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

Optimal. Leaf size=186 \[ -\frac{4 a^2 \sqrt{c-\frac{c}{a^2 x^2}}}{\sqrt{1-a^2 x^2}}+\frac{3 a \sqrt{c-\frac{c}{a^2 x^2}}}{2 x \sqrt{1-a^2 x^2}}-\frac{\sqrt{c-\frac{c}{a^2 x^2}}}{3 x^2 \sqrt{1-a^2 x^2}}-\frac{4 a^3 x \log (x) \sqrt{c-\frac{c}{a^2 x^2}}}{\sqrt{1-a^2 x^2}}+\frac{4 a^3 x \sqrt{c-\frac{c}{a^2 x^2}} \log (a x+1)}{\sqrt{1-a^2 x^2}} \]

[Out]

(-4*a^2*Sqrt[c - c/(a^2*x^2)])/Sqrt[1 - a^2*x^2] - Sqrt[c - c/(a^2*x^2)]/(3*x^2*Sqrt[1 - a^2*x^2]) + (3*a*Sqrt
[c - c/(a^2*x^2)])/(2*x*Sqrt[1 - a^2*x^2]) - (4*a^3*Sqrt[c - c/(a^2*x^2)]*x*Log[x])/Sqrt[1 - a^2*x^2] + (4*a^3
*Sqrt[c - c/(a^2*x^2)]*x*Log[1 + a*x])/Sqrt[1 - a^2*x^2]

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Rubi [A]  time = 0.281378, antiderivative size = 186, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {6160, 6150, 88} \[ -\frac{4 a^2 \sqrt{c-\frac{c}{a^2 x^2}}}{\sqrt{1-a^2 x^2}}+\frac{3 a \sqrt{c-\frac{c}{a^2 x^2}}}{2 x \sqrt{1-a^2 x^2}}-\frac{\sqrt{c-\frac{c}{a^2 x^2}}}{3 x^2 \sqrt{1-a^2 x^2}}-\frac{4 a^3 x \log (x) \sqrt{c-\frac{c}{a^2 x^2}}}{\sqrt{1-a^2 x^2}}+\frac{4 a^3 x \sqrt{c-\frac{c}{a^2 x^2}} \log (a x+1)}{\sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*a^2*Sqrt[c - c/(a^2*x^2)])/Sqrt[1 - a^2*x^2] - Sqrt[c - c/(a^2*x^2)]/(3*x^2*Sqrt[1 - a^2*x^2]) + (3*a*Sqrt
[c - c/(a^2*x^2)])/(2*x*Sqrt[1 - a^2*x^2]) - (4*a^3*Sqrt[c - c/(a^2*x^2)]*x*Log[x])/Sqrt[1 - a^2*x^2] + (4*a^3
*Sqrt[c - c/(a^2*x^2)]*x*Log[1 + a*x])/Sqrt[1 - a^2*x^2]

Rule 6160

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

Rule 6150

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

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin{align*} \int \frac{e^{-3 \tanh ^{-1}(a x)} \sqrt{c-\frac{c}{a^2 x^2}}}{x^3} \, dx &=\frac{\left (\sqrt{c-\frac{c}{a^2 x^2}} x\right ) \int \frac{e^{-3 \tanh ^{-1}(a x)} \sqrt{1-a^2 x^2}}{x^4} \, dx}{\sqrt{1-a^2 x^2}}\\ &=\frac{\left (\sqrt{c-\frac{c}{a^2 x^2}} x\right ) \int \frac{(1-a x)^2}{x^4 (1+a x)} \, dx}{\sqrt{1-a^2 x^2}}\\ &=\frac{\left (\sqrt{c-\frac{c}{a^2 x^2}} x\right ) \int \left (\frac{1}{x^4}-\frac{3 a}{x^3}+\frac{4 a^2}{x^2}-\frac{4 a^3}{x}+\frac{4 a^4}{1+a x}\right ) \, dx}{\sqrt{1-a^2 x^2}}\\ &=-\frac{4 a^2 \sqrt{c-\frac{c}{a^2 x^2}}}{\sqrt{1-a^2 x^2}}-\frac{\sqrt{c-\frac{c}{a^2 x^2}}}{3 x^2 \sqrt{1-a^2 x^2}}+\frac{3 a \sqrt{c-\frac{c}{a^2 x^2}}}{2 x \sqrt{1-a^2 x^2}}-\frac{4 a^3 \sqrt{c-\frac{c}{a^2 x^2}} x \log (x)}{\sqrt{1-a^2 x^2}}+\frac{4 a^3 \sqrt{c-\frac{c}{a^2 x^2}} x \log (1+a x)}{\sqrt{1-a^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0444212, size = 73, normalized size = 0.39 \[ \frac{x \sqrt{c-\frac{c}{a^2 x^2}} \left (-\frac{4 a^2}{x}-4 a^3 \log (x)+4 a^3 \log (a x+1)+\frac{3 a}{2 x^2}-\frac{1}{3 x^3}\right )}{\sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c - c/(a^2*x^2)]*x*(-1/(3*x^3) + (3*a)/(2*x^2) - (4*a^2)/x - 4*a^3*Log[x] + 4*a^3*Log[1 + a*x]))/Sqrt[1
- a^2*x^2]

