3.146 \(\int x^m \coth ^{-1}(\tanh (a+b x))^3 \, dx\)

Optimal. Leaf size=110 \[ \frac {6 b^2 x^{m+3} \coth ^{-1}(\tanh (a+b x))}{m^3+6 m^2+11 m+6}-\frac {3 b x^{m+2} \coth ^{-1}(\tanh (a+b x))^2}{m^2+3 m+2}+\frac {x^{m+1} \coth ^{-1}(\tanh (a+b x))^3}{m+1}-\frac {6 b^3 x^{m+4}}{(m+1) \left (m^3+9 m^2+26 m+24\right )} \]

[Out]

-6*b^3*x^(4+m)/(1+m)/(m^3+9*m^2+26*m+24)+6*b^2*x^(3+m)*arccoth(tanh(b*x+a))/(m^3+6*m^2+11*m+6)-3*b*x^(2+m)*arc
coth(tanh(b*x+a))^2/(m^2+3*m+2)+x^(1+m)*arccoth(tanh(b*x+a))^3/(1+m)

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Rubi [A]  time = 0.06, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2168, 30} \[ \frac {6 b^2 x^{m+3} \coth ^{-1}(\tanh (a+b x))}{m^3+6 m^2+11 m+6}-\frac {3 b x^{m+2} \coth ^{-1}(\tanh (a+b x))^2}{m^2+3 m+2}+\frac {x^{m+1} \coth ^{-1}(\tanh (a+b x))^3}{m+1}-\frac {6 b^3 x^{m+4}}{(m+1) \left (m^3+9 m^2+26 m+24\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^m*ArcCoth[Tanh[a + b*x]]^3,x]

[Out]

(-6*b^3*x^(4 + m))/((1 + m)*(24 + 26*m + 9*m^2 + m^3)) + (6*b^2*x^(3 + m)*ArcCoth[Tanh[a + b*x]])/(6 + 11*m +
6*m^2 + m^3) - (3*b*x^(2 + m)*ArcCoth[Tanh[a + b*x]]^2)/(2 + 3*m + m^2) + (x^(1 + m)*ArcCoth[Tanh[a + b*x]]^3)
/(1 + m)

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^m \coth ^{-1}(\tanh (a+b x))^3 \, dx &=\frac {x^{1+m} \coth ^{-1}(\tanh (a+b x))^3}{1+m}-\frac {(3 b) \int x^{1+m} \coth ^{-1}(\tanh (a+b x))^2 \, dx}{1+m}\\ &=-\frac {3 b x^{2+m} \coth ^{-1}(\tanh (a+b x))^2}{2+3 m+m^2}+\frac {x^{1+m} \coth ^{-1}(\tanh (a+b x))^3}{1+m}+\frac {\left (6 b^2\right ) \int x^{2+m} \coth ^{-1}(\tanh (a+b x)) \, dx}{2+3 m+m^2}\\ &=\frac {6 b^2 x^{3+m} \coth ^{-1}(\tanh (a+b x))}{6+11 m+6 m^2+m^3}-\frac {3 b x^{2+m} \coth ^{-1}(\tanh (a+b x))^2}{2+3 m+m^2}+\frac {x^{1+m} \coth ^{-1}(\tanh (a+b x))^3}{1+m}-\frac {\left (6 b^3\right ) \int x^{3+m} \, dx}{6+11 m+6 m^2+m^3}\\ &=-\frac {6 b^3 x^{4+m}}{(4+m) \left (6+11 m+6 m^2+m^3\right )}+\frac {6 b^2 x^{3+m} \coth ^{-1}(\tanh (a+b x))}{6+11 m+6 m^2+m^3}-\frac {3 b x^{2+m} \coth ^{-1}(\tanh (a+b x))^2}{2+3 m+m^2}+\frac {x^{1+m} \coth ^{-1}(\tanh (a+b x))^3}{1+m}\\ \end {align*}

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Mathematica [A]  time = 0.28, size = 97, normalized size = 0.88 \[ \frac {x^{m+1} \left (6 b^2 (m+4) x^2 \coth ^{-1}(\tanh (a+b x))-3 b \left (m^2+7 m+12\right ) x \coth ^{-1}(\tanh (a+b x))^2+\left (m^3+9 m^2+26 m+24\right ) \coth ^{-1}(\tanh (a+b x))^3-6 b^3 x^3\right )}{(m+1) (m+2) (m+3) (m+4)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^m*ArcCoth[Tanh[a + b*x]]^3,x]

