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

Optimal. Leaf size=102 \[ -\frac {2 b}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))}+\frac {1}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))}+\frac {2 b \log (x)}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}-\frac {2 b \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3} \]

[Out]

-2*b/(b*x-arctanh(tanh(b*x+a)))^2/arctanh(tanh(b*x+a))+1/x/(b*x-arctanh(tanh(b*x+a)))/arctanh(tanh(b*x+a))+2*b
*ln(x)/(b*x-arctanh(tanh(b*x+a)))^3-2*b*ln(arctanh(tanh(b*x+a)))/(b*x-arctanh(tanh(b*x+a)))^3

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Rubi [A]  time = 0.07, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {2171, 2163, 2160, 2157, 29} \[ -\frac {2 b}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))}+\frac {1}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))}+\frac {2 b \log (x)}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}-\frac {2 b \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b)/((b*x - ArcTanh[Tanh[a + b*x]])^2*ArcTanh[Tanh[a + b*x]]) + 1/(x*(b*x - ArcTanh[Tanh[a + b*x]])*ArcTanh
[Tanh[a + b*x]]) + (2*b*Log[x])/(b*x - ArcTanh[Tanh[a + b*x]])^3 - (2*b*Log[ArcTanh[Tanh[a + b*x]]])/(b*x - Ar
cTanh[Tanh[a + b*x]])^3

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]

Rule 2171

Int[(u_)^(m_)*(v_)^(n_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, -Simp[(u^(m + 1)*v^
(n + 1))/((m + 1)*(b*u - a*v)), x] + Dist[(b*(m + n + 2))/((m + 1)*(b*u - a*v)), Int[u^(m + 1)*v^n, x], x] /;
NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && NeQ[m + n + 2, 0] && LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {1}{x^2 \tanh ^{-1}(\tanh (a+b x))^2} \, dx &=\frac {1}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))}-\frac {(2 b) \int \frac {1}{x \tanh ^{-1}(\tanh (a+b x))^2} \, dx}{-b x+\tanh ^{-1}(\tanh (a+b x))}\\ &=-\frac {2 b}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))}+\frac {1}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))}-\frac {(2 b) \int \frac {1}{x \tanh ^{-1}(\tanh (a+b x))} \, dx}{\left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=-\frac {2 b}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))}+\frac {1}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))}+\frac {(2 b) \int \frac {1}{x} \, dx}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2}-\frac {\left (2 b^2\right ) \int \frac {1}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=-\frac {2 b}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))}+\frac {1}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))}+\frac {2 b \log (x)}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}-\frac {(2 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 ) \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=-\frac {2 b}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))}+\frac {1}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))}+\frac {2 b \log (x)}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}-\frac {2 b \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 70, normalized size = 0.69 \[ \frac {\tanh ^{-1}(\tanh (a+b x))^2+2 b x \tanh ^{-1}(\tanh (a+b x)) \left (\log (x)-\log \left (\tanh ^{-1}(\tanh (a+b x))\right )\right )-b^2 x^2}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(b^2*x^2) + ArcTanh[Tanh[a + b*x]]^2 + 2*b*x*ArcTanh[Tanh[a + b*x]]*(Log[x] - Log[ArcTanh[Tanh[a + b*x]]]))/
(x*(b*x - ArcTanh[Tanh[a + b*x]])^3*ArcTanh[Tanh[a + b*x]])

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fricas [A]  time = 0.54, size = 63, normalized size = 0.62 \[ -\frac {2 \, a b x + a^{2} - 2 \, {\left (b^{2} x^{2} + a b x\right )} \log \left (b x + a\right ) + 2 \, {\left (b^{2} x^{2} + a b x\right )} \log \relax (x)}{a^{3} b x^{2} + a^{4} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*a*b*x + a^2 - 2*(b^2*x^2 + a*b*x)*log(b*x + a) + 2*(b^2*x^2 + a*b*x)*log(x))/(a^3*b*x^2 + a^4*x)

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giac [A]  time = 0.16, size = 45, normalized size = 0.44 \[ \frac {2 \, b \log \left ({\left | b x + a \right |}\right )}{a^{3}} - \frac {2 \, b \log \left ({\left | x \right |}\right )}{a^{3}} - \frac {2 \, b x + a}{{\left (b x^{2} + a x\right )} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.15, size = 91, normalized size = 0.89 \[ -\frac {1}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} x}-\frac {2 b \ln \relax (x )}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3}}-\frac {b}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \arctanh \left (\tanh \left (b x +a \right )\right )}+\frac {2 b \ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right )}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(arctanh(tanh(b*x+a))-b*x)^2/x-2/(arctanh(tanh(b*x+a))-b*x)^3*b*ln(x)-1/(arctanh(tanh(b*x+a))-b*x)^2*b/arct
anh(tanh(b*x+a))+2/(arctanh(tanh(b*x+a))-b*x)^3*b*ln(arctanh(tanh(b*x+a)))

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maxima [A]  time = 0.74, size = 45, normalized size = 0.44 \[ -\frac {2 \, b x + a}{a^{2} b x^{2} + a^{3} x} + \frac {2 \, b \log \left (b x + a\right )}{a^{3}} - \frac {2 \, b \log \relax (x)}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.44, size = 432, normalized size = 4.24 \[ -\frac {4\,{\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 )\,\left (8\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+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 )\,32{}\mathrm {i}\right )+4\,{\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2-16\,b^2\,x^2+b\,x\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\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 )\,32{}\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 )\right )\,{\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 )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*atanh(tanh(a + b*x))^2),x)

[Out]

-(4*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))*(8*log(1/(exp(2*
a)*exp(2*b*x) + 1)) + b*x*atan((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)/(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))*32i) + 4*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))^2 - 16*b^2*x^2 + b*x*log(1/(exp(2
*a)*exp(2*b*x) + 1))*atan((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)/(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))*32i)/(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)))*(l
og(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)^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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