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

Optimal. Leaf size=170 \[ \frac{6 b^2}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^4 \coth ^{-1}(\tanh (a+b x))}-\frac{3 b^2}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3 \coth ^{-1}(\tanh (a+b x))^2}-\frac{6 b^2 \log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^5}+\frac{6 b^2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^5}+\frac{1}{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))^2}+\frac{2 b}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \coth ^{-1}(\tanh (a+b x))^2} \]

[Out]

(-3*b^2)/((b*x - ArcCoth[Tanh[a + b*x]])^3*ArcCoth[Tanh[a + b*x]]^2) + (2*b)/(x*(b*x - ArcCoth[Tanh[a + b*x]])
^2*ArcCoth[Tanh[a + b*x]]^2) + 1/(2*x^2*(b*x - ArcCoth[Tanh[a + b*x]])*ArcCoth[Tanh[a + b*x]]^2) + (6*b^2)/((b
*x - ArcCoth[Tanh[a + b*x]])^4*ArcCoth[Tanh[a + b*x]]) - (6*b^2*Log[x])/(b*x - ArcCoth[Tanh[a + b*x]])^5 + (6*
b^2*Log[ArcCoth[Tanh[a + b*x]]])/(b*x - ArcCoth[Tanh[a + b*x]])^5

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Rubi [A]  time = 0.126766, antiderivative size = 170, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ \frac{6 b^2}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^4 \coth ^{-1}(\tanh (a+b x))}-\frac{3 b^2}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3 \coth ^{-1}(\tanh (a+b x))^2}-\frac{6 b^2 \log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^5}+\frac{6 b^2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^5}+\frac{1}{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))^2}+\frac{2 b}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \coth ^{-1}(\tanh (a+b x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b^2)/((b*x - ArcCoth[Tanh[a + b*x]])^3*ArcCoth[Tanh[a + b*x]]^2) + (2*b)/(x*(b*x - ArcCoth[Tanh[a + b*x]])
^2*ArcCoth[Tanh[a + b*x]]^2) + 1/(2*x^2*(b*x - ArcCoth[Tanh[a + b*x]])*ArcCoth[Tanh[a + b*x]]^2) + (6*b^2)/((b
*x - ArcCoth[Tanh[a + b*x]])^4*ArcCoth[Tanh[a + b*x]]) - (6*b^2*Log[x])/(b*x - ArcCoth[Tanh[a + b*x]])^5 + (6*
b^2*Log[ArcCoth[Tanh[a + b*x]]])/(b*x - ArcCoth[Tanh[a + b*x]])^5

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]

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^3 \coth ^{-1}(\tanh (a+b x))^3} \, dx &=\frac{1}{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))^2}+\frac{(2 b) \int \frac{1}{x^2 \coth ^{-1}(\tanh (a+b x))^3} \, dx}{b x-\coth ^{-1}(\tanh (a+b x))}\\ &=\frac{2 b}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \coth ^{-1}(\tanh (a+b x))^2}+\frac{1}{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))^2}-\frac{\left (6 b^2\right ) \int \frac{1}{x \coth ^{-1}(\tanh (a+b x))^3} \, 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))^2}+\frac{2 b}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \coth ^{-1}(\tanh (a+b x))^2}+\frac{1}{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))^2}+\frac{\left (6 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 )^2 \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))^2}+\frac{2 b}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \coth ^{-1}(\tanh (a+b x))^2}+\frac{1}{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))^2}+\frac{6 b^2}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^4 \coth ^{-1}(\tanh (a+b x))}+\frac{\left (6 b^2\right ) \int \frac{1}{x \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))^2}+\frac{2 b}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \coth ^{-1}(\tanh (a+b x))^2}+\frac{1}{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))^2}+\frac{6 b^2}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^4 \coth ^{-1}(\tanh (a+b x))}-\frac{\left (6 b^2\right ) \int \frac{1}{x} \, dx}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3 \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )^2}+\frac{\left (6 b^3\right ) \int \frac{1}{\coth ^{-1}(\tanh (a+b x))} \, dx}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3 \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))^2}+\frac{2 b}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \coth ^{-1}(\tanh (a+b x))^2}+\frac{1}{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))^2}+\frac{6 b^2}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^4 \coth ^{-1}(\tanh (a+b x))}-\frac{6 b^2 \log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^5}+\frac{\left (6 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 )^3 \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))^2}+\frac{2 b}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \coth ^{-1}(\tanh (a+b x))^2}+\frac{1}{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))^2}+\frac{6 b^2}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^4 \coth ^{-1}(\tanh (a+b x))}-\frac{6 b^2 \log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^5}+\frac{6 b^2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^5}\\ \end{align*}

