3.1276 \(\int \frac {e^{-3 \tanh ^{-1}(a x)}}{(c-a^2 c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=46 \[ -\frac {\sqrt {1-a^2 x^2}}{2 a c (a x+1)^2 \sqrt {c-a^2 c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6143, 6140, 32} \[ -\frac {\sqrt {1-a^2 x^2}}{2 a c (a x+1)^2 \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sqrt[1 - a^2*x^2]/(2*a*c*(1 + a*x)^2*Sqrt[c - a^2*c*x^2])

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 6140

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[(1 - a*x)^(p - n/2)*
(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p] || GtQ[c, 0])

Rule 6143

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(c^IntPart[p]*(c + d*x^2)^Frac
Part[p])/(1 - a^2*x^2)^FracPart[p], Int[(1 - a^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a, c, d, n, p}, x
] && EqQ[a^2*c + d, 0] &&  !(IntegerQ[p] || GtQ[c, 0])

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 53, normalized size = 1.15 \[ \frac {\sqrt {1-a^2 x^2} \sqrt {c-a^2 c x^2}}{2 a c^2 (a x-1) (a x+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - a^2*x^2]*Sqrt[c - a^2*c*x^2])/(2*a*c^2*(-1 + a*x)*(1 + a*x)^3)

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fricas [A]  time = 0.56, size = 72, normalized size = 1.57 \[ -\frac {\sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} {\left (a x^{2} + 2 \, x\right )}}{2 \, {\left (a^{4} c^{2} x^{4} + 2 \, a^{3} c^{2} x^{3} - 2 \, a c^{2} x - c^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} {\left (a x + 1\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 38, normalized size = 0.83 \[ -\frac {\left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{2 \left (a x +1\right )^{2} a \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.43, size = 29, normalized size = 0.63 \[ -\frac {1}{2 \, {\left (a^{3} c^{\frac {3}{2}} x^{2} + 2 \, a^{2} c^{\frac {3}{2}} x + a c^{\frac {3}{2}}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.17, size = 58, normalized size = 1.26 \[ -\frac {\sqrt {c-a^2\,c\,x^2}\,\sqrt {1-a^2\,x^2}}{2\,a^5\,c^2\,\left (\frac {2\,x}{a^3}+\frac {1}{a^4}-x^4-\frac {2\,x^3}{a}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}{\left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}} \left (a x + 1\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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