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

Optimal. Leaf size=59 \[ \frac{\tan ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c-a^2 c x^2}}\right )}{a \sqrt{c}}-\frac{2 (a x+1)}{a \sqrt{c-a^2 c x^2}} \]

[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])

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Rubi [A]  time = 0.104667, antiderivative size = 59, normalized size of antiderivative = 1., 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, 6141, 653, 217, 203} \[ \frac{\tan ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c-a^2 c x^2}}\right )}{a \sqrt{c}}-\frac{2 (a x+1)}{a \sqrt{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[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 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]

Rule 6141

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[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]) && IGt
Q[n/2, 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 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 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])

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*}

Mathematica [A]  time = 0.0371664, size = 82, normalized size = 1.39 \[ -\frac{2 \sqrt{1-a^2 x^2} \left (\sqrt{a x+1}+\sqrt{1-a x} \sin ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{2}}\right )\right )}{a \sqrt{1-a x} \sqrt{c-a^2 c x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.045, size = 79, normalized size = 1.3 \begin{align*}{\arctan \left ({x\sqrt{{a}^{2}c}{\frac{1}{\sqrt{-{a}^{2}c{x}^{2}+c}}}} \right ){\frac{1}{\sqrt{{a}^{2}c}}}}+2\,{\frac{1}{{a}^{2}c}\sqrt{-c{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,ac \left ( x-{a}^{-1} \right ) } \left ( x-{a}^{-1} \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.79183, size = 339, normalized size = 5.75 \begin{align*} \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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a x + 1}{\sqrt{- c \left (a x - 1\right ) \left (a x + 1\right )} \left (a x - 1\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{undef} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

undef