3.55 \(\int x^3 \tanh ^{-1}(\tanh (a+b x))^3 \, dx\)

Optimal. Leaf size=61 \[ \frac{1}{20} b^2 x^6 \tanh ^{-1}(\tanh (a+b x))-\frac{3}{20} b x^5 \tanh ^{-1}(\tanh (a+b x))^2+\frac{1}{4} x^4 \tanh ^{-1}(\tanh (a+b x))^3-\frac{1}{140} b^3 x^7 \]

[Out]

-(b^3*x^7)/140 + (b^2*x^6*ArcTanh[Tanh[a + b*x]])/20 - (3*b*x^5*ArcTanh[Tanh[a + b*x]]^2)/20 + (x^4*ArcTanh[Ta
nh[a + b*x]]^3)/4

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Rubi [A]  time = 0.04446, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2168, 30} \[ \frac{1}{20} b^2 x^6 \tanh ^{-1}(\tanh (a+b x))-\frac{3}{20} b x^5 \tanh ^{-1}(\tanh (a+b x))^2+\frac{1}{4} x^4 \tanh ^{-1}(\tanh (a+b x))^3-\frac{1}{140} b^3 x^7 \]

Antiderivative was successfully verified.

[In]

Int[x^3*ArcTanh[Tanh[a + b*x]]^3,x]

[Out]

-(b^3*x^7)/140 + (b^2*x^6*ArcTanh[Tanh[a + b*x]])/20 - (3*b*x^5*ArcTanh[Tanh[a + b*x]]^2)/20 + (x^4*ArcTanh[Ta
nh[a + b*x]]^3)/4

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]))

Rule 30

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

Rubi steps

\begin{align*} \int x^3 \tanh ^{-1}(\tanh (a+b x))^3 \, dx &=\frac{1}{4} x^4 \tanh ^{-1}(\tanh (a+b x))^3-\frac{1}{4} (3 b) \int x^4 \tanh ^{-1}(\tanh (a+b x))^2 \, dx\\ &=-\frac{3}{20} b x^5 \tanh ^{-1}(\tanh (a+b x))^2+\frac{1}{4} x^4 \tanh ^{-1}(\tanh (a+b x))^3+\frac{1}{10} \left (3 b^2\right ) \int x^5 \tanh ^{-1}(\tanh (a+b x)) \, dx\\ &=\frac{1}{20} b^2 x^6 \tanh ^{-1}(\tanh (a+b x))-\frac{3}{20} b x^5 \tanh ^{-1}(\tanh (a+b x))^2+\frac{1}{4} x^4 \tanh ^{-1}(\tanh (a+b x))^3-\frac{1}{20} b^3 \int x^6 \, dx\\ &=-\frac{1}{140} b^3 x^7+\frac{1}{20} b^2 x^6 \tanh ^{-1}(\tanh (a+b x))-\frac{3}{20} b x^5 \tanh ^{-1}(\tanh (a+b x))^2+\frac{1}{4} x^4 \tanh ^{-1}(\tanh (a+b x))^3\\ \end{align*}

Mathematica [A]  time = 0.0232976, size = 54, normalized size = 0.89 \[ -\frac{1}{140} x^4 \left (-7 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))+21 b x \tanh ^{-1}(\tanh (a+b x))^2-35 \tanh ^{-1}(\tanh (a+b x))^3+b^3 x^3\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x^4*(b^3*x^3 - 7*b^2*x^2*ArcTanh[Tanh[a + b*x]] + 21*b*x*ArcTanh[Tanh[a + b*x]]^2 - 35*ArcTanh[Tanh[a + b*x]
]^3))/140

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Maple [A]  time = 0.038, size = 56, normalized size = 0.9 \begin{align*}{\frac{{x}^{4} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3}}{4}}-{\frac{3\,b}{4} \left ({\frac{{x}^{5} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}{5}}-{\frac{2\,b}{5} \left ({\frac{{x}^{6}{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{6}}-{\frac{{x}^{7}b}{42}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^4*arctanh(tanh(b*x+a))^3-3/4*b*(1/5*x^5*arctanh(tanh(b*x+a))^2-2/5*b*(1/6*x^6*arctanh(tanh(b*x+a))-1/42*
x^7*b))

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Maxima [A]  time = 1.56287, size = 73, normalized size = 1.2 \begin{align*} -\frac{3}{20} \, b x^{5} \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{2} + \frac{1}{4} \, x^{4} \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{3} - \frac{1}{140} \,{\left (b^{2} x^{7} - 7 \, b x^{6} \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/20*b*x^5*arctanh(tanh(b*x + a))^2 + 1/4*x^4*arctanh(tanh(b*x + a))^3 - 1/140*(b^2*x^7 - 7*b*x^6*arctanh(tan
h(b*x + a)))*b

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Fricas [A]  time = 1.42188, size = 80, normalized size = 1.31 \begin{align*} \frac{1}{7} \, b^{3} x^{7} + \frac{1}{2} \, a b^{2} x^{6} + \frac{3}{5} \, a^{2} b x^{5} + \frac{1}{4} \, a^{3} x^{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*b^3*x^7 + 1/2*a*b^2*x^6 + 3/5*a^2*b*x^5 + 1/4*a^3*x^4

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Sympy [A]  time = 6.78503, size = 80, normalized size = 1.31 \begin{align*} \begin{cases} \frac{x^{3} \operatorname{atanh}^{4}{\left (\tanh{\left (a + b x \right )} \right )}}{4 b} - \frac{3 x^{2} \operatorname{atanh}^{5}{\left (\tanh{\left (a + b x \right )} \right )}}{20 b^{2}} + \frac{x \operatorname{atanh}^{6}{\left (\tanh{\left (a + b x \right )} \right )}}{20 b^{3}} - \frac{\operatorname{atanh}^{7}{\left (\tanh{\left (a + b x \right )} \right )}}{140 b^{4}} & \text{for}\: b \neq 0 \\\frac{x^{4} \operatorname{atanh}^{3}{\left (\tanh{\left (a \right )} \right )}}{4} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x**3*atanh(tanh(a + b*x))**4/(4*b) - 3*x**2*atanh(tanh(a + b*x))**5/(20*b**2) + x*atanh(tanh(a + b*
x))**6/(20*b**3) - atanh(tanh(a + b*x))**7/(140*b**4), Ne(b, 0)), (x**4*atanh(tanh(a))**3/4, True))

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Giac [A]  time = 1.14231, size = 47, normalized size = 0.77 \begin{align*} \frac{1}{7} \, b^{3} x^{7} + \frac{1}{2} \, a b^{2} x^{6} + \frac{3}{5} \, a^{2} b x^{5} + \frac{1}{4} \, a^{3} x^{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*b^3*x^7 + 1/2*a*b^2*x^6 + 3/5*a^2*b*x^5 + 1/4*a^3*x^4