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

Optimal. Leaf size=14 \[ -\frac {x}{c^2 (1-a x)^2} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 14, 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 = {6167, 6129, 34} \[ -\frac {x}{c^2 (1-a x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x/(c^2*(1 - a*x)^2))

Rule 34

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

Rule 6129

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

Rule 6167

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Dist[(-1)^(n/2), Int[u*E^(n*ArcTanh[a*x]), x], x] /; Free
Q[a, x] && IntegerQ[n/2]

Rubi steps

\begin {align*} \int \frac {e^{2 \coth ^{-1}(a x)}}{(c-a c x)^2} \, dx &=-\int \frac {e^{2 \tanh ^{-1}(a x)}}{(c-a c x)^2} \, dx\\ &=-\frac {\int \frac {1+a x}{(1-a x)^3} \, dx}{c^2}\\ &=-\frac {x}{c^2 (1-a x)^2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 25, normalized size = 1.79 \[ -\frac {(a x+1)^2}{4 a c^2 (1-a x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(1 + a*x)^2/(a*c^2*(1 - a*x)^2)

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fricas [A]  time = 0.65, size = 26, normalized size = 1.86 \[ -\frac {x}{a^{2} c^{2} x^{2} - 2 \, a c^{2} x + c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x-1)*(a*x+1)/(-a*c*x+c)^2,x, algorithm="fricas")

[Out]

-x/(a^2*c^2*x^2 - 2*a*c^2*x + c^2)

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giac [B]  time = 0.14, size = 34, normalized size = 2.43 \[ -\frac {1}{{\left (a c x - c\right )}^{2} a} - \frac {1}{{\left (a c x - c\right )} a c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x-1)*(a*x+1)/(-a*c*x+c)^2,x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.04, size = 30, normalized size = 2.14 \[ \frac {-\frac {1}{a \left (a x -1\right )^{2}}-\frac {1}{a \left (a x -1\right )}}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.31, size = 26, normalized size = 1.86 \[ -\frac {x}{a^{2} c^{2} x^{2} - 2 \, a c^{2} x + c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x-1)*(a*x+1)/(-a*c*x+c)^2,x, algorithm="maxima")

[Out]

-x/(a^2*c^2*x^2 - 2*a*c^2*x + c^2)

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mupad [B]  time = 1.19, size = 13, normalized size = 0.93 \[ -\frac {x}{c^2\,{\left (a\,x-1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x/(c^2*(a*x - 1)^2)

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sympy [A]  time = 0.16, size = 24, normalized size = 1.71 \[ - \frac {x}{a^{2} c^{2} x^{2} - 2 a c^{2} x + c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x/(a**2*c**2*x**2 - 2*a*c**2*x + c**2)

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