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[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]

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Rubi [A]  time = 0.0638568, antiderivative size = 95, normalized size of antiderivative = 1., 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 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]))

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 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 29

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

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*}

Mathematica [A]  time = 0.0759046, 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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Maple [A]  time = 0.04, size = 112, normalized size = 1.2 \begin{align*} -{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{4}}{x}}+4\,\ln \left ( x \right ) \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3}b+12\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \ln \left ( x \right ){x}^{2}{b}^{3}-18\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ){x}^{2}{b}^{3}-12\,{b}^{2} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}\ln \left ( x \right ) x-4\,{b}^{4}{x}^{3}\ln \left ( x \right ) +{\frac{22\,{x}^{3}{b}^{4}}{3}}+12\,{b}^{2} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}x \end{align*}

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

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Maxima [A]  time = 2.61009, size = 104, normalized size = 1.09 \begin{align*} 4 \, b \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{3} \log \left (x\right ) - \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 \left (x\right ) - 6 \, \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{3} \log \left (x\right )\right )} b \end{align*}

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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Fricas [A]  time = 1.48507, size = 103, normalized size = 1.08 \begin{align*} \frac{b^{4} x^{4} + 6 \, a b^{3} x^{3} + 18 \, a^{2} b^{2} x^{2} + 12 \, a^{3} b x \log \left (x\right ) - 3 \, a^{4}}{3 \, x} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{atanh}^{4}{\left (\tanh{\left (a + b x \right )} \right )}}{x^{2}}\, dx \end{align*}

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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Giac [A]  time = 1.18977, size = 59, normalized size = 0.62 \begin{align*} \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} \end{align*}

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