### 3.202 $$\int \frac{e^{-\coth ^{-1}(a x)}}{(c-a c x)^2} \, dx$$

Optimal. Leaf size=28 $-\frac{\sqrt{1-\frac{1}{a^2 x^2}}}{c^2 \left (a-\frac{1}{x}\right )}$

[Out]

-(Sqrt[1 - 1/(a^2*x^2)]/(c^2*(a - x^(-1))))

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Rubi [A]  time = 0.101127, antiderivative size = 28, normalized size of antiderivative = 1., 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, 6178, 651} $-\frac{\sqrt{1-\frac{1}{a^2 x^2}}}{c^2 \left (a-\frac{1}{x}\right )}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/(E^ArcCoth[a*x]*(c - a*c*x)^2),x]

[Out]

-(Sqrt[1 - 1/(a^2*x^2)]/(c^2*(a - x^(-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 6178

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.)*(x_)^(m_.), x_Symbol] :> -Dist[c^n, Subst[Int[((c
+ d*x)^(p - n)*(1 - x^2/a^2)^(n/2))/x^(m + 2), x], x, 1/x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[c + a*d, 0] &&
IntegerQ[(n - 1)/2] && IntegerQ[m] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p, n/2 + 1] || LtQ[-5, m, -1]) && In
tegerQ[2*p]

Rule 651

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^m*(a + c*x^2)^(p + 1))
/(2*c*d*(p + 1)), x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + 2*p
+ 2, 0]

Rubi steps

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

Mathematica [A]  time = 0.050955, size = 27, normalized size = 0.96 $-\frac{x \sqrt{1-\frac{1}{a^2 x^2}}}{c^2 (a x-1)}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/(E^ArcCoth[a*x]*(c - a*c*x)^2),x]

[Out]

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

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Maple [A]  time = 0.043, size = 36, normalized size = 1.3 \begin{align*} -{\frac{ax+1}{ \left ( ax-1 \right ) a{c}^{2}}\sqrt{{\frac{ax-1}{ax+1}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.03071, size = 31, normalized size = 1.11 \begin{align*} -\frac{1}{a c^{2} \sqrt{\frac{a x - 1}{a x + 1}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(a*c^2*sqrt((a*x - 1)/(a*x + 1)))

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Fricas [A]  time = 1.58156, size = 78, normalized size = 2.79 \begin{align*} -\frac{{\left (a x + 1\right )} \sqrt{\frac{a x - 1}{a x + 1}}}{a^{2} c^{2} x - a c^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\sqrt{\frac{a x}{a x + 1} - \frac{1}{a x + 1}}}{a^{2} x^{2} - 2 a x + 1}\, dx}{c^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{undef} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

undef