3.92 \(\int \frac {1}{x^3 \tanh ^{-1}(\tanh (a+b x))} \, dx\)

Optimal. Leaf size=92 \[ -\frac {b^2 \log (x)}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {b^2 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {1}{2 x^2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {b}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \]

[Out]

b/x/(b*x-arctanh(tanh(b*x+a)))^2+1/2/x^2/(b*x-arctanh(tanh(b*x+a)))-b^2*ln(x)/(b*x-arctanh(tanh(b*x+a)))^3+b^2
*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 = 92, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2163, 2160, 2157, 29} \[ -\frac {b^2 \log (x)}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {b^2 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {1}{2 x^2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {b}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*ArcTanh[Tanh[a + b*x]]),x]

[Out]

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

Rubi steps

\begin {align*} \int \frac {1}{x^3 \tanh ^{-1}(\tanh (a+b x))} \, dx &=\frac {1}{2 x^2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {b \int \frac {1}{x^2 \tanh ^{-1}(\tanh (a+b x))} \, dx}{-b x+\tanh ^{-1}(\tanh (a+b x))}\\ &=\frac {b}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {1}{2 x^2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {b^2 \int \frac {1}{x \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 )}\\ &=\frac {b}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {1}{2 x^2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {b^2 \int \frac {1}{x} \, dx}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {b^3 \int \frac {1}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {b}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {1}{2 x^2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {b^2 \log (x)}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}-\frac {b^2 \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 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {b}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {1}{2 x^2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {b^2 \log (x)}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {b^2 \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 = 0.04, size = 66, normalized size = 0.72 \[ \frac {b^2 x^2 \left (2 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )-2 \log (x)+3\right )-4 b x \tanh ^{-1}(\tanh (a+b x))+\tanh ^{-1}(\tanh (a+b x))^2}{2 x^2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*ArcTanh[Tanh[a + b*x]]),x]

[Out]

(-4*b*x*ArcTanh[Tanh[a + b*x]] + ArcTanh[Tanh[a + b*x]]^2 + b^2*x^2*(3 - 2*Log[x] + 2*Log[ArcTanh[Tanh[a + b*x
]]]))/(2*x^2*(b*x - ArcTanh[Tanh[a + b*x]])^3)

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fricas [A]  time = 0.50, size = 41, normalized size = 0.45 \[ -\frac {2 \, b^{2} x^{2} \log \left (b x + a\right ) - 2 \, b^{2} x^{2} \log \relax (x) - 2 \, a b x + a^{2}}{2 \, a^{3} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/arctanh(tanh(b*x+a)),x, algorithm="fricas")

[Out]

-1/2*(2*b^2*x^2*log(b*x + a) - 2*b^2*x^2*log(x) - 2*a*b*x + a^2)/(a^3*x^2)

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giac [A]  time = 0.14, size = 45, normalized size = 0.49 \[ -\frac {b^{2} \log \left ({\left | b x + a \right |}\right )}{a^{3}} + \frac {b^{2} \log \left ({\left | x \right |}\right )}{a^{3}} + \frac {2 \, a b x - a^{2}}{2 \, a^{3} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/arctanh(tanh(b*x+a)),x, algorithm="giac")

[Out]

-b^2*log(abs(b*x + a))/a^3 + b^2*log(abs(x))/a^3 + 1/2*(2*a*b*x - a^2)/(a^3*x^2)

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maple [A]  time = 0.15, size = 87, normalized size = 0.95 \[ -\frac {1}{2 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{2}}+\frac {b^{2} \ln \relax (x )}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3}}+\frac {b}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} x}-\frac {b^{2} \ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right )}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/arctanh(tanh(b*x+a)),x)

[Out]

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

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maxima [A]  time = 0.52, size = 40, normalized size = 0.43 \[ -\frac {b^{2} \log \left (b x + a\right )}{a^{3}} + \frac {b^{2} \log \relax (x)}{a^{3}} + \frac {2 \, b x - a}{2 \, a^{2} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/arctanh(tanh(b*x+a)),x, algorithm="maxima")

[Out]

-b^2*log(b*x + a)/a^3 + b^2*log(x)/a^3 + 1/2*(2*b*x - a)/(a^2*x^2)

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mupad [B]  time = 2.89, size = 286, normalized size = 3.11 \[ \frac {{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2-\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\left (2\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+8\,b\,x\right )+{\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2+12\,b^2\,x^2+8\,b\,x\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+b^2\,x^2\,\mathrm {atan}\left (\frac {-\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,1{}\mathrm {i}+\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,1{}\mathrm {i}+b\,x\,2{}\mathrm {i}}{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x}\right )\,16{}\mathrm {i}}{x^2\,{\left (\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*atanh(tanh(a + b*x))),x)

[Out]

(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))*(2*log(1/(exp(2*a)*
exp(2*b*x) + 1)) + 8*b*x) + log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))^2 + 12*b^2*x^2 + 8*b*x*log(1/
(exp(2*a)*exp(2*b*x) + 1)) + b^2*x^2*atan((log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))*1i - log(1/(ex
p(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)*e
xp(2*b*x) + 1)) + 2*b*x))*16i)/(x^2*(log(1/(exp(2*a)*exp(2*b*x) + 1)) - log((exp(2*a)*exp(2*b*x))/(exp(2*a)*ex
p(2*b*x) + 1)) + 2*b*x)^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{3} \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/atanh(tanh(b*x+a)),x)

[Out]

Integral(1/(x**3*atanh(tanh(a + b*x))), x)

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