### 3.613 $$\int \frac{e^{-3 \coth ^{-1}(a x)}}{(c-a^2 c x^2)^4} \, dx$$

Optimal. Leaf size=127 $\frac{(2 a x+1) e^{-3 \coth ^{-1}(a x)}}{9 a c^4 \left (1-a^2 x^2\right )^3}-\frac{8 (2 a x+3) e^{-3 \coth ^{-1}(a x)}}{21 a c^4 \left (1-a^2 x^2\right )}+\frac{10 (4 a x+3) e^{-3 \coth ^{-1}(a x)}}{63 a c^4 \left (1-a^2 x^2\right )^2}+\frac{16 e^{-3 \coth ^{-1}(a x)}}{63 a c^4}$

[Out]

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

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Rubi [A]  time = 0.137822, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.091, Rules used = {6185, 6183} $\frac{(2 a x+1) e^{-3 \coth ^{-1}(a x)}}{9 a c^4 \left (1-a^2 x^2\right )^3}-\frac{8 (2 a x+3) e^{-3 \coth ^{-1}(a x)}}{21 a c^4 \left (1-a^2 x^2\right )}+\frac{10 (4 a x+3) e^{-3 \coth ^{-1}(a x)}}{63 a c^4 \left (1-a^2 x^2\right )^2}+\frac{16 e^{-3 \coth ^{-1}(a x)}}{63 a c^4}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/(E^(3*ArcCoth[a*x])*(c - a^2*c*x^2)^4),x]

[Out]

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

Rule 6185

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])

Rule 6183

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]

Rubi steps

\begin{align*} \int \frac{e^{-3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx &=\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]  time = 0.250705, size = 82, normalized size = 0.65 $\frac{x \sqrt{1-\frac{1}{a^2 x^2}} \left (16 a^6 x^6+48 a^5 x^5+24 a^4 x^4-56 a^3 x^3-66 a^2 x^2-6 a x+19\right )}{63 c^4 (a x-1)^2 (a x+1)^5}$

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(E^(3*ArcCoth[a*x])*(c - a^2*c*x^2)^4),x]

[Out]

(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))/(63*c^
4*(-1 + a*x)^2*(1 + a*x)^5)

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Maple [A]  time = 0.046, size = 81, normalized size = 0.6 \begin{align*}{\frac{16\,{x}^{6}{a}^{6}+48\,{x}^{5}{a}^{5}+24\,{x}^{4}{a}^{4}-56\,{x}^{3}{a}^{3}-66\,{a}^{2}{x}^{2}-6\,ax+19}{63\, \left ({a}^{2}{x}^{2}-1 \right ) ^{3}{c}^{4}a} \left ({\frac{ax-1}{ax+1}} \right ) ^{{\frac{3}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/63*((a*x-1)/(a*x+1))^(3/2)*(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

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Maxima [A]  time = 1.05721, size = 184, normalized size = 1.45 \begin{align*} \frac{1}{4032} \, a{\left (\frac{7 \, \left (\frac{a x - 1}{a x + 1}\right )^{\frac{9}{2}} - 54 \, \left (\frac{a x - 1}{a x + 1}\right )^{\frac{7}{2}} + 189 \, \left (\frac{a x - 1}{a x + 1}\right )^{\frac{5}{2}} - 420 \, \left (\frac{a x - 1}{a x + 1}\right )^{\frac{3}{2}} + 945 \, \sqrt{\frac{a x - 1}{a x + 1}}}{a^{2} c^{4}} + \frac{21 \,{\left (\frac{18 \,{\left (a x - 1\right )}}{a x + 1} - 1\right )}}{a^{2} c^{4} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{3}{2}}}\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4032*a*((7*((a*x - 1)/(a*x + 1))^(9/2) - 54*((a*x - 1)/(a*x + 1))^(7/2) + 189*((a*x - 1)/(a*x + 1))^(5/2) -
420*((a*x - 1)/(a*x + 1))^(3/2) + 945*sqrt((a*x - 1)/(a*x + 1)))/(a^2*c^4) + 21*(18*(a*x - 1)/(a*x + 1) - 1)/(
a^2*c^4*((a*x - 1)/(a*x + 1))^(3/2)))

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Fricas [A]  time = 1.63781, size = 278, normalized size = 2.19 \begin{align*} \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} + 2 \, a^{6} c^{4} x^{5} - a^{5} c^{4} x^{4} - 4 \, a^{4} c^{4} x^{3} - a^{3} c^{4} x^{2} + 2 \, a^{2} c^{4} x + a c^{4}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (\frac{a x - 1}{a x + 1}\right )^{\frac{3}{2}}}{{\left (a^{2} c x^{2} - c\right )}^{4}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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