3.174 \(\int \frac {1}{x^3 \coth ^{-1}(\tanh (a+b x))^2} \, dx\)

Optimal. Leaf size=143 \[ -\frac {3 b^2}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3 \coth ^{-1}(\tanh (a+b x))}+\frac {3 b^2 \log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^4}-\frac {3 b^2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^4}+\frac {1}{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))}+\frac {3 b}{2 x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \coth ^{-1}(\tanh (a+b x))} \]

[Out]

-3*b^2/(b*x-arccoth(tanh(b*x+a)))^3/arccoth(tanh(b*x+a))+3/2*b/x/(b*x-arccoth(tanh(b*x+a)))^2/arccoth(tanh(b*x
+a))+1/2/x^2/(b*x-arccoth(tanh(b*x+a)))/arccoth(tanh(b*x+a))+3*b^2*ln(x)/(b*x-arccoth(tanh(b*x+a)))^4-3*b^2*ln
(arccoth(tanh(b*x+a)))/(b*x-arccoth(tanh(b*x+a)))^4

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Rubi [A]  time = 0.09, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 7, 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} \[ -\frac {3 b^2}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3 \coth ^{-1}(\tanh (a+b x))}+\frac {3 b^2 \log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^4}-\frac {3 b^2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^4}+\frac {1}{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))}+\frac {3 b}{2 x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \coth ^{-1}(\tanh (a+b x))} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*ArcCoth[Tanh[a + b*x]]^2),x]

[Out]

(-3*b^2)/((b*x - ArcCoth[Tanh[a + b*x]])^3*ArcCoth[Tanh[a + b*x]]) + (3*b)/(2*x*(b*x - ArcCoth[Tanh[a + b*x]])
^2*ArcCoth[Tanh[a + b*x]]) + 1/(2*x^2*(b*x - ArcCoth[Tanh[a + b*x]])*ArcCoth[Tanh[a + b*x]]) + (3*b^2*Log[x])/
(b*x - ArcCoth[Tanh[a + b*x]])^4 - (3*b^2*Log[ArcCoth[Tanh[a + b*x]]])/(b*x - ArcCoth[Tanh[a + b*x]])^4

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^3 \coth ^{-1}(\tanh (a+b x))^2} \, dx &=\frac {1}{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))}+\frac {(3 b) \int \frac {1}{x^2 \coth ^{-1}(\tanh (a+b x))^2} \, dx}{2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {3 b}{2 x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \coth ^{-1}(\tanh (a+b x))}+\frac {1}{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))}-\frac {\left (3 b^2\right ) \int \frac {1}{x \coth ^{-1}(\tanh (a+b x))^2} \, dx}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )}\\ &=-\frac {3 b^2}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3 \coth ^{-1}(\tanh (a+b x))}+\frac {3 b}{2 x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \coth ^{-1}(\tanh (a+b x))}+\frac {1}{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))}-\frac {\left (3 b^2\right ) \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 )^2}\\ &=-\frac {3 b^2}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3 \coth ^{-1}(\tanh (a+b x))}+\frac {3 b}{2 x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \coth ^{-1}(\tanh (a+b x))}+\frac {1}{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))}+\frac {\left (3 b^2\right ) \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 )^2}-\frac {\left (3 b^3\right ) \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 )^2}\\ &=-\frac {3 b^2}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3 \coth ^{-1}(\tanh (a+b x))}+\frac {3 b}{2 x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \coth ^{-1}(\tanh (a+b x))}+\frac {1}{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))}+\frac {3 b^2 \log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^4}-\frac {\left (3 b^2\right ) \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 )^2}\\ &=-\frac {3 b^2}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3 \coth ^{-1}(\tanh (a+b x))}+\frac {3 b}{2 x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \coth ^{-1}(\tanh (a+b x))}+\frac {1}{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))}+\frac {3 b^2 \log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^4}-\frac {3 b^2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^4}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 92, normalized size = 0.64 \[ -\frac {-3 b^2 x^2 \coth ^{-1}(\tanh (a+b x)) \left (-2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )+2 \log (x)-1\right )-6 b x \coth ^{-1}(\tanh (a+b x))^2+\coth ^{-1}(\tanh (a+b x))^3+2 b^3 x^3}{2 x^2 \coth ^{-1}(\tanh (a+b x)) \left (\coth ^{-1}(\tanh (a+b x))-b x\right )^4} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*ArcCoth[Tanh[a + b*x]]^2),x]

