3.900 $$\int \frac{e^{3 \coth ^{-1}(a x)} \sqrt{c-\frac{c}{a^2 x^2}}}{x^2} \, dx$$

Optimal. Leaf size=147 $\frac{3 \sqrt{c-\frac{c}{a^2 x^2}}}{x \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{\sqrt{c-\frac{c}{a^2 x^2}}}{2 a x^2 \sqrt{1-\frac{1}{a^2 x^2}}}-\frac{4 a \log (x) \sqrt{c-\frac{c}{a^2 x^2}}}{\sqrt{1-\frac{1}{a^2 x^2}}}+\frac{4 a \sqrt{c-\frac{c}{a^2 x^2}} \log (1-a x)}{\sqrt{1-\frac{1}{a^2 x^2}}}$

[Out]

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

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Rubi [A]  time = 0.276782, antiderivative size = 147, 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 = {6197, 6193, 88} $\frac{3 \sqrt{c-\frac{c}{a^2 x^2}}}{x \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{\sqrt{c-\frac{c}{a^2 x^2}}}{2 a x^2 \sqrt{1-\frac{1}{a^2 x^2}}}-\frac{4 a \log (x) \sqrt{c-\frac{c}{a^2 x^2}}}{\sqrt{1-\frac{1}{a^2 x^2}}}+\frac{4 a \sqrt{c-\frac{c}{a^2 x^2}} \log (1-a x)}{\sqrt{1-\frac{1}{a^2 x^2}}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 6197

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

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

Mathematica [A]  time = 0.0464821, size = 66, normalized size = 0.45 $\frac{\sqrt{c-\frac{c}{a^2 x^2}} \left (-4 a^2 \log (x)+4 a^2 \log (1-a x)+\frac{3 a}{x}+\frac{1}{2 x^2}\right )}{a \sqrt{1-\frac{1}{a^2 x^2}}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(Sqrt[c - c/(a^2*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)])

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Maple [A]  time = 0.23, size = 82, normalized size = 0.6 \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\, \left ( ax+1 \right ) ^{2}x}\sqrt{{\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}{x}^{2}}}} \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)*(c-c/a^2/x^2)^(1/2)/x^2,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)/a^2/x^2)^(1/2)*(a*x-1)/(a*x+1)^2/x/((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{c - \frac{c}{a^{2} x^{2}}}}{x^{2} \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)*(c-c/a^2/x^2)^(1/2)/x^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.57251, size = 194, normalized size = 1.32 \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^{2} 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)*(c-c/a^2/x^2)^(1/2)/x^2,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)) + sq
rt(a^2*c)*(6*a*x + 1))/(a^2*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)*(c-c/a**2/x**2)**(1/2)/x**2,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c - \frac{c}{a^{2} x^{2}}}}{x^{2} \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)*(c-c/a^2/x^2)^(1/2)/x^2,x, algorithm="giac")

[Out]

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