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

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Rubi [A]  time = 0.07, antiderivative size = 97, 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} \[ \frac {1}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \coth ^{-1}(\tanh (a+b x))}-\frac {1}{2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))^2}-\frac {\log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3}+\frac {\log \left (\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3} \]

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 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 \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*}

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Mathematica [A]  time = 0.11, 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)

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fricas [B]  time = 0.61, size = 811, normalized size = 8.36 \[ -\frac {8 \, {\left (9 \, \pi ^{6} a + 60 \, \pi ^{4} a^{3} + 48 \, \pi ^{2} a^{5} - 192 \, a^{7} + 8 \, {\left (\pi ^{4} b^{3} - 16 \, a^{4} b^{3}\right )} x^{3} + 4 \, {\left (9 \, \pi ^{4} a b^{2} + 8 \, \pi ^{2} a^{3} b^{2} - 112 \, a^{5} b^{2}\right )} x^{2} + 4 \, {\left (\pi ^{6} b + 16 \, \pi ^{4} a^{2} b + 16 \, \pi ^{2} a^{4} b - 128 \, a^{6} b\right )} x - 2 \, {\left (\pi ^{7} - 4 \, \pi ^{5} a^{2} - 80 \, \pi ^{3} a^{4} - 192 \, \pi a^{6} + 16 \, {\left (\pi ^{3} b^{4} - 12 \, \pi a^{2} b^{4}\right )} x^{4} + 64 \, {\left (\pi ^{3} a b^{3} - 12 \, \pi a^{3} b^{3}\right )} x^{3} + 8 \, {\left (\pi ^{5} b^{2} - 144 \, \pi a^{4} b^{2}\right )} x^{2} + 16 \, {\left (\pi ^{5} a b - 8 \, \pi ^{3} a^{3} b - 48 \, \pi a^{5} b\right )} x\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 ) - {\left (3 \, \pi ^{6} a + 20 \, \pi ^{4} a^{3} + 16 \, \pi ^{2} a^{5} - 64 \, a^{7} + 16 \, {\left (3 \, \pi ^{2} a b^{4} - 4 \, a^{3} b^{4}\right )} x^{4} + 64 \, {\left (3 \, \pi ^{2} a^{2} b^{3} - 4 \, a^{4} b^{3}\right )} x^{3} + 8 \, {\left (3 \, \pi ^{4} a b^{2} + 32 \, \pi ^{2} a^{3} b^{2} - 48 \, a^{5} b^{2}\right )} x^{2} + 16 \, {\left (3 \, \pi ^{4} a^{2} b + 8 \, \pi ^{2} a^{4} b - 16 \, a^{6} b\right )} x\right )} \log \left (4 \, b^{2} x^{2} + 8 \, a b x + \pi ^{2} + 4 \, a^{2}\right ) + 2 \, {\left (3 \, \pi ^{6} a + 20 \, \pi ^{4} a^{3} + 16 \, \pi ^{2} a^{5} - 64 \, a^{7} + 16 \, {\left (3 \, \pi ^{2} a b^{4} - 4 \, a^{3} b^{4}\right )} x^{4} + 64 \, {\left (3 \, \pi ^{2} a^{2} b^{3} - 4 \, a^{4} b^{3}\right )} x^{3} + 8 \, {\left (3 \, \pi ^{4} a b^{2} + 32 \, \pi ^{2} a^{3} b^{2} - 48 \, a^{5} b^{2}\right )} x^{2} + 16 \, {\left (3 \, \pi ^{4} a^{2} b + 8 \, \pi ^{2} a^{4} b - 16 \, a^{6} b\right )} x\right )} \log \relax (x)\right )}}{\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 \, {\left (\pi ^{6} b^{4} + 12 \, \pi ^{4} a^{2} b^{4} + 48 \, \pi ^{2} a^{4} b^{4} + 64 \, a^{6} b^{4}\right )} x^{4} + 64 \, {\left (\pi ^{6} a b^{3} + 12 \, \pi ^{4} a^{3} b^{3} + 48 \, \pi ^{2} a^{5} b^{3} + 64 \, a^{7} b^{3}\right )} x^{3} + 8 \, {\left (\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}\right )} x^{2} + 16 \, {\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} \]

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)

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

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)

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

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)

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maxima [C]  time = 1.13, size = 173, normalized size = 1.78 \[ \frac {8 \, {\left (-3 i \, \pi + 4 \, b x + 6 \, a\right )}}{2 \, \pi ^{4} + 16 i \, \pi ^{3} a - 48 \, \pi ^{2} a^{2} - 64 i \, \pi a^{3} + 32 \, a^{4} - {\left (8 \, \pi ^{2} b^{2} + 32 i \, \pi a b^{2} - 32 \, a^{2} b^{2}\right )} x^{2} + {\left (8 i \, \pi ^{3} b - 48 \, \pi ^{2} a b - 96 i \, \pi a^{2} b + 64 \, a^{3} b\right )} x} + \frac {8 \, \log \left (-i \, \pi + 2 \, b x + 2 \, a\right )}{-i \, \pi ^{3} + 6 \, \pi ^{2} a + 12 i \, \pi a^{2} - 8 \, a^{3}} - \frac {8 \, \log \relax (x)}{-i \, \pi ^{3} + 6 \, \pi ^{2} a + 12 i \, \pi a^{2} - 8 \, a^{3}} \]

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)

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mupad [B]  time = 6.49, size = 902, normalized size = 9.30 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*acoth(tanh(a + b*x))^3),x)

[Out]

- (16*atanh((16*(4*b*x - ((2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + log(-2/(exp(2*a)*exp
(2*b*x) - 1)) + 2*b*x)^3 - 8*a^3 - 6*a*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + log(-2/
(exp(2*a)*exp(2*b*x) - 1)) + 2*b*x)^2 + 12*a^2*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) +
log(-2/(exp(2*a)*exp(2*b*x) - 1)) + 2*b*x))/((2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) +
log(-2/(exp(2*a)*exp(2*b*x) - 1)) + 2*b*x)^2 - 4*a*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1
)) + log(-2/(exp(2*a)*exp(2*b*x) - 1)) + 2*b*x) + 4*a^2))*((2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*
b*x) - 1)) + log(-2/(exp(2*a)*exp(2*b*x) - 1)) + 2*b*x)^2/16 - (a*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)
*exp(2*b*x) - 1)) + log(-2/(exp(2*a)*exp(2*b*x) - 1)) + 2*b*x))/4 + a^2/4))/(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)^3))/(log(-2/(exp(2*a)*exp(2*b*x) - 1)) - lo
g((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + 2*b*x)^3 - (12/(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) - (16*b*x)/((2*a - log((2*exp(2*a)*exp(2*b*x))/(e
xp(2*a)*exp(2*b*x) - 1)) + log(-2/(exp(2*a)*exp(2*b*x) - 1)) + 2*b*x)^2 - 4*a*(2*a - log((2*exp(2*a)*exp(2*b*x
))/(exp(2*a)*exp(2*b*x) - 1)) + log(-2/(exp(2*a)*exp(2*b*x) - 1)) + 2*b*x) + 4*a^2))/((2*a - log((2*exp(2*a)*e
xp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + log(-2/(exp(2*a)*exp(2*b*x) - 1)) + 2*b*x)^2 - 4*a*(2*a - log((2*exp(2
*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + log(-2/(exp(2*a)*exp(2*b*x) - 1)) + 2*b*x) + 4*a^2 + x*(8*a*b - 4
*b*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + log(-2/(exp(2*a)*exp(2*b*x) - 1)) + 2*b*x))
+ 4*b^2*x^2)

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

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)

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