[Out]

(x^(1 + m)*(-6*b^3*x^3 + 6*b^2*(4 + m)*x^2*ArcCoth[Tanh[a + b*x]] - 3*b*(12 + 7*m + m^2)*x*ArcCoth[Tanh[a + b*
x]]^2 + (24 + 26*m + 9*m^2 + m^3)*ArcCoth[Tanh[a + b*x]]^3))/((1 + m)*(2 + m)*(3 + m)*(4 + m))

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fricas [A]  time = 0.62, size = 209, normalized size = 1.90 \[ \frac {{\left (4 \, {\left (b^{3} m^{3} + 6 \, b^{3} m^{2} + 11 \, b^{3} m + 6 \, b^{3}\right )} x^{4} + 12 \, {\left (a b^{2} m^{3} + 7 \, a b^{2} m^{2} + 14 \, a b^{2} m + 8 \, a b^{2}\right )} x^{3} + 3 \, {\left (4 \, a^{2} b m^{3} + 32 \, a^{2} b m^{2} + 76 \, a^{2} b m - \pi ^{2} {\left (b m^{3} + 8 \, b m^{2} + 19 \, b m + 12 \, b\right )} + 48 \, a^{2} b\right )} x^{2} + {\left (4 \, a^{3} m^{3} + 36 \, a^{3} m^{2} + 104 \, a^{3} m - 3 \, \pi ^{2} {\left (a m^{3} + 9 \, a m^{2} + 26 \, a m + 24 \, a\right )} + 96 \, a^{3}\right )} x\right )} x^{m}}{4 \, {\left (m^{4} + 10 \, m^{3} + 35 \, m^{2} + 50 \, m + 24\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*arccoth(tanh(b*x+a))^3,x, algorithm="fricas")

[Out]

1/4*(4*(b^3*m^3 + 6*b^3*m^2 + 11*b^3*m + 6*b^3)*x^4 + 12*(a*b^2*m^3 + 7*a*b^2*m^2 + 14*a*b^2*m + 8*a*b^2)*x^3
+ 3*(4*a^2*b*m^3 + 32*a^2*b*m^2 + 76*a^2*b*m - pi^2*(b*m^3 + 8*b*m^2 + 19*b*m + 12*b) + 48*a^2*b)*x^2 + (4*a^3
*m^3 + 36*a^3*m^2 + 104*a^3*m - 3*pi^2*(a*m^3 + 9*a*m^2 + 26*a*m + 24*a) + 96*a^3)*x)*x^m/(m^4 + 10*m^3 + 35*m
^2 + 50*m + 24)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{m} \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*arccoth(tanh(b*x+a))^3,x, algorithm="giac")

[Out]

integrate(x^m*arccoth(tanh(b*x + a))^3, x)

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maple [C]  time = 9.36, size = 63382, normalized size = 576.20 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*arccoth(tanh(b*x+a))^3,x)

[Out]

result too large to display

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maxima [A]  time = 0.46, size = 109, normalized size = 0.99 \[ -\frac {3 \, b x^{2} x^{m} \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{2}}{{\left (m + 2\right )} {\left (m + 1\right )}} + \frac {x^{m + 1} \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{3}}{m + 1} - \frac {6 \, {\left (\frac {b^{2} x^{4} x^{m}}{{\left (m + 4\right )} {\left (m + 3\right )} {\left (m + 2\right )}} - \frac {b x^{3} x^{m} \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )}{{\left (m + 3\right )} {\left (m + 2\right )}}\right )} b}{m + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*arccoth(tanh(b*x+a))^3,x, algorithm="maxima")

[Out]

-3*b*x^2*x^m*arccoth(tanh(b*x + a))^2/((m + 2)*(m + 1)) + x^(m + 1)*arccoth(tanh(b*x + a))^3/(m + 1) - 6*(b^2*
x^4*x^m/((m + 4)*(m + 3)*(m + 2)) - b*x^3*x^m*arccoth(tanh(b*x + a))/((m + 3)*(m + 2)))*b/(m + 1)

