3.9.85 \(\int \frac {e^{\coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^2} \, dx\) [885]

Optimal. Leaf size=46 \[ -\frac {\sqrt {c-\frac {c}{a^2 x^2}} (1+a x)^2}{2 a \sqrt {1-\frac {1}{a^2 x^2}} x^2} \]

[Out]

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

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Rubi [A]
time = 0.17, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {6332, 6328, 37} \begin {gather*} -\frac {(a x+1)^2 \sqrt {c-\frac {c}{a^2 x^2}}}{2 a x^2 \sqrt {1-\frac {1}{a^2 x^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 6328

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symbol] :> Dist[c^p/a^(2*p), Int[(u/x^(
2*p))*(-1 + a*x)^(p - n/2)*(1 + a*x)^(p + n/2), 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 6332

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])

Rubi steps

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

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Mathematica [A]
time = 0.02, size = 47, normalized size = 1.02 \begin {gather*} \frac {\sqrt {c-\frac {c}{a^2 x^2}} \left (-\frac {1}{2 x^2}-\frac {a}{x}\right )}{a \sqrt {1-\frac {1}{a^2 x^2}}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]
time = 0.03, size = 53, normalized size = 1.15

method	result	size
gosper	\(-\frac {\left (2 a x +1\right ) \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}}{2 x \left (a x +1\right ) \sqrt {\frac {a x -1}{a x +1}}}\)	\(53\)
default	\(-\frac {\left (2 a x +1\right ) \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}}{2 x \left (a x +1\right ) \sqrt {\frac {a x -1}{a x +1}}}\)	\(53\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a*x-1)/(a*x+1))^(1/2)*(c-c/a^2/x^2)^(1/2)/x^2,x,method=_RETURNVERBOSE)

[Out]

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

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

[Out]

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

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Fricas [A]
time = 0.35, size = 21, normalized size = 0.46 \begin {gather*} -\frac {\sqrt {a^{2} c} {\left (2 \, a x + 1\right )}}{2 \, a^{2} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

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

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Mupad [B]
time = 1.40, size = 63, normalized size = 1.37 \begin {gather*} \frac {\left (x\,\sqrt {c-\frac {c}{a^2\,x^2}}+\frac {\sqrt {c-\frac {c}{a^2\,x^2}}}{2\,a}\right )\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{\frac {x}{a}-x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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