3.1.56 \(\int x^2 \tanh ^{-1}(\tanh (a+b x))^3 \, dx\) [56]

Optimal. Leaf size=53 \[ \frac {x^2 \tanh ^{-1}(\tanh (a+b x))^4}{4 b}-\frac {x \tanh ^{-1}(\tanh (a+b x))^5}{10 b^2}+\frac {\tanh ^{-1}(\tanh (a+b x))^6}{60 b^3} \]

[Out]

1/4*x^2*arctanh(tanh(b*x+a))^4/b-1/10*x*arctanh(tanh(b*x+a))^5/b^2+1/60*arctanh(tanh(b*x+a))^6/b^3

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Rubi [A]
time = 0.03, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2199, 2188, 30} \begin {gather*} \frac {\tanh ^{-1}(\tanh (a+b x))^6}{60 b^3}-\frac {x \tanh ^{-1}(\tanh (a+b x))^5}{10 b^2}+\frac {x^2 \tanh ^{-1}(\tanh (a+b x))^4}{4 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*ArcTanh[Tanh[a + b*x]]^3,x]

[Out]

(x^2*ArcTanh[Tanh[a + b*x]]^4)/(4*b) - (x*ArcTanh[Tanh[a + b*x]]^5)/(10*b^2) + ArcTanh[Tanh[a + b*x]]^6/(60*b^
3)

Rule 30

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

Rule 2188

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 2199

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^2 \tanh ^{-1}(\tanh (a+b x))^3 \, dx &=\frac {x^2 \tanh ^{-1}(\tanh (a+b x))^4}{4 b}-\frac {\int x \tanh ^{-1}(\tanh (a+b x))^4 \, dx}{2 b}\\ &=\frac {x^2 \tanh ^{-1}(\tanh (a+b x))^4}{4 b}-\frac {x \tanh ^{-1}(\tanh (a+b x))^5}{10 b^2}+\frac {\int \tanh ^{-1}(\tanh (a+b x))^5 \, dx}{10 b^2}\\ &=\frac {x^2 \tanh ^{-1}(\tanh (a+b x))^4}{4 b}-\frac {x \tanh ^{-1}(\tanh (a+b x))^5}{10 b^2}+\frac {\text {Subst}\left (\int x^5 \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{10 b^3}\\ &=\frac {x^2 \tanh ^{-1}(\tanh (a+b x))^4}{4 b}-\frac {x \tanh ^{-1}(\tanh (a+b x))^5}{10 b^2}+\frac {\tanh ^{-1}(\tanh (a+b x))^6}{60 b^3}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 54, normalized size = 1.02 \begin {gather*} -\frac {1}{60} x^3 \left (b^3 x^3-6 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))+15 b x \tanh ^{-1}(\tanh (a+b x))^2-20 \tanh ^{-1}(\tanh (a+b x))^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/60*(x^3*(b^3*x^3 - 6*b^2*x^2*ArcTanh[Tanh[a + b*x]] + 15*b*x*ArcTanh[Tanh[a + b*x]]^2 - 20*ArcTanh[Tanh[a +
b*x]]^3))

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Maple [A]
time = 0.02, size = 56, normalized size = 1.06 \[\frac {x^{3} \arctanh \left (\tanh \left (b x +a \right )\right )^{3}}{3}-b \left (\frac {x^{4} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}{4}-\frac {b \left (\frac {x^{5} \arctanh \left (\tanh \left (b x +a \right )\right )}{5}-\frac {b \,x^{6}}{30}\right )}{2}\right )\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*arctanh(tanh(b*x+a))^3-b*(1/4*x^4*arctanh(tanh(b*x+a))^2-1/2*b*(1/5*x^5*arctanh(tanh(b*x+a))-1/30*b*x^
6))

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Maxima [A]
time = 0.37, size = 54, normalized size = 1.02 \begin {gather*} -\frac {1}{4} \, b x^{4} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2} + \frac {1}{3} \, x^{3} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3} - \frac {1}{60} \, {\left (b^{2} x^{6} - 6 \, b x^{5} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )\right )} b \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*b*x^4*arctanh(tanh(b*x + a))^2 + 1/3*x^3*arctanh(tanh(b*x + a))^3 - 1/60*(b^2*x^6 - 6*b*x^5*arctanh(tanh(
b*x + a)))*b

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Fricas [A]
time = 0.33, size = 35, normalized size = 0.66 \begin {gather*} \frac {1}{6} \, b^{3} x^{6} + \frac {3}{5} \, a b^{2} x^{5} + \frac {3}{4} \, a^{2} b x^{4} + \frac {1}{3} \, a^{3} x^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*b^3*x^6 + 3/5*a*b^2*x^5 + 3/4*a^2*b*x^4 + 1/3*a^3*x^3

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Sympy [A]
time = 0.44, size = 56, normalized size = 1.06 \begin {gather*} - \frac {b^{3} x^{6}}{60} + \frac {b^{2} x^{5} \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}{10} - \frac {b x^{4} \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{4} + \frac {x^{3} \operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b**3*x**6/60 + b**2*x**5*atanh(tanh(a + b*x))/10 - b*x**4*atanh(tanh(a + b*x))**2/4 + x**3*atanh(tanh(a + b*x
))**3/3

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Giac [A]
time = 0.38, size = 35, normalized size = 0.66 \begin {gather*} \frac {1}{6} \, b^{3} x^{6} + \frac {3}{5} \, a b^{2} x^{5} + \frac {3}{4} \, a^{2} b x^{4} + \frac {1}{3} \, a^{3} x^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*b^3*x^6 + 3/5*a*b^2*x^5 + 3/4*a^2*b*x^4 + 1/3*a^3*x^3

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Mupad [B]
time = 0.14, size = 53, normalized size = 1.00 \begin {gather*} -\frac {b^3\,x^6}{60}+\frac {b^2\,x^5\,\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}{10}-\frac {b\,x^4\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2}{4}+\frac {x^3\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^3}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*atanh(tanh(a + b*x))^3,x)

[Out]

(x^3*atanh(tanh(a + b*x))^3)/3 - (b^3*x^6)/60 - (b*x^4*atanh(tanh(a + b*x))^2)/4 + (b^2*x^5*atanh(tanh(a + b*x
)))/10

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