3.1.37 \(\int x^2 \tanh ^{-1}(\tanh (a+b x)) \, dx\) [37]

Optimal. Leaf size=23 \[ -\frac {b x^4}{12}+\frac {1}{3} x^3 \tanh ^{-1}(\tanh (a+b x)) \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2199, 30} \begin {gather*} \frac {1}{3} x^3 \tanh ^{-1}(\tanh (a+b x))-\frac {b x^4}{12} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*ArcTanh[Tanh[a + b*x]],x]

[Out]

-1/12*(b*x^4) + (x^3*ArcTanh[Tanh[a + b*x]])/3

Rule 30

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

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

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Mathematica [A]
time = 0.01, size = 20, normalized size = 0.87 \begin {gather*} -\frac {1}{12} x^3 \left (b x-4 \tanh ^{-1}(\tanh (a+b x))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*(x^3*(b*x - 4*ArcTanh[Tanh[a + b*x]]))

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Maple [A]
time = 0.05, size = 20, normalized size = 0.87

method	result	size
default	\(-\frac {b \,x^{4}}{12}+\frac {x^{3} \arctanh \left (\tanh \left (b x +a \right )\right )}{3}\)	\(20\)
risch	\(\frac {x^{3} \ln \left ({\mathrm e}^{b x +a}\right )}{3}-\frac {b \,x^{4}}{12}+\frac {i \pi \,x^{3} \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}}{12}+\frac {i \pi \,x^{3} \mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2}}{6}-\frac {i \pi \,x^{3} \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3}}{12}-\frac {i \pi \,x^{3} \mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )}{12}+\frac {i \pi \,x^{3} \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}}{12}-\frac {i \pi \,x^{3} \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}}{12}-\frac {i \pi \,x^{3} \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )}{12}\)	\(304\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.30, size = 19, normalized size = 0.83 \begin {gather*} -\frac {1}{12} \, b x^{4} + \frac {1}{3} \, x^{3} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.34, size = 13, normalized size = 0.57 \begin {gather*} \frac {1}{4} \, b x^{4} + \frac {1}{3} \, a x^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b*x^4 + 1/3*a*x^3

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Sympy [A]
time = 0.12, size = 19, normalized size = 0.83 \begin {gather*} - \frac {b x^{4}}{12} + \frac {x^{3} \operatorname {atanh}{\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)),x)

[Out]

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

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Giac [A]
time = 0.39, size = 13, normalized size = 0.57 \begin {gather*} \frac {1}{4} \, b x^{4} + \frac {1}{3} \, a x^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b*x^4 + 1/3*a*x^3

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Mupad [B]
time = 1.00, size = 19, normalized size = 0.83 \begin {gather*} \frac {x^3\,\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}{3}-\frac {b\,x^4}{12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*atanh(tanh(a + b*x)),x)

[Out]

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

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