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

Optimal. Leaf size=131 \[ \frac {3 b}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3 \coth ^{-1}(\tanh (a+b x))}-\frac {3 b}{2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \coth ^{-1}(\tanh (a+b x))^2}+\frac {1}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))^2}-\frac {3 b \log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^4}+\frac {3 b \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^4} \]

[Out]

-3/2*b/(b*x-arccoth(tanh(b*x+a)))^2/arccoth(tanh(b*x+a))^2+1/x/(b*x-arccoth(tanh(b*x+a)))/arccoth(tanh(b*x+a))
^2+3*b/(b*x-arccoth(tanh(b*x+a)))^3/arccoth(tanh(b*x+a))-3*b*ln(x)/(b*x-arccoth(tanh(b*x+a)))^4+3*b*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 = 131, 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} \[ \frac {3 b}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3 \coth ^{-1}(\tanh (a+b x))}-\frac {3 b}{2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \coth ^{-1}(\tanh (a+b x))^2}+\frac {1}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))^2}-\frac {3 b \log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^4}+\frac {3 b \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b)/(2*(b*x - ArcCoth[Tanh[a + b*x]])^2*ArcCoth[Tanh[a + b*x]]^2) + 1/(x*(b*x - ArcCoth[Tanh[a + b*x]])*Arc
Coth[Tanh[a + b*x]]^2) + (3*b)/((b*x - ArcCoth[Tanh[a + b*x]])^3*ArcCoth[Tanh[a + b*x]]) - (3*b*Log[x])/(b*x -
ArcCoth[Tanh[a + b*x]])^4 + (3*b*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^2 \coth ^{-1}(\tanh (a+b x))^3} \, dx &=\frac {1}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))^2}-\frac {(3 b) \int \frac {1}{x \coth ^{-1}(\tanh (a+b x))^3} \, dx}{-b x+\coth ^{-1}(\tanh (a+b x))}\\ &=-\frac {3 b}{2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \coth ^{-1}(\tanh (a+b x))^2}+\frac {1}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))^2}+\frac {(3 b) \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 )^2 \coth ^{-1}(\tanh (a+b x))^2}+\frac {1}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))^2}+\frac {3 b}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3 \coth ^{-1}(\tanh (a+b x))}+\frac {(3 b) \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 )^2 \coth ^{-1}(\tanh (a+b x))^2}+\frac {1}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))^2}+\frac {3 b}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3 \coth ^{-1}(\tanh (a+b x))}-\frac {(3 b) \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^2\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 )^2 \coth ^{-1}(\tanh (a+b x))^2}+\frac {1}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))^2}+\frac {3 b}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3 \coth ^{-1}(\tanh (a+b x))}-\frac {3 b \log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^4}+\frac {(3 b) \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 )^2 \coth ^{-1}(\tanh (a+b x))^2}+\frac {1}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))^2}+\frac {3 b}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3 \coth ^{-1}(\tanh (a+b x))}-\frac {3 b \log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^4}+\frac {3 b \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 = 0.05, size = 93, normalized size = 0.71 \[ -\frac {-6 b^2 x^2 \coth ^{-1}(\tanh (a+b x))+2 \coth ^{-1}(\tanh (a+b x))^3+3 b x \coth ^{-1}(\tanh (a+b x))^2 \left (-2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )+2 \log (x)+1\right )+b^3 x^3}{2 x \coth ^{-1}(\tanh (a+b x))^2 \left (\coth ^{-1}(\tanh (a+b x))-b x\right )^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.54, size = 1078, normalized size = 8.23 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8*(6*pi^8*a + 64*pi^6*a^3 + 192*pi^4*a^5 - 512*a^9 + 96*(3*pi^4*a*b^4 + 8*pi^2*a^3*b^4 - 16*a^5*b^4)*x^4 - 12*
(pi^6*b^3 - 92*pi^4*a^2*b^3 - 272*pi^2*a^4*b^3 + 448*a^6*b^3)*x^3 + 8*(11*pi^6*a*b^2 + 228*pi^4*a^3*b^2 + 528*
pi^2*a^5*b^2 - 832*a^7*b^2)*x^2 - (5*pi^8*b - 176*pi^6*a^2*b - 1440*pi^4*a^4*b - 1792*pi^2*a^6*b + 3328*a^8*b)
*x - 96*(16*(pi^3*a*b^5 - 4*pi*a^3*b^5)*x^5 + 64*(pi^3*a^2*b^4 - 4*pi*a^4*b^4)*x^4 + 8*(pi^5*a*b^3 + 8*pi^3*a^
3*b^3 - 48*pi*a^5*b^3)*x^3 + 16*(pi^5*a^2*b^2 - 16*pi*a^6*b^2)*x^2 + (pi^7*a*b + 4*pi^5*a^3*b - 16*pi^3*a^5*b
- 64*pi*a^7*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*(16*(pi^4*b^5 - 24*
pi^2*a^2*b^5 + 16*a^4*b^5)*x^5 + 64*(pi^4*a*b^4 - 24*pi^2*a^3*b^4 + 16*a^5*b^4)*x^4 + 8*(pi^6*b^3 - 12*pi^4*a^
2*b^3 - 272*pi^2*a^4*b^3 + 192*a^6*b^3)*x^3 + 16*(pi^6*a*b^2 - 20*pi^4*a^3*b^2 - 80*pi^2*a^5*b^2 + 64*a^7*b^2)
*x^2 + (pi^8*b - 16*pi^6*a^2*b - 160*pi^4*a^4*b - 256*pi^2*a^6*b + 256*a^8*b)*x)*log(4*b^2*x^2 + 8*a*b*x + pi^
2 + 4*a^2) - 6*(16*(pi^4*b^5 - 24*pi^2*a^2*b^5 + 16*a^4*b^5)*x^5 + 64*(pi^4*a*b^4 - 24*pi^2*a^3*b^4 + 16*a^5*b
^4)*x^4 + 8*(pi^6*b^3 - 12*pi^4*a^2*b^3 - 272*pi^2*a^4*b^3 + 192*a^6*b^3)*x^3 + 16*(pi^6*a*b^2 - 20*pi^4*a^3*b
^2 - 80*pi^2*a^5*b^2 + 64*a^7*b^2)*x^2 + (pi^8*b - 16*pi^6*a^2*b - 160*pi^4*a^4*b - 256*pi^2*a^6*b + 256*a^8*b
)*x)*log(x))/(16*(pi^8*b^4 + 16*pi^6*a^2*b^4 + 96*pi^4*a^4*b^4 + 256*pi^2*a^6*b^4 + 256*a^8*b^4)*x^5 + 64*(pi^
8*a*b^3 + 16*pi^6*a^3*b^3 + 96*pi^4*a^5*b^3 + 256*pi^2*a^7*b^3 + 256*a^9*b^3)*x^4 + 8*(pi^10*b^2 + 28*pi^8*a^2
*b^2 + 288*pi^6*a^4*b^2 + 1408*pi^4*a^6*b^2 + 3328*pi^2*a^8*b^2 + 3072*a^10*b^2)*x^3 + 16*(pi^10*a*b + 20*pi^8
*a^3*b + 160*pi^6*a^5*b + 640*pi^4*a^7*b + 1280*pi^2*a^9*b + 1024*a^11*b)*x^2 + (pi^12 + 24*pi^10*a^2 + 240*pi
^8*a^4 + 1280*pi^6*a^6 + 3840*pi^4*a^8 + 6144*pi^2*a^10 + 4096*a^12)*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 1.15, size = 243, normalized size = 1.85 \[ \frac {48 \, b \log \left (-i \, \pi + 2 \, b x + 2 \, a\right )}{\pi ^{4} + 8 i \, \pi ^{3} a - 24 \, \pi ^{2} a^{2} - 32 i \, \pi a^{3} + 16 \, a^{4}} - \frac {48 \, b \log \relax (x)}{\pi ^{4} + 8 i \, \pi ^{3} a - 24 \, \pi ^{2} a^{2} - 32 i \, \pi a^{3} + 16 \, a^{4}} + \frac {8 \, {\left (12 \, b^{2} x^{2} - \pi ^{2} - 4 i \, \pi a + 4 \, a^{2} + {\left (-9 i \, \pi b + 18 \, a b\right )} x\right )}}{{\left (-4 i \, \pi ^{3} b^{2} + 24 \, \pi ^{2} a b^{2} + 48 i \, \pi a^{2} b^{2} - 32 \, a^{3} b^{2}\right )} x^{3} - {\left (4 \, \pi ^{4} b + 32 i \, \pi ^{3} a b - 96 \, \pi ^{2} a^{2} b - 128 i \, \pi a^{3} b + 64 \, a^{4} b\right )} x^{2} + {\left (i \, \pi ^{5} - 10 \, \pi ^{4} a - 40 i \, \pi ^{3} a^{2} + 80 \, \pi ^{2} a^{3} + 80 i \, \pi a^{4} - 32 \, a^{5}\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

