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

Optimal. Leaf size=65 \[ \frac {1}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {b \log (x)}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {b \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2163, 2160, 2157, 29} \[ \frac {1}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {b \log (x)}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {b \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*ArcTanh[Tanh[a + b*x]]),x]

[Out]

1/(x*(b*x - ArcTanh[Tanh[a + b*x]])) - (b*Log[x])/(b*x - ArcTanh[Tanh[a + b*x]])^2 + (b*Log[ArcTanh[Tanh[a + b
*x]]])/(b*x - ArcTanh[Tanh[a + b*x]])^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 2160

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]

Rule 2163

Int[(v_)^(n_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[v^(n + 1)/((n + 1)*
(b*u - a*v)), x] - Dist[(a*(n + 1))/((n + 1)*(b*u - a*v)), Int[v^(n + 1)/u, x], x] /; NeQ[b*u - a*v, 0]] /; Pi
ecewiseLinearQ[u, v, x] && LtQ[n, -1]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 45, normalized size = 0.69 \[ \frac {b x \left (\log \left (\tanh ^{-1}(\tanh (a+b x))\right )-\log (x)+1\right )-\tanh ^{-1}(\tanh (a+b x))}{x \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*ArcTanh[Tanh[a + b*x]]),x]

[Out]

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

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fricas [A]  time = 0.38, size = 26, normalized size = 0.40 \[ \frac {b x \log \left (b x + a\right ) - b x \log \relax (x) - a}{a^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/arctanh(tanh(b*x+a)),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.27, size = 30, normalized size = 0.46 \[ \frac {b \log \left ({\left | b x + a \right |}\right )}{a^{2}} - \frac {b \log \left ({\left | x \right |}\right )}{a^{2}} - \frac {1}{a x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/arctanh(tanh(b*x+a)),x, algorithm="giac")

[Out]

b*log(abs(b*x + a))/a^2 - b*log(abs(x))/a^2 - 1/(a*x)

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maple [A]  time = 0.14, size = 64, normalized size = 0.98 \[ -\frac {1}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x}-\frac {b \ln \relax (x )}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2}}+\frac {b \ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right )}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^2/arctanh(tanh(b*x+a)),x)

[Out]

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

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maxima [A]  time = 0.52, size = 28, normalized size = 0.43 \[ \frac {b \log \left (b x + a\right )}{a^{2}} - \frac {b \log \relax (x)}{a^{2}} - \frac {1}{a x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/arctanh(tanh(b*x+a)),x, algorithm="maxima")

[Out]

b*log(b*x + a)/a^2 - b*log(x)/a^2 - 1/(a*x)

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mupad [B]  time = 2.83, size = 210, normalized size = 3.23 \[ \frac {2\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-2\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+4\,b\,x+b\,x\,\mathrm {atan}\left (\frac {-\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,1{}\mathrm {i}+\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,1{}\mathrm {i}+b\,x\,2{}\mathrm {i}}{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x}\right )\,8{}\mathrm {i}}{x\,{\left (\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*atanh(tanh(a + b*x))),x)

[Out]

(2*log(1/(exp(2*a)*exp(2*b*x) + 1)) - 2*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 4*b*x + b*x*ata
n((log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))*1i - log(1/(exp(2*a)*exp(2*b*x) + 1))*1i + b*x*2i)/(lo
g(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))*8i)/(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)^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{2} \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**2/atanh(tanh(b*x+a)),x)

[Out]

Integral(1/(x**2*atanh(tanh(a + b*x))), x)

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