3.136 \(\int x^m \coth ^{-1}(\tanh (a+b x))^2 \, dx\)

Optimal. Leaf size=71 \[ -\frac {2 b x^{m+2} \coth ^{-1}(\tanh (a+b x))}{m^2+3 m+2}+\frac {x^{m+1} \coth ^{-1}(\tanh (a+b x))^2}{m+1}+\frac {2 b^2 x^{m+3}}{m^3+6 m^2+11 m+6} \]

[Out]

2*b^2*x^(3+m)/(m^3+6*m^2+11*m+6)-2*b*x^(2+m)*arccoth(tanh(b*x+a))/(m^2+3*m+2)+x^(1+m)*arccoth(tanh(b*x+a))^2/(
1+m)

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Rubi [A]  time = 0.04, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2168, 30} \[ -\frac {2 b x^{m+2} \coth ^{-1}(\tanh (a+b x))}{m^2+3 m+2}+\frac {x^{m+1} \coth ^{-1}(\tanh (a+b x))^2}{m+1}+\frac {2 b^2 x^{m+3}}{m^3+6 m^2+11 m+6} \]

Antiderivative was successfully verified.

[In]

Int[x^m*ArcCoth[Tanh[a + b*x]]^2,x]

[Out]

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

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Mathematica [A]  time = 0.17, size = 62, normalized size = 0.87 \[ \frac {x^{m+1} \left (\left (m^2+5 m+6\right ) \coth ^{-1}(\tanh (a+b x))^2-2 b (m+3) x \coth ^{-1}(\tanh (a+b x))+2 b^2 x^2\right )}{(m+1) (m+2) (m+3)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^m*ArcCoth[Tanh[a + b*x]]^2,x]

[Out]

(x^(1 + m)*(2*b^2*x^2 - 2*b*(3 + m)*x*ArcCoth[Tanh[a + b*x]] + (6 + 5*m + m^2)*ArcCoth[Tanh[a + b*x]]^2))/((1
+ m)*(2 + m)*(3 + m))

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fricas [A]  time = 0.50, size = 101, normalized size = 1.42 \[ \frac {{\left (4 \, {\left (b^{2} m^{2} + 3 \, b^{2} m + 2 \, b^{2}\right )} x^{3} + 8 \, {\left (a b m^{2} + 4 \, a b m + 3 \, a b\right )} x^{2} + {\left (4 \, a^{2} m^{2} - \pi ^{2} {\left (m^{2} + 5 \, m + 6\right )} + 20 \, a^{2} m + 24 \, a^{2}\right )} x\right )} x^{m}}{4 \, {\left (m^{3} + 6 \, m^{2} + 11 \, m + 6\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*arccoth(tanh(b*x+a))^2,x, algorithm="fricas")

[Out]

1/4*(4*(b^2*m^2 + 3*b^2*m + 2*b^2)*x^3 + 8*(a*b*m^2 + 4*a*b*m + 3*a*b)*x^2 + (4*a^2*m^2 - pi^2*(m^2 + 5*m + 6)
+ 20*a^2*m + 24*a^2)*x)*x^m/(m^3 + 6*m^2 + 11*m + 6)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{m} \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*arccoth(tanh(b*x+a))^2,x, algorithm="giac")

[Out]

integrate(x^m*arccoth(tanh(b*x + a))^2, x)

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maple [C]  time = 1.00, size = 9175, normalized size = 129.23 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*arccoth(tanh(b*x+a))^2,x)

[Out]

result too large to display

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maxima [A]  time = 0.39, size = 73, normalized size = 1.03 \[ \frac {2 \, b^{2} x^{3} x^{m}}{{\left (m + 3\right )} {\left (m + 2\right )} {\left (m + 1\right )}} - \frac {2 \, b x^{2} x^{m} \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )}{{\left (m + 2\right )} {\left (m + 1\right )}} + \frac {x^{m + 1} \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{2}}{m + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*arccoth(tanh(b*x+a))^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.32, size = 203, normalized size = 2.86 \[ \frac {4\,b^2\,x^m\,x^3\,\left (m^2+3\,m+2\right )}{4\,m^3+24\,m^2+44\,m+24}+\frac {x\,x^m\,{\left (\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^2\,\left (m^2+5\,m+6\right )}{4\,m^3+24\,m^2+44\,m+24}-\frac {4\,b\,x^m\,x^2\,\left (\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )\,\left (m^2+4\,m+3\right )}{4\,m^3+24\,m^2+44\,m+24} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*acoth(tanh(a + b*x))^2,x)

[Out]

(4*b^2*x^m*x^3*(3*m + m^2 + 2))/(44*m + 24*m^2 + 4*m^3 + 24) + (x*x^m*(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*(5*m + m^2 + 6))/(44*m + 24*m^2 + 4*m^3 + 24) -
(4*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)*(4*m + m^2 + 3))/(44*m + 24*m^2 + 4*m^3 + 24)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \begin {cases} b^{2} \log {\relax (x )} - \frac {b \operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )}}{x} - \frac {\operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{2 x^{2}} & \text {for}\: m = -3 \\\int \frac {\operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{x^{2}}\, dx & \text {for}\: m = -2 \\\int \frac {\operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{x}\, dx & \text {for}\: m = -1 \\\frac {2 b^{2} x^{3} x^{m}}{m^{3} + 6 m^{2} + 11 m + 6} - \frac {2 b m x^{2} x^{m} \operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )}}{m^{3} + 6 m^{2} + 11 m + 6} - \frac {6 b x^{2} x^{m} \operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )}}{m^{3} + 6 m^{2} + 11 m + 6} + \frac {m^{2} x x^{m} \operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{m^{3} + 6 m^{2} + 11 m + 6} + \frac {5 m x x^{m} \operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{m^{3} + 6 m^{2} + 11 m + 6} + \frac {6 x x^{m} \operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{m^{3} + 6 m^{2} + 11 m + 6} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m*acoth(tanh(b*x+a))**2,x)

[Out]

Piecewise((b**2*log(x) - b*acoth(tanh(a + b*x))/x - acoth(tanh(a + b*x))**2/(2*x**2), Eq(m, -3)), (Integral(ac
oth(tanh(a + b*x))**2/x**2, x), Eq(m, -2)), (Integral(acoth(tanh(a + b*x))**2/x, x), Eq(m, -1)), (2*b**2*x**3*
x**m/(m**3 + 6*m**2 + 11*m + 6) - 2*b*m*x**2*x**m*acoth(tanh(a + b*x))/(m**3 + 6*m**2 + 11*m + 6) - 6*b*x**2*x
**m*acoth(tanh(a + b*x))/(m**3 + 6*m**2 + 11*m + 6) + m**2*x*x**m*acoth(tanh(a + b*x))**2/(m**3 + 6*m**2 + 11*
m + 6) + 5*m*x*x**m*acoth(tanh(a + b*x))**2/(m**3 + 6*m**2 + 11*m + 6) + 6*x*x**m*acoth(tanh(a + b*x))**2/(m**
3 + 6*m**2 + 11*m + 6), True))

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