\(\int \frac {e^{2 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx\) [176]

Optimal result

Integrand size = 18, antiderivative size = 37 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=-\frac {1}{2 a c^4 (1-a x)^4}+\frac {1}{3 a c^4 (1-a x)^3} \]

[Out]

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

[In]

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

[Out]

-1/2*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^{2 \text {arctanh}(a x)}}{(c-a c x)^4} \, dx \\ & = -\frac {\int \frac {1+a x}{(1-a x)^5} \, dx}{c^4} \\ & = -\frac {\int \left (-\frac {2}{(-1+a x)^5}-\frac {1}{(-1+a x)^4}\right ) \, dx}{c^4} \\ & = -\frac {1}{2 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 = 23, normalized size of antiderivative = 0.62 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=-\frac {1+2 a x}{6 a c^4 (-1+a x)^4} \]

[In]

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

[Out]

-1/6*(1 + 2*a*x)/(a*c^4*(-1 + a*x)^4)

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.57

method	result	size
risch	\(\frac {-\frac {x}{3}-\frac {1}{6 a}}{c^{4} \left (a x -1\right )^{4}}\)	\(21\)
gosper	\(-\frac {2 a x +1}{6 a \,c^{4} \left (a x -1\right )^{4}}\)	\(22\)
default	\(\frac {-\frac {1}{2 a \left (a x -1\right )^{4}}-\frac {1}{3 a \left (a x -1\right )^{3}}}{c^{4}}\)	\(30\)
parallelrisch	\(\frac {a^{3} x^{4}-4 a^{2} x^{3}+6 a \,x^{2}-6 x}{6 c^{4} \left (a x -1\right )^{4}}\)	\(38\)
norman	\(\frac {\frac {a \,x^{2}}{c}-\frac {x}{c}-\frac {2 a^{2} x^{3}}{3 c}+\frac {a^{3} x^{4}}{6 c}}{\left (a x -1\right )^{4} c^{3}}\)	\(49\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.54 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=-\frac {2 \, a x + 1}{6 \, {\left (a^{5} c^{4} x^{4} - 4 \, a^{4} c^{4} x^{3} + 6 \, a^{3} c^{4} x^{2} - 4 \, a^{2} c^{4} x + a c^{4}\right )}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (29) = 58\).

Time = 0.19 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.62 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=\frac {- 2 a x - 1}{6 a^{5} c^{4} x^{4} - 24 a^{4} c^{4} x^{3} + 36 a^{3} c^{4} x^{2} - 24 a^{2} c^{4} x + 6 a c^{4}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.54 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=-\frac {2 \, a x + 1}{6 \, {\left (a^{5} c^{4} x^{4} - 4 \, a^{4} c^{4} x^{3} + 6 \, a^{3} c^{4} x^{2} - 4 \, a^{2} c^{4} x + a c^{4}\right )}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.57 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=-\frac {2 \, a x + 1}{6 \, {\left (a x - 1\right )}^{4} a c^{4}} \]

[In]

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

[Out]

-1/6*(2*a*x + 1)/((a*x - 1)^4*a*c^4)

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.51 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=-\frac {\frac {x}{3}+\frac {1}{6\,a}}{a^4\,c^4\,x^4-4\,a^3\,c^4\,x^3+6\,a^2\,c^4\,x^2-4\,a\,c^4\,x+c^4} \]

[In]

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

[Out]

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