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

Optimal. Leaf size=97 \[ \frac{1}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \coth ^{-1}(\tanh (a+b x))}-\frac{1}{2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))^2}-\frac{\log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3}+\frac{\log \left (\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0658827, antiderivative size = 97, normalized size of antiderivative = 1., 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} \[ \frac{1}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \coth ^{-1}(\tanh (a+b x))}-\frac{1}{2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))^2}-\frac{\log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3}+\frac{\log \left (\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*ArcCoth[Tanh[a + b*x]]^3),x]

[Out]

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

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 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 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 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rubi steps

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

Mathematica [A]  time = 0.106451, size = 74, normalized size = 0.76 \[ \frac{-4 b x \coth ^{-1}(\tanh (a+b x))+\coth ^{-1}(\tanh (a+b x))^2 \left (-2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )+2 \log (b x)+3\right )+b^2 x^2}{2 \coth ^{-1}(\tanh (a+b x))^2 \left (\coth ^{-1}(\tanh (a+b x))-b x\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 180., size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{x \left ({\rm arccoth} \left (\tanh \left ( bx+a \right ) \right ) \right ) ^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [C]  time = 3.55119, size = 234, normalized size = 2.41 \begin{align*} \frac{8 \,{\left (-3 i \, \pi + 4 \, b x + 6 \, a\right )}}{2 \, \pi ^{4} + 16 i \, \pi ^{3} a - 48 \, \pi ^{2} a^{2} - 64 i \, \pi a^{3} + 32 \, a^{4} -{\left (8 \, \pi ^{2} b^{2} + 32 i \, \pi a b^{2} - 32 \, a^{2} b^{2}\right )} x^{2} +{\left (8 i \, \pi ^{3} b - 48 \, \pi ^{2} a b - 96 i \, \pi a^{2} b + 64 \, a^{3} b\right )} x} + \frac{8 \, \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 \, \log \left (x\right )}{-i \, \pi ^{3} + 6 \, \pi ^{2} a + 12 i \, \pi a^{2} - 8 \, a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8*(-3*I*pi + 4*b*x + 6*a)/(2*pi^4 + 16*I*pi^3*a - 48*pi^2*a^2 - 64*I*pi*a^3 + 32*a^4 - (8*pi^2*b^2 + 32*I*pi*a
*b^2 - 32*a^2*b^2)*x^2 + (8*I*pi^3*b - 48*pi^2*a*b - 96*I*pi*a^2*b + 64*a^3*b)*x) + 8*log(-I*pi + 2*b*x + 2*a)
/(-I*pi^3 + 6*pi^2*a + 12*I*pi*a^2 - 8*a^3) - 8*log(x)/(-I*pi^3 + 6*pi^2*a + 12*I*pi*a^2 - 8*a^3)

________________________________________________________________________________________

Fricas [B]  time = 1.93464, size = 1841, normalized size = 18.98 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8*(9*pi^6*a + 60*pi^4*a^3 + 48*pi^2*a^5 - 192*a^7 + 8*(pi^4*b^3 - 16*a^4*b^3)*x^3 + 4*(9*pi^4*a*b^2 + 8*pi^2*
a^3*b^2 - 112*a^5*b^2)*x^2 + 4*(pi^6*b + 16*pi^4*a^2*b + 16*pi^2*a^4*b - 128*a^6*b)*x - 2*(pi^7 - 4*pi^5*a^2 -
80*pi^3*a^4 - 192*pi*a^6 + 16*(pi^3*b^4 - 12*pi*a^2*b^4)*x^4 + 64*(pi^3*a*b^3 - 12*pi*a^3*b^3)*x^3 + 8*(pi^5*
b^2 - 144*pi*a^4*b^2)*x^2 + 16*(pi^5*a*b - 8*pi^3*a^3*b - 48*pi*a^5*b)*x)*arctan(-(2*b*x + 2*a - sqrt(4*b^2*x^
2 + 8*a*b*x + pi^2 + 4*a^2))/pi) - (3*pi^6*a + 20*pi^4*a^3 + 16*pi^2*a^5 - 64*a^7 + 16*(3*pi^2*a*b^4 - 4*a^3*b
^4)*x^4 + 64*(3*pi^2*a^2*b^3 - 4*a^4*b^3)*x^3 + 8*(3*pi^4*a*b^2 + 32*pi^2*a^3*b^2 - 48*a^5*b^2)*x^2 + 16*(3*pi
^4*a^2*b + 8*pi^2*a^4*b - 16*a^6*b)*x)*log(4*b^2*x^2 + 8*a*b*x + pi^2 + 4*a^2) + 2*(3*pi^6*a + 20*pi^4*a^3 + 1
6*pi^2*a^5 - 64*a^7 + 16*(3*pi^2*a*b^4 - 4*a^3*b^4)*x^4 + 64*(3*pi^2*a^2*b^3 - 4*a^4*b^3)*x^3 + 8*(3*pi^4*a*b^
2 + 32*pi^2*a^3*b^2 - 48*a^5*b^2)*x^2 + 16*(3*pi^4*a^2*b + 8*pi^2*a^4*b - 16*a^6*b)*x)*log(x))/(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 + 16*(pi^6*b^4 + 12*pi^4*a^2*b^4 + 48*pi^2*a^4
*b^4 + 64*a^6*b^4)*x^4 + 64*(pi^6*a*b^3 + 12*pi^4*a^3*b^3 + 48*pi^2*a^5*b^3 + 64*a^7*b^3)*x^3 + 8*(pi^8*b^2 +
24*pi^6*a^2*b^2 + 192*pi^4*a^4*b^2 + 640*pi^2*a^6*b^2 + 768*a^8*b^2)*x^2 + 16*(pi^8*a*b + 16*pi^6*a^3*b + 96*p
i^4*a^5*b + 256*pi^2*a^7*b + 256*a^9*b)*x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x \operatorname{acoth}^{3}{\left (\tanh{\left (a + b x \right )} \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x \operatorname{arcoth}\left (\tanh \left (b x + a\right )\right )^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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