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

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

[Out]

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

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Rubi [A]  time = 0.06, 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} \[ -\frac {b^2 \log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3}+\frac {b^2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3}+\frac {1}{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}+\frac {b}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*ArcCoth[Tanh[a + b*x]]),x]

[Out]

b/(x*(b*x - ArcCoth[Tanh[a + b*x]])^2) + 1/(2*x^2*(b*x - ArcCoth[Tanh[a + b*x]])) - (b^2*Log[x])/(b*x - ArcCot
h[Tanh[a + b*x]])^3 + (b^2*Log[ArcCoth[Tanh[a + b*x]]])/(b*x - ArcCoth[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 \coth ^{-1}(\tanh (a+b x))} \, dx &=\frac {1}{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}-\frac {b \int \frac {1}{x^2 \coth ^{-1}(\tanh (a+b x))} \, dx}{-b x+\coth ^{-1}(\tanh (a+b x))}\\ &=\frac {b}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2}+\frac {1}{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}-\frac {b^2 \int \frac {1}{x \coth ^{-1}(\tanh (a+b x))} \, dx}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {b}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2}+\frac {1}{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}+\frac {b^2 \int \frac {1}{x} \, dx}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )}-\frac {b^3 \int \frac {1}{\coth ^{-1}(\tanh (a+b x))} \, dx}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {b}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2}+\frac {1}{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}-\frac {b^2 \log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3}-\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {b}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2}+\frac {1}{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}-\frac {b^2 \log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3}+\frac {b^2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 66, normalized size = 0.72 \[ \frac {b^2 x^2 \left (2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )-2 \log (x)+3\right )-4 b x \coth ^{-1}(\tanh (a+b x))+\coth ^{-1}(\tanh (a+b x))^2}{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.68, size = 199, normalized size = 2.16 \[ -\frac {2 \, {\left (\pi ^{4} a + 8 \, \pi ^{2} a^{3} + 16 \, a^{5} - 8 \, {\left (\pi ^{3} b^{2} - 12 \, \pi a^{2} b^{2}\right )} x^{2} \arctan \left (-\frac {2 \, b x + 2 \, a - \sqrt {4 \, b^{2} x^{2} + 8 \, a b x + \pi ^{2} + 4 \, a^{2}}}{\pi }\right ) - 4 \, {\left (3 \, \pi ^{2} a b^{2} - 4 \, a^{3} b^{2}\right )} x^{2} \log \left (4 \, b^{2} x^{2} + 8 \, a b x + \pi ^{2} + 4 \, a^{2}\right ) + 8 \, {\left (3 \, \pi ^{2} a b^{2} - 4 \, a^{3} b^{2}\right )} x^{2} \log \relax (x) + 2 \, {\left (\pi ^{4} b - 16 \, a^{4} b\right )} x\right )}}{{\left (\pi ^{6} + 12 \, \pi ^{4} a^{2} + 48 \, \pi ^{2} a^{4} + 64 \, a^{6}\right )} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(pi^4*a + 8*pi^2*a^3 + 16*a^5 - 8*(pi^3*b^2 - 12*pi*a^2*b^2)*x^2*arctan(-(2*b*x + 2*a - sqrt(4*b^2*x^2 + 8*
a*b*x + pi^2 + 4*a^2))/pi) - 4*(3*pi^2*a*b^2 - 4*a^3*b^2)*x^2*log(4*b^2*x^2 + 8*a*b*x + pi^2 + 4*a^2) + 8*(3*p
i^2*a*b^2 - 4*a^3*b^2)*x^2*log(x) + 2*(pi^4*b - 16*a^4*b)*x)/((pi^6 + 12*pi^4*a^2 + 48*pi^2*a^4 + 64*a^6)*x^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F(-1)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{3} \mathrm {arccoth}\left (\tanh \left (b x +a \right )\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 0.55, size = 108, normalized size = 1.17 \[ \frac {8 \, b^{2} \log \left (-i \, \pi + 2 \, b x + 2 \, a\right )}{-i \, \pi ^{3} + 6 \, \pi ^{2} a + 12 i \, \pi a^{2} - 8 \, a^{3}} - \frac {8 \, b^{2} \log \relax (x)}{-i \, \pi ^{3} + 6 \, \pi ^{2} a + 12 i \, \pi a^{2} - 8 \, a^{3}} - \frac {2 \, {\left (i \, \pi + 4 \, b x - 2 \, a\right )}}{{\left (2 \, \pi ^{2} + 8 i \, \pi a - 8 \, a^{2}\right )} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8*b^2*log(-I*pi + 2*b*x + 2*a)/(-I*pi^3 + 6*pi^2*a + 12*I*pi*a^2 - 8*a^3) - 8*b^2*log(x)/(-I*pi^3 + 6*pi^2*a +
12*I*pi*a^2 - 8*a^3) - 2*(I*pi + 4*b*x - 2*a)/((2*pi^2 + 8*I*pi*a - 8*a^2)*x^2)

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mupad [B]  time = 3.60, size = 300, normalized size = 3.26 \[ \frac {{\ln \left (-\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}^2-\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,\left (2\,\ln \left (-\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+8\,b\,x\right )+{\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}^2+12\,b^2\,x^2+8\,b\,x\,\ln \left (-\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+b^2\,x^2\,\mathrm {atan}\left (\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,1{}\mathrm {i}-\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,1{}\mathrm {i}+b\,x\,2{}\mathrm {i}}{\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x}\right )\,16{}\mathrm {i}}{x^2\,{\left (\ln \left (-\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*acoth(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 + b^2*x^2*at
an((log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1))*1i - log(-2/(exp(2*a)*exp(2*b*x) - 1))*1i + b*x*2i)
/(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))*16i + 8
*b*x*log(-1/(exp(2*a)*exp(2*b*x) - 1)))/(x^2*(log(-1/(exp(2*a)*exp(2*b*x) - 1)) - log((exp(2*a)*exp(2*b*x))/(e
xp(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 \[ \int \frac {1}{x^{3} \operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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