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

Optimal. Leaf size=80 \[ -\frac {1}{70} b^3 x^8 \tanh ^{-1}(\tanh (a+b x))+\frac {2}{35} b^2 x^7 \tanh ^{-1}(\tanh (a+b x))^2-\frac {2}{15} b x^6 \tanh ^{-1}(\tanh (a+b x))^3+\frac {1}{5} x^5 \tanh ^{-1}(\tanh (a+b x))^4+\frac {b^4 x^9}{630} \]

[Out]

1/630*b^4*x^9-1/70*b^3*x^8*arctanh(tanh(b*x+a))+2/35*b^2*x^7*arctanh(tanh(b*x+a))^2-2/15*b*x^6*arctanh(tanh(b*
x+a))^3+1/5*x^5*arctanh(tanh(b*x+a))^4

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Rubi [A]  time = 0.05, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2168, 30} \[ -\frac {1}{70} b^3 x^8 \tanh ^{-1}(\tanh (a+b x))+\frac {2}{35} b^2 x^7 \tanh ^{-1}(\tanh (a+b x))^2-\frac {2}{15} b x^6 \tanh ^{-1}(\tanh (a+b x))^3+\frac {1}{5} x^5 \tanh ^{-1}(\tanh (a+b x))^4+\frac {b^4 x^9}{630} \]

Antiderivative was successfully verified.

[In]

Int[x^4*ArcTanh[Tanh[a + b*x]]^4,x]

[Out]

(b^4*x^9)/630 - (b^3*x^8*ArcTanh[Tanh[a + b*x]])/70 + (2*b^2*x^7*ArcTanh[Tanh[a + b*x]]^2)/35 - (2*b*x^6*ArcTa
nh[Tanh[a + b*x]]^3)/15 + (x^5*ArcTanh[Tanh[a + b*x]]^4)/5

Rule 30

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

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^4 \tanh ^{-1}(\tanh (a+b x))^4 \, dx &=\frac {1}{5} x^5 \tanh ^{-1}(\tanh (a+b x))^4-\frac {1}{5} (4 b) \int x^5 \tanh ^{-1}(\tanh (a+b x))^3 \, dx\\ &=-\frac {2}{15} b x^6 \tanh ^{-1}(\tanh (a+b x))^3+\frac {1}{5} x^5 \tanh ^{-1}(\tanh (a+b x))^4+\frac {1}{5} \left (2 b^2\right ) \int x^6 \tanh ^{-1}(\tanh (a+b x))^2 \, dx\\ &=\frac {2}{35} b^2 x^7 \tanh ^{-1}(\tanh (a+b x))^2-\frac {2}{15} b x^6 \tanh ^{-1}(\tanh (a+b x))^3+\frac {1}{5} x^5 \tanh ^{-1}(\tanh (a+b x))^4-\frac {1}{35} \left (4 b^3\right ) \int x^7 \tanh ^{-1}(\tanh (a+b x)) \, dx\\ &=-\frac {1}{70} b^3 x^8 \tanh ^{-1}(\tanh (a+b x))+\frac {2}{35} b^2 x^7 \tanh ^{-1}(\tanh (a+b x))^2-\frac {2}{15} b x^6 \tanh ^{-1}(\tanh (a+b x))^3+\frac {1}{5} x^5 \tanh ^{-1}(\tanh (a+b x))^4+\frac {1}{70} b^4 \int x^8 \, dx\\ &=\frac {b^4 x^9}{630}-\frac {1}{70} b^3 x^8 \tanh ^{-1}(\tanh (a+b x))+\frac {2}{35} b^2 x^7 \tanh ^{-1}(\tanh (a+b x))^2-\frac {2}{15} b x^6 \tanh ^{-1}(\tanh (a+b x))^3+\frac {1}{5} x^5 \tanh ^{-1}(\tanh (a+b x))^4\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 71, normalized size = 0.89 \[ \frac {1}{630} x^5 \left (-9 b^3 x^3 \tanh ^{-1}(\tanh (a+b x))+36 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))^2-84 b x \tanh ^{-1}(\tanh (a+b x))^3+126 \tanh ^{-1}(\tanh (a+b x))^4+b^4 x^4\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^5*(b^4*x^4 - 9*b^3*x^3*ArcTanh[Tanh[a + b*x]] + 36*b^2*x^2*ArcTanh[Tanh[a + b*x]]^2 - 84*b*x*ArcTanh[Tanh[a
+ b*x]]^3 + 126*ArcTanh[Tanh[a + b*x]]^4))/630