48*b*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*log(x)/(pi^4 + 8
*I*pi^3*a - 24*pi^2*a^2 - 32*I*pi*a^3 + 16*a^4) + 8*(12*b^2*x^2 - pi^2 - 4*I*pi*a + 4*a^2 + (-9*I*pi*b + 18*a*
b)*x)/((-4*I*pi^3*b^2 + 24*pi^2*a*b^2 + 48*I*pi*a^2*b^2 - 32*a^3*b^2)*x^3 - (4*pi^4*b + 32*I*pi^3*a*b - 96*pi^
2*a^2*b - 128*I*pi*a^3*b + 64*a^4*b)*x^2 + (I*pi^5 - 10*pi^4*a - 40*I*pi^3*a^2 + 80*pi^2*a^3 + 80*I*pi*a^4 - 3
2*a^5)*x)

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mupad [B]  time = 5.26, size = 1074, normalized size = 8.20 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*acoth(tanh(a + b*x))^3),x)

[Out]

(8/(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) - (72*
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) + (96*b^2*x^2)/((log(-2/(exp(2*a)*exp(2*b*x) - 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*ex
p(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)))/(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) + x^2*(8*a*b - 4*b*(2*a - log((2*exp(2*a)*exp(2*b*x))/(ex
p(2*a)*exp(2*b*x) - 1)) + log(-2/(exp(2*a)*exp(2*b*x) - 1)) + 2*b*x)) + 4*b^2*x^3) + (96*b*atanh(((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 + 24*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^4 - 8*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)^3 - 32*a^3*(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))/(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)^4 - (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)^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))/(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)) - log((2*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 \[ \int \frac {1}{x^{2} \operatorname {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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