3.70 \(\int x^2 \tanh ^{-1}(\tanh (a+b x))^4 \, dx\)

Optimal. Leaf size=53 \[ \frac{\tanh ^{-1}(\tanh (a+b x))^7}{105 b^3}-\frac{x \tanh ^{-1}(\tanh (a+b x))^6}{15 b^2}+\frac{x^2 \tanh ^{-1}(\tanh (a+b x))^5}{5 b} \]

[Out]

(x^2*ArcTanh[Tanh[a + b*x]]^5)/(5*b) - (x*ArcTanh[Tanh[a + b*x]]^6)/(15*b^2) + ArcTanh[Tanh[a + b*x]]^7/(105*b
^3)

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Rubi [A]  time = 0.0297045, antiderivative size = 53, normalized size of antiderivative = 1., 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 = {2168, 2157, 30} \[ \frac{\tanh ^{-1}(\tanh (a+b x))^7}{105 b^3}-\frac{x \tanh ^{-1}(\tanh (a+b x))^6}{15 b^2}+\frac{x^2 \tanh ^{-1}(\tanh (a+b x))^5}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[x^2*ArcTanh[Tanh[a + b*x]]^4,x]

[Out]

(x^2*ArcTanh[Tanh[a + b*x]]^5)/(5*b) - (x*ArcTanh[Tanh[a + b*x]]^6)/(15*b^2) + ArcTanh[Tanh[a + b*x]]^7/(105*b
^3)

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 2157

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 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^2 \tanh ^{-1}(\tanh (a+b x))^4 \, dx &=\frac{x^2 \tanh ^{-1}(\tanh (a+b x))^5}{5 b}-\frac{2 \int x \tanh ^{-1}(\tanh (a+b x))^5 \, dx}{5 b}\\ &=\frac{x^2 \tanh ^{-1}(\tanh (a+b x))^5}{5 b}-\frac{x \tanh ^{-1}(\tanh (a+b x))^6}{15 b^2}+\frac{\int \tanh ^{-1}(\tanh (a+b x))^6 \, dx}{15 b^2}\\ &=\frac{x^2 \tanh ^{-1}(\tanh (a+b x))^5}{5 b}-\frac{x \tanh ^{-1}(\tanh (a+b x))^6}{15 b^2}+\frac{\operatorname{Subst}\left (\int x^6 \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{15 b^3}\\ &=\frac{x^2 \tanh ^{-1}(\tanh (a+b x))^5}{5 b}-\frac{x \tanh ^{-1}(\tanh (a+b x))^6}{15 b^2}+\frac{\tanh ^{-1}(\tanh (a+b x))^7}{105 b^3}\\ \end{align*}

Mathematica [A]  time = 0.0509028, size = 71, normalized size = 1.34 \[ \frac{1}{105} x^3 \left (-7 b^3 x^3 \tanh ^{-1}(\tanh (a+b x))+21 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))^2-35 b x \tanh ^{-1}(\tanh (a+b x))^3+35 \tanh ^{-1}(\tanh (a+b x))^4+b^4 x^4\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*ArcTanh[Tanh[a + b*x]]^4,x]

[Out]

(x^3*(b^4*x^4 - 7*b^3*x^3*ArcTanh[Tanh[a + b*x]] + 21*b^2*x^2*ArcTanh[Tanh[a + b*x]]^2 - 35*b*x*ArcTanh[Tanh[a
+ b*x]]^3 + 35*ArcTanh[Tanh[a + b*x]]^4))/105

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Maple [A]  time = 0.043, size = 74, normalized size = 1.4 \begin{align*}{\frac{{x}^{3} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{4}}{3}}-{\frac{4\,b}{3} \left ({\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 ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*arctanh(tanh(b*x+a))^4,x)

[Out]

1/3*x^3*arctanh(tanh(b*x+a))^4-4/3*b*(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 = 1.74675, size = 97, normalized size = 1.83 \begin{align*} -\frac{1}{3} \, b x^{4} \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{3} + \frac{1}{3} \, x^{3} \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{4} + \frac{1}{105} \,{\left (21 \, b x^{5} \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{2} +{\left (b^{2} x^{7} - 7 \, b x^{6} \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )\right )} b\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arctanh(tanh(b*x+a))^4,x, algorithm="maxima")

[Out]

-1/3*b*x^4*arctanh(tanh(b*x + a))^3 + 1/3*x^3*arctanh(tanh(b*x + a))^4 + 1/105*(21*b*x^5*arctanh(tanh(b*x + a)
)^2 + (b^2*x^7 - 7*b*x^6*arctanh(tanh(b*x + a)))*b)*b

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Fricas [A]  time = 1.49441, size = 99, normalized size = 1.87 \begin{align*} \frac{1}{7} \, b^{4} x^{7} + \frac{2}{3} \, a b^{3} x^{6} + \frac{6}{5} \, a^{2} b^{2} x^{5} + a^{3} b x^{4} + \frac{1}{3} \, a^{4} x^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arctanh(tanh(b*x+a))^4,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 7.45079, size = 60, normalized size = 1.13 \begin{align*} \begin{cases} \frac{x^{2} \operatorname{atanh}^{5}{\left (\tanh{\left (a + b x \right )} \right )}}{5 b} - \frac{x \operatorname{atanh}^{6}{\left (\tanh{\left (a + b x \right )} \right )}}{15 b^{2}} + \frac{\operatorname{atanh}^{7}{\left (\tanh{\left (a + b x \right )} \right )}}{105 b^{3}} & \text{for}\: b \neq 0 \\\frac{x^{3} \operatorname{atanh}^{4}{\left (\tanh{\left (a \right )} \right )}}{3} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*atanh(tanh(b*x+a))**4,x)

[Out]

Piecewise((x**2*atanh(tanh(a + b*x))**5/(5*b) - x*atanh(tanh(a + b*x))**6/(15*b**2) + atanh(tanh(a + b*x))**7/
(105*b**3), Ne(b, 0)), (x**3*atanh(tanh(a))**4/3, True))

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Giac [A]  time = 1.17556, size = 61, normalized size = 1.15 \begin{align*} \frac{1}{7} \, b^{4} x^{7} + \frac{2}{3} \, a b^{3} x^{6} + \frac{6}{5} \, a^{2} b^{2} x^{5} + a^{3} b x^{4} + \frac{1}{3} \, a^{4} x^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arctanh(tanh(b*x+a))^4,x, algorithm="giac")

[Out]

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