3.1.45 \(\int x^3 \tanh ^{-1}(\tanh (a+b x))^2 \, dx\) [45]

Optimal. Leaf size=42 \[ \frac {b^2 x^6}{60}-\frac {1}{10} b x^5 \tanh ^{-1}(\tanh (a+b x))+\frac {1}{4} x^4 \tanh ^{-1}(\tanh (a+b x))^2 \]

[Out]

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

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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}{10} b x^5 \tanh ^{-1}(\tanh (a+b x))+\frac {1}{4} x^4 \tanh ^{-1}(\tanh (a+b x))^2+\frac {b^2 x^6}{60} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*ArcTanh[Tanh[a + b*x]]^2,x]

[Out]

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

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

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Mathematica [A]
time = 0.02, size = 37, normalized size = 0.88 \begin {gather*} \frac {1}{60} x^4 \left (b^2 x^2-6 b x \tanh ^{-1}(\tanh (a+b x))+15 \tanh ^{-1}(\tanh (a+b x))^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^4*(b^2*x^2 - 6*b*x*ArcTanh[Tanh[a + b*x]] + 15*ArcTanh[Tanh[a + b*x]]^2))/60

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Maple [A]
time = 35.96, size = 38, normalized size = 0.90

method	result	size
default	\(\frac {x^{4} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}{4}-\frac {b \left (\frac {x^{5} \arctanh \left (\tanh \left (b x +a \right )\right )}{5}-\frac {b \,x^{6}}{30}\right )}{2}\)	\(38\)
risch	\(\text {Expression too large to display}\)	\(2083\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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 = 0.34, size = 36, normalized size = 0.86 \begin {gather*} \frac {1}{60} \, b^{2} x^{6} - \frac {1}{10} \, b x^{5} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right ) + \frac {1}{4} \, x^{4} \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^3*arctanh(tanh(b*x+a))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.34, size = 24, normalized size = 0.57 \begin {gather*} \frac {1}{6} \, b^{2} x^{6} + \frac {2}{5} \, a b x^{5} + \frac {1}{4} \, a^{2} x^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.26, size = 37, normalized size = 0.88 \begin {gather*} \frac {b^{2} x^{6}}{60} - \frac {b x^{5} \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}{10} + \frac {x^{4} \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.40, size = 24, normalized size = 0.57 \begin {gather*} \frac {1}{6} \, b^{2} x^{6} + \frac {2}{5} \, a b x^{5} + \frac {1}{4} \, a^{2} x^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.00, size = 36, normalized size = 0.86 \begin {gather*} \frac {b^2\,x^6}{60}-\frac {b\,x^5\,\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}{10}+\frac {x^4\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*atanh(tanh(a + b*x))^2,x)

[Out]

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

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