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

Optimal. Leaf size=34 \[ \frac {x \tanh ^{-1}(\tanh (a+b x))^3}{3 b}-\frac {\tanh ^{-1}(\tanh (a+b x))^4}{12 b^2} \]

[Out]

1/3*x*arctanh(tanh(b*x+a))^3/b-1/12*arctanh(tanh(b*x+a))^4/b^2

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Rubi [A]  time = 0.02, 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 = {2168, 2157, 30} \[ \frac {x \tanh ^{-1}(\tanh (a+b x))^3}{3 b}-\frac {\tanh ^{-1}(\tanh (a+b x))^4}{12 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x*ArcTanh[Tanh[a + b*x]]^2,x]

[Out]

(x*ArcTanh[Tanh[a + b*x]]^3)/(3*b) - ArcTanh[Tanh[a + b*x]]^4/(12*b^2)

Rule 30

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

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 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]))

Rubi steps

\begin {align*} \int x \tanh ^{-1}(\tanh (a+b x))^2 \, dx &=\frac {x \tanh ^{-1}(\tanh (a+b x))^3}{3 b}-\frac {\int \tanh ^{-1}(\tanh (a+b x))^3 \, dx}{3 b}\\ &=\frac {x \tanh ^{-1}(\tanh (a+b x))^3}{3 b}-\frac {\operatorname {Subst}\left (\int x^3 \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{3 b^2}\\ &=\frac {x \tanh ^{-1}(\tanh (a+b x))^3}{3 b}-\frac {\tanh ^{-1}(\tanh (a+b x))^4}{12 b^2}\\ \end {align*}

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Mathematica [B]  time = 0.08, size = 74, normalized size = 2.18 \[ \frac {(a+b x) \left (4 \left (2 a^2+a b x-b^2 x^2\right ) \tanh ^{-1}(\tanh (a+b x))-\left ((3 a-b x) (a+b x)^2\right )-6 (a-b x) \tanh ^{-1}(\tanh (a+b x))^2\right )}{12 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)*(-((3*a - b*x)*(a + b*x)^2) + 4*(2*a^2 + a*b*x - b^2*x^2)*ArcTanh[Tanh[a + b*x]] - 6*(a - b*x)*ArcT
anh[Tanh[a + b*x]]^2))/(12*b^2)

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fricas [A]  time = 0.50, size = 24, normalized size = 0.71 \[ \frac {1}{4} \, b^{2} x^{4} + \frac {2}{3} \, a b x^{3} + \frac {1}{2} \, a^{2} x^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b^2*x^4 + 2/3*a*b*x^3 + 1/2*a^2*x^2

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giac [A]  time = 0.16, size = 24, normalized size = 0.71 \[ \frac {1}{4} \, b^{2} x^{4} + \frac {2}{3} \, a b x^{3} + \frac {1}{2} \, a^{2} x^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b^2*x^4 + 2/3*a*b*x^3 + 1/2*a^2*x^2

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maple [A]  time = 0.14, size = 38, normalized size = 1.12 \[ \frac {x^{2} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}{2}-b \left (-\frac {b \,x^{4}}{12}+\frac {x^{3} \arctanh \left (\tanh \left (b x +a \right )\right )}{3}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*arctanh(tanh(b*x+a))^2-b*(-1/12*b*x^4+1/3*x^3*arctanh(tanh(b*x+a)))

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maxima [A]  time = 0.45, size = 36, normalized size = 1.06 \[ \frac {1}{12} \, b^{2} x^{4} - \frac {1}{3} \, b x^{3} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right ) + \frac {1}{2} \, x^{2} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*b^2*x^4 - 1/3*b*x^3*arctanh(tanh(b*x + a)) + 1/2*x^2*arctanh(tanh(b*x + a))^2

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mupad [B]  time = 0.94, size = 36, normalized size = 1.06 \[ \frac {b^2\,x^4}{12}-\frac {b\,x^3\,\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}{3}+\frac {x^2\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*atanh(tanh(a + b*x))^2,x)

[Out]

(x^2*atanh(tanh(a + b*x))^2)/2 + (b^2*x^4)/12 - (b*x^3*atanh(tanh(a + b*x)))/3

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sympy [A]  time = 0.63, size = 41, normalized size = 1.21 \[ \begin {cases} \frac {x \operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{3 b} - \frac {\operatorname {atanh}^{4}{\left (\tanh {\left (a + b x \right )} \right )}}{12 b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{2} \operatorname {atanh}^{2}{\left (\tanh {\relax (a )} \right )}}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*atanh(tanh(b*x+a))**2,x)

[Out]

Piecewise((x*atanh(tanh(a + b*x))**3/(3*b) - atanh(tanh(a + b*x))**4/(12*b**2), Ne(b, 0)), (x**2*atanh(tanh(a)
)**2/2, True))

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