\(\int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a^2 c x^2)^4} \, dx\) [580]

Optimal result

Integrand size = 22, antiderivative size = 127 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=-\frac {16 e^{3 \coth ^{-1}(a x)}}{63 a c^4}-\frac {e^{3 \coth ^{-1}(a x)} (1-2 a x)}{9 a c^4 \left (1-a^2 x^2\right )^3}-\frac {10 e^{3 \coth ^{-1}(a x)} (3-4 a x)}{63 a c^4 \left (1-a^2 x^2\right )^2}+\frac {8 e^{3 \coth ^{-1}(a x)} (3-2 a x)}{21 a c^4 \left (1-a^2 x^2\right )} \]

[Out]

-16/63/((a*x-1)/(a*x+1))^(3/2)/a/c^4-1/9/((a*x-1)/(a*x+1))^(3/2)*(-2*a*x+1)/a/c^4/(-a^2*x^2+1)^3-10/63/((a*x-1
)/(a*x+1))^(3/2)*(-4*a*x+3)/a/c^4/(-a^2*x^2+1)^2+8/21/((a*x-1)/(a*x+1))^(3/2)*(-2*a*x+3)/a/c^4/(-a^2*x^2+1)

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {6320, 6318} \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=-\frac {10 (3-4 a x) e^{3 \coth ^{-1}(a x)}}{63 a c^4 \left (1-a^2 x^2\right )^2}+\frac {8 (3-2 a x) e^{3 \coth ^{-1}(a x)}}{21 a c^4 \left (1-a^2 x^2\right )}-\frac {(1-2 a x) e^{3 \coth ^{-1}(a x)}}{9 a c^4 \left (1-a^2 x^2\right )^3}-\frac {16 e^{3 \coth ^{-1}(a x)}}{63 a c^4} \]

[In]

Int[E^(3*ArcCoth[a*x])/(c - a^2*c*x^2)^4,x]

[Out]

(-16*E^(3*ArcCoth[a*x]))/(63*a*c^4) - (E^(3*ArcCoth[a*x])*(1 - 2*a*x))/(9*a*c^4*(1 - a^2*x^2)^3) - (10*E^(3*Ar
cCoth[a*x])*(3 - 4*a*x))/(63*a*c^4*(1 - a^2*x^2)^2) + (8*E^(3*ArcCoth[a*x])*(3 - 2*a*x))/(21*a*c^4*(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)} (1-2 a x)}{9 a c^4 \left (1-a^2 x^2\right )^3}+\frac {10 \int \frac {e^{3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx}{9 c} \\ & = -\frac {e^{3 \coth ^{-1}(a x)} (1-2 a x)}{9 a c^4 \left (1-a^2 x^2\right )^3}-\frac {10 e^{3 \coth ^{-1}(a x)} (3-4 a x)}{63 a c^4 \left (1-a^2 x^2\right )^2}+\frac {40 \int \frac {e^{3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx}{21 c^2} \\ & = -\frac {e^{3 \coth ^{-1}(a x)} (1-2 a x)}{9 a c^4 \left (1-a^2 x^2\right )^3}-\frac {10 e^{3 \coth ^{-1}(a x)} (3-4 a x)}{63 a c^4 \left (1-a^2 x^2\right )^2}+\frac {8 e^{3 \coth ^{-1}(a x)} (3-2 a x)}{21 a c^4 \left (1-a^2 x^2\right )}-\frac {16 \int \frac {e^{3 \coth ^{-1}(a x)}}{c-a^2 c x^2} \, dx}{21 c^3} \\ & = -\frac {16 e^{3 \coth ^{-1}(a x)}}{63 a c^4}-\frac {e^{3 \coth ^{-1}(a x)} (1-2 a x)}{9 a c^4 \left (1-a^2 x^2\right )^3}-\frac {10 e^{3 \coth ^{-1}(a x)} (3-4 a x)}{63 a c^4 \left (1-a^2 x^2\right )^2}+\frac {8 e^{3 \coth ^{-1}(a x)} (3-2 a x)}{21 a c^4 \left (1-a^2 x^2\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.49 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.65 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=-\frac {\sqrt {1-\frac {1}{a^2 x^2}} x \left (19+6 a x-66 a^2 x^2+56 a^3 x^3+24 a^4 x^4-48 a^5 x^5+16 a^6 x^6\right )}{63 c^4 (-1+a x)^5 (1+a x)^2} \]

[In]

Integrate[E^(3*ArcCoth[a*x])/(c - a^2*c*x^2)^4,x]

[Out]

-1/63*(Sqrt[1 - 1/(a^2*x^2)]*x*(19 + 6*a*x - 66*a^2*x^2 + 56*a^3*x^3 + 24*a^4*x^4 - 48*a^5*x^5 + 16*a^6*x^6))/
(c^4*(-1 + a*x)^5*(1 + a*x)^2)

Maple [A] (verified)

