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

Optimal. Leaf size=113 \[ \frac {x \sqrt {1-\frac {1}{a^2 x^2}}}{\sqrt {c-\frac {c}{a^2 x^2}}}-\frac {2 \sqrt {1-\frac {1}{a^2 x^2}}}{a (a x+1) \sqrt {c-\frac {c}{a^2 x^2}}}-\frac {3 \sqrt {1-\frac {1}{a^2 x^2}} \log (a x+1)}{a \sqrt {c-\frac {c}{a^2 x^2}}} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6197, 6193, 77} \[ \frac {x \sqrt {1-\frac {1}{a^2 x^2}}}{\sqrt {c-\frac {c}{a^2 x^2}}}-\frac {2 \sqrt {1-\frac {1}{a^2 x^2}}}{a (a x+1) \sqrt {c-\frac {c}{a^2 x^2}}}-\frac {3 \sqrt {1-\frac {1}{a^2 x^2}} \log (a x+1)}{a \sqrt {c-\frac {c}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(3*ArcCoth[a*x])*Sqrt[c - c/(a^2*x^2)]),x]

[Out]

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

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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 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])

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 54, normalized size = 0.48 \[ \frac {\sqrt {1-\frac {1}{a^2 x^2}} \left (a x-\frac {2}{a x+1}-3 \log (a x+1)\right )}{a \sqrt {c-\frac {c}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(3*ArcCoth[a*x])*Sqrt[c - c/(a^2*x^2)]),x]

[Out]

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

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fricas [A]  time = 0.59, size = 47, normalized size = 0.42 \[ \frac {{\left (a^{2} x^{2} + a x - 3 \, {\left (a x + 1\right )} \log \left (a x + 1\right ) - 2\right )} \sqrt {a^{2} c}}{a^{3} c x + a^{2} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(a*x+1)]Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is re
al):Check [abs(t_nostep)]sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument
Value

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maple [A]  time = 0.06, size = 87, normalized size = 0.77 \[ -\frac {\left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} \left (a x +1\right ) \left (-a^{2} x^{2}+3 a x \ln \left (a x +1\right )-a x +3 \ln \left (a x +1\right )+2\right )}{\left (a x -1\right ) \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}\, x \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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