Mathematica [A]  time = 0.0433847, size = 107, normalized size = 0.63 \[ \frac{8 b^3 x^3 \coth ^{-1}(\tanh (a+b x))-12 b^2 x^2 \coth ^{-1}(\tanh (a+b x))^2 \left (\log (x)-\log \left (\coth ^{-1}(\tanh (a+b x))\right )\right )-8 b x \coth ^{-1}(\tanh (a+b x))^3+\coth ^{-1}(\tanh (a+b x))^4-b^4 x^4}{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^5 \coth ^{-1}(\tanh (a+b x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(b^4*x^4) + 8*b^3*x^3*ArcCoth[Tanh[a + b*x]] - 8*b*x*ArcCoth[Tanh[a + b*x]]^3 + ArcCoth[Tanh[a + b*x]]^4 - 1
2*b^2*x^2*ArcCoth[Tanh[a + b*x]]^2*(Log[x] - Log[ArcCoth[Tanh[a + b*x]]]))/(2*x^2*(b*x - ArcCoth[Tanh[a + b*x]
])^5*ArcCoth[Tanh[a + b*x]]^2)

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Maple [F]  time = 180., size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{3} \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^3/arccoth(tanh(b*x+a))^3,x)

[Out]

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

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Maxima [C]  time = 3.5402, size = 448, normalized size = 2.64 \begin{align*} \frac{192 \, b^{2} \log \left (-i \, \pi + 2 \, b x + 2 \, a\right )}{i \, \pi ^{5} - 10 \, \pi ^{4} a - 40 i \, \pi ^{3} a^{2} + 80 \, \pi ^{2} a^{3} + 80 i \, \pi a^{4} - 32 \, a^{5}} - \frac{192 \, b^{2} \log \left (x\right )}{i \, \pi ^{5} - 10 \, \pi ^{4} a - 40 i \, \pi ^{3} a^{2} + 80 \, \pi ^{2} a^{3} + 80 i \, \pi a^{4} - 32 \, a^{5}} + \frac{8 \,{\left (96 \, b^{3} x^{3} - i \, \pi ^{3} + 6 \, \pi ^{2} a + 12 i \, \pi a^{2} - 8 \, a^{3} +{\left (-72 i \, \pi b^{2} + 144 \, a b^{2}\right )} x^{2} -{\left (8 \, \pi ^{2} b + 32 i \, \pi a b - 32 \, a^{2} b\right )} x\right )}}{{\left (8 \, \pi ^{4} b^{2} + 64 i \, \pi ^{3} a b^{2} - 192 \, \pi ^{2} a^{2} b^{2} - 256 i \, \pi a^{3} b^{2} + 128 \, a^{4} b^{2}\right )} x^{4} +{\left (-8 i \, \pi ^{5} b + 80 \, \pi ^{4} a b + 320 i \, \pi ^{3} a^{2} b - 640 \, \pi ^{2} a^{3} b - 640 i \, \pi a^{4} b + 256 \, a^{5} b\right )} x^{3} -{\left (2 \, \pi ^{6} + 24 i \, \pi ^{5} a - 120 \, \pi ^{4} a^{2} - 320 i \, \pi ^{3} a^{3} + 480 \, \pi ^{2} a^{4} + 384 i \, \pi a^{5} - 128 \, a^{6}\right )} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

192*b^2*log(-I*pi + 2*b*x + 2*a)/(I*pi^5 - 10*pi^4*a - 40*I*pi^3*a^2 + 80*pi^2*a^3 + 80*I*pi*a^4 - 32*a^5) - 1
92*b^2*log(x)/(I*pi^5 - 10*pi^4*a - 40*I*pi^3*a^2 + 80*pi^2*a^3 + 80*I*pi*a^4 - 32*a^5) + 8*(96*b^3*x^3 - I*pi
^3 + 6*pi^2*a + 12*I*pi*a^2 - 8*a^3 + (-72*I*pi*b^2 + 144*a*b^2)*x^2 - (8*pi^2*b + 32*I*pi*a*b - 32*a^2*b)*x)/
((8*pi^4*b^2 + 64*I*pi^3*a*b^2 - 192*pi^2*a^2*b^2 - 256*I*pi*a^3*b^2 + 128*a^4*b^2)*x^4 + (-8*I*pi^5*b + 80*pi
^4*a*b + 320*I*pi^3*a^2*b - 640*pi^2*a^3*b - 640*I*pi*a^4*b + 256*a^5*b)*x^3 - (2*pi^6 + 24*I*pi^5*a - 120*pi^
4*a^2 - 320*I*pi^3*a^3 + 480*pi^2*a^4 + 384*I*pi*a^5 - 128*a^6)*x^2)

