3.1266 \(\int \frac {e^{-3 \tanh ^{-1}(a x)} \sqrt {c-a^2 c x^2}}{x} \, dx\)

Optimal. Leaf size=102 \[ \frac {a x \sqrt {c-a^2 c x^2}}{\sqrt {1-a^2 x^2}}+\frac {\log (x) \sqrt {c-a^2 c x^2}}{\sqrt {1-a^2 x^2}}-\frac {4 \sqrt {c-a^2 c x^2} \log (a x+1)}{\sqrt {1-a^2 x^2}} \]

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

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Rubi [A]  time = 0.20, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6153, 6150, 72} \[ \frac {a x \sqrt {c-a^2 c x^2}}{\sqrt {1-a^2 x^2}}+\frac {\log (x) \sqrt {c-a^2 c x^2}}{\sqrt {1-a^2 x^2}}-\frac {4 \sqrt {c-a^2 c x^2} \log (a x+1)}{\sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c - a^2*c*x^2]/(E^(3*ArcTanh[a*x])*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] - (4*Sqrt[c - a^2
*c*x^2]*Log[1 + a*x])/Sqrt[1 - a^2*x^2]

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

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^{-3 \tanh ^{-1}(a x)} \sqrt {c-a^2 c x^2}}{x} \, dx &=\frac {\sqrt {c-a^2 c x^2} \int \frac {e^{-3 \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)^2}{x (1+a x)} \, dx}{\sqrt {1-a^2 x^2}}\\ &=\frac {\sqrt {c-a^2 c x^2} \int \left (a+\frac {1}{x}-\frac {4 a}{1+a 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}}-\frac {4 \sqrt {c-a^2 c x^2} \log (1+a x)}{\sqrt {1-a^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 44, normalized size = 0.43 \[ \frac {\sqrt {c-a^2 c x^2} (a x-4 \log (a x+1)+\log (x))}{\sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c - a^2*c*x^2]/(E^(3*ArcTanh[a*x])*x),x]

[Out]

(Sqrt[c - a^2*c*x^2]*(a*x + Log[x] - 4*Log[1 + a*x]))/Sqrt[1 - a^2*x^2]

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fricas [F]  time = 0.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} {\left (a x - 1\right )}}{a^{2} x^{3} + 2 \, a x^{2} + x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*c*x^2+c)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/x,x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {-a^{2} c x^{2} + c} {\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (a x + 1\right )}^{3} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*c*x^2+c)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/x,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.04, size = 56, normalized size = 0.55 \[ \frac {\left (-a x -\ln \relax (x )+4 \ln \left (a x +1\right )\right ) \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \sqrt {-a^{2} x^{2}+1}}{a^{2} x^{2}-1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {-a^{2} c x^{2} + c} {\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (a x + 1\right )}^{3} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*c*x^2+c)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/x,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {c-a^2\,c\,x^2}\,{\left (1-a^2\,x^2\right )}^{3/2}}{x\,{\left (a\,x+1\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}} \sqrt {- c \left (a x - 1\right ) \left (a x + 1\right )}}{x \left (a x + 1\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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