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Maple [A]  time = 0.15, size = 86, normalized size = 0.5 \begin{align*}{\frac{24\,{a}^{3}\ln \left ( x \right ){x}^{3}-24\,{a}^{3}{x}^{3}\ln \left ( ax+1 \right ) +24\,{a}^{2}{x}^{2}-9\,ax+2}{6\,{x}^{2} \left ({a}^{2}{x}^{2}-1 \right ) }\sqrt{{\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}{x}^{2}}}}\sqrt{-{a}^{2}{x}^{2}+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(c*(a^2*x^2-1)/a^2/x^2)^(1/2)/x^2*(-a^2*x^2+1)^(1/2)*(24*a^3*ln(x)*x^3-24*a^3*x^3*ln(a*x+1)+24*a^2*x^2-9*a
*x+2)/(a^2*x^2-1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.68808, size = 1102, normalized size = 5.92 \begin{align*} \left [\frac{12 \,{\left (a^{4} x^{4} - a^{2} x^{2}\right )} \sqrt{-c} \log \left (\frac{4 \, a^{5} c x^{5} +{\left (2 \, a^{6} + 4 \, a^{5} + 6 \, a^{4} + 4 \, a^{3} + a^{2}\right )} c x^{6} +{\left (4 \, a^{4} - 4 \, a^{3} - 6 \, a^{2} - 4 \, a - 1\right )} c x^{4} - 5 \, a^{2} c x^{2} - 4 \, a c x +{\left (4 \, a^{4} x^{4} + 6 \, a^{3} x^{3} -{\left (4 \, a^{4} + 6 \, a^{3} + 4 \, a^{2} + a\right )} x^{5} + 4 \, a^{2} x^{2} + a x\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{-c} \sqrt{\frac{a^{2} c x^{2} - c}{a^{2} x^{2}}} - c}{a^{4} x^{6} + 2 \, a^{3} x^{5} - 2 \, a x^{3} - x^{2}}\right ) +{\left (24 \, a^{2} x^{2} -{\left (24 \, a^{2} - 9 \, a + 2\right )} x^{3} - 9 \, a x + 2\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{\frac{a^{2} c x^{2} - c}{a^{2} x^{2}}}}{6 \,{\left (a^{2} x^{4} - x^{2}\right )}}, \frac{24 \,{\left (a^{4} x^{4} - a^{2} x^{2}\right )} \sqrt{c} \arctan \left (-\frac{{\left (2 \, a^{2} x^{2} +{\left (2 \, a^{3} + 2 \, a^{2} + a\right )} x^{3} + a x\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{c} \sqrt{\frac{a^{2} c x^{2} - c}{a^{2} x^{2}}}}{2 \, a^{3} c x^{3} -{\left (2 \, a^{3} + a^{2}\right )} c x^{4} +{\left (a^{2} + 2 \, a + 1\right )} c x^{2} - 2 \, a c x - c}\right ) +{\left (24 \, a^{2} x^{2} -{\left (24 \, a^{2} - 9 \, a + 2\right )} x^{3} - 9 \, a x + 2\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{\frac{a^{2} c x^{2} - c}{a^{2} x^{2}}}}{6 \,{\left (a^{2} x^{4} - x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(12*(a^4*x^4 - a^2*x^2)*sqrt(-c)*log((4*a^5*c*x^5 + (2*a^6 + 4*a^5 + 6*a^4 + 4*a^3 + a^2)*c*x^6 + (4*a^4
- 4*a^3 - 6*a^2 - 4*a - 1)*c*x^4 - 5*a^2*c*x^2 - 4*a*c*x + (4*a^4*x^4 + 6*a^3*x^3 - (4*a^4 + 6*a^3 + 4*a^2 + a
)*x^5 + 4*a^2*x^2 + a*x)*sqrt(-a^2*x^2 + 1)*sqrt(-c)*sqrt((a^2*c*x^2 - c)/(a^2*x^2)) - c)/(a^4*x^6 + 2*a^3*x^5
- 2*a*x^3 - x^2)) + (24*a^2*x^2 - (24*a^2 - 9*a + 2)*x^3 - 9*a*x + 2)*sqrt(-a^2*x^2 + 1)*sqrt((a^2*c*x^2 - c)
/(a^2*x^2)))/(a^2*x^4 - x^2), 1/6*(24*(a^4*x^4 - a^2*x^2)*sqrt(c)*arctan(-(2*a^2*x^2 + (2*a^3 + 2*a^2 + a)*x^3
+ a*x)*sqrt(-a^2*x^2 + 1)*sqrt(c)*sqrt((a^2*c*x^2 - c)/(a^2*x^2))/(2*a^3*c*x^3 - (2*a^3 + a^2)*c*x^4 + (a^2 +
2*a + 1)*c*x^2 - 2*a*c*x - c)) + (24*a^2*x^2 - (24*a^2 - 9*a + 2)*x^3 - 9*a*x + 2)*sqrt(-a^2*x^2 + 1)*sqrt((a
^2*c*x^2 - c)/(a^2*x^2)))/(a^2*x^4 - x^2)]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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