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Fricas [B]  time = 2.23993, size = 3069, normalized size = 18.05 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8*(3*pi^10*a + 44*pi^8*a^3 + 224*pi^6*a^5 + 384*pi^4*a^7 - 256*pi^2*a^9 - 1024*a^11 + 192*(pi^6*b^5 - 20*pi^4*
a^2*b^5 - 80*pi^2*a^4*b^5 + 64*a^6*b^5)*x^5 + 96*(11*pi^6*a*b^4 - 140*pi^4*a^3*b^4 - 624*pi^2*a^5*b^4 + 448*a^
7*b^4)*x^4 + 16*(5*pi^8*b^3 + 16*pi^6*a^2*b^3 - 1440*pi^4*a^4*b^3 - 4864*pi^2*a^6*b^3 + 3328*a^8*b^3)*x^3 + 4*
(65*pi^8*a*b^2 - 272*pi^6*a^3*b^2 - 4896*pi^4*a^5*b^2 - 9472*pi^2*a^7*b^2 + 6400*a^9*b^2)*x^2 + 2*(3*pi^10*b -
12*pi^8*a^2*b - 416*pi^6*a^4*b - 1920*pi^4*a^6*b - 2304*pi^2*a^8*b + 1024*a^10*b)*x - 48*(16*(pi^5*b^6 - 40*p
i^3*a^2*b^6 + 80*pi*a^4*b^6)*x^6 + 64*(pi^5*a*b^5 - 40*pi^3*a^3*b^5 + 80*pi*a^5*b^5)*x^5 + 8*(pi^7*b^4 - 28*pi
^5*a^2*b^4 - 400*pi^3*a^4*b^4 + 960*pi*a^6*b^4)*x^4 + 16*(pi^7*a*b^3 - 36*pi^5*a^3*b^3 - 80*pi^3*a^5*b^3 + 320
*pi*a^7*b^3)*x^3 + (pi^9*b^2 - 32*pi^7*a^2*b^2 - 224*pi^5*a^4*b^2 + 1280*pi*a^8*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) - 24*(16*(5*pi^4*a*b^6 - 40*pi^2*a^3*b^6 + 16*a^5*b^6)*x^6 +
64*(5*pi^4*a^2*b^5 - 40*pi^2*a^4*b^5 + 16*a^6*b^5)*x^5 + 8*(5*pi^6*a*b^4 + 20*pi^4*a^3*b^4 - 464*pi^2*a^5*b^4
+ 192*a^7*b^4)*x^4 + 16*(5*pi^6*a^2*b^3 - 20*pi^4*a^4*b^3 - 144*pi^2*a^6*b^3 + 64*a^8*b^3)*x^3 + (5*pi^8*a*b^2
- 224*pi^4*a^5*b^2 - 512*pi^2*a^7*b^2 + 256*a^9*b^2)*x^2)*log(4*b^2*x^2 + 8*a*b*x + pi^2 + 4*a^2) + 48*(16*(5
*pi^4*a*b^6 - 40*pi^2*a^3*b^6 + 16*a^5*b^6)*x^6 + 64*(5*pi^4*a^2*b^5 - 40*pi^2*a^4*b^5 + 16*a^6*b^5)*x^5 + 8*(
5*pi^6*a*b^4 + 20*pi^4*a^3*b^4 - 464*pi^2*a^5*b^4 + 192*a^7*b^4)*x^4 + 16*(5*pi^6*a^2*b^3 - 20*pi^4*a^4*b^3 -
144*pi^2*a^6*b^3 + 64*a^8*b^3)*x^3 + (5*pi^8*a*b^2 - 224*pi^4*a^5*b^2 - 512*pi^2*a^7*b^2 + 256*a^9*b^2)*x^2)*l
og(x))/(16*(pi^10*b^4 + 20*pi^8*a^2*b^4 + 160*pi^6*a^4*b^4 + 640*pi^4*a^6*b^4 + 1280*pi^2*a^8*b^4 + 1024*a^10*
b^4)*x^6 + 64*(pi^10*a*b^3 + 20*pi^8*a^3*b^3 + 160*pi^6*a^5*b^3 + 640*pi^4*a^7*b^3 + 1280*pi^2*a^9*b^3 + 1024*
a^11*b^3)*x^5 + 8*(pi^12*b^2 + 32*pi^10*a^2*b^2 + 400*pi^8*a^4*b^2 + 2560*pi^6*a^6*b^2 + 8960*pi^4*a^8*b^2 + 1
6384*pi^2*a^10*b^2 + 12288*a^12*b^2)*x^4 + 16*(pi^12*a*b + 24*pi^10*a^3*b + 240*pi^8*a^5*b + 1280*pi^6*a^7*b +
3840*pi^4*a^9*b + 6144*pi^2*a^11*b + 4096*a^13*b)*x^3 + (pi^14 + 28*pi^12*a^2 + 336*pi^10*a^4 + 2240*pi^8*a^6
+ 8960*pi^6*a^8 + 21504*pi^4*a^10 + 28672*pi^2*a^12 + 16384*a^14)*x^2)

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{3} \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^3/arccoth(tanh(b*x+a))^3,x, algorithm="giac")

[Out]

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