3.369 \(\int \frac {e^{n \coth ^{-1}(a x)}}{(c-a c x)^2} \, dx\)

Optimal. Leaf size=48 \[ -\frac {\left (1-\frac {1}{a x}\right )^{-\frac {n}{2}-1} \left (\frac {1}{a x}+1\right )^{\frac {n+2}{2}}}{a c^2 (n+2)} \]

[Out]

-(1-1/a/x)^(-1-1/2*n)*(1+1/a/x)^(1+1/2*n)/a/c^2/(2+n)

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Rubi [A]  time = 0.11, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6175, 6180, 37} \[ -\frac {\left (1-\frac {1}{a x}\right )^{-\frac {n}{2}-1} \left (\frac {1}{a x}+1\right )^{\frac {n+2}{2}}}{a c^2 (n+2)} \]

Antiderivative was successfully verified.

[In]

Int[E^(n*ArcCoth[a*x])/(c - a*c*x)^2,x]

[Out]

-(((1 - 1/(a*x))^(-1 - n/2)*(1 + 1/(a*x))^((2 + n)/2))/(a*c^2*(2 + n)))

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 6175

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_))^(p_.), x_Symbol] :> Dist[d^p, Int[u*x^p*(1 + c/(d*
x))^p*E^(n*ArcCoth[a*x]), x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[a^2*c^2 - d^2, 0] &&  !IntegerQ[n/2] && Inte
gerQ[p]

Rule 6180

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.)*(x_)^(m_.), x_Symbol] :> -Dist[c^p, Subst[Int[((1
+ (d*x)/c)^p*(1 + x/a)^(n/2))/(x^(m + 2)*(1 - x/a)^(n/2)), x], x, 1/x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ
[c^2 - a^2*d^2, 0] &&  !IntegerQ[n/2] && (IntegerQ[p] || GtQ[c, 0]) && IntegerQ[m]

Rubi steps

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

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Mathematica [A]  time = 0.18, size = 33, normalized size = 0.69 \[ -\frac {(a x+1) e^{n \coth ^{-1}(a x)}}{a c^2 (n+2) (a x-1)} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(n*ArcCoth[a*x])/(c - a*c*x)^2,x]

[Out]

-((E^(n*ArcCoth[a*x])*(1 + a*x))/(a*c^2*(2 + n)*(-1 + a*x)))

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fricas [A]  time = 0.44, size = 59, normalized size = 1.23 \[ -\frac {{\left (a x + 1\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{2} \, n}}{a c^{2} n - 2 \, a c^{2} - {\left (a^{2} c^{2} n - 2 \, a^{2} c^{2}\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arccoth(a*x))/(-a*c*x+c)^2,x, algorithm="fricas")

[Out]

-(a*x + 1)*((a*x - 1)/(a*x + 1))^(1/2*n)/(a*c^2*n - 2*a*c^2 - (a^2*c^2*n - 2*a^2*c^2)*x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{2} \, n}}{{\left (a c x - c\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arccoth(a*x))/(-a*c*x+c)^2,x, algorithm="giac")

[Out]

integrate(((a*x - 1)/(a*x + 1))^(1/2*n)/(a*c*x - c)^2, x)

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maple [A]  time = 0.04, size = 33, normalized size = 0.69 \[ -\frac {{\mathrm e}^{n \,\mathrm {arccoth}\left (a x \right )} \left (a x +1\right )}{\left (a x -1\right ) c^{2} \left (2+n \right ) a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*arccoth(a*x))/(-a*c*x+c)^2,x)

[Out]

-exp(n*arccoth(a*x))*(a*x+1)/(a*x-1)/c^2/(2+n)/a

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{2} \, n}}{{\left (a c x - c\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arccoth(a*x))/(-a*c*x+c)^2,x, algorithm="maxima")

[Out]

integrate(((a*x - 1)/(a*x + 1))^(1/2*n)/(a*c*x - c)^2, x)

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mupad [B]  time = 1.53, size = 32, normalized size = 0.67 \[ -\frac {{\mathrm {e}}^{n\,\mathrm {acoth}\left (a\,x\right )}\,\left (a\,x+1\right )}{a\,c^2\,\left (a\,x-1\right )\,\left (n+2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*acoth(a*x))/(c - a*c*x)^2,x)

[Out]

-(exp(n*acoth(a*x))*(a*x + 1))/(a*c^2*(a*x - 1)*(n + 2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \begin {cases} \text {NaN} & \text {for}\: a = \frac {1}{x} \wedge c = 0 \wedge n = -2 \\\tilde {\infty } x e^{\infty n} & \text {for}\: a = \frac {1}{x} \\\tilde {\infty } \int e^{n \operatorname {acoth}{\left (a x \right )}}\, dx & \text {for}\: c = 0 \\- \frac {a x \operatorname {acoth}{\left (a x \right )}}{a^{2} c^{2} x e^{2 \operatorname {acoth}{\left (a x \right )}} - a c^{2} e^{2 \operatorname {acoth}{\left (a x \right )}}} - \frac {\operatorname {acoth}{\left (a x \right )}}{a^{2} c^{2} x e^{2 \operatorname {acoth}{\left (a x \right )}} - a c^{2} e^{2 \operatorname {acoth}{\left (a x \right )}}} & \text {for}\: n = -2 \\- \frac {a x e^{n \operatorname {acoth}{\left (a x \right )}}}{a^{2} c^{2} n x + 2 a^{2} c^{2} x - a c^{2} n - 2 a c^{2}} - \frac {e^{n \operatorname {acoth}{\left (a x \right )}}}{a^{2} c^{2} n x + 2 a^{2} c^{2} x - a c^{2} n - 2 a c^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*acoth(a*x))/(-a*c*x+c)**2,x)

[Out]

Piecewise((nan, Eq(c, 0) & Eq(n, -2) & Eq(a, 1/x)), (zoo*x*exp(oo*n), Eq(a, 1/x)), (zoo*Integral(exp(n*acoth(a
*x)), x), Eq(c, 0)), (-a*x*acoth(a*x)/(a**2*c**2*x*exp(2*acoth(a*x)) - a*c**2*exp(2*acoth(a*x))) - acoth(a*x)/
(a**2*c**2*x*exp(2*acoth(a*x)) - a*c**2*exp(2*acoth(a*x))), Eq(n, -2)), (-a*x*exp(n*acoth(a*x))/(a**2*c**2*n*x
+ 2*a**2*c**2*x - a*c**2*n - 2*a*c**2) - exp(n*acoth(a*x))/(a**2*c**2*n*x + 2*a**2*c**2*x - a*c**2*n - 2*a*c*
*2), True))

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