3.74 \(\int \frac {\tanh ^{-1}(\tanh (a+b x))^4}{x^2} \, dx\)

Optimal. Leaf size=95 \[ 4 b^2 x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2-\frac {\tanh ^{-1}(\tanh (a+b x))^4}{x}+\frac {4}{3} b \tanh ^{-1}(\tanh (a+b x))^3-2 b \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^2-4 b \log (x) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2168, 2159, 2158, 29} \[ 4 b^2 x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2-\frac {\tanh ^{-1}(\tanh (a+b x))^4}{x}+\frac {4}{3} b \tanh ^{-1}(\tanh (a+b x))^3-2 b \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^2-4 b \log (x) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^4/x^2,x]

[Out]

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

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Mathematica [A]  time = 0.09, size = 85, normalized size = 0.89 \[ 6 b^3 x^2 (2 \log (x)-1) \tanh ^{-1}(\tanh (a+b x))-12 b^2 x \log (x) \tanh ^{-1}(\tanh (a+b x))^2-\frac {\tanh ^{-1}(\tanh (a+b x))^4}{x}+4 b (\log (x)+1) \tanh ^{-1}(\tanh (a+b x))^3+\frac {2}{3} b^4 x^3 (5-6 \log (x)) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^4/x^2,x]

[Out]

-(ArcTanh[Tanh[a + b*x]]^4/x) + (2*b^4*x^3*(5 - 6*Log[x]))/3 - 12*b^2*x*ArcTanh[Tanh[a + b*x]]^2*Log[x] + 4*b*
ArcTanh[Tanh[a + b*x]]^3*(1 + Log[x]) + 6*b^3*x^2*ArcTanh[Tanh[a + b*x]]*(-1 + 2*Log[x])

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fricas [A]  time = 0.51, size = 47, normalized size = 0.49 \[ \frac {b^{4} x^{4} + 6 \, a b^{3} x^{3} + 18 \, a^{2} b^{2} x^{2} + 12 \, a^{3} b x \log \relax (x) - 3 \, a^{4}}{3 \, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^4/x^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.33, size = 44, normalized size = 0.46 \[ \frac {1}{3} \, b^{4} x^{3} + 2 \, a b^{3} x^{2} + 6 \, a^{2} b^{2} x + 4 \, a^{3} b \log \left ({\left | x \right |}\right ) - \frac {a^{4}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^4/x^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.18, size = 112, normalized size = 1.18 \[ -\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{4}}{x}+4 \ln \relax (x ) \arctanh \left (\tanh \left (b x +a \right )\right )^{3} b +12 \arctanh \left (\tanh \left (b x +a \right )\right ) \ln \relax (x ) x^{2} b^{3}+12 b^{2} \arctanh \left (\tanh \left (b x +a \right )\right )^{2} x +\frac {22 x^{3} b^{4}}{3}-4 b^{4} x^{3} \ln \relax (x )-12 b^{2} \arctanh \left (\tanh \left (b x +a \right )\right )^{2} \ln \relax (x ) x -18 \arctanh \left (\tanh \left (b x +a \right )\right ) x^{2} b^{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^4/x^2,x)

[Out]

-arctanh(tanh(b*x+a))^4/x+4*ln(x)*arctanh(tanh(b*x+a))^3*b+12*arctanh(tanh(b*x+a))*ln(x)*x^2*b^3+12*b^2*arctan
h(tanh(b*x+a))^2*x+22/3*x^3*b^4-4*b^4*x^3*ln(x)-12*b^2*arctanh(tanh(b*x+a))^2*ln(x)*x-18*arctanh(tanh(b*x+a))*
x^2*b^3

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maxima [A]  time = 0.78, size = 77, normalized size = 0.81 \[ 4 \, b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3} \log \relax (x) - \frac {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{4}}{x} + \frac {2}{3} \, {\left (2 \, b^{3} x^{3} + 9 \, a b^{2} x^{2} + 18 \, a^{2} b x + 6 \, a^{3} \log \relax (x) - 6 \, \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3} \log \relax (x)\right )} b \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^4/x^2,x, algorithm="maxima")

[Out]

4*b*arctanh(tanh(b*x + a))^3*log(x) - arctanh(tanh(b*x + a))^4/x + 2/3*(2*b^3*x^3 + 9*a*b^2*x^2 + 18*a^2*b*x +
6*a^3*log(x) - 6*arctanh(tanh(b*x + a))^3*log(x))*b

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mupad [B]  time = 0.14, size = 553, normalized size = 5.82 \[ \ln \relax (x)\,\left (4\,a^3\,b-\frac {b\,{\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}{2}+3\,a\,b\,{\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2-6\,a^2\,b\,\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )\right )-\frac {{\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^4+24\,a^2\,{\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2+16\,a^4-8\,a\,{\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3-32\,a^3\,\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{16\,x}+\frac {b^4\,x^3}{3}+\frac {3\,b^2\,x\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{2}-b^3\,x^2\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^4/x^2,x)

[Out]

log(x)*(4*a^3*b - (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)^3)/2 + 3*a*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)^2 - 6*a^2*b*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + log(2/(e
xp(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/(e
xp(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)) + l
og(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))/(16*x) + (b^4*x^3)/3 + (3*b^2*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)^2)/2 - b^3*x^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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {atanh}^{4}{\left (\tanh {\left (a + b x \right )} \right )}}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))**4/x**2,x)

[Out]

Integral(atanh(tanh(a + b*x))**4/x**2, x)

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