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

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

[Out]

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

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Rubi [A]  time = 0.16, 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, 72} \[ \frac {\sqrt {c-a^2 c x^2}}{\sqrt {1-\frac {1}{a^2 x^2}}}-\frac {\log (x) \sqrt {c-a^2 c x^2}}{a x \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {4 \sqrt {c-a^2 c x^2} \log (1-a x)}{a x \sqrt {1-\frac {1}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.03, size = 53, normalized size = 0.46 \[ \frac {\sqrt {c-a^2 c x^2} (a x+4 \log (1-a x)-\log (x))}{a x \sqrt {1-\frac {1}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.68, size = 28, normalized size = 0.25 \[ \frac {\sqrt {-a^{2} c} {\left (a x + 4 \, \log \left (a x - 1\right ) - \log \relax (x)\right )}}{a} \]

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,x, algorithm="fricas")

[Out]

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

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

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,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 59, normalized size = 0.52 \[ -\frac {\sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (-a x +\ln \relax (x )-4 \ln \left (a x -1\right )\right ) \left (a x -1\right )}{\left (a x +1\right )^{2} \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}} \]

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,x)

[Out]

-(-c*(a^2*x^2-1))^(1/2)*(-a*x+ln(x)-4*ln(a*x-1))*(a*x-1)/(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 \[ \int \frac {\sqrt {-a^{2} c x^{2} + c}}{x \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}\,{d x} \]

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,x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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,x)

[Out]

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

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