3.67 \(\int x^5 \tanh ^{-1}(\tanh (a+b x))^4 \, dx\)

Optimal. Leaf size=80 \[ -\frac {1}{126} b^3 x^9 \tanh ^{-1}(\tanh (a+b x))+\frac {1}{28} b^2 x^8 \tanh ^{-1}(\tanh (a+b x))^2-\frac {2}{21} b x^7 \tanh ^{-1}(\tanh (a+b x))^3+\frac {1}{6} x^6 \tanh ^{-1}(\tanh (a+b x))^4+\frac {b^4 x^{10}}{1260} \]

[Out]

1/1260*b^4*x^10-1/126*b^3*x^9*arctanh(tanh(b*x+a))+1/28*b^2*x^8*arctanh(tanh(b*x+a))^2-2/21*b*x^7*arctanh(tanh
(b*x+a))^3+1/6*x^6*arctanh(tanh(b*x+a))^4

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Rubi [A]  time = 0.06, 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} \[ -\frac {1}{126} b^3 x^9 \tanh ^{-1}(\tanh (a+b x))+\frac {1}{28} b^2 x^8 \tanh ^{-1}(\tanh (a+b x))^2-\frac {2}{21} b x^7 \tanh ^{-1}(\tanh (a+b x))^3+\frac {1}{6} x^6 \tanh ^{-1}(\tanh (a+b x))^4+\frac {b^4 x^{10}}{1260} \]

Antiderivative was successfully verified.

[In]

Int[x^5*ArcTanh[Tanh[a + b*x]]^4,x]

[Out]

(b^4*x^10)/1260 - (b^3*x^9*ArcTanh[Tanh[a + b*x]])/126 + (b^2*x^8*ArcTanh[Tanh[a + b*x]]^2)/28 - (2*b*x^7*ArcT
anh[Tanh[a + b*x]]^3)/21 + (x^6*ArcTanh[Tanh[a + b*x]]^4)/6

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^5 \tanh ^{-1}(\tanh (a+b x))^4 \, dx &=\frac {1}{6} x^6 \tanh ^{-1}(\tanh (a+b x))^4-\frac {1}{3} (2 b) \int x^6 \tanh ^{-1}(\tanh (a+b x))^3 \, dx\\ &=-\frac {2}{21} b x^7 \tanh ^{-1}(\tanh (a+b x))^3+\frac {1}{6} x^6 \tanh ^{-1}(\tanh (a+b x))^4+\frac {1}{7} \left (2 b^2\right ) \int x^7 \tanh ^{-1}(\tanh (a+b x))^2 \, dx\\ &=\frac {1}{28} b^2 x^8 \tanh ^{-1}(\tanh (a+b x))^2-\frac {2}{21} b x^7 \tanh ^{-1}(\tanh (a+b x))^3+\frac {1}{6} x^6 \tanh ^{-1}(\tanh (a+b x))^4-\frac {1}{14} b^3 \int x^8 \tanh ^{-1}(\tanh (a+b x)) \, dx\\ &=-\frac {1}{126} b^3 x^9 \tanh ^{-1}(\tanh (a+b x))+\frac {1}{28} b^2 x^8 \tanh ^{-1}(\tanh (a+b x))^2-\frac {2}{21} b x^7 \tanh ^{-1}(\tanh (a+b x))^3+\frac {1}{6} x^6 \tanh ^{-1}(\tanh (a+b x))^4+\frac {1}{126} b^4 \int x^9 \, dx\\ &=\frac {b^4 x^{10}}{1260}-\frac {1}{126} b^3 x^9 \tanh ^{-1}(\tanh (a+b x))+\frac {1}{28} b^2 x^8 \tanh ^{-1}(\tanh (a+b x))^2-\frac {2}{21} b x^7 \tanh ^{-1}(\tanh (a+b x))^3+\frac {1}{6} x^6 \tanh ^{-1}(\tanh (a+b x))^4\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 71, normalized size = 0.89 \[ \frac {x^6 \left (-10 b^3 x^3 \tanh ^{-1}(\tanh (a+b x))+45 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))^2-120 b x \tanh ^{-1}(\tanh (a+b x))^3+210 \tanh ^{-1}(\tanh (a+b x))^4+b^4 x^4\right )}{1260} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5*ArcTanh[Tanh[a + b*x]]^4,x]

[Out]

(x^6*(b^4*x^4 - 10*b^3*x^3*ArcTanh[Tanh[a + b*x]] + 45*b^2*x^2*ArcTanh[Tanh[a + b*x]]^2 - 120*b*x*ArcTanh[Tanh
[a + b*x]]^3 + 210*ArcTanh[Tanh[a + b*x]]^4))/1260

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fricas [A]  time = 0.52, size = 46, normalized size = 0.58 \[ \frac {1}{10} \, b^{4} x^{10} + \frac {4}{9} \, a b^{3} x^{9} + \frac {3}{4} \, a^{2} b^{2} x^{8} + \frac {4}{7} \, a^{3} b x^{7} + \frac {1}{6} \, a^{4} x^{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*arctanh(tanh(b*x+a))^4,x, algorithm="fricas")

