3.1161 \(\int \frac{e^{3 \tanh ^{-1}(a x)} \sqrt{c-a^2 c x^2}}{x^2} \, dx\)

Optimal. Leaf size=108 \[ -\frac{\sqrt{c-a^2 c x^2}}{x \sqrt{1-a^2 x^2}}+\frac{3 a \log (x) \sqrt{c-a^2 c x^2}}{\sqrt{1-a^2 x^2}}-\frac{4 a \sqrt{c-a^2 c x^2} \log (1-a x)}{\sqrt{1-a^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.200332, antiderivative size = 108, 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} \[ -\frac{\sqrt{c-a^2 c x^2}}{x \sqrt{1-a^2 x^2}}+\frac{3 a \log (x) \sqrt{c-a^2 c x^2}}{\sqrt{1-a^2 x^2}}-\frac{4 a \sqrt{c-a^2 c x^2} \log (1-a x)}{\sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0336291, size = 51, normalized size = 0.47 \[ \frac{\sqrt{c-a^2 c x^2} \left (3 a \log (x)-4 a \log (1-a x)-\frac{1}{x}\right )}{\sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.089, size = 60, normalized size = 0.6 \begin{align*}{\frac{-3\,a\ln \left ( x \right ) x+4\,\ln \left ( ax-1 \right ) xa+1}{x \left ({a}^{2}{x}^{2}-1 \right ) }\sqrt{-{a}^{2}{x}^{2}+1}\sqrt{-c \left ({a}^{2}{x}^{2}-1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-3*a*ln(x)*x+4*ln(a*x-1)*x*a+1)*(-a^2*x^2+1)^(1/2)*(-c*(a^2*x^2-1))^(1/2)/(a^2*x^2-1)/x

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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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^{4} - 2 \, a x^{3} + x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*(a*x + 1)/(a^2*x^4 - 2*a*x^3 + x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(-c*(a*x - 1)*(a*x + 1))*(a*x + 1)**3/(x**2*(-(a*x - 1)*(a*x + 1))**(3/2)), 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 x + 1\right )}^{3}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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