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

Optimal. Leaf size=43 \[ -\frac {(a x+1)^2 \sqrt {c-\frac {c}{a^2 x^2}}}{2 x \sqrt {1-a^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {6160, 6150, 37} \[ -\frac {(a x+1)^2 \sqrt {c-\frac {c}{a^2 x^2}}}{2 x \sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 6160

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

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 42, normalized size = 0.98 \[ -\frac {(2 a x+1) \sqrt {c-\frac {c}{a^2 x^2}}}{2 x \sqrt {1-a^2 x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [A]  time = 0.56, size = 63, normalized size = 1.47 \[ -\frac {\sqrt {-a^{2} x^{2} + 1} {\left ({\left (2 \, a + 1\right )} x^{2} - 2 \, a x - 1\right )} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{2 \, {\left (a^{2} x^{3} - x\right )}} \]

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^2/x^2)^(1/2)/x^2,x, algorithm="fricas")

[Out]

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

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

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^2/x^2)^(1/2)/x^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.03, size = 43, normalized size = 1.00 \[ -\frac {\left (2 a x +1\right ) \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}}{2 x \sqrt {-a^{2} x^{2}+1}} \]

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^2/x^2)^(1/2)/x^2,x)

[Out]

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

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maxima [C]  time = 0.38, size = 20, normalized size = 0.47 \[ \frac {i \, \sqrt {c}}{x} + \frac {i \, \sqrt {c}}{2 \, a x^{2}} \]

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^2/x^2)^(1/2)/x^2,x, algorithm="maxima")

[Out]

I*sqrt(c)/x + 1/2*I*sqrt(c)/(a*x^2)

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mupad [B]  time = 1.01, size = 35, normalized size = 0.81 \[ -\frac {\sqrt {c-\frac {c}{a^2\,x^2}}\,\left (a\,x+\frac {1}{2}\right )}{x\,\sqrt {1-a^2\,x^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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**2/x**2)**(1/2)/x**2,x)

[Out]

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

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