\(\int \frac {e^{-2 \coth ^{-1}(a x)}}{c-a^2 c x^2} \, dx\) [602]

Optimal result

Integrand size = 22, antiderivative size = 14 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{c-a^2 c x^2} \, dx=\frac {1}{a c (1+a x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 14, 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, 32} \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{c-a^2 c x^2} \, dx=\frac {1}{a c (a x+1)} \]

[In]

Int[1/(E^(2*ArcCoth[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 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^{-2 \text {arctanh}(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*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 3 vs. order 1 in optimal.

Time = 0.06 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.29 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{c-a^2 c x^2} \, dx=-\frac {e^{-2 \coth ^{-1}(a x)}}{2 a c} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.59 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00

method	result	size
norman	\(-\frac {x}{c \left (a x +1\right )}\)	\(14\)
parallelrisch	\(-\frac {x}{c \left (a x +1\right )}\)	\(14\)
gosper	\(\frac {1}{a c \left (a x +1\right )}\)	\(15\)
default	\(\frac {1}{a c \left (a x +1\right )}\)	\(15\)
risch	\(\frac {1}{a c \left (a x +1\right )}\)	\(15\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{c-a^2 c x^2} \, dx=\frac {1}{a^{2} c x + a c} \]

[In]

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

[Out]

1/(a^2*c*x + a*c)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.71 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{c-a^2 c x^2} \, dx=\frac {1}{a^{2} c x + a c} \]

[In]

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

[Out]

1/(a**2*c*x + a*c)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{c-a^2 c x^2} \, dx=\frac {1}{a^{2} c x + a c} \]

[In]

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

[Out]

1/(a^2*c*x + a*c)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{c-a^2 c x^2} \, dx=\frac {1}{{\left (a x + 1\right )} a c} \]

[In]

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

[Out]

1/((a*x + 1)*a*c)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{c-a^2 c x^2} \, dx=\frac {1}{a\,\left (c+a\,c\,x\right )} \]

[In]

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

[Out]

1/(a*(c + a*c*x))