\(\int \frac {e^{2 \coth ^{-1}(a x)}}{(c-a c x)^3} \, dx\) [175]

Optimal result

Integrand size = 18, antiderivative size = 37 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{(c-a c x)^3} \, dx=-\frac {2}{3 a c^3 (1-a x)^3}+\frac {1}{2 a c^3 (1-a x)^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6302, 6264, 45} \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{(c-a c x)^3} \, dx=\frac {1}{2 a c^3 (1-a x)^2}-\frac {2}{3 a c^3 (1-a x)^3} \]

[In]

Int[E^(2*ArcCoth[a*x])/(c - a*c*x)^3,x]

[Out]

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.62 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{(c-a c x)^3} \, dx=\frac {1+3 a x}{6 a c^3 (-1+a x)^3} \]

[In]

Integrate[E^(2*ArcCoth[a*x])/(c - a*c*x)^3,x]

[Out]

(1 + 3*a*x)/(6*a*c^3*(-1 + a*x)^3)

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.57

method	result	size
risch	\(\frac {\frac {x}{2}+\frac {1}{6 a}}{\left (a x -1\right )^{3} c^{3}}\)	\(21\)
gosper	\(\frac {3 a x +1}{6 a \,c^{3} \left (a x -1\right )^{3}}\)	\(22\)
default	\(\frac {\frac {1}{2 \left (a x -1\right )^{2} a}+\frac {2}{3 a \left (a x -1\right )^{3}}}{c^{3}}\)	\(30\)
parallelrisch	\(\frac {a^{2} x^{3}-3 a \,x^{2}+6 x}{6 c^{3} \left (a x -1\right )^{3}}\)	\(30\)
norman	\(\frac {\frac {x}{c}-\frac {a \,x^{2}}{2 c}+\frac {a^{2} x^{3}}{6 c}}{\left (a x -1\right )^{3} c^{2}}\)	\(38\)

[In]

int(1/(a*x-1)*(a*x+1)/(-a*c*x+c)^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.27 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{(c-a c x)^3} \, dx=\frac {3 \, a x + 1}{6 \, {\left (a^{4} c^{3} x^{3} - 3 \, a^{3} c^{3} x^{2} + 3 \, a^{2} c^{3} x - a c^{3}\right )}} \]

[In]

integrate(1/(a*x-1)*(a*x+1)/(-a*c*x+c)^3,x, algorithm="fricas")

[Out]

1/6*(3*a*x + 1)/(a^4*c^3*x^3 - 3*a^3*c^3*x^2 + 3*a^2*c^3*x - a*c^3)

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.32 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{(c-a c x)^3} \, dx=- \frac {- 3 a x - 1}{6 a^{4} c^{3} x^{3} - 18 a^{3} c^{3} x^{2} + 18 a^{2} c^{3} x - 6 a c^{3}} \]

[In]

integrate(1/(a*x-1)*(a*x+1)/(-a*c*x+c)**3,x)

[Out]

-(-3*a*x - 1)/(6*a**4*c**3*x**3 - 18*a**3*c**3*x**2 + 18*a**2*c**3*x - 6*a*c**3)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.27 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{(c-a c x)^3} \, dx=\frac {3 \, a x + 1}{6 \, {\left (a^{4} c^{3} x^{3} - 3 \, a^{3} c^{3} x^{2} + 3 \, a^{2} c^{3} x - a c^{3}\right )}} \]

[In]

integrate(1/(a*x-1)*(a*x+1)/(-a*c*x+c)^3,x, algorithm="maxima")

[Out]

1/6*(3*a*x + 1)/(a^4*c^3*x^3 - 3*a^3*c^3*x^2 + 3*a^2*c^3*x - a*c^3)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.57 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{(c-a c x)^3} \, dx=\frac {3 \, a x + 1}{6 \, {\left (a x - 1\right )}^{3} a c^{3}} \]

[In]

integrate(1/(a*x-1)*(a*x+1)/(-a*c*x+c)^3,x, algorithm="giac")

[Out]

1/6*(3*a*x + 1)/((a*x - 1)^3*a*c^3)

Mupad [B] (verification not implemented)

Time = 4.32 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.24 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{(c-a c x)^3} \, dx=-\frac {\frac {x}{2}+\frac {1}{6\,a}}{-a^3\,c^3\,x^3+3\,a^2\,c^3\,x^2-3\,a\,c^3\,x+c^3} \]

[In]

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

[Out]

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