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

Optimal. Leaf size=65 \[ \frac {a x \sqrt {c-a^2 c x^2}}{\sqrt {1-a^2 x^2}}+\frac {\log (x) \sqrt {c-a^2 c x^2}}{\sqrt {1-a^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {6153, 6150, 43} \[ \frac {a x \sqrt {c-a^2 c x^2}}{\sqrt {1-a^2 x^2}}+\frac {\log (x) \sqrt {c-a^2 c x^2}}{\sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 36, normalized size = 0.55 \[ \frac {\sqrt {c-a^2 c x^2} (a x+\log (x))}{\sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.87, size = 262, normalized size = 4.03 \[ \left [\frac {{\left (a^{2} x^{2} - 1\right )} \sqrt {c} \log \left (\frac {a^{2} c x^{6} + a^{2} c x^{2} - c x^{4} - \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} {\left (x^{4} - 1\right )} \sqrt {c} - c}{a^{2} x^{4} - x^{2}}\right ) - 2 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} {\left (a x - a\right )}}{2 \, {\left (a^{2} x^{2} - 1\right )}}, \frac {{\left (a^{2} x^{2} - 1\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} {\left (x^{2} + 1\right )} \sqrt {-c}}{a^{2} c x^{4} - {\left (a^{2} + 1\right )} c x^{2} + c}\right ) - \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} {\left (a x - a\right )}}{a^{2} x^{2} - 1}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*((a^2*x^2 - 1)*sqrt(c)*log((a^2*c*x^6 + a^2*c*x^2 - c*x^4 - sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*(x^4
- 1)*sqrt(c) - c)/(a^2*x^4 - x^2)) - 2*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*(a*x - a))/(a^2*x^2 - 1), ((a^2
*x^2 - 1)*sqrt(-c)*arctan(sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*(x^2 + 1)*sqrt(-c)/(a^2*c*x^4 - (a^2 + 1)*c*
x^2 + c)) - sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*(a*x - a))/(a^2*x^2 - 1)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 48, normalized size = 0.74 \[ \frac {\left (-a x -\ln \relax (x )\right ) \sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a^{2} x^{2}-1\right )}}{a^{2} x^{2}-1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.34, size = 56, normalized size = 0.86 \[ a \sqrt {c} x - \frac {1}{2} \, \left (-1\right )^{-2 \, a^{2} c x^{2} + 2 \, c} \sqrt {c} \log \left (-2 \, a^{2} c + \frac {2 \, c}{x^{2}}\right ) - \frac {1}{2} \, \sqrt {c} \log \left (x^{2} - \frac {1}{a^{2}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\sqrt {c-a^2\,c\,x^2}\,\left (a\,x+1\right )}{x\,\sqrt {1-a^2\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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