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

Optimal. Leaf size=77 \[ \frac {2 x \sqrt {1-\frac {1}{a^2 x^2}}}{(a x+1) \sqrt {c-a^2 c x^2}}+\frac {x \sqrt {1-\frac {1}{a^2 x^2}} \log (a x+1)}{\sqrt {c-a^2 c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 77, 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 = {6192, 6193, 43} \[ \frac {2 x \sqrt {1-\frac {1}{a^2 x^2}}}{(a x+1) \sqrt {c-a^2 c x^2}}+\frac {x \sqrt {1-\frac {1}{a^2 x^2}} \log (a x+1)}{\sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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

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Mathematica [A]  time = 0.04, size = 52, normalized size = 0.68 \[ \frac {x \sqrt {1-\frac {1}{a^2 x^2}} ((a x+1) \log (a x+1)+2)}{(a x+1) \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.62, size = 38, normalized size = 0.49 \[ -\frac {\sqrt {-a^{2} c} {\left ({\left (a x + 1\right )} \log \left (a x + 1\right ) + 2\right )}}{a^{3} c x + a^{2} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 62, normalized size = 0.81 \[ -\frac {\sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (a x \ln \left (a x +1\right )+\ln \left (a x +1\right )+2\right ) \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{a c \left (a x -1\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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