\(\int e^{-3 \coth ^{-1}(a x)} (c-a^2 c x^2)^{3/2} \, dx\) [662]

Optimal result

Integrand size = 24, antiderivative size = 47 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{3/2} \, dx=\frac {(1-a x)^4 \left (c-a^2 c x^2\right )^{3/2}}{4 a^4 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6327, 6328, 32} \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{3/2} \, dx=\frac {(1-a x)^4 \left (c-a^2 c x^2\right )^{3/2}}{4 a^4 x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}} \]

[In]

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

[Out]

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

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 6327

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 6328

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symbol] :> Dist[c^p/a^(2*p), Int[(u/x^(
2*p))*(-1 + a*x)^(p - n/2)*(1 + a*x)^(p + n/2), 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*} \text {integral}& = \frac {\left (c-a^2 c x^2\right )^{3/2} \int e^{-3 \coth ^{-1}(a x)} \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3 \, dx}{\left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3} \\ & = \frac {\left (c-a^2 c x^2\right )^{3/2} \int (-1+a x)^3 \, dx}{a^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3} \\ & = \frac {(1-a x)^4 \left (c-a^2 c x^2\right )^{3/2}}{4 a^4 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.23 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{3/2} \, dx=-\frac {c \sqrt {c-a^2 c x^2} \left (-4+6 a x-4 a^2 x^2+a^3 x^3\right )}{4 a \sqrt {1-\frac {1}{a^2 x^2}}} \]

[In]

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

[Out]

-1/4*(c*Sqrt[c - a^2*c*x^2]*(-4 + 6*a*x - 4*a^2*x^2 + a^3*x^3))/(a*Sqrt[1 - 1/(a^2*x^2)])

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.02

method	result	size
default	\(-\frac {\left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} \left (a x +1\right ) \left (a x -1\right )^{2} \sqrt {-c \left (a^{2} x^{2}-1\right )}\, c}{4 a}\)	\(48\)
gosper	\(\frac {x \left (a^{3} x^{3}-4 a^{2} x^{2}+6 a x -4\right ) \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}} \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{4 \left (a x -1\right )^{3}}\)	\(60\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.89 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{3/2} \, dx=-\frac {{\left (a^{3} c x^{4} - 4 \, a^{2} c x^{3} + 6 \, a c x^{2} - 4 \, c x\right )} \sqrt {-a^{2} c}}{4 \, a} \]

[In]

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

[Out]

-1/4*(a^3*c*x^4 - 4*a^2*c*x^3 + 6*a*c*x^2 - 4*c*x)*sqrt(-a^2*c)/a

Sympy [F(-1)]

Timed out. \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{3/2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 97 vs. \(2 (40) = 80\).

Time = 0.21 (sec) , antiderivative size = 97, normalized size of antiderivative = 2.06 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{3/2} \, dx=-\frac {{\left (a^{5} \sqrt {-c} c x^{5} - 3 \, a^{4} \sqrt {-c} c x^{4} + 2 \, a^{3} \sqrt {-c} c x^{3} + 2 \, a^{2} \sqrt {-c} c x^{2} + 4 \, \sqrt {-c} c\right )} {\left (a x - 1\right )}^{2}}{4 \, {\left (a^{3} x^{2} - 2 \, a^{2} x + a\right )} {\left (a x + 1\right )}} \]

[In]

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

[Out]

-1/4*(a^5*sqrt(-c)*c*x^5 - 3*a^4*sqrt(-c)*c*x^4 + 2*a^3*sqrt(-c)*c*x^3 + 2*a^2*sqrt(-c)*c*x^2 + 4*sqrt(-c)*c)*
(a*x - 1)^2/((a^3*x^2 - 2*a^2*x + a)*(a*x + 1))

Giac [F]

\[ \int e^{-3 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{3/2} \, dx=\int { {\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{3/2} \, dx=\int {\left (c-a^2\,c\,x^2\right )}^{3/2}\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2} \,d x \]

[In]

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

[Out]

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