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]

1/5*x^2*arctanh(tanh(b*x+a))^5/b-1/15*x*arctanh(tanh(b*x+a))^6/b^2+1/105*arctanh(tanh(b*x+a))^7/b^3

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Rubi [A]  time = 0.03, antiderivative size = 53, normalized size of antiderivative = 1.00, 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 30

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

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 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]))

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*}

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Mathematica [A]  time = 0.06, 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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fricas [A]  time = 0.46, size = 45, normalized size = 0.85 \[ \frac {1}{7} \, b^{4} x^{7} + \frac {2}{3} \, a b^{3} x^{6} + \frac {6}{5} \, a^{2} b^{2} x^{5} + a^{3} b x^{4} + \frac {1}{3} \, a^{4} x^{3} \]

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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giac [A]  time = 0.19, size = 45, normalized size = 0.85 \[ \frac {1}{7} \, b^{4} x^{7} + \frac {2}{3} \, a b^{3} x^{6} + \frac {6}{5} \, a^{2} b^{2} x^{5} + a^{3} b x^{4} + \frac {1}{3} \, a^{4} x^{3} \]

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

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maple [A]  time = 0.15, size = 74, normalized size = 1.40 \[ \frac {x^{3} \arctanh \left (\tanh \left (b x +a \right )\right )^{4}}{3}-\frac {4 b \left (\frac {x^{4} \arctanh \left (\tanh \left (b x +a \right )\right )^{3}}{4}-\frac {3 b \left (\frac {x^{5} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}{5}-\frac {2 b \left (\frac {x^{6} \arctanh \left (\tanh \left (b x +a \right )\right )}{6}-\frac {x^{7} b}{42}\right )}{5}\right )}{4}\right )}{3} \]

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 = 0.58, size = 72, normalized size = 1.36 \[ -\frac {1}{3} \, b x^{4} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3} + \frac {1}{3} \, x^{3} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{4} + \frac {1}{105} \, {\left (21 \, b x^{5} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2} + {\left (b^{2} x^{7} - 7 \, b x^{6} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )\right )} b\right )} b \]

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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mupad [B]  time = 0.15, size = 70, normalized size = 1.32 \[ \frac {b^4\,x^7}{105}-\frac {b^3\,x^6\,\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}{15}+\frac {b^2\,x^5\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2}{5}-\frac {b\,x^4\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^3}{3}+\frac {x^3\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^4}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*atanh(tanh(a + b*x))^4,x)

[Out]

(x^3*atanh(tanh(a + b*x))^4)/3 + (b^4*x^7)/105 + (b^2*x^5*atanh(tanh(a + b*x))^2)/5 - (b*x^4*atanh(tanh(a + b*
x))^3)/3 - (b^3*x^6*atanh(tanh(a + b*x)))/15

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sympy [A]  time = 4.01, size = 60, normalized size = 1.13 \[ \begin {cases} \frac {x^{2} \operatorname {atanh}^{5}{\left (\tanh {\left (a + b x \right )} \right )}}{5 b} - \frac {x \operatorname {atanh}^{6}{\left (\tanh {\left (a + b x \right )} \right )}}{15 b^{2}} + \frac {\operatorname {atanh}^{7}{\left (\tanh {\left (a + b x \right )} \right )}}{105 b^{3}} & \text {for}\: b \neq 0 \\\frac {x^{3} \operatorname {atanh}^{4}{\left (\tanh {\relax (a )} \right )}}{3} & \text {otherwise} \end {cases} \]

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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