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

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

[Out]

-3*b^2/(b*x-arctanh(tanh(b*x+a)))^3/arctanh(tanh(b*x+a))^2+2*b/x/(b*x-arctanh(tanh(b*x+a)))^2/arctanh(tanh(b*x
+a))^2+1/2/x^2/(b*x-arctanh(tanh(b*x+a)))/arctanh(tanh(b*x+a))^2+6*b^2/(b*x-arctanh(tanh(b*x+a)))^4/arctanh(ta
nh(b*x+a))-6*b^2*ln(x)/(b*x-arctanh(tanh(b*x+a)))^5+6*b^2*ln(arctanh(tanh(b*x+a)))/(b*x-arctanh(tanh(b*x+a)))^
5

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Rubi [A]  time = 0.12, antiderivative size = 170, normalized size of antiderivative = 1.00, 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} \[ \frac {6 b^2}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \tanh ^{-1}(\tanh (a+b x))}-\frac {3 b^2}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^2}-\frac {6 b^2 \log (x)}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^5}+\frac {6 b^2 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^5}+\frac {1}{2 x^2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^2}+\frac {2 b}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.05, size = 107, normalized size = 0.63 \[ \frac {8 b^3 x^3 \tanh ^{-1}(\tanh (a+b x))-12 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))^2 \left (\log (x)-\log \left (\tanh ^{-1}(\tanh (a+b x))\right )\right )-8 b x \tanh ^{-1}(\tanh (a+b x))^3+\tanh ^{-1}(\tanh (a+b x))^4-b^4 x^4}{2 x^2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^5 \tanh ^{-1}(\tanh (a+b x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.46, size = 130, normalized size = 0.76 \[ \frac {12 \, a b^{3} x^{3} + 18 \, a^{2} b^{2} x^{2} + 4 \, a^{3} b x - a^{4} - 12 \, {\left (b^{4} x^{4} + 2 \, a b^{3} x^{3} + a^{2} b^{2} x^{2}\right )} \log \left (b x + a\right ) + 12 \, {\left (b^{4} x^{4} + 2 \, a b^{3} x^{3} + a^{2} b^{2} x^{2}\right )} \log \relax (x)}{2 \, {\left (a^{5} b^{2} x^{4} + 2 \, a^{6} b x^{3} + a^{7} x^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(12*a*b^3*x^3 + 18*a^2*b^2*x^2 + 4*a^3*b*x - a^4 - 12*(b^4*x^4 + 2*a*b^3*x^3 + a^2*b^2*x^2)*log(b*x + a) +
12*(b^4*x^4 + 2*a*b^3*x^3 + a^2*b^2*x^2)*log(x))/(a^5*b^2*x^4 + 2*a^6*b*x^3 + a^7*x^2)

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giac [A]  time = 0.18, size = 73, normalized size = 0.43 \[ -\frac {6 \, b^{2} \log \left ({\left | b x + a \right |}\right )}{a^{5}} + \frac {6 \, b^{2} \log \left ({\left | x \right |}\right )}{a^{5}} + \frac {12 \, b^{3} x^{3} + 18 \, a b^{2} x^{2} + 4 \, a^{2} b x - a^{3}}{2 \, {\left (b x^{2} + a x\right )}^{2} a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-6*b^2*log(abs(b*x + a))/a^5 + 6*b^2*log(abs(x))/a^5 + 1/2*(12*b^3*x^3 + 18*a*b^2*x^2 + 4*a^2*b*x - a^3)/((b*x
^2 + a*x)^2*a^4)

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maple [A]  time = 0.16, size = 145, normalized size = 0.85 \[ -\frac {1}{2 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3} x^{2}}+\frac {6 b^{2} \ln \relax (x )}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{5}}+\frac {3 b}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{4} x}-\frac {6 b^{2} \ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right )}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{5}}+\frac {3 b^{2}}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{4} \arctanh \left (\tanh \left (b x +a \right )\right )}+\frac {b^{2}}{2 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/(arctanh(tanh(b*x+a))-b*x)^3/x^2+6/(arctanh(tanh(b*x+a))-b*x)^5*b^2*ln(x)+3/(arctanh(tanh(b*x+a))-b*x)^4*
b/x-6/(arctanh(tanh(b*x+a))-b*x)^5*b^2*ln(arctanh(tanh(b*x+a)))+3/(arctanh(tanh(b*x+a))-b*x)^4*b^2/arctanh(tan
h(b*x+a))+1/2/(arctanh(tanh(b*x+a))-b*x)^3*b^2/arctanh(tanh(b*x+a))^2

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maxima [A]  time = 3.14, size = 86, normalized size = 0.51 \[ \frac {12 \, b^{3} x^{3} + 18 \, a b^{2} x^{2} + 4 \, a^{2} b x - a^{3}}{2 \, {\left (a^{4} b^{2} x^{4} + 2 \, a^{5} b x^{3} + a^{6} x^{2}\right )}} - \frac {6 \, b^{2} \log \left (b x + a\right )}{a^{5}} + \frac {6 \, b^{2} \log \relax (x)}{a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(12*b^3*x^3 + 18*a*b^2*x^2 + 4*a^2*b*x - a^3)/(a^4*b^2*x^4 + 2*a^5*b*x^3 + a^6*x^2) - 6*b^2*log(b*x + a)/a
^5 + 6*b^2*log(x)/a^5

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mupad [B]  time = 2.73, size = 909, normalized size = 5.35 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*atanh(tanh(a + b*x))^3),x)

[Out]

(4*log(1/(exp(2*a)*exp(2*b*x) + 1))^4 - 16*log(1/(exp(2*a)*exp(2*b*x) + 1))^3*log((exp(2*a)*exp(2*b*x))/(exp(2
*a)*exp(2*b*x) + 1)) - 16*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)
)^3 + 4*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))^4 + 24*log(1/(exp(2*a)*exp(2*b*x) + 1))^2*log((ex
p(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))^2 - 64*b^4*x^4 - 64*b*x*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(
2*b*x) + 1))^3 - 256*b^3*x^3*log(1/(exp(2*a)*exp(2*b*x) + 1)) + 256*b^3*x^3*log((exp(2*a)*exp(2*b*x))/(exp(2*a
)*exp(2*b*x) + 1)) + 64*b*x*log(1/(exp(2*a)*exp(2*b*x) + 1))^3 + 192*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))^2 - 192*b*x*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)) + b^2*x^2*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))*384i + b^2*x^2*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)*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))*384i - b^2*x^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))*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)
)*768i)/(x^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*(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)^5)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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