3.188 \(\int x \coth ^{-1}(\tanh (a+b x))^n \, dx\)

Optimal. Leaf size=48 \[ \frac {x \coth ^{-1}(\tanh (a+b x))^{n+1}}{b (n+1)}-\frac {\coth ^{-1}(\tanh (a+b x))^{n+2}}{b^2 (n+1) (n+2)} \]

[Out]

x*arccoth(tanh(b*x+a))^(1+n)/b/(1+n)-arccoth(tanh(b*x+a))^(2+n)/b^2/(1+n)/(2+n)

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Rubi [A]  time = 0.02, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2168, 2157, 30} \[ \frac {x \coth ^{-1}(\tanh (a+b x))^{n+1}}{b (n+1)}-\frac {\coth ^{-1}(\tanh (a+b x))^{n+2}}{b^2 (n+1) (n+2)} \]

Antiderivative was successfully verified.

[In]

Int[x*ArcCoth[Tanh[a + b*x]]^n,x]

[Out]

(x*ArcCoth[Tanh[a + b*x]]^(1 + n))/(b*(1 + n)) - ArcCoth[Tanh[a + b*x]]^(2 + n)/(b^2*(1 + n)*(2 + n))

Rule 30

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

Rule 2157

Int[(u_)^(m_.), x_Symbol] :> With[{c = Simplify[D[u, x]]}, Dist[1/c, Subst[Int[x^m, x], x, u], x]] /; FreeQ[m,
x] && PiecewiseLinearQ[u, x]

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 \coth ^{-1}(\tanh (a+b x))^n \, dx &=\frac {x \coth ^{-1}(\tanh (a+b x))^{1+n}}{b (1+n)}-\frac {\int \coth ^{-1}(\tanh (a+b x))^{1+n} \, dx}{b (1+n)}\\ &=\frac {x \coth ^{-1}(\tanh (a+b x))^{1+n}}{b (1+n)}-\frac {\operatorname {Subst}\left (\int x^{1+n} \, dx,x,\coth ^{-1}(\tanh (a+b x))\right )}{b^2 (1+n)}\\ &=\frac {x \coth ^{-1}(\tanh (a+b x))^{1+n}}{b (1+n)}-\frac {\coth ^{-1}(\tanh (a+b x))^{2+n}}{b^2 (1+n) (2+n)}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 41, normalized size = 0.85 \[ \frac {\left (b (n+2) x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))^{n+1}}{b^2 (n+1) (n+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*(2 + n)*x - ArcCoth[Tanh[a + b*x]])*ArcCoth[Tanh[a + b*x]]^(1 + n))/(b^2*(1 + n)*(2 + n))

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fricas [B]  time = 0.57, size = 210, normalized size = 4.38 \[ \frac {{\left (4 \, a b n x + 4 \, {\left (b^{2} n + b^{2}\right )} x^{2} + \pi ^{2} - 4 \, a^{2}\right )} {\left (b^{2} x^{2} + 2 \, a b x + \frac {1}{4} \, \pi ^{2} + a^{2}\right )}^{\frac {1}{2} \, n} \cos \left (2 \, n \arctan \left (-\frac {2 \, b x}{\pi } - \frac {2 \, a}{\pi } + \frac {\sqrt {4 \, b^{2} x^{2} + 8 \, a b x + \pi ^{2} + 4 \, a^{2}}}{\pi }\right )\right ) - 2 \, {\left (\pi b n x - 2 \, \pi a\right )} {\left (b^{2} x^{2} + 2 \, a b x + \frac {1}{4} \, \pi ^{2} + a^{2}\right )}^{\frac {1}{2} \, n} \sin \left (2 \, n \arctan \left (-\frac {2 \, b x}{\pi } - \frac {2 \, a}{\pi } + \frac {\sqrt {4 \, b^{2} x^{2} + 8 \, a b x + \pi ^{2} + 4 \, a^{2}}}{\pi }\right )\right )}{4 \, {\left (b^{2} n^{2} + 3 \, b^{2} n + 2 \, b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((4*a*b*n*x + 4*(b^2*n + b^2)*x^2 + pi^2 - 4*a^2)*(b^2*x^2 + 2*a*b*x + 1/4*pi^2 + a^2)^(1/2*n)*cos(2*n*arc
tan(-2*b*x/pi - 2*a/pi + sqrt(4*b^2*x^2 + 8*a*b*x + pi^2 + 4*a^2)/pi)) - 2*(pi*b*n*x - 2*pi*a)*(b^2*x^2 + 2*a*
b*x + 1/4*pi^2 + a^2)^(1/2*n)*sin(2*n*arctan(-2*b*x/pi - 2*a/pi + sqrt(4*b^2*x^2 + 8*a*b*x + pi^2 + 4*a^2)/pi)
))/(b^2*n^2 + 3*b^2*n + 2*b^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 13.67, size = 480, normalized size = 10.00 \[ \frac {\left (\frac {1}{2}\right )^{n} \left (2 \ln \left ({\mathrm e}^{b x +a}\right )-\frac {i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \left (-\mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )+\mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right )\right )^{2}}{2}-\frac {i \pi \,\mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right ) \left (-\mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )+\mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )\right ) \left (-\mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )+\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )\right )}{2}-i \pi -i \pi \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2} \left (\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )-1\right )\right )^{1+n} x}{2 b \left (1+n \right )}-\frac {\left (\frac {1}{2}\right )^{n} \left (2 \ln \left ({\mathrm e}^{b x +a}\right )-\frac {i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \left (-\mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )+\mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right )\right )^{2}}{2}-\frac {i \pi \,\mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right ) \left (-\mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )+\mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )\right ) \left (-\mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )+\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )\right )}{2}-i \pi -i \pi \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2} \left (\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )-1\right )\right )^{2+n}}{4 b^{2} \left (1+n \right ) \left (2+n \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/b*(1/2)^n*(2*ln(exp(b*x+a))-1/2*I*Pi*csgn(I*exp(2*b*x+2*a))*(-csgn(I*exp(2*b*x+2*a))+csgn(I*exp(b*x+a)))^2
-1/2*I*Pi*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))*(-csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))+csgn(I*exp(2*b
*x+2*a)))*(-csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))+csgn(I/(exp(2*b*x+2*a)+1)))-I*Pi-I*Pi*csgn(I/(exp(2*b*x+
2*a)+1))^2*(csgn(I/(exp(2*b*x+2*a)+1))-1))^(1+n)/(1+n)*x-1/4/b^2*(1/2)^n/(1+n)*(2*ln(exp(b*x+a))-1/2*I*Pi*csgn
(I*exp(2*b*x+2*a))*(-csgn(I*exp(2*b*x+2*a))+csgn(I*exp(b*x+a)))^2-1/2*I*Pi*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*
a)+1))*(-csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))+csgn(I*exp(2*b*x+2*a)))*(-csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+
2*a)+1))+csgn(I/(exp(2*b*x+2*a)+1)))-I*Pi-I*Pi*csgn(I/(exp(2*b*x+2*a)+1))^2*(csgn(I/(exp(2*b*x+2*a)+1))-1))^(2
+n)/(2+n)

