3.1187 \(\int \frac {e^{4 \tanh ^{-1}(a x)}}{c-a^2 c x^2} \, dx\)

Optimal. Leaf size=13 \[ \frac {x}{c (1-a x)^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 13, 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, 34} \[ \frac {x}{c (1-a x)^2} \]

Antiderivative was successfully verified.

[In]

Int[E^(4*ArcTanh[a*x])/(c - a^2*c*x^2),x]

[Out]

x/(c*(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 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^{4 \tanh ^{-1}(a x)}}{c-a^2 c x^2} \, dx &=\frac {\int \frac {1+a x}{(1-a x)^3} \, dx}{c}\\ &=\frac {x}{c (1-a x)^2}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 25, normalized size = 1.92 \[ \frac {(a x+1)^2}{4 a c (1-a x)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(4*ArcTanh[a*x])/(c - a^2*c*x^2),x]

[Out]

(1 + a*x)^2/(4*a*c*(1 - a*x)^2)

________________________________________________________________________________________

fricas [A]  time = 0.54, size = 19, normalized size = 1.46 \[ \frac {x}{a^{2} c x^{2} - 2 \, a c x + c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^4/(-a^2*x^2+1)^2/(-a^2*c*x^2+c),x, algorithm="fricas")

[Out]

x/(a^2*c*x^2 - 2*a*c*x + c)

________________________________________________________________________________________

giac [A]  time = 0.16, size = 12, normalized size = 0.92 \[ \frac {x}{{\left (a x - 1\right )}^{2} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^4/(-a^2*x^2+1)^2/(-a^2*c*x^2+c),x, algorithm="giac")

[Out]

x/((a*x - 1)^2*c)

________________________________________________________________________________________

maple [B]  time = 0.03, size = 28, normalized size = 2.15 \[ \frac {\frac {1}{a \left (a x -1\right )}+\frac {1}{a \left (a x -1\right )^{2}}}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)^4/(-a^2*x^2+1)^2/(-a^2*c*x^2+c),x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.32, size = 19, normalized size = 1.46 \[ \frac {x}{a^{2} c x^{2} - 2 \, a c x + c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^4/(-a^2*x^2+1)^2/(-a^2*c*x^2+c),x, algorithm="maxima")

[Out]

x/(a^2*c*x^2 - 2*a*c*x + c)

________________________________________________________________________________________

mupad [B]  time = 0.92, size = 12, normalized size = 0.92 \[ \frac {x}{c\,{\left (a\,x-1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x + 1)^4/((c - a^2*c*x^2)*(a^2*x^2 - 1)^2),x)

[Out]

x/(c*(a*x - 1)^2)

________________________________________________________________________________________

sympy [B]  time = 0.24, size = 17, normalized size = 1.31 \[ \frac {x}{a^{2} c x^{2} - 2 a c x + c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)**4/(-a**2*x**2+1)**2/(-a**2*c*x**2+c),x)

[Out]

x/(a**2*c*x**2 - 2*a*c*x + c)

________________________________________________________________________________________