[Out]

-1/2*(2*b^3*x^3 - 6*b*x*ArcCoth[Tanh[a + b*x]]^2 + ArcCoth[Tanh[a + b*x]]^3 - 3*b^2*x^2*ArcCoth[Tanh[a + b*x]]
*(-1 + 2*Log[x] - 2*Log[ArcCoth[Tanh[a + b*x]]]))/(x^2*ArcCoth[Tanh[a + b*x]]*(-(b*x) + ArcCoth[Tanh[a + b*x]]
)^4)

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fricas [B]  time = 0.55, size = 644, normalized size = 4.50 \[ \frac {2 \, {\left (\pi ^{8} + 8 \, \pi ^{6} a^{2} - 128 \, \pi ^{2} a^{6} - 256 \, a^{8} - 96 \, {\left (3 \, \pi ^{4} a b^{3} + 8 \, \pi ^{2} a^{3} b^{3} - 16 \, a^{5} b^{3}\right )} x^{3} + 12 \, {\left (\pi ^{6} b^{2} - 44 \, \pi ^{4} a^{2} b^{2} - 144 \, \pi ^{2} a^{4} b^{2} + 192 \, a^{6} b^{2}\right )} x^{2} - 8 \, {\left (5 \, \pi ^{6} a b + 36 \, \pi ^{4} a^{3} b + 48 \, \pi ^{2} a^{5} b - 64 \, a^{7} b\right )} x + 384 \, {\left (4 \, {\left (\pi ^{3} a b^{4} - 4 \, \pi a^{3} b^{4}\right )} x^{4} + 8 \, {\left (\pi ^{3} a^{2} b^{3} - 4 \, \pi a^{4} b^{3}\right )} x^{3} + {\left (\pi ^{5} a b^{2} - 16 \, \pi a^{5} b^{2}\right )} x^{2}\right )} \arctan \left (-\frac {2 \, b x + 2 \, a - \sqrt {4 \, b^{2} x^{2} + 8 \, a b x + \pi ^{2} + 4 \, a^{2}}}{\pi }\right ) - 12 \, {\left (4 \, {\left (\pi ^{4} b^{4} - 24 \, \pi ^{2} a^{2} b^{4} + 16 \, a^{4} b^{4}\right )} x^{4} + 8 \, {\left (\pi ^{4} a b^{3} - 24 \, \pi ^{2} a^{3} b^{3} + 16 \, a^{5} b^{3}\right )} x^{3} + {\left (\pi ^{6} b^{2} - 20 \, \pi ^{4} a^{2} b^{2} - 80 \, \pi ^{2} a^{4} b^{2} + 64 \, a^{6} b^{2}\right )} x^{2}\right )} \log \left (4 \, b^{2} x^{2} + 8 \, a b x + \pi ^{2} + 4 \, a^{2}\right ) + 24 \, {\left (4 \, {\left (\pi ^{4} b^{4} - 24 \, \pi ^{2} a^{2} b^{4} + 16 \, a^{4} b^{4}\right )} x^{4} + 8 \, {\left (\pi ^{4} a b^{3} - 24 \, \pi ^{2} a^{3} b^{3} + 16 \, a^{5} b^{3}\right )} x^{3} + {\left (\pi ^{6} b^{2} - 20 \, \pi ^{4} a^{2} b^{2} - 80 \, \pi ^{2} a^{4} b^{2} + 64 \, a^{6} b^{2}\right )} x^{2}\right )} \log \relax (x)\right )}}{4 \, {\left (\pi ^{8} b^{2} + 16 \, \pi ^{6} a^{2} b^{2} + 96 \, \pi ^{4} a^{4} b^{2} + 256 \, \pi ^{2} a^{6} b^{2} + 256 \, a^{8} b^{2}\right )} x^{4} + 8 \, {\left (\pi ^{8} a b + 16 \, \pi ^{6} a^{3} b + 96 \, \pi ^{4} a^{5} b + 256 \, \pi ^{2} a^{7} b + 256 \, a^{9} b\right )} x^{3} + {\left (\pi ^{10} + 20 \, \pi ^{8} a^{2} + 160 \, \pi ^{6} a^{4} + 640 \, \pi ^{4} a^{6} + 1280 \, \pi ^{2} a^{8} + 1024 \, a^{10}\right )} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/arccoth(tanh(b*x+a))^2,x, algorithm="fricas")

