3.94 \(\int \frac{x^4}{\tanh ^{-1}(\tanh (a+b x))^2} \, dx\)

Optimal. Leaf size=98 \[ \frac{2 x^2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^3}+\frac{4 x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{b^4}+\frac{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^5}-\frac{x^4}{b \tanh ^{-1}(\tanh (a+b x))}+\frac{4 x^3}{3 b^2} \]

[Out]

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

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Rubi [A]  time = 0.0806588, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ \frac{2 x^2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^3}+\frac{4 x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{b^4}+\frac{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^5}-\frac{x^4}{b \tanh ^{-1}(\tanh (a+b x))}+\frac{4 x^3}{3 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^4/ArcTanh[Tanh[a + b*x]]^2,x]

[Out]

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

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^4}{\tanh ^{-1}(\tanh (a+b x))^2} \, dx &=-\frac{x^4}{b \tanh ^{-1}(\tanh (a+b x))}+\frac{4 \int \frac{x^3}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{b}\\ &=\frac{4 x^3}{3 b^2}-\frac{x^4}{b \tanh ^{-1}(\tanh (a+b x))}-\frac{\left (4 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )\right ) \int \frac{x^2}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{b^2}\\ &=\frac{4 x^3}{3 b^2}+\frac{2 x^2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^3}-\frac{x^4}{b \tanh ^{-1}(\tanh (a+b x))}+\frac{\left (4 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2\right ) \int \frac{x}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{b^3}\\ &=\frac{4 x^3}{3 b^2}+\frac{2 x^2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^3}+\frac{4 x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{b^4}-\frac{x^4}{b \tanh ^{-1}(\tanh (a+b x))}-\frac{\left (4 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^3\right ) \int \frac{1}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{b^4}\\ &=\frac{4 x^3}{3 b^2}+\frac{2 x^2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^3}+\frac{4 x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{b^4}-\frac{x^4}{b \tanh ^{-1}(\tanh (a+b x))}-\frac{\left (4 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^3\right ) \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{b^5}\\ &=\frac{4 x^3}{3 b^2}+\frac{2 x^2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^3}+\frac{4 x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{b^4}-\frac{x^4}{b \tanh ^{-1}(\tanh (a+b x))}+\frac{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^5}\\ \end{align*}

Mathematica [A]  time = 0.0972113, size = 106, normalized size = 1.08 \[ -\frac{x^2 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )}{b^3}-\frac{\left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^4}{b^5 \tanh ^{-1}(\tanh (a+b x))}+\frac{3 x \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2}{b^4}-\frac{4 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^3 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^5}+\frac{x^3}{3 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/ArcTanh[Tanh[a + b*x]]^2,x]

[Out]

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

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Maple [B]  time = 0.042, size = 350, normalized size = 3.6 \begin{align*}{\frac{{x}^{3}}{3\,{b}^{2}}}-{\frac{a{x}^{2}}{{b}^{3}}}-{\frac{{x}^{2} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) }{{b}^{3}}}+3\,{\frac{{a}^{2}x}{{b}^{4}}}+6\,{\frac{a \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) x}{{b}^{4}}}+3\,{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{2}x}{{b}^{4}}}-4\,{\frac{\ln \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ){a}^{3}}{{b}^{5}}}-12\,{\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}^{5}}}-12\,{\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}^{5}}}-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}^{5}}}-{\frac{{a}^{4}}{{b}^{5}{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}-4\,{\frac{{a}^{3} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) }{{b}^{5}{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}-6\,{\frac{{a}^{2} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{2}}{{b}^{5}{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}-4\,{\frac{a \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{3}}{{b}^{5}{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}-{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{4}}{{b}^{5}{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/arctanh(tanh(b*x+a))^2,x)

[Out]

1/3*x^3/b^2-1/b^3*a*x^2-1/b^3*x^2*(arctanh(tanh(b*x+a))-b*x-a)+3/b^4*x*a^2+6/b^4*a*(arctanh(tanh(b*x+a))-b*x-a
)*x+3/b^4*(arctanh(tanh(b*x+a))-b*x-a)^2*x-4/b^5*ln(arctanh(tanh(b*x+a)))*a^3-12/b^5*ln(arctanh(tanh(b*x+a)))*
a^2*(arctanh(tanh(b*x+a))-b*x-a)-12/b^5*ln(arctanh(tanh(b*x+a)))*a*(arctanh(tanh(b*x+a))-b*x-a)^2-4/b^5*ln(arc
tanh(tanh(b*x+a)))*(arctanh(tanh(b*x+a))-b*x-a)^3-1/b^5/arctanh(tanh(b*x+a))*a^4-4/b^5/arctanh(tanh(b*x+a))*a^
3*(arctanh(tanh(b*x+a))-b*x-a)-6/b^5/arctanh(tanh(b*x+a))*a^2*(arctanh(tanh(b*x+a))-b*x-a)^2-4/b^5/arctanh(tan
h(b*x+a))*a*(arctanh(tanh(b*x+a))-b*x-a)^3-1/b^5/arctanh(tanh(b*x+a))*(arctanh(tanh(b*x+a))-b*x-a)^4

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Maxima [A]  time = 2.43619, size = 95, normalized size = 0.97 \begin{align*} \frac{b^{4} x^{4} - 2 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} + 9 \, a^{3} b x - 3 \, a^{4}}{3 \,{\left (b^{6} x + a b^{5}\right )}} - \frac{4 \, a^{3} \log \left (b x + a\right )}{b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/arctanh(tanh(b*x+a))^2,x, algorithm="maxima")

[Out]

1/3*(b^4*x^4 - 2*a*b^3*x^3 + 6*a^2*b^2*x^2 + 9*a^3*b*x - 3*a^4)/(b^6*x + a*b^5) - 4*a^3*log(b*x + a)/b^5

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Fricas [A]  time = 1.51892, size = 155, normalized size = 1.58 \begin{align*} \frac{b^{4} x^{4} - 2 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} + 9 \, a^{3} b x - 3 \, a^{4} - 12 \,{\left (a^{3} b x + a^{4}\right )} \log \left (b x + a\right )}{3 \,{\left (b^{6} x + a b^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/arctanh(tanh(b*x+a))^2,x, algorithm="fricas")

[Out]

1/3*(b^4*x^4 - 2*a*b^3*x^3 + 6*a^2*b^2*x^2 + 9*a^3*b*x - 3*a^4 - 12*(a^3*b*x + a^4)*log(b*x + a))/(b^6*x + a*b
^5)

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

[Out]

Integral(x**4/atanh(tanh(a + b*x))**2, x)

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Giac [A]  time = 1.16089, size = 84, normalized size = 0.86 \begin{align*} -\frac{4 \, a^{3} \log \left ({\left | b x + a \right |}\right )}{b^{5}} - \frac{a^{4}}{{\left (b x + a\right )} b^{5}} + \frac{b^{4} x^{3} - 3 \, a b^{3} x^{2} + 9 \, a^{2} b^{2} x}{3 \, b^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/arctanh(tanh(b*x+a))^2,x, algorithm="giac")

[Out]

-4*a^3*log(abs(b*x + a))/b^5 - a^4/((b*x + a)*b^5) + 1/3*(b^4*x^3 - 3*a*b^3*x^2 + 9*a^2*b^2*x)/b^6