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]

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

________________________________________________________________________________________

Rubi [A]  time = 0.23, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, 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 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]

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 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 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]

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 = 0.02, size = 41, normalized size = 1.00 \[ -\frac {2 (a x+1)^{3/2} \sqrt {c-\frac {c}{a x}}}{3 x \sqrt {1-a x}} \]

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])

________________________________________________________________________________________

fricas [A]  time = 0.50, size = 47, normalized size = 1.15 \[ \frac {2 \, \sqrt {-a^{2} x^{2} + 1} {\left (a x + 1\right )} \sqrt {\frac {a c x - c}{a x}}}{3 \, {\left (a x^{2} - x\right )}} \]

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)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a x + 1\right )} \sqrt {c - \frac {c}{a x}}}{\sqrt {-a^{2} x^{2} + 1} x^{2}}\,{d x} \]

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)

________________________________________________________________________________________

maple [A]  time = 0.03, size = 40, normalized size = 0.98 \[ -\frac {2 \left (a x +1\right )^{2} \sqrt {\frac {c \left (a x -1\right )}{a x}}}{3 x \sqrt {-a^{2} x^{2}+1}} \]

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)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a x + 1\right )} \sqrt {c - \frac {c}{a x}}}{\sqrt {-a^{2} x^{2} + 1} x^{2}}\,{d x} \]

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)

________________________________________________________________________________________

mupad [B]  time = 1.12, size = 44, normalized size = 1.07 \[ -\frac {\sqrt {c-\frac {c}{a\,x}}\,\left (\frac {2\,a^2\,x^2}{3}+\frac {4\,a\,x}{3}+\frac {2}{3}\right )}{x\,\sqrt {1-a^2\,x^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {- c \left (-1 + \frac {1}{a x}\right )} \left (a x + 1\right )}{x^{2} \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \]

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)

________________________________________________________________________________________