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

Optimal. Leaf size=114 \[ \frac {\sqrt {c-a^2 c x^2}}{a x^2 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {3 \log (x) \sqrt {c-a^2 c x^2}}{x \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {4 \sqrt {c-a^2 c x^2} \log (1-a x)}{x \sqrt {1-\frac {1}{a^2 x^2}}} \]

[Out]

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

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Rubi [A]  time = 0.24, antiderivative size = 114, 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 = {6192, 6193, 88} \[ \frac {\sqrt {c-a^2 c x^2}}{a x^2 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {3 \log (x) \sqrt {c-a^2 c x^2}}{x \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {4 \sqrt {c-a^2 c x^2} \log (1-a x)}{x \sqrt {1-\frac {1}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 88

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 6192

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

Rule 6193

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

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 55, normalized size = 0.48 \[ \frac {\sqrt {c-a^2 c x^2} \left (-3 a \log (x)+4 a \log (1-a x)+\frac {1}{x}\right )}{a x \sqrt {1-\frac {1}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.55, size = 33, normalized size = 0.29 \[ \frac {\sqrt {-a^{2} c} {\left (4 \, a x \log \left (a x - 1\right ) - 3 \, a x \log \relax (x) + 1\right )}}{a x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 65, normalized size = 0.57 \[ -\frac {\sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (3 a \ln \relax (x ) x -4 \ln \left (a x -1\right ) x a -1\right ) \left (a x -1\right )}{x \left (a x +1\right )^{2} \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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