3.629 $$\int \frac{e^{2 \coth ^{-1}(a x)}}{(c-a^2 c x^2)^{3/2}} \, dx$$

Optimal. Leaf size=51 $-\frac{x}{3 c \sqrt{c-a^2 c x^2}}-\frac{2 (a x+1)}{3 a \left (c-a^2 c x^2\right )^{3/2}}$

[Out]

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

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Rubi [A]  time = 0.104968, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.167, Rules used = {6167, 6141, 653, 191} $-\frac{x}{3 c \sqrt{c-a^2 c x^2}}-\frac{2 (a x+1)}{3 a \left (c-a^2 c x^2\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rubi steps

\begin{align*} \int \frac{e^{2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{3/2}} \, dx &=-\int \frac{e^{2 \tanh ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{3/2}} \, dx\\ &=-\left (c \int \frac{(1+a x)^2}{\left (c-a^2 c x^2\right )^{5/2}} \, dx\right )\\ &=-\frac{2 (1+a x)}{3 a \left (c-a^2 c x^2\right )^{3/2}}-\frac{1}{3} \int \frac{1}{\left (c-a^2 c x^2\right )^{3/2}} \, dx\\ &=-\frac{2 (1+a x)}{3 a \left (c-a^2 c x^2\right )^{3/2}}-\frac{x}{3 c \sqrt{c-a^2 c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0349434, size = 64, normalized size = 1.25 $-\frac{(2-a x) \sqrt{a x+1} \sqrt{1-a^2 x^2}}{3 a c (1-a x)^{3/2} \sqrt{c-a^2 c x^2}}$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.04, size = 31, normalized size = 0.6 \begin{align*}{\frac{ \left ( ax+1 \right ) ^{2} \left ( ax-2 \right ) }{3\,a} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{-{\frac{3}{2}}}} \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)^(3/2),x)

[Out]

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

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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)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.65754, size = 97, normalized size = 1.9 \begin{align*} \frac{\sqrt{-a^{2} c x^{2} + c}{\left (a x - 2\right )}}{3 \,{\left (a^{3} c^{2} x^{2} - 2 \, a^{2} c^{2} x + a c^{2}\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)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a x + 1}{\left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac{3}{2}} \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)**(3/2),x)

[Out]

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

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Giac [B]  time = 1.20202, size = 200, normalized size = 3.92 \begin{align*} \frac{{\left (a c - 3 \, \sqrt{-a^{2} c} \sqrt{c}\right )} \mathrm{sgn}\left (x\right )}{3 \,{\left (a^{2} c^{\frac{5}{2}} - \sqrt{-a^{2} c} a c^{2}\right )}} - \frac{2 \,{\left (2 \, a^{2} c + 3 \, a \sqrt{c}{\left (\sqrt{-a^{2} c + \frac{c}{x^{2}}} - \frac{\sqrt{c}}{x}\right )} + 3 \,{\left (\sqrt{-a^{2} c + \frac{c}{x^{2}}} - \frac{\sqrt{c}}{x}\right )}^{2}\right )}}{3 \,{\left (a \sqrt{c} + \sqrt{-a^{2} c + \frac{c}{x^{2}}} - \frac{\sqrt{c}}{x}\right )}^{3} c \mathrm{sgn}\left (x\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)^(3/2),x, algorithm="giac")

[Out]

1/3*(a*c - 3*sqrt(-a^2*c)*sqrt(c))*sgn(x)/(a^2*c^(5/2) - sqrt(-a^2*c)*a*c^2) - 2/3*(2*a^2*c + 3*a*sqrt(c)*(sqr
t(-a^2*c + c/x^2) - sqrt(c)/x) + 3*(sqrt(-a^2*c + c/x^2) - sqrt(c)/x)^2)/((a*sqrt(c) + sqrt(-a^2*c + c/x^2) -
sqrt(c)/x)^3*c*sgn(x))