### 3.648 $$\int \frac{e^{-\coth ^{-1}(a x)}}{(c-a^2 c x^2)^{3/2}} \, dx$$

Optimal. Leaf size=90 $\frac{a^2 x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}{2 (a x+1) \left (c-a^2 c x^2\right )^{3/2}}-\frac{a^2 x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} \tanh ^{-1}(a x)}{2 \left (c-a^2 c x^2\right )^{3/2}}$

[Out]

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

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Rubi [A]  time = 0.183534, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.167, Rules used = {6192, 6193, 44, 207} $\frac{a^2 x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}{2 (a x+1) \left (c-a^2 c x^2\right )^{3/2}}-\frac{a^2 x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} \tanh ^{-1}(a x)}{2 \left (c-a^2 c x^2\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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]

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 207

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

Rubi steps

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

Mathematica [A]  time = 0.0489857, size = 54, normalized size = 0.6 $\frac{x \sqrt{1-\frac{1}{a^2 x^2}} \left ((a x+1) \tanh ^{-1}(a x)-1\right )}{2 (a c x+c) \sqrt{c-a^2 c x^2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.145, size = 84, normalized size = 0.9 \begin{align*} -{\frac{ax\ln \left ( ax+1 \right ) -\ln \left ( ax-1 \right ) xa+\ln \left ( ax+1 \right ) -\ln \left ( ax-1 \right ) -2}{ \left ( 4\,{a}^{2}{x}^{2}-4 \right ){c}^{2}a}\sqrt{{\frac{ax-1}{ax+1}}}\sqrt{-c \left ({a}^{2}{x}^{2}-1 \right ) }} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*((a*x-1)/(a*x+1))^(1/2)*(-c*(a^2*x^2-1))^(1/2)*(a*x*ln(a*x+1)-ln(a*x-1)*x*a+ln(a*x+1)-ln(a*x-1)-2)/(a^2*x
^2-1)/c^2/a

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.84255, size = 177, normalized size = 1.97 \begin{align*} -\frac{{\left (a^{2} x + a\right )} \sqrt{-c} \log \left (\frac{a^{2} c x^{2} - 2 \, \sqrt{-a^{2} c} \sqrt{-c} x + c}{a^{2} x^{2} - 1}\right ) - 2 \, \sqrt{-a^{2} c}}{4 \,{\left (a^{3} c^{2} x + a^{2} c^{2}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*((a^2*x + a)*sqrt(-c)*log((a^2*c*x^2 - 2*sqrt(-a^2*c)*sqrt(-c)*x + c)/(a^2*x^2 - 1)) - 2*sqrt(-a^2*c))/(a
^3*c^2*x + a^2*c^2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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