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

Optimal. Leaf size=68 \[ -3 b^2 x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )-\frac {\tanh ^{-1}(\tanh (a+b x))^3}{x}+\frac {3}{2} b \tanh ^{-1}(\tanh (a+b x))^2+3 b \log (x) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2168, 2159, 2158, 29} \[ -3 b^2 x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )-\frac {\tanh ^{-1}(\tanh (a+b x))^3}{x}+\frac {3}{2} b \tanh ^{-1}(\tanh (a+b x))^2+3 b \log (x) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^3/x^2,x]

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2158

Int[(v_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(b*x)/a, x] - Dist[(b*u
- a*v)/a, Int[1/u, x], x] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x]

Rule 2159

Int[(v_)^(n_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[v^n/(a*n), x] - Dis
t[(b*u - a*v)/a, Int[v^(n - 1)/u, x], x] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && GtQ[n, 0] && Ne
Q[n, 1]

Rule 2168

Int[(u_)^(m_)*(v_)^(n_.), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(u^(m + 1)*v^
n)/(a*(m + 1)), x] - Dist[(b*n)/(a*(m + 1)), Int[u^(m + 1)*v^(n - 1), x], x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m
, n}, x] && PiecewiseLinearQ[u, v, x] && NeQ[m, -1] && ((LtQ[m, -1] && GtQ[n, 0] &&  !(ILtQ[m + n, -2] && (Fra
ctionQ[m] || GeQ[2*n + m + 1, 0]))) || (IGtQ[n, 0] && IGtQ[m, 0] && LeQ[n, m]) || (IGtQ[n, 0] &&  !IntegerQ[m]
) || (ILtQ[m, 0] &&  !IntegerQ[n]))

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 62, normalized size = 0.91 \[ -6 b^2 x \log (x) \tanh ^{-1}(\tanh (a+b x))-\frac {\tanh ^{-1}(\tanh (a+b x))^3}{x}+3 b (\log (x)+1) \tanh ^{-1}(\tanh (a+b x))^2+\frac {3}{2} b^3 x^2 (2 \log (x)-1) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.74, size = 36, normalized size = 0.53 \[ \frac {b^{3} x^{3} + 6 \, a b^{2} x^{2} + 6 \, a^{2} b x \log \relax (x) - 2 \, a^{3}}{2 \, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.19, size = 33, normalized size = 0.49 \[ \frac {1}{2} \, b^{3} x^{2} + 3 \, a b^{2} x + 3 \, a^{2} b \log \left ({\left | x \right |}\right ) - \frac {a^{3}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.18, size = 76, normalized size = 1.12 \[ -\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{3}}{x}+3 \ln \relax (x ) \arctanh \left (\tanh \left (b x +a \right )\right )^{2} b +3 b^{3} x^{2} \ln \relax (x )-\frac {9 x^{2} b^{3}}{2}-6 b^{2} \arctanh \left (\tanh \left (b x +a \right )\right ) \ln \relax (x ) x +6 b^{2} \arctanh \left (\tanh \left (b x +a \right )\right ) x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arctanh(tanh(b*x+a))^3/x+3*ln(x)*arctanh(tanh(b*x+a))^2*b+3*b^3*x^2*ln(x)-9/2*x^2*b^3-6*b^2*arctanh(tanh(b*x+
a))*ln(x)*x+6*b^2*arctanh(tanh(b*x+a))*x

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maxima [A]  time = 0.78, size = 65, normalized size = 0.96 \[ 3 \, b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2} \log \relax (x) + \frac {3}{2} \, {\left (b^{2} x^{2} + 4 \, a b x + 2 \, a^{2} \log \relax (x) - 2 \, \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2} \log \relax (x)\right )} b - \frac {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*b*arctanh(tanh(b*x + a))^2*log(x) + 3/2*(b^2*x^2 + 4*a*b*x + 2*a^2*log(x) - 2*arctanh(tanh(b*x + a))^2*log(x
))*b - arctanh(tanh(b*x + a))^3/x

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mupad [B]  time = 1.07, size = 415, normalized size = 6.10 \[ \frac {3\,b\,{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2}{4}-\frac {{\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^3}{8\,x}+\frac {3\,b\,{\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2}{4}+\frac {{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^3}{8\,x}-\frac {3\,b^3\,x^2}{2}+\frac {3\,b\,{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2\,\ln \relax (x)}{4}+\frac {3\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,{\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2}{8\,x}-\frac {3\,{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{8\,x}+\frac {3\,b\,{\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2\,\ln \relax (x)}{4}-\frac {3\,b\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}+3\,b^3\,x^2\,\ln \relax (x)-\frac {3\,b\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\ln \relax (x)}{2}+3\,b^2\,x\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\ln \relax (x)-3\,b^2\,x\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\ln \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^3/x^2,x)

[Out]

(3*b*log(1/(exp(2*a)*exp(2*b*x) + 1))^2)/4 - log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))^3/(8*x) + (3
*b*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))^2)/4 + log(1/(exp(2*a)*exp(2*b*x) + 1))^3/(8*x) - (3*b
^3*x^2)/2 + (3*b*log(1/(exp(2*a)*exp(2*b*x) + 1))^2*log(x))/4 + (3*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*x) - (3*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)))/(8*x) + (3*b*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))^2*log(x)
)/4 - (3*b*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 + 3*b^3*x^
2*log(x) - (3*b*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))*log(x))/
2 + 3*b^2*x*log(1/(exp(2*a)*exp(2*b*x) + 1))*log(x) - 3*b^2*x*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) +
1))*log(x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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