3.1.66 \(\int x^6 \tanh ^{-1}(\tanh (a+b x))^4 \, dx\) [66]

Optimal. Leaf size=80 \[ \frac {b^4 x^{11}}{2310}-\frac {1}{210} b^3 x^{10} \tanh ^{-1}(\tanh (a+b x))+\frac {1}{42} b^2 x^9 \tanh ^{-1}(\tanh (a+b x))^2-\frac {1}{14} b x^8 \tanh ^{-1}(\tanh (a+b x))^3+\frac {1}{7} x^7 \tanh ^{-1}(\tanh (a+b x))^4 \]

[Out]

1/2310*b^4*x^11-1/210*b^3*x^10*arctanh(tanh(b*x+a))+1/42*b^2*x^9*arctanh(tanh(b*x+a))^2-1/14*b*x^8*arctanh(tan
h(b*x+a))^3+1/7*x^7*arctanh(tanh(b*x+a))^4

________________________________________________________________________________________

Rubi [A]
time = 0.04, 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 = {2199, 30} \begin {gather*} -\frac {1}{210} b^3 x^{10} \tanh ^{-1}(\tanh (a+b x))+\frac {1}{42} b^2 x^9 \tanh ^{-1}(\tanh (a+b x))^2-\frac {1}{14} b x^8 \tanh ^{-1}(\tanh (a+b x))^3+\frac {1}{7} x^7 \tanh ^{-1}(\tanh (a+b x))^4+\frac {b^4 x^{11}}{2310} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^6*ArcTanh[Tanh[a + b*x]]^4,x]

[Out]

(b^4*x^11)/2310 - (b^3*x^10*ArcTanh[Tanh[a + b*x]])/210 + (b^2*x^9*ArcTanh[Tanh[a + b*x]]^2)/42 - (b*x^8*ArcTa
nh[Tanh[a + b*x]]^3)/14 + (x^7*ArcTanh[Tanh[a + b*x]]^4)/7

Rule 30

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

Rule 2199

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^6 \tanh ^{-1}(\tanh (a+b x))^4 \, dx &=\frac {1}{7} x^7 \tanh ^{-1}(\tanh (a+b x))^4-\frac {1}{7} (4 b) \int x^7 \tanh ^{-1}(\tanh (a+b x))^3 \, dx\\ &=-\frac {1}{14} b x^8 \tanh ^{-1}(\tanh (a+b x))^3+\frac {1}{7} x^7 \tanh ^{-1}(\tanh (a+b x))^4+\frac {1}{14} \left (3 b^2\right ) \int x^8 \tanh ^{-1}(\tanh (a+b x))^2 \, dx\\ &=\frac {1}{42} b^2 x^9 \tanh ^{-1}(\tanh (a+b x))^2-\frac {1}{14} b x^8 \tanh ^{-1}(\tanh (a+b x))^3+\frac {1}{7} x^7 \tanh ^{-1}(\tanh (a+b x))^4-\frac {1}{21} b^3 \int x^9 \tanh ^{-1}(\tanh (a+b x)) \, dx\\ &=-\frac {1}{210} b^3 x^{10} \tanh ^{-1}(\tanh (a+b x))+\frac {1}{42} b^2 x^9 \tanh ^{-1}(\tanh (a+b x))^2-\frac {1}{14} b x^8 \tanh ^{-1}(\tanh (a+b x))^3+\frac {1}{7} x^7 \tanh ^{-1}(\tanh (a+b x))^4+\frac {1}{210} b^4 \int x^{10} \, dx\\ &=\frac {b^4 x^{11}}{2310}-\frac {1}{210} b^3 x^{10} \tanh ^{-1}(\tanh (a+b x))+\frac {1}{42} b^2 x^9 \tanh ^{-1}(\tanh (a+b x))^2-\frac {1}{14} b x^8 \tanh ^{-1}(\tanh (a+b x))^3+\frac {1}{7} x^7 \tanh ^{-1}(\tanh (a+b x))^4\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.04, size = 71, normalized size = 0.89 \begin {gather*} \frac {x^7 \left (b^4 x^4-11 b^3 x^3 \tanh ^{-1}(\tanh (a+b x))+55 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))^2-165 b x \tanh ^{-1}(\tanh (a+b x))^3+330 \tanh ^{-1}(\tanh (a+b x))^4\right )}{2310} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^6*ArcTanh[Tanh[a + b*x]]^4,x]

[Out]

(x^7*(b^4*x^4 - 11*b^3*x^3*ArcTanh[Tanh[a + b*x]] + 55*b^2*x^2*ArcTanh[Tanh[a + b*x]]^2 - 165*b*x*ArcTanh[Tanh
[a + b*x]]^3 + 330*ArcTanh[Tanh[a + b*x]]^4))/2310

