3.1.46 \(\int x^2 \tanh ^{-1}(\tanh (a+b x))^2 \, dx\) [46]

Optimal. Leaf size=42 \[ \frac {b^2 x^5}{30}-\frac {1}{6} b x^4 \tanh ^{-1}(\tanh (a+b x))+\frac {1}{3} x^3 \tanh ^{-1}(\tanh (a+b x))^2 \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.02, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2199, 30} \begin {gather*} -\frac {1}{6} b x^4 \tanh ^{-1}(\tanh (a+b x))+\frac {1}{3} x^3 \tanh ^{-1}(\tanh (a+b x))^2+\frac {b^2 x^5}{30} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*ArcTanh[Tanh[a + b*x]]^2,x]

[Out]

(b^2*x^5)/30 - (b*x^4*ArcTanh[Tanh[a + b*x]])/6 + (x^3*ArcTanh[Tanh[a + b*x]]^2)/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))^2 \, dx &=\frac {1}{3} x^3 \tanh ^{-1}(\tanh (a+b x))^2-\frac {1}{3} (2 b) \int x^3 \tanh ^{-1}(\tanh (a+b x)) \, dx\\ &=-\frac {1}{6} b x^4 \tanh ^{-1}(\tanh (a+b x))+\frac {1}{3} x^3 \tanh ^{-1}(\tanh (a+b x))^2+\frac {1}{6} b^2 \int x^4 \, dx\\ &=\frac {b^2 x^5}{30}-\frac {1}{6} b x^4 \tanh ^{-1}(\tanh (a+b x))+\frac {1}{3} x^3 \tanh ^{-1}(\tanh (a+b x))^2\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.04, size = 37, normalized size = 0.88 \begin {gather*} \frac {1}{30} x^3 \left (b^2 x^2-5 b x \tanh ^{-1}(\tanh (a+b x))+10 \tanh ^{-1}(\tanh (a+b x))^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^3*(b^2*x^2 - 5*b*x*ArcTanh[Tanh[a + b*x]] + 10*ArcTanh[Tanh[a + b*x]]^2))/30

________________________________________________________________________________________

Maple [A]
time = 30.39, size = 38, normalized size = 0.90

method	result	size
default	\(\frac {x^{3} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}{3}-\frac {2 b \left (\frac {x^{4} \arctanh \left (\tanh \left (b x +a \right )\right )}{4}-\frac {b \,x^{5}}{20}\right )}{3}\)	\(38\)
risch	\(\text {Expression too large to display}\)	\(2083\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.33, size = 36, normalized size = 0.86 \begin {gather*} \frac {1}{30} \, b^{2} x^{5} - \frac {1}{6} \, b x^{4} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right ) + \frac {1}{3} \, x^{3} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 0.32, size = 24, normalized size = 0.57 \begin {gather*} \frac {1}{5} \, b^{2} x^{5} + \frac {1}{2} \, a b x^{4} + \frac {1}{3} \, a^{2} x^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]
time = 0.17, size = 37, normalized size = 0.88 \begin {gather*} \frac {b^{2} x^{5}}{30} - \frac {b x^{4} \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}{6} + \frac {x^{3} \operatorname {atanh}^{2}{\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))**2,x)

[Out]

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

________________________________________________________________________________________

Giac [A]
time = 0.40, size = 24, normalized size = 0.57 \begin {gather*} \frac {1}{5} \, b^{2} x^{5} + \frac {1}{2} \, a b x^{4} + \frac {1}{3} \, a^{2} x^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Mupad [B]
time = 0.97, size = 36, normalized size = 0.86 \begin {gather*} \frac {b^2\,x^5}{30}-\frac {b\,x^4\,\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}{6}+\frac {x^3\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*atanh(tanh(a + b*x))^2,x)

[Out]

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

________________________________________________________________________________________