\(\int e^{-3 \coth ^{-1}(a x)} x \sqrt {c-a^2 c x^2} \, dx\) [723]

Optimal result

Integrand size = 25, antiderivative size = 151 \[ \int e^{-3 \coth ^{-1}(a x)} x \sqrt {c-a^2 c x^2} \, dx=\frac {4 \sqrt {c-a^2 c x^2}}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {3 x \sqrt {c-a^2 c x^2}}{2 a \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {x^2 \sqrt {c-a^2 c x^2}}{3 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {4 \sqrt {c-a^2 c x^2} \log (1+a x)}{a^3 \sqrt {1-\frac {1}{a^2 x^2}} x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {6327, 6328, 78} \[ \int e^{-3 \coth ^{-1}(a x)} x \sqrt {c-a^2 c x^2} \, dx=\frac {x^2 \sqrt {c-a^2 c x^2}}{3 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {3 x \sqrt {c-a^2 c x^2}}{2 a \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {4 \sqrt {c-a^2 c x^2}}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {4 \sqrt {c-a^2 c x^2} \log (a x+1)}{a^3 x \sqrt {1-\frac {1}{a^2 x^2}}} \]

[In]

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

[Out]

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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 {\sqrt {c-a^2 c x^2} \int e^{-3 \coth ^{-1}(a x)} \sqrt {1-\frac {1}{a^2 x^2}} x^2 \, dx}{\sqrt {1-\frac {1}{a^2 x^2}} x} \\ & = \frac {\sqrt {c-a^2 c x^2} \int \frac {x (-1+a x)^2}{1+a x} \, dx}{a \sqrt {1-\frac {1}{a^2 x^2}} x} \\ & = \frac {\sqrt {c-a^2 c x^2} \int \left (\frac {4}{a}-3 x+a x^2-\frac {4}{a (1+a x)}\right ) \, dx}{a \sqrt {1-\frac {1}{a^2 x^2}} x} \\ & = \frac {4 \sqrt {c-a^2 c x^2}}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {3 x \sqrt {c-a^2 c x^2}}{2 a \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {x^2 \sqrt {c-a^2 c x^2}}{3 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {4 \sqrt {c-a^2 c x^2} \log (1+a x)}{a^3 \sqrt {1-\frac {1}{a^2 x^2}} x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.43 \[ \int e^{-3 \coth ^{-1}(a x)} x \sqrt {c-a^2 c x^2} \, dx=\frac {\sqrt {c-a^2 c x^2} \left (a x \left (24-9 a x+2 a^2 x^2\right )-24 \log (1+a x)\right )}{6 a^3 \sqrt {1-\frac {1}{a^2 x^2}} x} \]

[In]

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

[Out]

(Sqrt[c - a^2*c*x^2]*(a*x*(24 - 9*a*x + 2*a^2*x^2) - 24*Log[1 + a*x]))/(6*a^3*Sqrt[1 - 1/(a^2*x^2)]*x)

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.50

method	result	size
default	\(-\frac {\left (-2 a^{3} x^{3}+9 a^{2} x^{2}-24 a x +24 \ln \left (a x +1\right )\right ) \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (a x +1\right ) \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{6 a^{2} \left (a x -1\right )^{2}}\)	\(76\)

[In]

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

[Out]

-1/6*(-2*a^3*x^3+9*a^2*x^2-24*a*x+24*ln(a*x+1))*(-c*(a^2*x^2-1))^(1/2)*(a*x+1)*((a*x-1)/(a*x+1))^(3/2)/a^2/(a*
x-1)^2

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.28 \[ \int e^{-3 \coth ^{-1}(a x)} x \sqrt {c-a^2 c x^2} \, dx=\frac {{\left (2 \, a^{3} x^{3} - 9 \, a^{2} x^{2} + 24 \, a x - 24 \, \log \left (a x + 1\right )\right )} \sqrt {-a^{2} c}}{6 \, a^{3}} \]

[In]

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

[Out]

1/6*(2*a^3*x^3 - 9*a^2*x^2 + 24*a*x - 24*log(a*x + 1))*sqrt(-a^2*c)/a^3

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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