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

Optimal. Leaf size=33 \[ -\frac {2 \sqrt {c-a c x}}{a c^2 \sqrt {1-a^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {6127, 649} \[ -\frac {2 \sqrt {c-a c x}}{a c^2 \sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 649

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)*(a + c*x^2)^(p
 + 1))/(c*(p + 1)), x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p,
 0]

Rule 6127

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

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 35, normalized size = 1.06 \[ -\frac {2 (1-a x)^{3/2}}{a \sqrt {a x+1} (c-a c x)^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [A]  time = 0.48, size = 42, normalized size = 1.27 \[ \frac {2 \, \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}}{a^{3} c^{2} x^{2} - a c^{2}} \]

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*c*x+c)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.20, size = 30, normalized size = 0.91 \[ \frac {{\left (\frac {\sqrt {2}}{a \sqrt {c}} - \frac {2}{\sqrt {a c x + c} a}\right )} {\left | c \right |}}{c^{2}} \]

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*c*x+c)^(3/2),x, algorithm="giac")

[Out]

(sqrt(2)/(a*sqrt(c)) - 2/(sqrt(a*c*x + c)*a))*abs(c)/c^2

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maple [A]  time = 0.02, size = 34, normalized size = 1.03 \[ -\frac {2 \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{\left (-a c x +c \right )^{\frac {3}{2}} \left (a x +1\right )^{2} a} \]

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*c*x+c)^(3/2),x)

[Out]

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

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maxima [A]  time = 0.33, size = 28, normalized size = 0.85 \[ -\frac {2 \, \sqrt {a x + 1} \sqrt {c}}{a^{2} c^{2} x + a c^{2}} \]

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*c*x+c)^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.02, size = 43, normalized size = 1.30 \[ -\frac {2\,\sqrt {1-a^2\,x^2}\,\sqrt {c-a\,c\,x}}{a\,\left (c^2-a^2\,c^2\,x^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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*c*x+c)**(3/2),x)

[Out]

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

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