3.103 \(\int \frac{x^4}{\tanh ^{-1}(\tanh (a+b x))^3} \, dx\)

Optimal. Leaf size=92 \[ -\frac{2 x^3}{b^2 \tanh ^{-1}(\tanh (a+b x))}+\frac{6 x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^4}+\frac{6 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^5}-\frac{x^4}{2 b \tanh ^{-1}(\tanh (a+b x))^2}+\frac{3 x^2}{b^3} \]

[Out]

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

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Rubi [A]  time = 0.0727309, antiderivative size = 92, 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} \[ -\frac{2 x^3}{b^2 \tanh ^{-1}(\tanh (a+b x))}+\frac{6 x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^4}+\frac{6 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^5}-\frac{x^4}{2 b \tanh ^{-1}(\tanh (a+b x))^2}+\frac{3 x^2}{b^3} \]

Antiderivative was successfully verified.

[In]

Int[x^4/ArcTanh[Tanh[a + b*x]]^3,x]

[Out]

(3*x^2)/b^3 + (6*x*(b*x - ArcTanh[Tanh[a + b*x]]))/b^4 - x^4/(2*b*ArcTanh[Tanh[a + b*x]]^2) - (2*x^3)/(b^2*Arc
Tanh[Tanh[a + b*x]]) + (6*(b*x - ArcTanh[Tanh[a + b*x]])^2*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))^3} \, dx &=-\frac{x^4}{2 b \tanh ^{-1}(\tanh (a+b x))^2}+\frac{2 \int \frac{x^3}{\tanh ^{-1}(\tanh (a+b x))^2} \, dx}{b}\\ &=-\frac{x^4}{2 b \tanh ^{-1}(\tanh (a+b x))^2}-\frac{2 x^3}{b^2 \tanh ^{-1}(\tanh (a+b x))}+\frac{6 \int \frac{x^2}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{b^2}\\ &=\frac{3 x^2}{b^3}-\frac{x^4}{2 b \tanh ^{-1}(\tanh (a+b x))^2}-\frac{2 x^3}{b^2 \tanh ^{-1}(\tanh (a+b x))}-\frac{\left (6 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )\right ) \int \frac{x}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{b^3}\\ &=\frac{3 x^2}{b^3}+\frac{6 x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^4}-\frac{x^4}{2 b \tanh ^{-1}(\tanh (a+b x))^2}-\frac{2 x^3}{b^2 \tanh ^{-1}(\tanh (a+b x))}+\frac{\left (6 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2\right ) \int \frac{1}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{b^4}\\ &=\frac{3 x^2}{b^3}+\frac{6 x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^4}-\frac{x^4}{2 b \tanh ^{-1}(\tanh (a+b x))^2}-\frac{2 x^3}{b^2 \tanh ^{-1}(\tanh (a+b x))}+\frac{\left (6 \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^5}\\ &=\frac{3 x^2}{b^3}+\frac{6 x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^4}-\frac{x^4}{2 b \tanh ^{-1}(\tanh (a+b x))^2}-\frac{2 x^3}{b^2 \tanh ^{-1}(\tanh (a+b x))}+\frac{6 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^5}\\ \end{align*}

Mathematica [A]  time = 0.0418385, size = 114, normalized size = 1.24 \[ -\frac{\left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^4}{2 b^5 \tanh ^{-1}(\tanh (a+b x))^2}+\frac{4 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^3}{b^5 \tanh ^{-1}(\tanh (a+b x))}-\frac{3 x \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )}{b^4}+\frac{6 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^5}+\frac{x^2}{2 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/ArcTanh[Tanh[a + b*x]]^3,x]

[Out]

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

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Maple [B]  time = 0.041, size = 371, normalized size = 4. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/arctanh(tanh(b*x+a))^3,x)

[Out]

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

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Maxima [A]  time = 3.54089, size = 109, normalized size = 1.18 \begin{align*} \frac{b^{4} x^{4} - 4 \, a b^{3} x^{3} - 11 \, a^{2} b^{2} x^{2} + 2 \, a^{3} b x + 7 \, a^{4}}{2 \,{\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}} + \frac{6 \, a^{2} \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))^3,x, algorithm="maxima")

[Out]

1/2*(b^4*x^4 - 4*a*b^3*x^3 - 11*a^2*b^2*x^2 + 2*a^3*b*x + 7*a^4)/(b^7*x^2 + 2*a*b^6*x + a^2*b^5) + 6*a^2*log(b
*x + a)/b^5

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Fricas [A]  time = 1.50021, size = 200, normalized size = 2.17 \begin{align*} \frac{b^{4} x^{4} - 4 \, a b^{3} x^{3} - 11 \, a^{2} b^{2} x^{2} + 2 \, a^{3} b x + 7 \, a^{4} + 12 \,{\left (a^{2} b^{2} x^{2} + 2 \, a^{3} b x + a^{4}\right )} \log \left (b x + a\right )}{2 \,{\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/arctanh(tanh(b*x+a))^3,x, algorithm="fricas")

[Out]

1/2*(b^4*x^4 - 4*a*b^3*x^3 - 11*a^2*b^2*x^2 + 2*a^3*b*x + 7*a^4 + 12*(a^2*b^2*x^2 + 2*a^3*b*x + a^4)*log(b*x +
a))/(b^7*x^2 + 2*a*b^6*x + a^2*b^5)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\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**4/atanh(tanh(b*x+a))**3,x)

[Out]

Integral(x**4/atanh(tanh(a + b*x))**3, x)

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Giac [A]  time = 1.1599, size = 82, normalized size = 0.89 \begin{align*} \frac{6 \, a^{2} \log \left ({\left | b x + a \right |}\right )}{b^{5}} + \frac{b^{3} x^{2} - 6 \, a b^{2} x}{2 \, b^{6}} + \frac{8 \, a^{3} b x + 7 \, a^{4}}{2 \,{\left (b x + a\right )}^{2} b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/arctanh(tanh(b*x+a))^3,x, algorithm="giac")

[Out]

6*a^2*log(abs(b*x + a))/b^5 + 1/2*(b^3*x^2 - 6*a*b^2*x)/b^6 + 1/2*(8*a^3*b*x + 7*a^4)/((b*x + a)^2*b^5)