\(\int \frac {e^{4 \coth ^{-1}(a x)}}{(c-a^2 c x^2)^2} \, dx\) [587]

Optimal result

Integrand size = 22, antiderivative size = 18 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx=\frac {1}{3 a c^2 (1-a x)^3} \]

[Out]

1/3/a/c^2/(-a*x+1)^3

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 18, 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} \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx=\frac {1}{3 a c^2 (1-a x)^3} \]

[In]

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

[Out]

1/(3*a*c^2*(1 - a*x)^3)

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^{4 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx \\ & = \frac {\int \frac {1}{(1-a x)^4} \, dx}{c^2} \\ & = \frac {1}{3 a c^2 (1-a x)^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx=\frac {1}{3 a c^2 (1-a x)^3} \]

[In]

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

[Out]

1/(3*a*c^2*(1 - a*x)^3)

Maple [A] (verified)

Time = 0.62 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89

method	result	size
gosper	\(-\frac {1}{3 c^{2} \left (a x -1\right )^{3} a}\)	\(16\)
default	\(-\frac {1}{3 c^{2} \left (a x -1\right )^{3} a}\)	\(16\)
risch	\(-\frac {1}{3 c^{2} \left (a x -1\right )^{3} a}\)	\(16\)
parallelrisch	\(\frac {-a^{2} x^{3}+3 a \,x^{2}-3 x}{3 \left (a x -1\right )^{3} c^{2}}\)	\(31\)
norman	\(\frac {-\frac {x}{c}-\frac {a^{3} x^{4}}{3 c}+\frac {2 a^{2} x^{3}}{3 c}}{\left (a x -1\right )^{3} \left (a x +1\right ) c}\)	\(48\)

[In]

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

[Out]

-1/3/c^2/(a*x-1)^3/a

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (15) = 30\).

Time = 0.23 (sec) , antiderivative size = 41, normalized size of antiderivative = 2.28 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx=-\frac {1}{3 \, {\left (a^{4} c^{2} x^{3} - 3 \, a^{3} c^{2} x^{2} + 3 \, a^{2} c^{2} x - a c^{2}\right )}} \]

[In]

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

[Out]

-1/3/(a^4*c^2*x^3 - 3*a^3*c^2*x^2 + 3*a^2*c^2*x - a*c^2)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (14) = 28\).

Time = 0.11 (sec) , antiderivative size = 42, normalized size of antiderivative = 2.33 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx=- \frac {1}{3 a^{4} c^{2} x^{3} - 9 a^{3} c^{2} x^{2} + 9 a^{2} c^{2} x - 3 a c^{2}} \]

[In]

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

[Out]

-1/(3*a**4*c**2*x**3 - 9*a**3*c**2*x**2 + 9*a**2*c**2*x - 3*a*c**2)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (15) = 30\).

Time = 0.19 (sec) , antiderivative size = 41, normalized size of antiderivative = 2.28 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx=-\frac {1}{3 \, {\left (a^{4} c^{2} x^{3} - 3 \, a^{3} c^{2} x^{2} + 3 \, a^{2} c^{2} x - a c^{2}\right )}} \]

[In]

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

[Out]

-1/3/(a^4*c^2*x^3 - 3*a^3*c^2*x^2 + 3*a^2*c^2*x - a*c^2)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx=-\frac {1}{3 \, {\left (a x - 1\right )}^{3} a c^{2}} \]

[In]

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

[Out]

-1/3/((a*x - 1)^3*a*c^2)

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 40, normalized size of antiderivative = 2.22 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx=\frac {1}{-3\,a^4\,c^2\,x^3+9\,a^3\,c^2\,x^2-9\,a^2\,c^2\,x+3\,a\,c^2} \]

[In]

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

[Out]

1/(3*a*c^2 - 9*a^2*c^2*x + 9*a^3*c^2*x^2 - 3*a^4*c^2*x^3)