3.1266 \(\int \frac{e^{-3 \tanh ^{-1}(a x)} \sqrt{c-a^2 c x^2}}{x} \, dx\)

Optimal. Leaf size=102 \[ \frac{a x \sqrt{c-a^2 c x^2}}{\sqrt{1-a^2 x^2}}+\frac{\log (x) \sqrt{c-a^2 c x^2}}{\sqrt{1-a^2 x^2}}-\frac{4 \sqrt{c-a^2 c x^2} \log (a x+1)}{\sqrt{1-a^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.196604, antiderivative size = 102, 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 = {6153, 6150, 72} \[ \frac{a x \sqrt{c-a^2 c x^2}}{\sqrt{1-a^2 x^2}}+\frac{\log (x) \sqrt{c-a^2 c x^2}}{\sqrt{1-a^2 x^2}}-\frac{4 \sqrt{c-a^2 c x^2} \log (a x+1)}{\sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6153

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(c^IntPart[p]*(c +
d*x^2)^FracPart[p])/(1 - a^2*x^2)^FracPart[p], Int[x^m*(1 - a^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a,
c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] &&  !(IntegerQ[p] || GtQ[c, 0]) &&  !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 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rubi steps

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

Mathematica [A]  time = 0.0253727, size = 44, normalized size = 0.43 \[ \frac{\sqrt{c-a^2 c x^2} (a x-4 \log (a x+1)+\log (x))}{\sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.088, size = 56, normalized size = 0.6 \begin{align*}{\frac{-ax-\ln \left ( x \right ) +4\,\ln \left ( ax+1 \right ) }{{a}^{2}{x}^{2}-1}\sqrt{-c \left ({a}^{2}{x}^{2}-1 \right ) }\sqrt{-{a}^{2}{x}^{2}+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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