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

Optimal. Leaf size=45 \[ -\frac {\sqrt {1-a^2 x^2}}{a c^3}-\frac {1}{a c^3 \sqrt {1-a^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {6131, 6128, 266, 43} \[ -\frac {\sqrt {1-a^2 x^2}}{a c^3}-\frac {1}{a c^3 \sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 6128

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

Rule 6131

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

Rubi steps

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

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Mathematica [A]  time = 0.14, size = 30, normalized size = 0.67 \[ \frac {a^2 x^2-2}{a c^3 \sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.37, size = 53, normalized size = 1.18 \[ -\frac {2 \, a^{2} x^{2} + {\left (a^{2} x^{2} - 2\right )} \sqrt {-a^{2} x^{2} + 1} - 2}{a^{3} c^{3} x^{2} - a c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.26, size = 38, normalized size = 0.84 \[ -\frac {\frac {\sqrt {-a^{2} x^{2} + 1}}{c^{3}} + \frac {1}{\sqrt {-a^{2} x^{2} + 1} c^{3}}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 43, normalized size = 0.96 \[ \frac {\left (-a^{2} x^{2}+1\right )^{\frac {3}{2}} \left (a^{2} x^{2}-2\right )}{a \left (a x -1\right )^{2} c^{3} \left (a x +1\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.33, size = 45, normalized size = 1.00 \[ -\frac {{\left (a^{2} x^{2} - 2\right )} \sqrt {a x + 1} \sqrt {-a x + 1}}{a^{3} c^{3} x^{2} - a c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.86, size = 116, normalized size = 2.58 \[ \frac {\sqrt {1-a^2\,x^2}}{2\,c^3\,\left (x\,\sqrt {-a^2}+\frac {\sqrt {-a^2}}{a}\right )\,\sqrt {-a^2}}-\frac {\sqrt {1-a^2\,x^2}}{a\,c^3}-\frac {\sqrt {1-a^2\,x^2}}{2\,c^3\,\left (x\,\sqrt {-a^2}-\frac {\sqrt {-a^2}}{a}\right )\,\sqrt {-a^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 23.32, size = 34, normalized size = 0.76 \[ - \frac {2 \left (\frac {\sqrt {- a^{2} x^{2} + 1}}{2 c^{3}} + \frac {1}{2 c^{3} \sqrt {- a^{2} x^{2} + 1}}\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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