3.95 \(\int \frac{x^3}{\tanh ^{-1}(\tanh (a+b x))^2} \, dx\)

Optimal. Leaf size=75 \[ \frac{3 x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^3}+\frac{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^4}-\frac{x^3}{b \tanh ^{-1}(\tanh (a+b x))}+\frac{3 x^2}{2 b^2} \]

[Out]

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

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Rubi [A]  time = 0.0529525, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {2168, 2159, 2158, 2157, 29} \[ \frac{3 x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^3}+\frac{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^4}-\frac{x^3}{b \tanh ^{-1}(\tanh (a+b x))}+\frac{3 x^2}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^3/ArcTanh[Tanh[a + b*x]]^2,x]

[Out]

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

Mathematica [A]  time = 0.052245, size = 83, normalized size = 1.11 \[ \frac{\left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^3}{b^4 \tanh ^{-1}(\tanh (a+b x))}-\frac{2 x \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )}{b^3}+\frac{3 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^4}+\frac{x^2}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/ArcTanh[Tanh[a + b*x]]^2,x]

[Out]

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

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Maple [B]  time = 0.041, size = 223, normalized size = 3. \begin{align*}{\frac{{x}^{2}}{2\,{b}^{2}}}-2\,{\frac{ax}{{b}^{3}}}-2\,{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) x}{{b}^{3}}}+{\frac{{a}^{3}}{{b}^{4}{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}+3\,{\frac{{a}^{2} \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{a \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 ) }}+{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{3}}{{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}^{2}}{{b}^{4}}}+6\,{\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 ) }{{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 ) ^{2}}{{b}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/arctanh(tanh(b*x+a))^2,x)

[Out]

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

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Maxima [A]  time = 2.43982, size = 80, normalized size = 1.07 \begin{align*} \frac{b^{3} x^{3} - 3 \, a b^{2} x^{2} - 4 \, a^{2} b x + 2 \, a^{3}}{2 \,{\left (b^{5} x + a b^{4}\right )}} + \frac{3 \, a^{2} \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))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.51337, size = 132, normalized size = 1.76 \begin{align*} \frac{b^{3} x^{3} - 3 \, a b^{2} x^{2} - 4 \, a^{2} b x + 2 \, a^{3} + 6 \,{\left (a^{2} b x + a^{3}\right )} \log \left (b x + a\right )}{2 \,{\left (b^{5} x + a b^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/arctanh(tanh(b*x+a))^2,x, algorithm="fricas")

[Out]

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

[Out]

Integral(x**3/atanh(tanh(a + b*x))**2, x)

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Giac [A]  time = 1.11688, size = 65, normalized size = 0.87 \begin{align*} \frac{3 \, a^{2} \log \left ({\left | b x + a \right |}\right )}{b^{4}} + \frac{a^{3}}{{\left (b x + a\right )} b^{4}} + \frac{b^{2} x^{2} - 4 \, a b x}{2 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/arctanh(tanh(b*x+a))^2,x, algorithm="giac")

[Out]

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