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

Optimal. Leaf size=29 \[ \frac {\sqrt {1-a^2 x^2}}{a c^2 (1-a x)} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6127, 651} \[ \frac {\sqrt {1-a^2 x^2}}{a c^2 (1-a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[1 - a^2*x^2]/(a*c^2*(1 - a*x))

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]

Rule 6127

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

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 26, normalized size = 0.90 \[ \frac {\sqrt {a x+1}}{a c^2 \sqrt {1-a x}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Sqrt[1 + a*x]/(a*c^2*Sqrt[1 - a*x])

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fricas [A]  time = 0.51, size = 37, normalized size = 1.28 \[ \frac {a x - \sqrt {-a^{2} x^{2} + 1} - 1}{a^{2} c^{2} x - a c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [C]  time = 0.24, size = 70, normalized size = 2.41 \[ -c^{2} {\left (\frac {\sqrt {-\frac {2 \, c}{a c x - c} - 1} \mathrm {sgn}\left (\frac {1}{a c x - c}\right ) \mathrm {sgn}\relax (a) \mathrm {sgn}\relax (c)}{a^{2} c^{4}} - \frac {i \, \mathrm {sgn}\left (\frac {1}{a c x - c}\right ) \mathrm {sgn}\relax (a) \mathrm {sgn}\relax (c)}{a^{2} c^{4}}\right )} {\left | a \right |} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c^2*(sqrt(-2*c/(a*c*x - c) - 1)*sgn(1/(a*c*x - c))*sgn(a)*sgn(c)/(a^2*c^4) - I*sgn(1/(a*c*x - c))*sgn(a)*sgn(
c)/(a^2*c^4))*abs(a)

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maple [A]  time = 0.03, size = 28, normalized size = 0.97 \[ -\frac {\sqrt {-a^{2} x^{2}+1}}{\left (a x -1\right ) a \,c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.04, size = 47, normalized size = 1.62 \[ \frac {\sqrt {1-a^2\,x^2}}{c^2\,\left (x\,\sqrt {-a^2}-\frac {\sqrt {-a^2}}{a}\right )\,\sqrt {-a^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sqrt {- a^{2} x^{2} + 1}}{a^{3} x^{3} - a^{2} x^{2} - a x + 1}\, dx}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(-a**2*x**2 + 1)/(a**3*x**3 - a**2*x**2 - a*x + 1), x)/c**2

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