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maxima [C]  time = 0.54, size = 101, normalized size = 2.10 \[ \frac {{\left (4 \, b^{2} {\left (n + 1\right )} x^{2} + \pi ^{2} + 4 i \, \pi a - 4 \, a^{2} + {\left (-2 i \, \pi b n + 4 \, a b n\right )} x\right )} {\left (\cosh \left (-n \log \left (-i \, \pi + 2 \, b x + 2 \, a\right )\right ) - \sinh \left (-n \log \left (-i \, \pi + 2 \, b x + 2 \, a\right )\right )\right )}}{{\left (2^{n + 2} n^{2} + 3 \cdot 2^{n + 2} n + 2^{n + 3}\right )} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*b^2*(n + 1)*x^2 + pi^2 + 4*I*pi*a - 4*a^2 + (-2*I*pi*b*n + 4*a*b*n)*x)*(cosh(-n*log(-I*pi + 2*b*x + 2*a)) -
sinh(-n*log(-I*pi + 2*b*x + 2*a)))/((2^(n + 2)*n^2 + 3*2^(n + 2)*n + 2^(n + 3))*b^2)

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mupad [B]  time = 1.31, size = 205, normalized size = 4.27 \[ -{\left (\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}{2}-\frac {\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}{2}\right )}^n\,\left (\frac {{\left (\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^2}{4\,b^2\,\left (n^2+3\,n+2\right )}-\frac {x^2\,\left (n+1\right )}{n^2+3\,n+2}+\frac {n\,x\,\left (\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}{2\,b\,\left (n^2+3\,n+2\right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*acoth(tanh(a + b*x))^n,x)

[Out]

-(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1))/2 - log(-2/(exp(2*a)*exp(2*b*x) - 1))/2)^n*((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/(4*b^2*(3*n + n^
2 + 2)) - (x^2*(n + 1))/(3*n + n^2 + 2) + (n*x*(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*b*(3*n + n^2 + 2)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*acoth(tanh(b*x+a))**n,x)

[Out]

Piecewise((x**2*acoth(tanh(a))**n/2, Eq(b, 0)), (-x/(b*acoth(tanh(a + b*x))) + log(acoth(tanh(a + b*x)))/b**2,
Eq(n, -2)), (Integral(x/acoth(tanh(a + b*x)), x), Eq(n, -1)), (b*n*x*acoth(tanh(a + b*x))*acoth(tanh(a + b*x)
)**n/(b**2*n**2 + 3*b**2*n + 2*b**2) + 2*b*x*acoth(tanh(a + b*x))*acoth(tanh(a + b*x))**n/(b**2*n**2 + 3*b**2*
n + 2*b**2) - acoth(tanh(a + b*x))**2*acoth(tanh(a + b*x))**n/(b**2*n**2 + 3*b**2*n + 2*b**2), True))

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