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

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

[Out]

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

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b)/(2*(b*x - ArcTanh[Tanh[a + b*x]])^2*ArcTanh[Tanh[a + b*x]]^2) + 1/(x*(b*x - ArcTanh[Tanh[a + b*x]])*Arc
Tanh[Tanh[a + b*x]]^2) + (3*b)/((b*x - ArcTanh[Tanh[a + b*x]])^3*ArcTanh[Tanh[a + b*x]]) - (3*b*Log[x])/(b*x -
ArcTanh[Tanh[a + b*x]])^4 + (3*b*Log[ArcTanh[Tanh[a + b*x]]])/(b*x - ArcTanh[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 \tanh ^{-1}(\tanh (a+b x))^3} \, dx &=\frac {1}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^2}-\frac {(3 b) \int \frac {1}{x \tanh ^{-1}(\tanh (a+b x))^3} \, dx}{-b x+\tanh ^{-1}(\tanh (a+b x))}\\ &=-\frac {3 b}{2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^2}+\frac {1}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^2}+\frac {(3 b) \int \frac {1}{x \tanh ^{-1}(\tanh (a+b x))^2} \, dx}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=-\frac {3 b}{2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^2}+\frac {1}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^2}+\frac {3 b}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))}+\frac {(3 b) \int \frac {1}{x \tanh ^{-1}(\tanh (a+b x))} \, dx}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=-\frac {3 b}{2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^2}+\frac {1}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^2}+\frac {3 b}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))}-\frac {(3 b) \int \frac {1}{x} \, dx}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {\left (3 b^2\right ) \int \frac {1}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=-\frac {3 b}{2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^2}+\frac {1}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^2}+\frac {3 b}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))}-\frac {3 b \log (x)}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}+\frac {(3 b) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=-\frac {3 b}{2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^2}+\frac {1}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^2}+\frac {3 b}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))}-\frac {3 b \log (x)}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}+\frac {3 b \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 93, normalized size = 0.71 \[ -\frac {-6 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))+2 \tanh ^{-1}(\tanh (a+b x))^3+3 b x \tanh ^{-1}(\tanh (a+b x))^2 \left (-2 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )+2 \log (x)+1\right )+b^3 x^3}{2 x \tanh ^{-1}(\tanh (a+b x))^2 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.52, size = 109, normalized size = 0.83 \[ -\frac {6 \, a b^{2} x^{2} + 9 \, a^{2} b x + 2 \, a^{3} - 6 \, {\left (b^{3} x^{3} + 2 \, a b^{2} x^{2} + a^{2} b x\right )} \log \left (b x + a\right ) + 6 \, {\left (b^{3} x^{3} + 2 \, a b^{2} x^{2} + a^{2} b x\right )} \log \relax (x)}{2 \, {\left (a^{4} b^{2} x^{3} + 2 \, a^{5} b x^{2} + a^{6} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(6*a*b^2*x^2 + 9*a^2*b*x + 2*a^3 - 6*(b^3*x^3 + 2*a*b^2*x^2 + a^2*b*x)*log(b*x + a) + 6*(b^3*x^3 + 2*a*b^
2*x^2 + a^2*b*x)*log(x))/(a^4*b^2*x^3 + 2*a^5*b*x^2 + a^6*x)

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giac [A]  time = 0.17, size = 60, normalized size = 0.46 \[ \frac {3 \, b \log \left ({\left | b x + a \right |}\right )}{a^{4}} - \frac {3 \, b \log \left ({\left | x \right |}\right )}{a^{4}} - \frac {6 \, a b^{2} x^{2} + 9 \, a^{2} b x + 2 \, a^{3}}{2 \, {\left (b x + a\right )}^{2} a^{4} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*b*log(abs(b*x + a))/a^4 - 3*b*log(abs(x))/a^4 - 1/2*(6*a*b^2*x^2 + 9*a^2*b*x + 2*a^3)/((b*x + a)^2*a^4*x)

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maple [A]  time = 0.15, size = 117, normalized size = 0.89 \[ -\frac {1}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3} x}-\frac {3 b \ln \relax (x )}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{4}}-\frac {b}{2 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}+\frac {3 b \ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right )}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{4}}-\frac {2 b}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \arctanh \left (\tanh \left (b x +a \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 3.13, size = 69, normalized size = 0.53 \[ -\frac {6 \, b^{2} x^{2} + 9 \, a b x + 2 \, a^{2}}{2 \, {\left (a^{3} b^{2} x^{3} + 2 \, a^{4} b x^{2} + a^{5} x\right )}} + \frac {3 \, b \log \left (b x + a\right )}{a^{4}} - \frac {3 \, b \log \relax (x)}{a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(6*b^2*x^2 + 9*a*b*x + 2*a^2)/(a^3*b^2*x^3 + 2*a^4*b*x^2 + a^5*x) + 3*b*log(b*x + a)/a^4 - 3*b*log(x)/a^4

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mupad [B]  time = 3.67, size = 804, normalized size = 6.14 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*atanh(tanh(a + b*x))^3),x)

[Out]

-(24*log(1/(exp(2*a)*exp(2*b*x) + 1))^2*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) - 24*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 - 8*log(1/(exp(2*a)*exp(2*b*x) + 1
))^3 + 8*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))^3 + 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 + 96*b^2*x^2*log(1/(exp(2*a)*exp(2*b*x) + 1)) - 96*b^2*x^2*log((exp(2*a)*exp(2*
b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 24*b*x*log(1/(exp(2*a)*exp(2*b*x) + 1))^2 - b*x*log((exp(2*a)*exp(2*b*x))/(
exp(2*a)*exp(2*b*x) + 1))^2*atan((log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))*1i - log(1/(exp(2*a)*ex
p(2*b*x) + 1))*1i + b*x*2i)/(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))*96i - b*x*log(1/(exp(2*a)*exp(2*b*x) + 1))^2*atan((log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*
b*x) + 1))*1i - log(1/(exp(2*a)*exp(2*b*x) + 1))*1i + b*x*2i)/(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))*96i - 48*b*x*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)) + b*x*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))*atan((log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))*1i - log(1/(exp(2*a)*exp(2*b*
x) + 1))*1i + b*x*2i)/(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))*192i)/(x*(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*(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 \[ \int \frac {1}{x^{2} \operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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