3.192 \(\int \frac {\coth ^{-1}(\tanh (a+b x))^n}{x^3} \, dx\)

Optimal. Leaf size=101 \[ \frac {b^2 n \coth ^{-1}(\tanh (a+b x))^{n-1} \, _2F_1\left (1,n-1;n;-\frac {\coth ^{-1}(\tanh (a+b x))}{b x-\coth ^{-1}(\tanh (a+b x))}\right )}{2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}-\frac {\coth ^{-1}(\tanh (a+b x))^n}{2 x^2}-\frac {b n \coth ^{-1}(\tanh (a+b x))^{n-1}}{2 x} \]

[Out]

-1/2*b*n*arccoth(tanh(b*x+a))^(-1+n)/x-1/2*arccoth(tanh(b*x+a))^n/x^2+1/2*b^2*n*arccoth(tanh(b*x+a))^(-1+n)*hy
pergeom([1, -1+n],[n],-arccoth(tanh(b*x+a))/(b*x-arccoth(tanh(b*x+a))))/(b*x-arccoth(tanh(b*x+a)))

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Rubi [A]  time = 0.07, antiderivative size = 101, 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, 2164} \[ \frac {b^2 n \coth ^{-1}(\tanh (a+b x))^{n-1} \, _2F_1\left (1,n-1;n;-\frac {\coth ^{-1}(\tanh (a+b x))}{b x-\coth ^{-1}(\tanh (a+b x))}\right )}{2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}-\frac {\coth ^{-1}(\tanh (a+b x))^n}{2 x^2}-\frac {b n \coth ^{-1}(\tanh (a+b x))^{n-1}}{2 x} \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[Tanh[a + b*x]]^n/x^3,x]

[Out]

-(b*n*ArcCoth[Tanh[a + b*x]]^(-1 + n))/(2*x) - ArcCoth[Tanh[a + b*x]]^n/(2*x^2) + (b^2*n*ArcCoth[Tanh[a + b*x]
]^(-1 + n)*Hypergeometric2F1[1, -1 + n, n, -(ArcCoth[Tanh[a + b*x]]/(b*x - ArcCoth[Tanh[a + b*x]]))])/(2*(b*x
- ArcCoth[Tanh[a + b*x]]))

Rule 2164

Int[(v_)^(n_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(v^(n + 1)*Hypergeo
metric2F1[1, n + 1, n + 2, -((a*v)/(b*u - a*v))])/((n + 1)*(b*u - a*v)), x] /; NeQ[b*u - a*v, 0]] /; Piecewise
LinearQ[u, v, x] &&  !IntegerQ[n]

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 \frac {\coth ^{-1}(\tanh (a+b x))^n}{x^3} \, dx &=-\frac {\coth ^{-1}(\tanh (a+b x))^n}{2 x^2}+\frac {1}{2} (b n) \int \frac {\coth ^{-1}(\tanh (a+b x))^{-1+n}}{x^2} \, dx\\ &=-\frac {b n \coth ^{-1}(\tanh (a+b x))^{-1+n}}{2 x}-\frac {\coth ^{-1}(\tanh (a+b x))^n}{2 x^2}-\frac {1}{2} \left (b^2 (1-n) n\right ) \int \frac {\coth ^{-1}(\tanh (a+b x))^{-2+n}}{x} \, dx\\ &=-\frac {b n \coth ^{-1}(\tanh (a+b x))^{-1+n}}{2 x}-\frac {\coth ^{-1}(\tanh (a+b x))^n}{2 x^2}+\frac {b^2 n \coth ^{-1}(\tanh (a+b x))^{-1+n} \, _2F_1\left (1,-1+n;n;-\frac {\coth ^{-1}(\tanh (a+b x))}{b x-\coth ^{-1}(\tanh (a+b x))}\right )}{2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 67, normalized size = 0.66 \[ \frac {\coth ^{-1}(\tanh (a+b x))^n \left (\frac {\coth ^{-1}(\tanh (a+b x))}{b x}\right )^{-n} \, _2F_1\left (2-n,-n;3-n;1-\frac {\coth ^{-1}(\tanh (a+b x))}{b x}\right )}{(n-2) x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(ArcCoth[Tanh[a + b*x]]^n*Hypergeometric2F1[2 - n, -n, 3 - n, 1 - ArcCoth[Tanh[a + b*x]]/(b*x)])/((-2 + n)*x^2
*(ArcCoth[Tanh[a + b*x]]/(b*x))^n)

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fricas [F]  time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{n}}{x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arccoth(tanh(b*x + a))^n/x^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 13.91, size = 0, normalized size = 0.00 \[ \int \frac {\mathrm {arccoth}\left (\tanh \left (b x +a \right )\right )^{n}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^n}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acoth(tanh(a + b*x))^n/x^3,x)

[Out]

int(acoth(tanh(a + b*x))^n/x^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {acoth}^{n}{\left (\tanh {\left (a + b x \right )} \right )}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(acoth(tanh(a + b*x))**n/x**3, x)

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