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mupad [B]  time = 1.45, size = 332, normalized size = 3.02 \[ \frac {8\,b^3\,x^m\,x^4\,\left (m^3+6\,m^2+11\,m+6\right )}{8\,m^4+80\,m^3+280\,m^2+400\,m+192}-\frac {x\,x^m\,{\left (\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^3\,\left (m^3+9\,m^2+26\,m+24\right )}{8\,m^4+80\,m^3+280\,m^2+400\,m+192}-\frac {12\,b^2\,x^m\,x^3\,\left (\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )\,\left (m^3+7\,m^2+14\,m+8\right )}{8\,m^4+80\,m^3+280\,m^2+400\,m+192}+\frac {6\,b\,x^m\,x^2\,{\left (\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^2\,\left (m^3+8\,m^2+19\,m+12\right )}{8\,m^4+80\,m^3+280\,m^2+400\,m+192} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*acoth(tanh(a + b*x))^3,x)

[Out]

(8*b^3*x^m*x^4*(11*m + 6*m^2 + m^3 + 6))/(400*m + 280*m^2 + 80*m^3 + 8*m^4 + 192) - (x*x^m*(log(-2/(exp(2*a)*e
xp(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*(26*m + 9*m^2 + m^3 + 24))
/(400*m + 280*m^2 + 80*m^3 + 8*m^4 + 192) - (12*b^2*x^m*x^3*(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)*(14*m + 7*m^2 + m^3 + 8))/(400*m + 280*m^2 + 80*m^3 + 8*m^4
+ 192) + (6*b*x^m*x^2*(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*(19*m + 8*m^2 + m^3 + 12))/(400*m + 280*m^2 + 80*m^3 + 8*m^4 + 192)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \begin {cases} b^{3} \log {\relax (x )} - \frac {b^{2} \operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )}}{x} - \frac {b \operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{2 x^{2}} - \frac {\operatorname {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{3 x^{3}} & \text {for}\: m = -4 \\\int \frac {\operatorname {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{x^{3}}\, dx & \text {for}\: m = -3 \\\int \frac {\operatorname {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{x^{2}}\, dx & \text {for}\: m = -2 \\\int \frac {\operatorname {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{x}\, dx & \text {for}\: m = -1 \\- \frac {6 b^{3} x^{4} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {6 b^{2} m x^{3} x^{m} \operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {24 b^{2} x^{3} x^{m} \operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} - \frac {3 b m^{2} x^{2} x^{m} \operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} - \frac {21 b m x^{2} x^{m} \operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} - \frac {36 b x^{2} x^{m} \operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {m^{3} x x^{m} \operatorname {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {9 m^{2} x x^{m} \operatorname {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {26 m x x^{m} \operatorname {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {24 x x^{m} \operatorname {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m*acoth(tanh(b*x+a))**3,x)

[Out]

Piecewise((b**3*log(x) - b**2*acoth(tanh(a + b*x))/x - b*acoth(tanh(a + b*x))**2/(2*x**2) - acoth(tanh(a + b*x
))**3/(3*x**3), Eq(m, -4)), (Integral(acoth(tanh(a + b*x))**3/x**3, x), Eq(m, -3)), (Integral(acoth(tanh(a + b
*x))**3/x**2, x), Eq(m, -2)), (Integral(acoth(tanh(a + b*x))**3/x, x), Eq(m, -1)), (-6*b**3*x**4*x**m/(m**4 +
10*m**3 + 35*m**2 + 50*m + 24) + 6*b**2*m*x**3*x**m*acoth(tanh(a + b*x))/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24
) + 24*b**2*x**3*x**m*acoth(tanh(a + b*x))/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) - 3*b*m**2*x**2*x**m*acoth(t
anh(a + b*x))**2/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) - 21*b*m*x**2*x**m*acoth(tanh(a + b*x))**2/(m**4 + 10*
m**3 + 35*m**2 + 50*m + 24) - 36*b*x**2*x**m*acoth(tanh(a + b*x))**2/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) +
m**3*x*x**m*acoth(tanh(a + b*x))**3/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 9*m**2*x*x**m*acoth(tanh(a + b*x)
)**3/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 26*m*x*x**m*acoth(tanh(a + b*x))**3/(m**4 + 10*m**3 + 35*m**2 +
50*m + 24) + 24*x*x**m*acoth(tanh(a + b*x))**3/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24), True))

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