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fricas [A]  time = 0.52, size = 46, normalized size = 0.58 \[ \frac {1}{9} \, b^{4} x^{9} + \frac {1}{2} \, a b^{3} x^{8} + \frac {6}{7} \, a^{2} b^{2} x^{7} + \frac {2}{3} \, a^{3} b x^{6} + \frac {1}{5} \, a^{4} x^{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*b^4*x^9 + 1/2*a*b^3*x^8 + 6/7*a^2*b^2*x^7 + 2/3*a^3*b*x^6 + 1/5*a^4*x^5

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giac [A]  time = 0.22, size = 46, normalized size = 0.58 \[ \frac {1}{9} \, b^{4} x^{9} + \frac {1}{2} \, a b^{3} x^{8} + \frac {6}{7} \, a^{2} b^{2} x^{7} + \frac {2}{3} \, a^{3} b x^{6} + \frac {1}{5} \, a^{4} x^{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*b^4*x^9 + 1/2*a*b^3*x^8 + 6/7*a^2*b^2*x^7 + 2/3*a^3*b*x^6 + 1/5*a^4*x^5

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maple [A]  time = 0.15, size = 74, normalized size = 0.92 \[ \frac {x^{5} \arctanh \left (\tanh \left (b x +a \right )\right )^{4}}{5}-\frac {4 b \left (\frac {x^{6} \arctanh \left (\tanh \left (b x +a \right )\right )^{3}}{6}-\frac {b \left (\frac {x^{7} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}{7}-\frac {2 b \left (\frac {x^{8} \arctanh \left (\tanh \left (b x +a \right )\right )}{8}-\frac {x^{9} b}{72}\right )}{7}\right )}{2}\right )}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*x^5*arctanh(tanh(b*x+a))^4-4/5*b*(1/6*x^6*arctanh(tanh(b*x+a))^3-1/2*b*(1/7*x^7*arctanh(tanh(b*x+a))^2-2/7
*b*(1/8*x^8*arctanh(tanh(b*x+a))-1/72*x^9*b)))

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maxima [A]  time = 0.57, size = 72, normalized size = 0.90 \[ -\frac {2}{15} \, b x^{6} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3} + \frac {1}{5} \, x^{5} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{4} + \frac {1}{630} \, {\left (36 \, b x^{7} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2} + {\left (b^{2} x^{9} - 9 \, b x^{8} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )\right )} b\right )} b \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*b*x^6*arctanh(tanh(b*x + a))^3 + 1/5*x^5*arctanh(tanh(b*x + a))^4 + 1/630*(36*b*x^7*arctanh(tanh(b*x + a
))^2 + (b^2*x^9 - 9*b*x^8*arctanh(tanh(b*x + a)))*b)*b

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mupad [B]  time = 1.06, size = 77, normalized size = 0.96 \[ \frac {{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^5\,\left (126\,b^4\,x^4-84\,b^3\,x^3\,\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )+36\,b^2\,x^2\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2-9\,b\,x\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^3+{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^4\right )}{630\,b^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*atanh(tanh(a + b*x))^4,x)

[Out]

(atanh(tanh(a + b*x))^5*(atanh(tanh(a + b*x))^4 + 126*b^4*x^4 + 36*b^2*x^2*atanh(tanh(a + b*x))^2 - 9*b*x*atan
h(tanh(a + b*x))^3 - 84*b^3*x^3*atanh(tanh(a + b*x))))/(630*b^5)

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sympy [A]  time = 11.72, size = 100, normalized size = 1.25 \[ \begin {cases} \frac {x^{4} \operatorname {atanh}^{5}{\left (\tanh {\left (a + b x \right )} \right )}}{5 b} - \frac {2 x^{3} \operatorname {atanh}^{6}{\left (\tanh {\left (a + b x \right )} \right )}}{15 b^{2}} + \frac {2 x^{2} \operatorname {atanh}^{7}{\left (\tanh {\left (a + b x \right )} \right )}}{35 b^{3}} - \frac {x \operatorname {atanh}^{8}{\left (\tanh {\left (a + b x \right )} \right )}}{70 b^{4}} + \frac {\operatorname {atanh}^{9}{\left (\tanh {\left (a + b x \right )} \right )}}{630 b^{5}} & \text {for}\: b \neq 0 \\\frac {x^{5} \operatorname {atanh}^{4}{\left (\tanh {\relax (a )} \right )}}{5} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x**4*atanh(tanh(a + b*x))**5/(5*b) - 2*x**3*atanh(tanh(a + b*x))**6/(15*b**2) + 2*x**2*atanh(tanh(a
+ b*x))**7/(35*b**3) - x*atanh(tanh(a + b*x))**8/(70*b**4) + atanh(tanh(a + b*x))**9/(630*b**5), Ne(b, 0)), (
x**5*atanh(tanh(a))**4/5, True))

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