3.1053 \(\int \frac {e^{2 \tanh ^{-1}(a x)}}{c-a^2 c x^2} \, dx\)

Optimal. Leaf size=15 \[ \frac {1}{a c (1-a x)} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {6140, 32} \[ \frac {1}{a c (1-a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

1/(a*c*(1 - a*x))

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 6140

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

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 18, normalized size = 1.20 \[ \frac {e^{2 \tanh ^{-1}(a x)}}{2 a c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

E^(2*ArcTanh[a*x])/(2*a*c)

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fricas [A]  time = 0.45, size = 15, normalized size = 1.00 \[ -\frac {1}{a^{2} c x - a c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(a^2*c*x - a*c)

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giac [A]  time = 0.17, size = 15, normalized size = 1.00 \[ -\frac {1}{{\left (a x - 1\right )} a c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/((a*x - 1)*a*c)

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maple [A]  time = 0.02, size = 16, normalized size = 1.07 \[ -\frac {1}{c a \left (a x -1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.34, size = 15, normalized size = 1.00 \[ -\frac {1}{a^{2} c x - a c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(a^2*c*x - a*c)

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mupad [B]  time = 0.89, size = 13, normalized size = 0.87 \[ \frac {1}{a\,\left (c-a\,c\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(a*(c - a*c*x))

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sympy [A]  time = 0.14, size = 12, normalized size = 0.80 \[ - \frac {1}{a^{2} c x - a c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(a**2*c*x - a*c)

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