\(\int \frac {e^{4 \coth ^{-1}(a x)}}{(c-a c x)^3} \, dx\) [195]

Optimal result

Integrand size = 18, antiderivative size = 52 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{(c-a c x)^3} \, dx=\frac {1}{a c^3 (1-a x)^4}-\frac {4}{3 a c^3 (1-a x)^3}+\frac {1}{2 a c^3 (1-a x)^2} \]

[Out]

1/a/c^3/(-a*x+1)^4-4/3/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 = 52, 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} \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{(c-a c x)^3} \, dx=\frac {1}{2 a c^3 (1-a x)^2}-\frac {4}{3 a c^3 (1-a x)^3}+\frac {1}{a c^3 (1-a x)^4} \]

[In]

Int[E^(4*ArcCoth[a*x])/(c - a*c*x)^3,x]

[Out]

1/(a*c^3*(1 - a*x)^4) - 4/(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^{4 \text {arctanh}(a x)}}{(c-a c x)^3} \, dx \\ & = \frac {\int \frac {(1+a x)^2}{(1-a x)^5} \, dx}{c^3} \\ & = \frac {\int \left (-\frac {4}{(-1+a x)^5}-\frac {4}{(-1+a x)^4}-\frac {1}{(-1+a x)^3}\right ) \, dx}{c^3} \\ & = \frac {1}{a c^3 (1-a x)^4}-\frac {4}{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 = 31, normalized size of antiderivative = 0.60 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{(c-a c x)^3} \, dx=\frac {1+2 a x+3 a^2 x^2}{6 a c^3 (-1+a x)^4} \]

[In]

Integrate[E^(4*ArcCoth[a*x])/(c - a*c*x)^3,x]

[Out]

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

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.52

method	result	size
risch	\(\frac {\frac {a \,x^{2}}{2}+\frac {x}{3}+\frac {1}{6 a}}{\left (a x -1\right )^{4} c^{3}}\)	\(27\)
gosper	\(\frac {3 a^{2} x^{2}+2 a x +1}{6 \left (a x -1\right )^{4} a \,c^{3}}\)	\(30\)
parallelrisch	\(\frac {-a^{3} x^{4}+4 a^{2} x^{3}-3 a \,x^{2}+6 x}{6 \left (a x -1\right )^{4} c^{3}}\)	\(39\)
default	\(\frac {\frac {1}{a \left (a x -1\right )^{4}}+\frac {1}{2 \left (a x -1\right )^{2} a}+\frac {4}{3 a \left (a x -1\right )^{3}}}{c^{3}}\)	\(41\)
norman	\(\frac {\frac {x}{c}-\frac {a \,x^{2}}{2 c}+\frac {2 a^{2} x^{3}}{3 c}-\frac {a^{3} x^{4}}{6 c}}{\left (a x -1\right )^{4} c^{2}}\)	\(49\)

[In]

int(1/(a*x-1)^2*(a*x+1)^2/(-a*c*x+c)^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.25 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{(c-a c x)^3} \, dx=\frac {3 \, a^{2} x^{2} + 2 \, a x + 1}{6 \, {\left (a^{5} c^{3} x^{4} - 4 \, a^{4} c^{3} x^{3} + 6 \, a^{3} c^{3} x^{2} - 4 \, a^{2} c^{3} x + a c^{3}\right )}} \]

[In]

integrate(1/(a*x-1)^2*(a*x+1)^2/(-a*c*x+c)^3,x, algorithm="fricas")

[Out]

1/6*(3*a^2*x^2 + 2*a*x + 1)/(a^5*c^3*x^4 - 4*a^4*c^3*x^3 + 6*a^3*c^3*x^2 - 4*a^2*c^3*x + a*c^3)

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.35 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{(c-a c x)^3} \, dx=- \frac {- 3 a^{2} x^{2} - 2 a x - 1}{6 a^{5} c^{3} x^{4} - 24 a^{4} c^{3} x^{3} + 36 a^{3} c^{3} x^{2} - 24 a^{2} c^{3} x + 6 a c^{3}} \]

[In]

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

[Out]

-(-3*a**2*x**2 - 2*a*x - 1)/(6*a**5*c**3*x**4 - 24*a**4*c**3*x**3 + 36*a**3*c**3*x**2 - 24*a**2*c**3*x + 6*a*c
**3)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.25 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{(c-a c x)^3} \, dx=\frac {3 \, a^{2} x^{2} + 2 \, a x + 1}{6 \, {\left (a^{5} c^{3} x^{4} - 4 \, a^{4} c^{3} x^{3} + 6 \, a^{3} c^{3} x^{2} - 4 \, a^{2} c^{3} x + a c^{3}\right )}} \]

[In]

integrate(1/(a*x-1)^2*(a*x+1)^2/(-a*c*x+c)^3,x, algorithm="maxima")

[Out]

1/6*(3*a^2*x^2 + 2*a*x + 1)/(a^5*c^3*x^4 - 4*a^4*c^3*x^3 + 6*a^3*c^3*x^2 - 4*a^2*c^3*x + a*c^3)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.81 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{(c-a c x)^3} \, dx=\frac {\frac {3}{{\left (a x - 1\right )}^{2} a} + \frac {8}{{\left (a x - 1\right )}^{3} a} + \frac {6}{{\left (a x - 1\right )}^{4} a}}{6 \, c^{3}} \]

[In]

integrate(1/(a*x-1)^2*(a*x+1)^2/(-a*c*x+c)^3,x, algorithm="giac")

[Out]

1/6*(3/((a*x - 1)^2*a) + 8/((a*x - 1)^3*a) + 6/((a*x - 1)^4*a))/c^3

Mupad [B] (verification not implemented)

Time = 4.44 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.56 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{(c-a c x)^3} \, dx=\frac {3\,a^2\,x^2+2\,a\,x+1}{6\,a\,c^3\,{\left (a\,x-1\right )}^4} \]

[In]

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

[Out]

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