________________________________________________________________________________________

Maple [A]
time = 0.03, size = 74, normalized size = 0.92 \[\frac {x^{7} \arctanh \left (\tanh \left (b x +a \right )\right )^{4}}{7}-\frac {4 b \left (\frac {x^{8} \arctanh \left (\tanh \left (b x +a \right )\right )^{3}}{8}-\frac {3 b \left (\frac {x^{9} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}{9}-\frac {2 b \left (\frac {x^{10} \arctanh \left (\tanh \left (b x +a \right )\right )}{10}-\frac {x^{11} b}{110}\right )}{9}\right )}{8}\right )}{7}\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6*arctanh(tanh(b*x+a))^4,x)

[Out]

1/7*x^7*arctanh(tanh(b*x+a))^4-4/7*b*(1/8*x^8*arctanh(tanh(b*x+a))^3-3/8*b*(1/9*x^9*arctanh(tanh(b*x+a))^2-2/9
*b*(1/10*x^10*arctanh(tanh(b*x+a))-1/110*x^11*b)))

________________________________________________________________________________________

Maxima [A]
time = 0.40, size = 72, normalized size = 0.90 \begin {gather*} -\frac {1}{14} \, b x^{8} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3} + \frac {1}{7} \, x^{7} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{4} + \frac {1}{2310} \, {\left (55 \, b x^{9} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2} + {\left (b^{2} x^{11} - 11 \, b x^{10} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )\right )} b\right )} b \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*arctanh(tanh(b*x+a))^4,x, algorithm="maxima")

[Out]

-1/14*b*x^8*arctanh(tanh(b*x + a))^3 + 1/7*x^7*arctanh(tanh(b*x + a))^4 + 1/2310*(55*b*x^9*arctanh(tanh(b*x +
a))^2 + (b^2*x^11 - 11*b*x^10*arctanh(tanh(b*x + a)))*b)*b

________________________________________________________________________________________

Fricas [A]
time = 0.34, size = 46, normalized size = 0.58 \begin {gather*} \frac {1}{11} \, b^{4} x^{11} + \frac {2}{5} \, a b^{3} x^{10} + \frac {2}{3} \, a^{2} b^{2} x^{9} + \frac {1}{2} \, a^{3} b x^{8} + \frac {1}{7} \, a^{4} x^{7} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*arctanh(tanh(b*x+a))^4,x, algorithm="fricas")

[Out]

1/11*b^4*x^11 + 2/5*a*b^3*x^10 + 2/3*a^2*b^2*x^9 + 1/2*a^3*b*x^8 + 1/7*a^4*x^7

________________________________________________________________________________________

Sympy [A]
time = 2.49, size = 75, normalized size = 0.94 \begin {gather*} \frac {b^{4} x^{11}}{2310} - \frac {b^{3} x^{10} \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}{210} + \frac {b^{2} x^{9} \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{42} - \frac {b x^{8} \operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{14} + \frac {x^{7} \operatorname {atanh}^{4}{\left (\tanh {\left (a + b x \right )} \right )}}{7} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6*atanh(tanh(b*x+a))**4,x)

[Out]

b**4*x**11/2310 - b**3*x**10*atanh(tanh(a + b*x))/210 + b**2*x**9*atanh(tanh(a + b*x))**2/42 - b*x**8*atanh(ta
nh(a + b*x))**3/14 + x**7*atanh(tanh(a + b*x))**4/7

________________________________________________________________________________________

Giac [A]
time = 0.38, size = 46, normalized size = 0.58 \begin {gather*} \frac {1}{11} \, b^{4} x^{11} + \frac {2}{5} \, a b^{3} x^{10} + \frac {2}{3} \, a^{2} b^{2} x^{9} + \frac {1}{2} \, a^{3} b x^{8} + \frac {1}{7} \, a^{4} x^{7} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*arctanh(tanh(b*x+a))^4,x, algorithm="giac")

[Out]

1/11*b^4*x^11 + 2/5*a*b^3*x^10 + 2/3*a^2*b^2*x^9 + 1/2*a^3*b*x^8 + 1/7*a^4*x^7

________________________________________________________________________________________

Mupad [B]
time = 1.03, size = 242, normalized size = 3.02 \begin {gather*} \frac {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 )}^4}{112}+\frac {b^4\,x^{11}}{11}-\frac {b\,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 )}^3}{16}-\frac {b^3\,x^{10}\,\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 )}{5}+\frac {b^2\,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 )}^2}{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6*atanh(tanh(a + b*x))^4,x)

[Out]

(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)^4)/11
2 + (b^4*x^11)/11 - (b*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)^3)/16 - (b^3*x^10*(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))/5 + (b^2*x^9*(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)/6

________________________________________________________________________________________