[Out]

2*(pi^8 + 8*pi^6*a^2 - 128*pi^2*a^6 - 256*a^8 - 96*(3*pi^4*a*b^3 + 8*pi^2*a^3*b^3 - 16*a^5*b^3)*x^3 + 12*(pi^6
*b^2 - 44*pi^4*a^2*b^2 - 144*pi^2*a^4*b^2 + 192*a^6*b^2)*x^2 - 8*(5*pi^6*a*b + 36*pi^4*a^3*b + 48*pi^2*a^5*b -
64*a^7*b)*x + 384*(4*(pi^3*a*b^4 - 4*pi*a^3*b^4)*x^4 + 8*(pi^3*a^2*b^3 - 4*pi*a^4*b^3)*x^3 + (pi^5*a*b^2 - 16
*pi*a^5*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) - 12*(4*(pi^4*b^4 - 24*
pi^2*a^2*b^4 + 16*a^4*b^4)*x^4 + 8*(pi^4*a*b^3 - 24*pi^2*a^3*b^3 + 16*a^5*b^3)*x^3 + (pi^6*b^2 - 20*pi^4*a^2*b
^2 - 80*pi^2*a^4*b^2 + 64*a^6*b^2)*x^2)*log(4*b^2*x^2 + 8*a*b*x + pi^2 + 4*a^2) + 24*(4*(pi^4*b^4 - 24*pi^2*a^
2*b^4 + 16*a^4*b^4)*x^4 + 8*(pi^4*a*b^3 - 24*pi^2*a^3*b^3 + 16*a^5*b^3)*x^3 + (pi^6*b^2 - 20*pi^4*a^2*b^2 - 80
*pi^2*a^4*b^2 + 64*a^6*b^2)*x^2)*log(x))/(4*(pi^8*b^2 + 16*pi^6*a^2*b^2 + 96*pi^4*a^4*b^2 + 256*pi^2*a^6*b^2 +
256*a^8*b^2)*x^4 + 8*(pi^8*a*b + 16*pi^6*a^3*b + 96*pi^4*a^5*b + 256*pi^2*a^7*b + 256*a^9*b)*x^3 + (pi^10 + 2
0*pi^8*a^2 + 160*pi^6*a^4 + 640*pi^4*a^6 + 1280*pi^2*a^8 + 1024*a^10)*x^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{3} \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/arccoth(tanh(b*x+a))^2,x, algorithm="giac")

[Out]

integrate(1/(x^3*arccoth(tanh(b*x + a))^2), x)

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maple [F(-1)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{3} \mathrm {arccoth}\left (\tanh \left (b x +a \right )\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/arccoth(tanh(b*x+a))^2,x)

[Out]

int(1/x^3/arccoth(tanh(b*x+a))^2,x)

