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

Optimal. Leaf size=71 \[ -\frac{3 x^2}{2 b^2 \tanh ^{-1}(\tanh (a+b x))}+\frac{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^4}-\frac{x^3}{2 b \tanh ^{-1}(\tanh (a+b x))^2}+\frac{3 x}{b^3} \]

[Out]

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

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Rubi [A]  time = 0.0483379, antiderivative size = 71, 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, 2158, 2157, 29} \[ -\frac{3 x^2}{2 b^2 \tanh ^{-1}(\tanh (a+b x))}+\frac{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^4}-\frac{x^3}{2 b \tanh ^{-1}(\tanh (a+b x))^2}+\frac{3 x}{b^3} \]

Antiderivative was successfully verified.

[In]

Int[x^3/ArcTanh[Tanh[a + b*x]]^3,x]

[Out]

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

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

Mathematica [A]  time = 0.0654912, size = 86, normalized size = 1.21 \[ -\frac{3 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))-b x \tanh ^{-1}(\tanh (a+b x))^2 \left (6 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )+11\right )+\tanh ^{-1}(\tanh (a+b x))^3 \left (6 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )+5\right )+b^3 x^3}{2 b^4 \tanh ^{-1}(\tanh (a+b x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b^3*x^3 + 3*b^2*x^2*ArcTanh[Tanh[a + b*x]] + ArcTanh[Tanh[a + b*x]]^3*(5 + 6*Log[ArcTanh[Tanh[a + b*x]]]) -
b*x*ArcTanh[Tanh[a + b*x]]^2*(11 + 6*Log[ArcTanh[Tanh[a + b*x]]]))/(2*b^4*ArcTanh[Tanh[a + b*x]]^2)

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Maple [B]  time = 0.04, size = 239, normalized size = 3.4 \begin{align*}{\frac{x}{{b}^{3}}}-3\,{\frac{{a}^{2}}{{b}^{4}{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}-6\,{\frac{a \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) }{{b}^{4}{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}-3\,{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{2}}{{b}^{4}{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}-3\,{\frac{\ln \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) a}{{b}^{4}}}-3\,{\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 ) }{{b}^{4}}}+{\frac{{a}^{3}}{2\,{b}^{4} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}}+{\frac{3\,{a}^{2} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) }{2\,{b}^{4} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}}+{\frac{3\,a \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{2}}{2\,{b}^{4} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}}+{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{3}}{2\,{b}^{4} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/b^3-3/b^4/arctanh(tanh(b*x+a))*a^2-6/b^4/arctanh(tanh(b*x+a))*a*(arctanh(tanh(b*x+a))-b*x-a)-3/b^4/arctanh(t
anh(b*x+a))*(arctanh(tanh(b*x+a))-b*x-a)^2-3/b^4*ln(arctanh(tanh(b*x+a)))*a-3/b^4*ln(arctanh(tanh(b*x+a)))*(ar
ctanh(tanh(b*x+a))-b*x-a)+1/2/b^4/arctanh(tanh(b*x+a))^2*a^3+3/2/b^4/arctanh(tanh(b*x+a))^2*a^2*(arctanh(tanh(
b*x+a))-b*x-a)+3/2/b^4/arctanh(tanh(b*x+a))^2*a*(arctanh(tanh(b*x+a))-b*x-a)^2+1/2/b^4/arctanh(tanh(b*x+a))^2*
(arctanh(tanh(b*x+a))-b*x-a)^3

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Maxima [A]  time = 3.48334, size = 93, normalized size = 1.31 \begin{align*} \frac{2 \, b^{3} x^{3} + 4 \, a b^{2} x^{2} - 4 \, a^{2} b x - 5 \, a^{3}}{2 \,{\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} - \frac{3 \, a \log \left (b x + a\right )}{b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*b^3*x^3 + 4*a*b^2*x^2 - 4*a^2*b*x - 5*a^3)/(b^6*x^2 + 2*a*b^5*x + a^2*b^4) - 3*a*log(b*x + a)/b^4

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Fricas [A]  time = 1.47845, size = 176, normalized size = 2.48 \begin{align*} \frac{2 \, b^{3} x^{3} + 4 \, a b^{2} x^{2} - 4 \, a^{2} b x - 5 \, a^{3} - 6 \,{\left (a b^{2} x^{2} + 2 \, a^{2} b x + a^{3}\right )} \log \left (b x + a\right )}{2 \,{\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*b^3*x^3 + 4*a*b^2*x^2 - 4*a^2*b*x - 5*a^3 - 6*(a*b^2*x^2 + 2*a^2*b*x + a^3)*log(b*x + a))/(b^6*x^2 + 2*
a*b^5*x + a^2*b^4)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{\operatorname{atanh}^{3}{\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))**3,x)

[Out]

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

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Giac [A]  time = 1.18488, size = 59, normalized size = 0.83 \begin{align*} \frac{x}{b^{3}} - \frac{3 \, a \log \left ({\left | b x + a \right |}\right )}{b^{4}} - \frac{6 \, a^{2} b x + 5 \, a^{3}}{2 \,{\left (b x + a\right )}^{2} b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/b^3 - 3*a*log(abs(b*x + a))/b^4 - 1/2*(6*a^2*b*x + 5*a^3)/((b*x + a)^2*b^4)