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

Optimal. Leaf size=105 \[ -b x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3+\frac{1}{2} \tanh ^{-1}(\tanh (a+b x))^2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2-\frac{1}{3} \tanh ^{-1}(\tanh (a+b x))^3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )+\frac{1}{4} \tanh ^{-1}(\tanh (a+b x))^4+\log (x) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \]

[Out]

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

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Rubi [A]  time = 0.064792, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {2159, 2158, 29} \[ -b x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3+\frac{1}{2} \tanh ^{-1}(\tanh (a+b x))^2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2-\frac{1}{3} \tanh ^{-1}(\tanh (a+b x))^3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )+\frac{1}{4} \tanh ^{-1}(\tanh (a+b x))^4+\log (x) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^4/x,x]

[Out]

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

Mathematica [A]  time = 0.0914317, size = 175, normalized size = 1.67 \[ \frac{1}{2} (a+b x)^2 \left (a^2-4 a \left (-\tanh ^{-1}(\tanh (a+b x))+a+b x\right )+6 \left (-\tanh ^{-1}(\tanh (a+b x))+a+b x\right )^2\right )+(a+b x) \left (-4 a^2 \left (-\tanh ^{-1}(\tanh (a+b x))+a+b x\right )+a^3+6 a \left (-\tanh ^{-1}(\tanh (a+b x))+a+b x\right )^2-4 \left (-\tanh ^{-1}(\tanh (a+b x))+a+b x\right )^3\right )+\frac{1}{4} (a+b x)^4-\frac{1}{3} (a+b x)^3 \left (-4 \tanh ^{-1}(\tanh (a+b x))+3 a+4 b x\right )+\log (b x) \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^4 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.038, size = 127, normalized size = 1.2 \begin{align*} \ln \left ( x \right ) \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{4}-4\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \ln \left ( x \right ){x}^{3}{b}^{3}+6\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}\ln \left ( x \right ){x}^{2}{b}^{2}+{\frac{22\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ){x}^{3}{b}^{3}}{3}}-9\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}{x}^{2}{b}^{2}-4\,b \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3}\ln \left ( x \right ) x+{b}^{4}{x}^{4}\ln \left ( x \right ) -{\frac{25\,{x}^{4}{b}^{4}}{12}}+4\,b \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3}x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(x)*arctanh(tanh(b*x+a))^4-4*arctanh(tanh(b*x+a))*ln(x)*x^3*b^3+6*arctanh(tanh(b*x+a))^2*ln(x)*x^2*b^2+22/3*
arctanh(tanh(b*x+a))*x^3*b^3-9*arctanh(tanh(b*x+a))^2*x^2*b^2-4*b*arctanh(tanh(b*x+a))^3*ln(x)*x+b^4*x^4*ln(x)
-25/12*x^4*b^4+4*b*arctanh(tanh(b*x+a))^3*x

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Maxima [A]  time = 2.40378, size = 57, normalized size = 0.54 \begin{align*} \frac{1}{4} \, b^{4} x^{4} + \frac{4}{3} \, a b^{3} x^{3} + 3 \, a^{2} b^{2} x^{2} + 4 \, a^{3} b x + a^{4} \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.61591, size = 95, normalized size = 0.9 \begin{align*} \frac{1}{4} \, b^{4} x^{4} + \frac{4}{3} \, a b^{3} x^{3} + 3 \, a^{2} b^{2} x^{2} + 4 \, a^{3} b x + a^{4} \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b^4*x^4 + 4/3*a*b^3*x^3 + 3*a^2*b^2*x^2 + 4*a^3*b*x + a^4*log(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}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.1105, size = 58, normalized size = 0.55 \begin{align*} \frac{1}{4} \, b^{4} x^{4} + \frac{4}{3} \, a b^{3} x^{3} + 3 \, a^{2} b^{2} x^{2} + 4 \, a^{3} b x + a^{4} \log \left ({\left | x \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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