\(\int \frac {e^{4 \coth ^{-1}(a x)}}{(c-a c x)^2} \, dx\) [194]

Optimal result

Integrand size = 18, antiderivative size = 25 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{(c-a c x)^2} \, dx=\frac {(1+a x)^3}{6 a c^2 (1-a x)^3} \]

[Out]

1/6*(a*x+1)^3/a/c^2/(-a*x+1)^3

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6302, 6264, 37} \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{(c-a c x)^2} \, dx=\frac {(a x+1)^3}{6 a c^2 (1-a x)^3} \]

[In]

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

[Out]

(1 + a*x)^3/(6*a*c^2*(1 - a*x)^3)

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 6264

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_))^(p_.), x_Symbol] :> Dist[c^p, Int[u*(1 + d*(x/c))^
p*((1 + a*x)^(n/2)/(1 - a*x)^(n/2)), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[a^2*c^2 - d^2, 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 c x)^2} \, dx \\ & = \frac {\int \frac {(1+a x)^2}{(1-a x)^4} \, dx}{c^2} \\ & = \frac {(1+a x)^3}{6 a c^2 (1-a x)^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{(c-a c x)^2} \, dx=\frac {(1+a x)^3}{6 a c^2 (1-a x)^3} \]

[In]

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

[Out]

(1 + a*x)^3/(6*a*c^2*(1 - a*x)^3)

Maple [A] (verified)

Time = 0.56 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96

method	result	size
risch	\(\frac {-a \,x^{2}-\frac {1}{3 a}}{\left (a x -1\right )^{3} c^{2}}\)	\(24\)
parallelrisch	\(\frac {-a^{2} x^{3}-3 x}{3 \left (a x -1\right )^{3} c^{2}}\)	\(25\)
gosper	\(-\frac {3 a^{2} x^{2}+1}{3 \left (a x -1\right )^{3} a \,c^{2}}\)	\(26\)
norman	\(\frac {-\frac {x}{c}-\frac {a^{2} x^{3}}{3 c}}{\left (a x -1\right )^{3} c}\)	\(30\)
default	\(\frac {-\frac {2}{\left (a x -1\right )^{2} a}-\frac {1}{a \left (a x -1\right )}-\frac {4}{3 a \left (a x -1\right )^{3}}}{c^{2}}\)	\(42\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (22) = 44\).

Time = 0.25 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.04 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{(c-a c x)^2} \, dx=-\frac {3 \, a^{2} x^{2} + 1}{3 \, {\left (a^{4} c^{2} x^{3} - 3 \, a^{3} c^{2} x^{2} + 3 \, a^{2} c^{2} x - a c^{2}\right )}} \]

[In]

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

[Out]

-1/3*(3*a^2*x^2 + 1)/(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. 51 vs. \(2 (19) = 38\).

Time = 0.18 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.04 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{(c-a c x)^2} \, dx=\frac {- 3 a^{2} x^{2} - 1}{3 a^{4} c^{2} x^{3} - 9 a^{3} c^{2} x^{2} + 9 a^{2} c^{2} x - 3 a c^{2}} \]

[In]

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

[Out]

(-3*a**2*x**2 - 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. 51 vs. \(2 (22) = 44\).

Time = 0.21 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.04 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{(c-a c x)^2} \, dx=-\frac {3 \, a^{2} x^{2} + 1}{3 \, {\left (a^{4} c^{2} x^{3} - 3 \, a^{3} c^{2} x^{2} + 3 \, a^{2} c^{2} x - a c^{2}\right )}} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 50 vs. \(2 (22) = 44\).

Time = 0.27 (sec) , antiderivative size = 50, normalized size of antiderivative = 2.00 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{(c-a c x)^2} \, dx=-\frac {2}{{\left (a c x - c\right )}^{2} a} - \frac {1}{{\left (a c x - c\right )} a c} - \frac {4 \, c}{3 \, {\left (a c x - c\right )}^{3} a} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.57 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{(c-a c x)^2} \, dx=-\frac {3\,a^2\,x^2+1}{3\,a\,c^2\,{\left (a\,x-1\right )}^3} \]

[In]

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

[Out]

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