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

Optimal. Leaf size=84 \[ \frac{10 a \sqrt{a x+1} \sqrt{c-\frac{c}{a x}}}{3 \sqrt{1-a x}}-\frac{2 \sqrt{1-a^2 x^2} \sqrt{c-\frac{c}{a x}}}{3 x (1-a x)} \]

[Out]

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

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Rubi [A]  time = 0.259863, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {6134, 6128, 879, 848, 37} \[ \frac{10 a \sqrt{a x+1} \sqrt{c-\frac{c}{a x}}}{3 \sqrt{1-a x}}-\frac{2 \sqrt{1-a^2 x^2} \sqrt{c-\frac{c}{a x}}}{3 x (1-a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(10*a*Sqrt[c - c/(a*x)]*Sqrt[1 + a*x])/(3*Sqrt[1 - a*x]) - (2*Sqrt[c - c/(a*x)]*Sqrt[1 - a^2*x^2])/(3*x*(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 879

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e^2*(e*f
 - d*g)*(d + e*x)^(m - 2)*(f + g*x)^(n + 1)*(a + c*x^2)^(p + 1))/(c*g*(n + 1)*(e*f + d*g)), x] - Dist[(e*(e*f*
(p + 1) - d*g*(2*n + p + 3)))/(g*(n + 1)*(e*f + d*g)), Int[(d + e*x)^(m - 1)*(f + g*x)^(n + 1)*(a + c*x^2)^p,
x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[e*f - d*g, 0] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] &&
 EqQ[m + p - 1, 0] && LtQ[n, -1] && 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{(1-a x)^{3/2}}{x^{5/2} \sqrt{1-a^2 x^2}} \, dx}{\sqrt{1-a x}}\\ &=-\frac{2 \sqrt{c-\frac{c}{a x}} \sqrt{1-a^2 x^2}}{3 x (1-a x)}-\frac{\left (5 a \sqrt{c-\frac{c}{a x}} \sqrt{x}\right ) \int \frac{\sqrt{1-a x}}{x^{3/2} \sqrt{1-a^2 x^2}} \, dx}{3 \sqrt{1-a x}}\\ &=-\frac{2 \sqrt{c-\frac{c}{a x}} \sqrt{1-a^2 x^2}}{3 x (1-a x)}-\frac{\left (5 a \sqrt{c-\frac{c}{a x}} \sqrt{x}\right ) \int \frac{1}{x^{3/2} \sqrt{1+a x}} \, dx}{3 \sqrt{1-a x}}\\ &=\frac{10 a \sqrt{c-\frac{c}{a x}} \sqrt{1+a x}}{3 \sqrt{1-a x}}-\frac{2 \sqrt{c-\frac{c}{a x}} \sqrt{1-a^2 x^2}}{3 x (1-a x)}\\ \end{align*}

Mathematica [A]  time = 0.0306416, size = 47, normalized size = 0.56 \[ \frac{2 \sqrt{a x+1} (5 a x-1) \sqrt{c-\frac{c}{a x}}}{3 x \sqrt{1-a x}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*Sqrt[c - c/(a*x)]*Sqrt[1 + a*x]*(-1 + 5*a*x))/(3*x*Sqrt[1 - a*x])

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Maple [A]  time = 0.093, size = 46, normalized size = 0.6 \begin{align*} -{\frac{10\,ax-2}{ \left ( 3\,ax-3 \right ) x}\sqrt{{\frac{c \left ( ax-1 \right ) }{ax}}}\sqrt{-{a}^{2}{x}^{2}+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(5*a*x-1)*(c*(a*x-1)/a/x)^(1/2)*(-a^2*x^2+1)^(1/2)/(a*x-1)/x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.33845, size = 99, normalized size = 1.18 \begin{align*} -\frac{2 \, \sqrt{-a^{2} x^{2} + 1}{\left (5 \, 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((c-c/a/x)^(1/2)/(a*x+1)*(-a^2*x^2+1)^(1/2)/x^2,x, algorithm="fricas")

[Out]

-2/3*sqrt(-a^2*x^2 + 1)*(5*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 )} \sqrt{- \left (a x - 1\right ) \left (a x + 1\right )}}{x^{2} \left (a x + 1\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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