3.13.67 \(\int \frac {e^{-3 \tanh ^{-1}(a x)} \sqrt {c-a^2 c x^2}}{x^2} \, dx\) [1267]

Optimal. Leaf size=107 \[ -\frac {\sqrt {c-a^2 c x^2}}{x \sqrt {1-a^2 x^2}}-\frac {3 a \sqrt {c-a^2 c x^2} \log (x)}{\sqrt {1-a^2 x^2}}+\frac {4 a \sqrt {c-a^2 c x^2} \log (1+a x)}{\sqrt {1-a^2 x^2}} \]

[Out]

-(-a^2*c*x^2+c)^(1/2)/x/(-a^2*x^2+1)^(1/2)-3*a*ln(x)*(-a^2*c*x^2+c)^(1/2)/(-a^2*x^2+1)^(1/2)+4*a*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 = 0.14, antiderivative size = 107, 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 = {6288, 6285, 90} \begin {gather*} -\frac {\sqrt {c-a^2 c x^2}}{x \sqrt {1-a^2 x^2}}-\frac {3 a \log (x) \sqrt {c-a^2 c x^2}}{\sqrt {1-a^2 x^2}}+\frac {4 a \sqrt {c-a^2 c x^2} \log (a x+1)}{\sqrt {1-a^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 6285

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 6288

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^2} \, dx &=\frac {\sqrt {c-a^2 c x^2} \int \frac {e^{-3 \tanh ^{-1}(a x)} \sqrt {1-a^2 x^2}}{x^2} \, dx}{\sqrt {1-a^2 x^2}}\\ &=\frac {\sqrt {c-a^2 c x^2} \int \frac {(1-a x)^2}{x^2 (1+a x)} \, dx}{\sqrt {1-a^2 x^2}}\\ &=\frac {\sqrt {c-a^2 c x^2} \int \left (\frac {1}{x^2}-\frac {3 a}{x}+\frac {4 a^2}{1+a x}\right ) \, dx}{\sqrt {1-a^2 x^2}}\\ &=-\frac {\sqrt {c-a^2 c x^2}}{x \sqrt {1-a^2 x^2}}-\frac {3 a \sqrt {c-a^2 c x^2} \log (x)}{\sqrt {1-a^2 x^2}}+\frac {4 a \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 = 0.02, size = 50, normalized size = 0.47 \begin {gather*} \frac {\sqrt {c-a^2 c x^2} \left (-\frac {1}{x}-3 a \log (x)+4 a \log (1+a x)\right )}{\sqrt {1-a^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.06, size = 60, normalized size = 0.56

method	result	size
default	\(\frac {\left (-4 \ln \left (a x +1\right ) a x +3 a \ln \left (x \right ) x +1\right ) \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \sqrt {-a^{2} x^{2}+1}}{\left (a^{2} x^{2}-1\right ) x}\)	\(60\)

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^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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^2,x, algorithm="maxima")

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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^2,x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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^{2} \left (a x + 1\right )^{3}}\, dx \end {gather*}

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**2,x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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^2,x, algorithm="giac")

[Out]

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

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

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^2*(a*x + 1)^3),x)

[Out]

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

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