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

Optimal. Leaf size=81 \[ \frac{x^2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{2 b^2}+\frac{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{b^3}+\frac{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^4}+\frac{x^3}{3 b} \]

[Out]

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

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Rubi [A]  time = 0.058285, antiderivative size = 81, 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 = {2159, 2158, 2157, 29} \[ \frac{x^2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{2 b^2}+\frac{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{b^3}+\frac{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^4}+\frac{x^3}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[x^3/ArcTanh[Tanh[a + b*x]],x]

[Out]

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

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 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 29

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

Rubi steps

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

Mathematica [A]  time = 0.04354, size = 79, normalized size = 0.98 \[ -\frac{x^2 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )}{2 b^2}+\frac{x \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2}{b^3}-\frac{\left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^3 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^4}+\frac{x^3}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.034, size = 202, normalized size = 2.5 \begin{align*}{\frac{{x}^{3}}{3\,b}}-{\frac{a{x}^{2}}{2\,{b}^{2}}}-{\frac{{x}^{2} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) }{2\,{b}^{2}}}+{\frac{{a}^{2}x}{{b}^{3}}}+2\,{\frac{a \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) x}{{b}^{3}}}+{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{2}x}{{b}^{3}}}-{\frac{\ln \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ){a}^{3}}{{b}^{4}}}-3\,{\frac{\ln \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ){a}^{2} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) }{{b}^{4}}}-3\,{\frac{\ln \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) a \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{2}}{{b}^{4}}}-{\frac{\ln \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{3}}{{b}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3/b-1/2/b^2*a*x^2-1/2/b^2*x^2*(arctanh(tanh(b*x+a))-b*x-a)+1/b^3*x*a^2+2/b^3*a*(arctanh(tanh(b*x+a))-b*x
-a)*x+1/b^3*(arctanh(tanh(b*x+a))-b*x-a)^2*x-1/b^4*ln(arctanh(tanh(b*x+a)))*a^3-3/b^4*ln(arctanh(tanh(b*x+a)))
*a^2*(arctanh(tanh(b*x+a))-b*x-a)-3/b^4*ln(arctanh(tanh(b*x+a)))*a*(arctanh(tanh(b*x+a))-b*x-a)^2-1/b^4*ln(arc
tanh(tanh(b*x+a)))*(arctanh(tanh(b*x+a))-b*x-a)^3

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Maxima [A]  time = 1.78117, size = 57, normalized size = 0.7 \begin{align*} -\frac{a^{3} \log \left (b x + a\right )}{b^{4}} + \frac{2 \, b^{2} x^{3} - 3 \, a b x^{2} + 6 \, a^{2} x}{6 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.48546, size = 92, normalized size = 1.14 \begin{align*} \frac{2 \, b^{3} x^{3} - 3 \, a b^{2} x^{2} + 6 \, a^{2} b x - 6 \, a^{3} \log \left (b x + a\right )}{6 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{\operatorname{atanh}{\left (\tanh{\left (a + b x \right )} \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.1273, size = 58, normalized size = 0.72 \begin{align*} -\frac{a^{3} \log \left ({\left | b x + a \right |}\right )}{b^{4}} + \frac{2 \, b^{2} x^{3} - 3 \, a b x^{2} + 6 \, a^{2} x}{6 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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