Time = 0.49 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.64

method	result	size
gosper	\(-\frac {16 a^{6} x^{6}-48 a^{5} x^{5}+24 a^{4} x^{4}+56 a^{3} x^{3}-66 a^{2} x^{2}+6 a x +19}{63 \left (a^{2} x^{2}-1\right )^{3} c^{4} \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} a}\)	\(81\)
default	\(-\frac {16 a^{6} x^{6}-48 a^{5} x^{5}+24 a^{4} x^{4}+56 a^{3} x^{3}-66 a^{2} x^{2}+6 a x +19}{63 \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} \left (a x +1\right )^{3} c^{4} \left (a x -1\right )^{3} a}\)	\(84\)
trager	\(-\frac {\left (16 a^{6} x^{6}-48 a^{5} x^{5}+24 a^{4} x^{4}+56 a^{3} x^{3}-66 a^{2} x^{2}+6 a x +19\right ) \sqrt {-\frac {-a x +1}{a x +1}}}{63 a \,c^{4} \left (a x +1\right ) \left (a x -1\right )^{5}}\)	\(86\)

[In]

int(1/((a*x-1)/(a*x+1))^(3/2)/(-a^2*c*x^2+c)^4,x,method=_RETURNVERBOSE)

[Out]

-1/63*(16*a^6*x^6-48*a^5*x^5+24*a^4*x^4+56*a^3*x^3-66*a^2*x^2+6*a*x+19)/(a^2*x^2-1)^3/c^4/((a*x-1)/(a*x+1))^(3
/2)/a

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.98 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=-\frac {{\left (16 \, a^{6} x^{6} - 48 \, a^{5} x^{5} + 24 \, a^{4} x^{4} + 56 \, a^{3} x^{3} - 66 \, a^{2} x^{2} + 6 \, a x + 19\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{63 \, {\left (a^{7} c^{4} x^{6} - 4 \, a^{6} c^{4} x^{5} + 5 \, a^{5} c^{4} x^{4} - 5 \, a^{3} c^{4} x^{2} + 4 \, a^{2} c^{4} x - a c^{4}\right )}} \]

[In]

integrate(1/((a*x-1)/(a*x+1))^(3/2)/(-a^2*c*x^2+c)^4,x, algorithm="fricas")

[Out]

-1/63*(16*a^6*x^6 - 48*a^5*x^5 + 24*a^4*x^4 + 56*a^3*x^3 - 66*a^2*x^2 + 6*a*x + 19)*sqrt((a*x - 1)/(a*x + 1))/
(a^7*c^4*x^6 - 4*a^6*c^4*x^5 + 5*a^5*c^4*x^4 - 5*a^3*c^4*x^2 + 4*a^2*c^4*x - a*c^4)

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\text {Timed out} \]

[In]

integrate(1/((a*x-1)/(a*x+1))**(3/2)/(-a**2*c*x**2+c)**4,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.03 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\frac {1}{4032} \, a {\left (\frac {21 \, {\left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} - 18 \, \sqrt {\frac {a x - 1}{a x + 1}}\right )}}{a^{2} c^{4}} + \frac {\frac {54 \, {\left (a x - 1\right )}}{a x + 1} - \frac {189 \, {\left (a x - 1\right )}^{2}}{{\left (a x + 1\right )}^{2}} + \frac {420 \, {\left (a x - 1\right )}^{3}}{{\left (a x + 1\right )}^{3}} - \frac {945 \, {\left (a x - 1\right )}^{4}}{{\left (a x + 1\right )}^{4}} - 7}{a^{2} c^{4} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {9}{2}}}\right )} \]

[In]

integrate(1/((a*x-1)/(a*x+1))^(3/2)/(-a^2*c*x^2+c)^4,x, algorithm="maxima")

[Out]

1/4032*a*(21*(((a*x - 1)/(a*x + 1))^(3/2) - 18*sqrt((a*x - 1)/(a*x + 1)))/(a^2*c^4) + (54*(a*x - 1)/(a*x + 1)
- 189*(a*x - 1)^2/(a*x + 1)^2 + 420*(a*x - 1)^3/(a*x + 1)^3 - 945*(a*x - 1)^4/(a*x + 1)^4 - 7)/(a^2*c^4*((a*x
- 1)/(a*x + 1))^(9/2)))

Giac [F]

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

[In]

integrate(1/((a*x-1)/(a*x+1))^(3/2)/(-a^2*c*x^2+c)^4,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 3.98 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.60 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=-\frac {16\,a^6\,x^6-48\,a^5\,x^5+24\,a^4\,x^4+56\,a^3\,x^3-66\,a^2\,x^2+6\,a\,x+19}{63\,a\,c^4\,{\left (a\,x+1\right )}^6\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{9/2}} \]

[In]

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

[Out]

-(6*a*x - 66*a^2*x^2 + 56*a^3*x^3 + 24*a^4*x^4 - 48*a^5*x^5 + 16*a^6*x^6 + 19)/(63*a*c^4*(a*x + 1)^6*((a*x - 1
)/(a*x + 1))^(9/2))