\(\int \frac {e^{-3 \coth ^{-1}(a x)}}{(c-a^2 c x^2)^3} \, dx\) [612]

Optimal result

Integrand size = 22, antiderivative size = 91 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {8 e^{-3 \coth ^{-1}(a x)}}{35 a c^3}+\frac {e^{-3 \coth ^{-1}(a x)} (3+4 a x)}{7 a c^3 \left (1-a^2 x^2\right )^2}-\frac {12 e^{-3 \coth ^{-1}(a x)} (3+2 a x)}{35 a c^3 \left (1-a^2 x^2\right )} \]

[Out]

8/35/a/c^3*((a*x-1)/(a*x+1))^(3/2)+1/7*(4*a*x+3)/a/c^3*((a*x-1)/(a*x+1))^(3/2)/(-a^2*x^2+1)^2-12/35*(2*a*x+3)/
a/c^3*((a*x-1)/(a*x+1))^(3/2)/(-a^2*x^2+1)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {6320, 6318} \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=-\frac {12 (2 a x+3) e^{-3 \coth ^{-1}(a x)}}{35 a c^3 \left (1-a^2 x^2\right )}+\frac {(4 a x+3) e^{-3 \coth ^{-1}(a x)}}{7 a c^3 \left (1-a^2 x^2\right )^2}+\frac {8 e^{-3 \coth ^{-1}(a x)}}{35 a c^3} \]

[In]

Int[1/(E^(3*ArcCoth[a*x])*(c - a^2*c*x^2)^3),x]

[Out]

8/(35*a*c^3*E^(3*ArcCoth[a*x])) + (3 + 4*a*x)/(7*a*c^3*E^(3*ArcCoth[a*x])*(1 - a^2*x^2)^2) - (12*(3 + 2*a*x))/
(35*a*c^3*E^(3*ArcCoth[a*x])*(1 - a^2*x^2))

Rule 6318

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))/((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[E^(n*ArcCoth[a*x])/(a*c*n), x] /; F
reeQ[{a, c, d, n}, x] && EqQ[a^2*c + d, 0] &&  !IntegerQ[n/2]

Rule 6320

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(n + 2*a*(p + 1)*x)*(c + d*x^2
)^(p + 1)*(E^(n*ArcCoth[a*x])/(a*c*(n^2 - 4*(p + 1)^2))), x] - Dist[2*(p + 1)*((2*p + 3)/(c*(n^2 - 4*(p + 1)^2
))), Int[(c + d*x^2)^(p + 1)*E^(n*ArcCoth[a*x]), x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[a^2*c + d, 0] &&  !In
tegerQ[n/2] && LtQ[p, -1] && NeQ[p, -3/2] && NeQ[n^2 - 4*(p + 1)^2, 0] && (IntegerQ[p] ||  !IntegerQ[n])

Rubi steps \begin{align*} \text {integral}& = \frac {e^{-3 \coth ^{-1}(a x)} (3+4 a x)}{7 a c^3 \left (1-a^2 x^2\right )^2}+\frac {12 \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx}{7 c} \\ & = \frac {e^{-3 \coth ^{-1}(a x)} (3+4 a x)}{7 a c^3 \left (1-a^2 x^2\right )^2}-\frac {12 e^{-3 \coth ^{-1}(a x)} (3+2 a x)}{35 a c^3 \left (1-a^2 x^2\right )}-\frac {24 \int \frac {e^{-3 \coth ^{-1}(a x)}}{c-a^2 c x^2} \, dx}{35 c^2} \\ & = \frac {8 e^{-3 \coth ^{-1}(a x)}}{35 a c^3}+\frac {e^{-3 \coth ^{-1}(a x)} (3+4 a x)}{7 a c^3 \left (1-a^2 x^2\right )^2}-\frac {12 e^{-3 \coth ^{-1}(a x)} (3+2 a x)}{35 a c^3 \left (1-a^2 x^2\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.73 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {\sqrt {1-\frac {1}{a^2 x^2}} x \left (-13-4 a x+20 a^2 x^2+24 a^3 x^3+8 a^4 x^4\right )}{35 c^3 (-1+a x) (1+a x)^4} \]

[In]

Integrate[1/(E^(3*ArcCoth[a*x])*(c - a^2*c*x^2)^3),x]

[Out]

(Sqrt[1 - 1/(a^2*x^2)]*x*(-13 - 4*a*x + 20*a^2*x^2 + 24*a^3*x^3 + 8*a^4*x^4))/(35*c^3*(-1 + a*x)*(1 + a*x)^4)

Maple [A] (verified)

