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

Optimal. Leaf size=41 \[ -\frac{2 (a x+1)^{3/2} \sqrt{c-\frac{c}{a x}}}{3 x \sqrt{1-a x}} \]

[Out]

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

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Rubi [A]  time = 0.226816, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {6134, 6128, 848, 37} \[ -\frac{2 (a x+1)^{3/2} \sqrt{c-\frac{c}{a x}}}{3 x \sqrt{1-a x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6134

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

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 848

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

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

\begin{align*} \int \frac{e^{\tanh ^{-1}(a x)} \sqrt{c-\frac{c}{a x}}}{x^2} \, dx &=\frac{\left (\sqrt{c-\frac{c}{a x}} \sqrt{x}\right ) \int \frac{e^{\tanh ^{-1}(a x)} \sqrt{1-a x}}{x^{5/2}} \, dx}{\sqrt{1-a x}}\\ &=\frac{\left (\sqrt{c-\frac{c}{a x}} \sqrt{x}\right ) \int \frac{\sqrt{1-a^2 x^2}}{x^{5/2} \sqrt{1-a x}} \, dx}{\sqrt{1-a x}}\\ &=\frac{\left (\sqrt{c-\frac{c}{a x}} \sqrt{x}\right ) \int \frac{\sqrt{1+a x}}{x^{5/2}} \, dx}{\sqrt{1-a x}}\\ &=-\frac{2 \sqrt{c-\frac{c}{a x}} (1+a x)^{3/2}}{3 x \sqrt{1-a x}}\\ \end{align*}

Mathematica [A]  time = 0.0212255, size = 41, normalized size = 1. \[ -\frac{2 (a x+1)^{3/2} \sqrt{c-\frac{c}{a x}}}{3 x \sqrt{1-a x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.083, size = 40, normalized size = 1. \begin{align*} -{\frac{2\, \left ( ax+1 \right ) ^{2}}{3\,x}\sqrt{{\frac{c \left ( ax-1 \right ) }{ax}}}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.90373, size = 95, normalized size = 2.32 \begin{align*} \frac{2 \, \sqrt{-a^{2} x^{2} + 1}{\left (a x + 1\right )} \sqrt{\frac{a c x - c}{a x}}}{3 \,{\left (a x^{2} - x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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