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

Optimal. Leaf size=60 \[ \frac {2 (1-a x)}{a \sqrt {c-a^2 c x^2}}+\frac {\tan ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{a \sqrt {c}} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {6167, 6142, 653, 217, 203} \[ \frac {2 (1-a x)}{a \sqrt {c-a^2 c x^2}}+\frac {\tan ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{a \sqrt {c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 653

Int[((d_) + (e_.)*(x_))^2*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)*(a + c*x^2)^(p + 1))/(c*(
p + 1)), x] - Dist[(e^2*(p + 2))/(c*(p + 1)), Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, p}, x] &&
EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && LtQ[p, -1]

Rule 6142

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

Rule 6167

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Dist[(-1)^(n/2), Int[u*E^(n*ArcTanh[a*x]), x], x] /; Free
Q[a, x] && IntegerQ[n/2]

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 100, normalized size = 1.67 \[ -\frac {2 \sqrt {1-a^2 x^2} \left (\sqrt {a x+1} (a x-1)+\sqrt {1-a x} (a x+1) \sin ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )\right )}{a \sqrt {1-a x} (a x+1) \sqrt {c-a^2 c x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [A]  time = 0.57, size = 151, normalized size = 2.52 \[ \left [-\frac {{\left (a x + 1\right )} \sqrt {-c} \log \left (2 \, a^{2} c x^{2} - 2 \, \sqrt {-a^{2} c x^{2} + c} a \sqrt {-c} x - c\right ) - 4 \, \sqrt {-a^{2} c x^{2} + c}}{2 \, {\left (a^{2} c x + a c\right )}}, -\frac {{\left (a x + 1\right )} \sqrt {c} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} a \sqrt {c} x}{a^{2} c x^{2} - c}\right ) - 2 \, \sqrt {-a^{2} c x^{2} + c}}{a^{2} c x + a c}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*((a*x + 1)*sqrt(-c)*log(2*a^2*c*x^2 - 2*sqrt(-a^2*c*x^2 + c)*a*sqrt(-c)*x - c) - 4*sqrt(-a^2*c*x^2 + c))
/(a^2*c*x + a*c), -((a*x + 1)*sqrt(c)*arctan(sqrt(-a^2*c*x^2 + c)*a*sqrt(c)*x/(a^2*c*x^2 - c)) - 2*sqrt(-a^2*c
*x^2 + c))/(a^2*c*x + a*c)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {undef} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

undef

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maple [A]  time = 0.04, size = 73, normalized size = 1.22 \[ \frac {\arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right )}{\sqrt {a^{2} c}}+\frac {2 \sqrt {-\left (x +\frac {1}{a}\right )^{2} a^{2} c +2 \left (x +\frac {1}{a}\right ) a c}}{a^{2} c \left (x +\frac {1}{a}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.40, size = 39, normalized size = 0.65 \[ \frac {2 \, \sqrt {-a^{2} c x^{2} + c}}{a^{2} c x + a c} + \frac {\arcsin \left (a x\right )}{a \sqrt {c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {a\,x-1}{\sqrt {c-a^2\,c\,x^2}\,\left (a\,x+1\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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