3.97 \(\int \frac {x}{\tanh ^{-1}(\tanh (a+b x))^2} \, dx\)

Optimal. Leaf size=28 \[ \frac {\log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^2}-\frac {x}{b \tanh ^{-1}(\tanh (a+b x))} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 28, 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 = {2168, 2157, 29} \[ \frac {\log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^2}-\frac {x}{b \tanh ^{-1}(\tanh (a+b x))} \]

Antiderivative was successfully verified.

[In]

Int[x/ArcTanh[Tanh[a + b*x]]^2,x]

[Out]

-(x/(b*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 2157

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 2168

Int[(u_)^(m_)*(v_)^(n_.), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(u^(m + 1)*v^
n)/(a*(m + 1)), x] - Dist[(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] && (Fra
ctionQ[m] || GeQ[2*n + m + 1, 0]))) || (IGtQ[n, 0] && IGtQ[m, 0] && LeQ[n, m]) || (IGtQ[n, 0] &&  !IntegerQ[m]
) || (ILtQ[m, 0] &&  !IntegerQ[n]))

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 27, normalized size = 0.96 \[ \frac {-\frac {b x}{\tanh ^{-1}(\tanh (a+b x))}+\log \left (\tanh ^{-1}(\tanh (a+b x))\right )+1}{b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x/ArcTanh[Tanh[a + b*x]]^2,x]

[Out]

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

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fricas [A]  time = 0.43, size = 28, normalized size = 1.00 \[ \frac {{\left (b x + a\right )} \log \left (b x + a\right ) + a}{b^{3} x + a b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/arctanh(tanh(b*x+a))^2,x, algorithm="fricas")

[Out]

((b*x + a)*log(b*x + a) + a)/(b^3*x + a*b^2)

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giac [A]  time = 0.16, size = 24, normalized size = 0.86 \[ \frac {\log \left ({\left | b x + a \right |}\right )}{b^{2}} + \frac {a}{{\left (b x + a\right )} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/arctanh(tanh(b*x+a))^2,x, algorithm="giac")

[Out]

log(abs(b*x + a))/b^2 + a/((b*x + a)*b^2)

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maple [A]  time = 0.14, size = 56, normalized size = 2.00 \[ \frac {\ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right )}{b^{2}}+\frac {a}{b^{2} \arctanh \left (\tanh \left (b x +a \right )\right )}+\frac {\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a}{b^{2} \arctanh \left (\tanh \left (b x +a \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/arctanh(tanh(b*x+a))^2,x)

[Out]

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

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maxima [A]  time = 0.74, size = 26, normalized size = 0.93 \[ \frac {a}{b^{3} x + a b^{2}} + \frac {\log \left (b x + a\right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/arctanh(tanh(b*x+a))^2,x, algorithm="maxima")

[Out]

a/(b^3*x + a*b^2) + log(b*x + a)/b^2

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mupad [B]  time = 0.08, size = 28, normalized size = 1.00 \[ \frac {\ln \left (\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )\right )}{b^2}-\frac {x}{b\,\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/atanh(tanh(a + b*x))^2,x)

[Out]

log(atanh(tanh(a + b*x)))/b^2 - x/(b*atanh(tanh(a + b*x)))

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/atanh(tanh(b*x+a))**2,x)

[Out]

Exception raised: TypeError

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