Time = 0.51 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.71

method	result	size
gosper	\(\frac {\left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} \left (8 a^{4} x^{4}+24 a^{3} x^{3}+20 a^{2} x^{2}-4 a x -13\right )}{35 \left (a^{2} x^{2}-1\right )^{2} c^{3} a}\)	\(65\)
default	\(\frac {\left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} \left (8 a^{4} x^{4}+24 a^{3} x^{3}+20 a^{2} x^{2}-4 a x -13\right )}{35 \left (a x -1\right )^{2} c^{3} a \left (a x +1\right )^{2}}\)	\(68\)
trager	\(\frac {\left (8 a^{4} x^{4}+24 a^{3} x^{3}+20 a^{2} x^{2}-4 a x -13\right ) \sqrt {-\frac {-a x +1}{a x +1}}}{35 a \,c^{3} \left (a x -1\right ) \left (a x +1\right )^{3}}\)	\(70\)

[In]

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

[Out]

1/35*((a*x-1)/(a*x+1))^(3/2)*(8*a^4*x^4+24*a^3*x^3+20*a^2*x^2-4*a*x-13)/(a^2*x^2-1)^2/c^3/a

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.95 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {{\left (8 \, a^{4} x^{4} + 24 \, a^{3} x^{3} + 20 \, a^{2} x^{2} - 4 \, a x - 13\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{35 \, {\left (a^{5} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{3} - 2 \, a^{2} c^{3} x - a c^{3}\right )}} \]

[In]

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

[Out]

1/35*(8*a^4*x^4 + 24*a^3*x^3 + 20*a^2*x^2 - 4*a*x - 13)*sqrt((a*x - 1)/(a*x + 1))/(a^5*c^3*x^4 + 2*a^4*c^3*x^3
- 2*a^2*c^3*x - a*c^3)

Sympy [F]

\[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=- \frac {\int \left (- \frac {\sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a^{7} x^{7} + a^{6} x^{6} - 3 a^{5} x^{5} - 3 a^{4} x^{4} + 3 a^{3} x^{3} + 3 a^{2} x^{2} - a x - 1}\right )\, dx + \int \frac {a x \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a^{7} x^{7} + a^{6} x^{6} - 3 a^{5} x^{5} - 3 a^{4} x^{4} + 3 a^{3} x^{3} + 3 a^{2} x^{2} - a x - 1}\, dx}{c^{3}} \]

[In]

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

[Out]

-(Integral(-sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/(a**7*x**7 + a**6*x**6 - 3*a**5*x**5 - 3*a**4*x**4 + 3*a**3*x**3
+ 3*a**2*x**2 - a*x - 1), x) + Integral(a*x*sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/(a**7*x**7 + a**6*x**6 - 3*a**5
*x**5 - 3*a**4*x**4 + 3*a**3*x**3 + 3*a**2*x**2 - a*x - 1), x))/c**3

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.13 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=-\frac {1}{560} \, a {\left (\frac {5 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {7}{2}} - 28 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{2}} + 70 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} - 140 \, \sqrt {\frac {a x - 1}{a x + 1}}}{a^{2} c^{3}} - \frac {35}{a^{2} c^{3} \sqrt {\frac {a x - 1}{a x + 1}}}\right )} \]

[In]

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

[Out]

-1/560*a*((5*((a*x - 1)/(a*x + 1))^(7/2) - 28*((a*x - 1)/(a*x + 1))^(5/2) + 70*((a*x - 1)/(a*x + 1))^(3/2) - 1
40*sqrt((a*x - 1)/(a*x + 1)))/(a^2*c^3) - 35/(a^2*c^3*sqrt((a*x - 1)/(a*x + 1))))

Giac [F]

\[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\int { -\frac {\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}{{\left (a^{2} c x^{2} - c\right )}^{3}} \,d x } \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 3.85 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.27 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {1}{16\,a\,c^3\,\sqrt {\frac {a\,x-1}{a\,x+1}}}+\frac {\sqrt {\frac {a\,x-1}{a\,x+1}}}{4\,a\,c^3}-\frac {{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}}{8\,a\,c^3}+\frac {{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/2}}{20\,a\,c^3}-\frac {{\left (\frac {a\,x-1}{a\,x+1}\right )}^{7/2}}{112\,a\,c^3} \]

[In]

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

[Out]

1/(16*a*c^3*((a*x - 1)/(a*x + 1))^(1/2)) + ((a*x - 1)/(a*x + 1))^(1/2)/(4*a*c^3) - ((a*x - 1)/(a*x + 1))^(3/2)
/(8*a*c^3) + ((a*x - 1)/(a*x + 1))^(5/2)/(20*a*c^3) - ((a*x - 1)/(a*x + 1))^(7/2)/(112*a*c^3)