[Out]

1/10*b^4*x^10 + 4/9*a*b^3*x^9 + 3/4*a^2*b^2*x^8 + 4/7*a^3*b*x^7 + 1/6*a^4*x^6

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giac [A]  time = 0.20, size = 46, normalized size = 0.58 \[ \frac {1}{10} \, b^{4} x^{10} + \frac {4}{9} \, a b^{3} x^{9} + \frac {3}{4} \, a^{2} b^{2} x^{8} + \frac {4}{7} \, a^{3} b x^{7} + \frac {1}{6} \, a^{4} x^{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*arctanh(tanh(b*x+a))^4,x, algorithm="giac")

[Out]

1/10*b^4*x^10 + 4/9*a*b^3*x^9 + 3/4*a^2*b^2*x^8 + 4/7*a^3*b*x^7 + 1/6*a^4*x^6

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maple [A]  time = 0.15, size = 74, normalized size = 0.92 \[ \frac {x^{6} \arctanh \left (\tanh \left (b x +a \right )\right )^{4}}{6}-\frac {2 b \left (\frac {x^{7} \arctanh \left (\tanh \left (b x +a \right )\right )^{3}}{7}-\frac {3 b \left (\frac {x^{8} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}{8}-\frac {b \left (\frac {x^{9} \arctanh \left (\tanh \left (b x +a \right )\right )}{9}-\frac {x^{10} b}{90}\right )}{4}\right )}{7}\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*arctanh(tanh(b*x+a))^4,x)

[Out]

1/6*x^6*arctanh(tanh(b*x+a))^4-2/3*b*(1/7*x^7*arctanh(tanh(b*x+a))^3-3/7*b*(1/8*x^8*arctanh(tanh(b*x+a))^2-1/4
*b*(1/9*x^9*arctanh(tanh(b*x+a))-1/90*x^10*b)))

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maxima [A]  time = 0.58, size = 72, normalized size = 0.90 \[ -\frac {2}{21} \, b x^{7} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3} + \frac {1}{6} \, x^{6} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{4} + \frac {1}{1260} \, {\left (45 \, b x^{8} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2} + {\left (b^{2} x^{10} - 10 \, b x^{9} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )\right )} b\right )} b \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*arctanh(tanh(b*x+a))^4,x, algorithm="maxima")

[Out]

-2/21*b*x^7*arctanh(tanh(b*x + a))^3 + 1/6*x^6*arctanh(tanh(b*x + a))^4 + 1/1260*(45*b*x^8*arctanh(tanh(b*x +
a))^2 + (b^2*x^10 - 10*b*x^9*arctanh(tanh(b*x + a)))*b)*b

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mupad [B]  time = 0.15, size = 242, normalized size = 3.02 \[ \frac {x^6\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^4}{96}+\frac {b^4\,x^{10}}{10}-\frac {b\,x^7\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}{14}-\frac {2\,b^3\,x^9\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{9}+\frac {3\,b^2\,x^8\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{16} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*atanh(tanh(a + b*x))^4,x)

[Out]

(x^6*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^4)/96
+ (b^4*x^10)/10 - (b*x^7*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x)
+ 1)) + 2*b*x)^3)/14 - (2*b^3*x^9*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*e
xp(2*b*x) + 1)) + 2*b*x))/9 + (3*b^2*x^8*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(
2*a)*exp(2*b*x) + 1)) + 2*b*x)^2)/16

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sympy [A]  time = 18.91, size = 117, normalized size = 1.46 \[ \begin {cases} \frac {x^{5} \operatorname {atanh}^{5}{\left (\tanh {\left (a + b x \right )} \right )}}{5 b} - \frac {x^{4} \operatorname {atanh}^{6}{\left (\tanh {\left (a + b x \right )} \right )}}{6 b^{2}} + \frac {2 x^{3} \operatorname {atanh}^{7}{\left (\tanh {\left (a + b x \right )} \right )}}{21 b^{3}} - \frac {x^{2} \operatorname {atanh}^{8}{\left (\tanh {\left (a + b x \right )} \right )}}{28 b^{4}} + \frac {x \operatorname {atanh}^{9}{\left (\tanh {\left (a + b x \right )} \right )}}{126 b^{5}} - \frac {\operatorname {atanh}^{10}{\left (\tanh {\left (a + b x \right )} \right )}}{1260 b^{6}} & \text {for}\: b \neq 0 \\\frac {x^{6} \operatorname {atanh}^{4}{\left (\tanh {\relax (a )} \right )}}{6} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*atanh(tanh(b*x+a))**4,x)

[Out]

Piecewise((x**5*atanh(tanh(a + b*x))**5/(5*b) - x**4*atanh(tanh(a + b*x))**6/(6*b**2) + 2*x**3*atanh(tanh(a +
b*x))**7/(21*b**3) - x**2*atanh(tanh(a + b*x))**8/(28*b**4) + x*atanh(tanh(a + b*x))**9/(126*b**5) - atanh(tan
h(a + b*x))**10/(1260*b**6), Ne(b, 0)), (x**6*atanh(tanh(a))**4/6, True))

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