________________________________________________________________________________________

maxima [C]  time = 0.76, size = 191, normalized size = 1.34 \[ -\frac {48 \, b^{2} \log \left (-i \, \pi + 2 \, b x + 2 \, a\right )}{\pi ^{4} + 8 i \, \pi ^{3} a - 24 \, \pi ^{2} a^{2} - 32 i \, \pi a^{3} + 16 \, a^{4}} + \frac {48 \, b^{2} \log \relax (x)}{\pi ^{4} + 8 i \, \pi ^{3} a - 24 \, \pi ^{2} a^{2} - 32 i \, \pi a^{3} + 16 \, a^{4}} - \frac {4 \, {\left (24 \, b^{2} x^{2} + \pi ^{2} + 4 i \, \pi a - 4 \, a^{2} + {\left (-6 i \, \pi b + 12 \, a b\right )} x\right )}}{{\left (-4 i \, \pi ^{3} b + 24 \, \pi ^{2} a b + 48 i \, \pi a^{2} b - 32 \, a^{3} b\right )} x^{3} - {\left (2 \, \pi ^{4} + 16 i \, \pi ^{3} a - 48 \, \pi ^{2} a^{2} - 64 i \, \pi a^{3} + 32 \, a^{4}\right )} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/arccoth(tanh(b*x+a))^2,x, algorithm="maxima")

[Out]

-48*b^2*log(-I*pi + 2*b*x + 2*a)/(pi^4 + 8*I*pi^3*a - 24*pi^2*a^2 - 32*I*pi*a^3 + 16*a^4) + 48*b^2*log(x)/(pi^
4 + 8*I*pi^3*a - 24*pi^2*a^2 - 32*I*pi*a^3 + 16*a^4) - 4*(24*b^2*x^2 + pi^2 + 4*I*pi*a - 4*a^2 + (-6*I*pi*b +
12*a*b)*x)/((-4*I*pi^3*b + 24*pi^2*a*b + 48*I*pi*a^2*b - 32*a^3*b)*x^3 - (2*pi^4 + 16*I*pi^3*a - 48*pi^2*a^2 -
64*I*pi*a^3 + 32*a^4)*x^2)

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mupad [B]  time = 4.90, size = 689, normalized size = 4.82 \[ \frac {2\,{\ln \left (-\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}^3-2\,{\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}^3-6\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,{\ln \left (-\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}^2+6\,{\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}^2\,\ln \left (-\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-32\,b^3\,x^3+24\,b\,x\,{\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}^2+24\,b\,x\,{\ln \left (-\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}^2-24\,b^2\,x^2\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+24\,b^2\,x^2\,\ln \left (-\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-b^2\,x^2\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,\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 )\,96{}\mathrm {i}-48\,b\,x\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,\ln \left (-\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+b^2\,x^2\,\ln \left (-\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,\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 )\,96{}\mathrm {i}}{x^2\,\left (\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (-\frac {1}{{\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 )}^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*acoth(tanh(a + b*x))^2),x)

[Out]

(2*log(-1/(exp(2*a)*exp(2*b*x) - 1))^3 - 2*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1))^3 - 6*log((exp
(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1))*log(-1/(exp(2*a)*exp(2*b*x) - 1))^2 + 6*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)) - 32*b^3*x^3 + 24*b*x*log((exp(2*a)*exp(2*b*x)
)/(exp(2*a)*exp(2*b*x) - 1))^2 + 24*b*x*log(-1/(exp(2*a)*exp(2*b*x) - 1))^2 - 24*b^2*x^2*log((exp(2*a)*exp(2*b
*x))/(exp(2*a)*exp(2*b*x) - 1)) + 24*b^2*x^2*log(-1/(exp(2*a)*exp(2*b*x) - 1)) - b^2*x^2*log((exp(2*a)*exp(2*b
*x))/(exp(2*a)*exp(2*b*x) - 1))*atan((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))*96i - 48*b*x*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1))*log(-1/(exp(2*a)*
exp(2*b*x) - 1)) + b^2*x^2*log(-1/(exp(2*a)*exp(2*b*x) - 1))*atan((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*ex
p(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + 2*b*x))*96i)/(x^2*(log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b
*x) - 1)) - log(-1/(exp(2*a)*exp(2*b*x) - 1)))*(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)^4)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{3} \operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/acoth(tanh(b*x+a))**2,x)

[Out]

Integral(1/(x**3*acoth(tanh(a + b*x))**2), x)

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