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

Optimal. Leaf size=91 \[ -\frac {(3-4 a x) e^{3 \coth ^{-1}(a x)}}{7 a c^3 \left (1-a^2 x^2\right )^2}+\frac {12 (3-2 a x) e^{3 \coth ^{-1}(a x)}}{35 a c^3 \left (1-a^2 x^2\right )}-\frac {8 e^{3 \coth ^{-1}(a x)}}{35 a c^3} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {6185, 6183} \[ -\frac {(3-4 a x) e^{3 \coth ^{-1}(a x)}}{7 a c^3 \left (1-a^2 x^2\right )^2}+\frac {12 (3-2 a x) e^{3 \coth ^{-1}(a x)}}{35 a c^3 \left (1-a^2 x^2\right )}-\frac {8 e^{3 \coth ^{-1}(a x)}}{35 a c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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])

Rubi steps

\begin {align*} \int \frac {e^{3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx &=-\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*}

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Mathematica [A]  time = 0.22, size = 66, normalized size = 0.73 \[ -\frac {x \sqrt {1-\frac {1}{a^2 x^2}} \left (8 a^4 x^4-24 a^3 x^3+20 a^2 x^2+4 a x-13\right )}{35 c^3 (a x-1)^4 (a x+1)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/35*(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))/(c^3*(-1 + a*x)^4*(1 + a*x
))

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fricas [A]  time = 0.47, size = 96, normalized size = 1.05 \[ -\frac {{\left (8 \, a^{4} x^{4} - 24 \, a^{3} x^{3} + 20 \, a^{2} x^{2} + 4 \, a x - 13\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{35 \, {\left (a^{5} c^{3} x^{4} - 4 \, a^{4} c^{3} x^{3} + 6 \, a^{3} c^{3} x^{2} - 4 \, a^{2} c^{3} x + a c^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [A]  time = 0.18, size = 104, normalized size = 1.14 \[ -\frac {\frac {{\left (a x + 1\right )}^{3} {\left (\frac {28 \, {\left (a x - 1\right )}}{a x + 1} - \frac {70 \, {\left (a x - 1\right )}^{2}}{{\left (a x + 1\right )}^{2}} + \frac {140 \, {\left (a x - 1\right )}^{3}}{{\left (a x + 1\right )}^{3}} - 5\right )}}{{\left (a x - 1\right )}^{3} \sqrt {\frac {a x - 1}{a x + 1}}} + 35 \, \sqrt {\frac {a x - 1}{a x + 1}}}{560 \, a c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/560*((a*x + 1)^3*(28*(a*x - 1)/(a*x + 1) - 70*(a*x - 1)^2/(a*x + 1)^2 + 140*(a*x - 1)^3/(a*x + 1)^3 - 5)/((
a*x - 1)^3*sqrt((a*x - 1)/(a*x + 1))) + 35*sqrt((a*x - 1)/(a*x + 1)))/(a*c^3)

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maple [A]  time = 0.04, size = 65, normalized size = 0.71 \[ -\frac {8 x^{4} a^{4}-24 x^{3} a^{3}+20 a^{2} x^{2}+4 a x -13}{35 \left (a^{2} x^{2}-1\right )^{2} c^{3} \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/35*(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*x-1)/(a*x+1))^(3/2)/a

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maxima [A]  time = 0.30, size = 97, normalized size = 1.07 \[ -\frac {1}{560} \, a {\left (\frac {35 \, \sqrt {\frac {a x - 1}{a x + 1}}}{a^{2} c^{3}} + \frac {\frac {28 \, {\left (a x - 1\right )}}{a x + 1} - \frac {70 \, {\left (a x - 1\right )}^{2}}{{\left (a x + 1\right )}^{2}} + \frac {140 \, {\left (a x - 1\right )}^{3}}{{\left (a x + 1\right )}^{3}} - 5}{a^{2} c^{3} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {7}{2}}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/560*a*(35*sqrt((a*x - 1)/(a*x + 1))/(a^2*c^3) + (28*(a*x - 1)/(a*x + 1) - 70*(a*x - 1)^2/(a*x + 1)^2 + 140*
(a*x - 1)^3/(a*x + 1)^3 - 5)/(a^2*c^3*((a*x - 1)/(a*x + 1))^(7/2)))

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mupad [B]  time = 1.26, size = 60, normalized size = 0.66 \[ -\frac {8\,a^4\,x^4-24\,a^3\,x^3+20\,a^2\,x^2+4\,a\,x-13}{35\,a\,c^3\,{\left (a\,x+1\right )}^4\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(4*a*x + 20*a^2*x^2 - 24*a^3*x^3 + 8*a^4*x^4 - 13)/(35*a*c^3*(a*x + 1)^4*((a*x - 1)/(a*x + 1))^(7/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {1}{\frac {a^{7} x^{7} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1} - \frac {a^{6} x^{6} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1} - \frac {3 a^{5} x^{5} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1} + \frac {3 a^{4} x^{4} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1} + \frac {3 a^{3} x^{3} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1} - \frac {3 a^{2} x^{2} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1} - \frac {a x \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1} + \frac {\sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1}}\, dx}{c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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