\(\int \frac {e^{4 \coth ^{-1}(a x)}}{c-a^2 c x^2} \, dx\) [586]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [B] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [B] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 22, antiderivative size = 13 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{c-a^2 c x^2} \, dx=\frac {x}{c (1-a x)^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {6302, 6275, 34} \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{c-a^2 c x^2} \, dx=\frac {x}{c (1-a x)^2} \]

[In]

Int[E^(4*ArcCoth[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 6275

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])

Rule 6302

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*} \text {integral}& = \int \frac {e^{4 \text {arctanh}(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] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.92 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{c-a^2 c x^2} \, dx=\frac {(1+a x)^2}{4 a c (1-a x)^2} \]

[In]

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

[Out]

(1 + a*x)^2/(4*a*c*(1 - a*x)^2)

Maple [A] (verified)

Time = 0.65 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00

method result size
gosper \(\frac {x}{c \left (a x -1\right )^{2}}\) \(13\)
norman \(\frac {x}{c \left (a x -1\right )^{2}}\) \(13\)
risch \(\frac {x}{c \left (a x -1\right )^{2}}\) \(13\)
parallelrisch \(\frac {x}{c \left (a x -1\right )^{2}}\) \(13\)
default \(\frac {\frac {1}{\left (a x -1\right )^{2} a}+\frac {1}{a \left (a x -1\right )}}{c}\) \(28\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.46 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{c-a^2 c x^2} \, dx=\frac {x}{a^{2} c x^{2} - 2 \, a c x + c} \]

[In]

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

[Out]

x/(a^2*c*x^2 - 2*a*c*x + c)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 17 vs. \(2 (8) = 16\).

Time = 0.09 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.31 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{c-a^2 c x^2} \, dx=\frac {x}{a^{2} c x^{2} - 2 a c x + c} \]

[In]

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

[Out]

x/(a**2*c*x**2 - 2*a*c*x + c)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.46 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{c-a^2 c x^2} \, dx=\frac {x}{a^{2} c x^{2} - 2 \, a c x + c} \]

[In]

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

[Out]

x/(a^2*c*x^2 - 2*a*c*x + c)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 27 vs. \(2 (12) = 24\).

Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 2.08 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{c-a^2 c x^2} \, dx=\frac {\frac {1}{{\left (a x - 1\right )} a} + \frac {1}{{\left (a x - 1\right )}^{2} a}}{c} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.92 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{c-a^2 c x^2} \, dx=\frac {x}{c\,{\left (a\,x-1\right )}^2} \]

[In]

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

[Out]

x/(c*(a*x - 1)^2)