\(\int \frac {e^{4 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx\) [196]

Optimal result

Integrand size = 18, antiderivative size = 53 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=\frac {4}{5 a c^4 (1-a x)^5}-\frac {1}{a c^4 (1-a x)^4}+\frac {1}{3 a c^4 (1-a x)^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 53, 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} \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=\frac {1}{3 a c^4 (1-a x)^3}-\frac {1}{a c^4 (1-a x)^4}+\frac {4}{5 a c^4 (1-a x)^5} \]

[In]

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

[Out]

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

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)^4} \, dx \\ & = \frac {\int \frac {(1+a x)^2}{(1-a x)^6} \, dx}{c^4} \\ & = \frac {\int \left (\frac {4}{(-1+a x)^6}+\frac {4}{(-1+a x)^5}+\frac {1}{(-1+a x)^4}\right ) \, dx}{c^4} \\ & = \frac {4}{5 a c^4 (1-a x)^5}-\frac {1}{a c^4 (1-a x)^4}+\frac {1}{3 a c^4 (1-a x)^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.58 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=-\frac {2+5 a x+5 a^2 x^2}{15 a c^4 (-1+a x)^5} \]

[In]

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

[Out]

-1/15*(2 + 5*a*x + 5*a^2*x^2)/(a*c^4*(-1 + a*x)^5)

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.51

method	result	size
risch	\(\frac {-\frac {a \,x^{2}}{3}-\frac {x}{3}-\frac {2}{15 a}}{\left (a x -1\right )^{5} c^{4}}\)	\(27\)
gosper	\(-\frac {5 a^{2} x^{2}+5 a x +2}{15 \left (a x -1\right )^{5} a \,c^{4}}\)	\(30\)
default	\(\frac {-\frac {1}{a \left (a x -1\right )^{4}}-\frac {1}{3 a \left (a x -1\right )^{3}}-\frac {4}{5 a \left (a x -1\right )^{5}}}{c^{4}}\)	\(42\)
parallelrisch	\(\frac {-2 a^{4} x^{5}+10 a^{3} x^{4}-20 a^{2} x^{3}+15 a \,x^{2}-15 x}{15 \left (a x -1\right )^{5} c^{4}}\)	\(47\)
norman	\(\frac {\frac {a \,x^{2}}{c}-\frac {x}{c}-\frac {4 a^{2} x^{3}}{3 c}+\frac {2 a^{3} x^{4}}{3 c}-\frac {2 a^{4} x^{5}}{15 c}}{\left (a x -1\right )^{5} c^{3}}\)	\(60\)

[In]

int(1/(a*x-1)^2*(a*x+1)^2/(-a*c*x+c)^4,x,method=_RETURNVERBOSE)

[Out]

(-1/3*a*x^2-1/3*x-2/15/a)/(a*x-1)^5/c^4

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.45 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=-\frac {5 \, a^{2} x^{2} + 5 \, a x + 2}{15 \, {\left (a^{6} c^{4} x^{5} - 5 \, a^{5} c^{4} x^{4} + 10 \, a^{4} c^{4} x^{3} - 10 \, a^{3} c^{4} x^{2} + 5 \, a^{2} c^{4} x - a c^{4}\right )}} \]

[In]

integrate(1/(a*x-1)^2*(a*x+1)^2/(-a*c*x+c)^4,x, algorithm="fricas")

[Out]

-1/15*(5*a^2*x^2 + 5*a*x + 2)/(a^6*c^4*x^5 - 5*a^5*c^4*x^4 + 10*a^4*c^4*x^3 - 10*a^3*c^4*x^2 + 5*a^2*c^4*x - a
*c^4)

Sympy [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.51 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=\frac {- 5 a^{2} x^{2} - 5 a x - 2}{15 a^{6} c^{4} x^{5} - 75 a^{5} c^{4} x^{4} + 150 a^{4} c^{4} x^{3} - 150 a^{3} c^{4} x^{2} + 75 a^{2} c^{4} x - 15 a c^{4}} \]

[In]

integrate(1/(a*x-1)**2*(a*x+1)**2/(-a*c*x+c)**4,x)

[Out]

(-5*a**2*x**2 - 5*a*x - 2)/(15*a**6*c**4*x**5 - 75*a**5*c**4*x**4 + 150*a**4*c**4*x**3 - 150*a**3*c**4*x**2 +
75*a**2*c**4*x - 15*a*c**4)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.45 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=-\frac {5 \, a^{2} x^{2} + 5 \, a x + 2}{15 \, {\left (a^{6} c^{4} x^{5} - 5 \, a^{5} c^{4} x^{4} + 10 \, a^{4} c^{4} x^{3} - 10 \, a^{3} c^{4} x^{2} + 5 \, a^{2} c^{4} x - a c^{4}\right )}} \]

[In]

integrate(1/(a*x-1)^2*(a*x+1)^2/(-a*c*x+c)^4,x, algorithm="maxima")

[Out]

-1/15*(5*a^2*x^2 + 5*a*x + 2)/(a^6*c^4*x^5 - 5*a^5*c^4*x^4 + 10*a^4*c^4*x^3 - 10*a^3*c^4*x^2 + 5*a^2*c^4*x - a
*c^4)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.79 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=-\frac {\frac {5}{{\left (a x - 1\right )}^{3} a} + \frac {15}{{\left (a x - 1\right )}^{4} a} + \frac {12}{{\left (a x - 1\right )}^{5} a}}{15 \, c^{4}} \]

[In]

integrate(1/(a*x-1)^2*(a*x+1)^2/(-a*c*x+c)^4,x, algorithm="giac")

[Out]

-1/15*(5/((a*x - 1)^3*a) + 15/((a*x - 1)^4*a) + 12/((a*x - 1)^5*a))/c^4

Mupad [B] (verification not implemented)

Time = 4.53 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.55 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=-\frac {5\,a^2\,x^2+5\,a\,x+2}{15\,a\,c^4\,{\left (a\,x-1\right )}^5} \]

[In]

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

[Out]

-(5*a*x + 5*a^2*x^2 + 2)/(15*a*c^4*(a*x - 1)^5)