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

Optimal. Leaf size=153 \[ \frac{3 \sqrt{c-a^2 c x^2}}{x^2 \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{\sqrt{c-a^2 c x^2}}{2 a x^3 \sqrt{1-\frac{1}{a^2 x^2}}}-\frac{4 a \log (x) \sqrt{c-a^2 c x^2}}{x \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{4 a \sqrt{c-a^2 c x^2} \log (1-a x)}{x \sqrt{1-\frac{1}{a^2 x^2}}} \]

[Out]

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

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Rubi [A]  time = 0.240446, antiderivative size = 153, normalized size of antiderivative = 1., 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} \[ \frac{3 \sqrt{c-a^2 c x^2}}{x^2 \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{\sqrt{c-a^2 c x^2}}{2 a x^3 \sqrt{1-\frac{1}{a^2 x^2}}}-\frac{4 a \log (x) \sqrt{c-a^2 c x^2}}{x \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{4 a \sqrt{c-a^2 c x^2} \log (1-a x)}{x \sqrt{1-\frac{1}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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]))

Rubi steps

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

Mathematica [A]  time = 0.0376005, size = 69, normalized size = 0.45 \[ \frac{\sqrt{c-a^2 c x^2} \left (-4 a^2 \log (x)+4 a^2 \log (1-a x)+\frac{3 a}{x}+\frac{1}{2 x^2}\right )}{a x \sqrt{1-\frac{1}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.191, size = 77, normalized size = 0.5 \begin{align*} -{\frac{ \left ( 8\,{a}^{2}\ln \left ( x \right ){x}^{2}-8\,\ln \left ( ax-1 \right ){a}^{2}{x}^{2}-6\,ax-1 \right ) \left ( ax-1 \right ) }{2\,{x}^{2} \left ( ax+1 \right ) ^{2}}\sqrt{-c \left ({a}^{2}{x}^{2}-1 \right ) } \left ({\frac{ax-1}{ax+1}} \right ) ^{-{\frac{3}{2}}}} \end{align*}

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^3,x)

[Out]

-1/2*(8*a^2*ln(x)*x^2-8*ln(a*x-1)*a^2*x^2-6*a*x-1)*(-c*(a^2*x^2-1))^(1/2)*(a*x-1)/x^2/(a*x+1)^2/((a*x-1)/(a*x+
1))^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{-a^{2} c x^{2} + c}}{x^{3} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{3}{2}}}\,{d x} \end{align*}

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^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.56174, size = 197, normalized size = 1.29 \begin{align*} \frac{8 \, a^{3} \sqrt{-c} x^{2} \log \left (\frac{2 \, a^{3} c x^{2} - 2 \, a^{2} c x + \sqrt{-a^{2} c}{\left (2 \, a x - 1\right )} \sqrt{-c} + a c}{a x^{2} - x}\right ) + \sqrt{-a^{2} c}{\left (6 \, a x + 1\right )}}{2 \, a x^{2}} \end{align*}

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^3,x, algorithm="fricas")

[Out]

1/2*(8*a^3*sqrt(-c)*x^2*log((2*a^3*c*x^2 - 2*a^2*c*x + sqrt(-a^2*c)*(2*a*x - 1)*sqrt(-c) + a*c)/(a*x^2 - x)) +
sqrt(-a^2*c)*(6*a*x + 1))/(a*x^2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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**3,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{-a^{2} c x^{2} + c}}{x^{3} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{3}{2}}}\,{d x} \end{align*}

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^3,x, algorithm="giac")

[Out]

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