3.1268 \(\int \frac{e^{-3 \tanh ^{-1}(a x)} \sqrt{c-a^2 c x^2}}{x^3} \, dx\)

Optimal. Leaf size=148 \[ \frac{3 a \sqrt{c-a^2 c x^2}}{x \sqrt{1-a^2 x^2}}-\frac{\sqrt{c-a^2 c x^2}}{2 x^2 \sqrt{1-a^2 x^2}}+\frac{4 a^2 \log (x) \sqrt{c-a^2 c x^2}}{\sqrt{1-a^2 x^2}}-\frac{4 a^2 \sqrt{c-a^2 c x^2} \log (a x+1)}{\sqrt{1-a^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.20577, antiderivative size = 148, 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, 88} \[ \frac{3 a \sqrt{c-a^2 c x^2}}{x \sqrt{1-a^2 x^2}}-\frac{\sqrt{c-a^2 c x^2}}{2 x^2 \sqrt{1-a^2 x^2}}+\frac{4 a^2 \log (x) \sqrt{c-a^2 c x^2}}{\sqrt{1-a^2 x^2}}-\frac{4 a^2 \sqrt{c-a^2 c x^2} \log (a x+1)}{\sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0419929, size = 62, normalized size = 0.42 \[ \frac{\sqrt{c-a^2 c x^2} \left (4 a^2 \log (x)-4 a^2 \log (a x+1)+\frac{3 a}{x}-\frac{1}{2 x^2}\right )}{\sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.094, size = 73, normalized size = 0.5 \begin{align*} -{\frac{8\,{a}^{2}\ln \left ( x \right ){x}^{2}-8\,\ln \left ( ax+1 \right ){a}^{2}{x}^{2}+6\,ax-1}{ \left ( 2\,{a}^{2}{x}^{2}-2 \right ){x}^{2}}\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^3,x)

[Out]

-1/2*(-c*(a^2*x^2-1))^(1/2)*(-a^2*x^2+1)^(1/2)*(8*a^2*ln(x)*x^2-8*ln(a*x+1)*a^2*x^2+6*a*x-1)/(a^2*x^2-1)/x^2

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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^{3}}\,{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^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 3.23728, size = 959, normalized size = 6.48 \begin{align*} \left [\frac{4 \,{\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^{3} x^{3} -{\left (4 \, a^{3} + 6 \, a^{2} + 4 \, a + 1\right )} x^{4} + 6 \, a^{2} x^{2} + 4 \, a x + 1\right )} \sqrt{-a^{2} c x^{2} + c} \sqrt{-a^{2} x^{2} + 1} \sqrt{c} - c}{a^{4} x^{6} + 2 \, a^{3} x^{5} - 2 \, a x^{3} - x^{2}}\right ) + \sqrt{-a^{2} c x^{2} + c} \sqrt{-a^{2} x^{2} + 1}{\left ({\left (6 \, a - 1\right )} x^{2} - 6 \, a x + 1\right )}}{2 \,{\left (a^{2} x^{4} - x^{2}\right )}}, \frac{8 \,{\left (a^{4} x^{4} - a^{2} x^{2}\right )} \sqrt{-c} \arctan \left (-\frac{\sqrt{-a^{2} c x^{2} + c} \sqrt{-a^{2} x^{2} + 1}{\left ({\left (2 \, a^{2} + 2 \, a + 1\right )} x^{2} + 2 \, a x + 1\right )} \sqrt{-c}}{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 ) + \sqrt{-a^{2} c x^{2} + c} \sqrt{-a^{2} x^{2} + 1}{\left ({\left (6 \, a - 1\right )} x^{2} - 6 \, a x + 1\right )}}{2 \,{\left (a^{2} x^{4} - x^{2}\right )}}\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^3,x, algorithm="fricas")

[Out]

[1/2*(4*(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^3*x^3 - (4*a^3 + 6*a^2 + 4*a + 1)*x^4 + 6*a^2*x^
2 + 4*a*x + 1)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*sqrt(c) - c)/(a^4*x^6 + 2*a^3*x^5 - 2*a*x^3 - x^2)) + s
qrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*((6*a - 1)*x^2 - 6*a*x + 1))/(a^2*x^4 - x^2), 1/2*(8*(a^4*x^4 - a^2*x^2
)*sqrt(-c)*arctan(-sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*((2*a^2 + 2*a + 1)*x^2 + 2*a*x + 1)*sqrt(-c)/(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)) + sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)
*((6*a - 1)*x^2 - 6*a*x + 1))/(a^2*x^4 - x^2)]

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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^{3} \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**3,x)

[Out]

Integral((-(a*x - 1)*(a*x + 1))**(3/2)*sqrt(-c*(a*x - 1)*(a*x + 1))/(x**3*(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^{3}}\,{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^3,x, algorithm="giac")

[Out]

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