∫ x tanh−1(tanh(a + bx))3 dx [57]

Optimal. Leaf size=34 \[ \frac {x \tanh ^{-1}(\tanh (a+b x))^4}{4 b}-\frac {\tanh ^{-1}(\tanh (a+b x))^5}{20 b^2} \]

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Rubi [A]
time = 0.01, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2199, 2188, 30} \begin {gather*} \frac {x \tanh ^{-1}(\tanh (a+b x))^4}{4 b}-\frac {\tanh ^{-1}(\tanh (a+b x))^5}{20 b^2} \end {gather*}

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Int[(u_)^(m_.), x_Symbol] :> With[{c = Simplify[D[u, x]]}, Dist[1/c, Subst[Int[x^m, x], x, u], x]] /; FreeQ[m,

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]

\begin {align*} \int x \tanh ^{-1}(\tanh (a+b x))^3 \, dx &=\frac {x \tanh ^{-1}(\tanh (a+b x))^4}{4 b}-\frac {\int \tanh ^{-1}(\tanh (a+b x))^4 \, dx}{4 b}\\ &=\frac {x \tanh ^{-1}(\tanh (a+b x))^4}{4 b}-\frac {\text {Subst}\left (\int x^4 \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{4 b^2}\\ &=\frac {x \tanh ^{-1}(\tanh (a+b x))^4}{4 b}-\frac {\tanh ^{-1}(\tanh (a+b x))^5}{20 b^2}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(99\) vs. \(2(34)=68\).
time = 0.05, size = 99, normalized size = 2.91 \begin {gather*} \frac {(a+b x) \left ((4 a-b x) (a+b x)^3-5 (3 a-b x) (a+b x)^2 \tanh ^{-1}(\tanh (a+b x))+10 \left (2 a^2+a b x-b^2 x^2\right ) \tanh ^{-1}(\tanh (a+b x))^2-10 (a-b x) \tanh ^{-1}(\tanh (a+b x))^3\right )}{20 b^2} \end {gather*}

((a + b*x)*((4*a - b*x)*(a + b*x)^3 - 5*(3*a - b*x)*(a + b*x)^2*ArcTanh[Tanh[a + b*x]] + 10*(2*a^2 + a*b*x - b

^2*x^2)*ArcTanh[Tanh[a + b*x]]^2 - 10*(a - b*x)*ArcTanh[Tanh[a + b*x]]^3))/(20*b^2)

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method	result	size

default	\(\frac {x^{2} \arctanh \left (\tanh \left (b x +a \right )\right )^{3}}{2}-\frac {3 b \left (\frac {x^{3} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}{3}-\frac {2 b \left (\frac {x^{4} \arctanh \left (\tanh \left (b x +a \right )\right )}{4}-\frac {b \,x^{5}}{20}\right )}{3}\right )}{2}\)	\(56\)
risch	\(\text {Expression too large to display}\)	\(8165\)

1/2*x^2*arctanh(tanh(b*x+a))^3-3/2*b*(1/3*x^3*arctanh(tanh(b*x+a))^2-2/3*b*(1/4*x^4*arctanh(tanh(b*x+a))-1/20*

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Maxima [A]
time = 0.37, size = 54, normalized size = 1.59 \begin {gather*} -\frac {1}{2} \, b x^{3} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2} + \frac {1}{2} \, x^{2} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3} - \frac {1}{20} \, {\left (b^{2} x^{5} - 5 \, b x^{4} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )\right )} b \end {gather*}

-1/2*b*x^3*arctanh(tanh(b*x + a))^2 + 1/2*x^2*arctanh(tanh(b*x + a))^3 - 1/20*(b^2*x^5 - 5*b*x^4*arctanh(tanh(

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Fricas [A]
time = 0.33, size = 34, normalized size = 1.00 \begin {gather*} \frac {1}{5} \, b^{3} x^{5} + \frac {3}{4} \, a b^{2} x^{4} + a^{2} b x^{3} + \frac {1}{2} \, a^{3} x^{2} \end {gather*}

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Sympy [A]
time = 0.23, size = 41, normalized size = 1.21 \begin {gather*} \begin {cases} \frac {x \operatorname {atanh}^{4}{\left (\tanh {\left (a + b x \right )} \right )}}{4 b} - \frac {\operatorname {atanh}^{5}{\left (\tanh {\left (a + b x \right )} \right )}}{20 b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{2} \operatorname {atanh}^{3}{\left (\tanh {\left (a \right )} \right )}}{2} & \text {otherwise} \end {cases} \end {gather*}

Piecewise((x*atanh(tanh(a + b*x))**4/(4*b) - atanh(tanh(a + b*x))**5/(20*b**2), Ne(b, 0)), (x**2*atanh(tanh(a)

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Giac [A]
time = 0.38, size = 34, normalized size = 1.00 \begin {gather*} \frac {1}{5} \, b^{3} x^{5} + \frac {3}{4} \, a b^{2} x^{4} + a^{2} b x^{3} + \frac {1}{2} \, a^{3} x^{2} \end {gather*}

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Mupad [B]
time = 0.98, size = 53, normalized size = 1.56 \begin {gather*} -\frac {b^3\,x^5}{20}+\frac {b^2\,x^4\,\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}{4}-\frac {b\,x^3\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2}{2}+\frac {x^2\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^3}{2} \end {gather*}

(x^2*atanh(tanh(a + b*x))^3)/2 - (b^3*x^5)/20 - (b*x^3*atanh(tanh(a + b*x))^2)/2 + (b^2*x^4*atanh(tanh(a + b*x

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3.1.57 \(\int x \tanh ^{-1}(\tanh (a+b x))^3 \, dx\) [57]