.40, size = 19, normalized size = 0.61 \begin {gather*} \frac {x}{b} - \frac {a \log \left ({\left | b x + a \right |}\right )}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/b - a*log(abs(b*x + a))/b^2

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Mupad [B]
time = 0.15, size = 108, normalized size = 3.48 \begin {gather*} \frac {x}{b}+\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 (\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 )}{2\,b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/atanh(tanh(a + b*x)),x)

[Out]

x/b + (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)))*(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))/(2*b^2)

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