\(\int e^{\coth ^{-1}(a x)} x \sqrt {c-a^2 c x^2} \, dx\) [669]

Optimal result

Integrand size = 23, antiderivative size = 74 \[ \int e^{\coth ^{-1}(a x)} x \sqrt {c-a^2 c x^2} \, dx=\frac {x \sqrt {c-a^2 c x^2}}{2 a \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {x^2 \sqrt {c-a^2 c x^2}}{3 \sqrt {1-\frac {1}{a^2 x^2}}} \]

[Out]

1/2*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)

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {6327, 6328, 45} \[ \int e^{\coth ^{-1}(a x)} x \sqrt {c-a^2 c x^2} \, dx=\frac {x^2 \sqrt {c-a^2 c x^2}}{3 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {x \sqrt {c-a^2 c x^2}}{2 a \sqrt {1-\frac {1}{a^2 x^2}}} \]

[In]

Int[E^ArcCoth[a*x]*x*Sqrt[c - a^2*c*x^2],x]

[Out]

(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)])

Rule 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.58 \[ \int e^{\coth ^{-1}(a x)} x \sqrt {c-a^2 c x^2} \, dx=\frac {x (3+2 a x) \sqrt {c-a^2 c x^2}}{6 a \sqrt {1-\frac {1}{a^2 x^2}}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.64

method	result	size
gosper	\(\frac {x^{2} \left (2 a x +3\right ) \sqrt {-a^{2} c \,x^{2}+c}}{6 \left (a x +1\right ) \sqrt {\frac {a x -1}{a x +1}}}\)	\(47\)
default	\(\frac {\left (2 a x +3\right ) x^{2} \sqrt {-c \left (a^{2} x^{2}-1\right )}}{6 \left (a x +1\right ) \sqrt {\frac {a x -1}{a x +1}}}\)	\(48\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.34 \[ \int e^{\coth ^{-1}(a x)} x \sqrt {c-a^2 c x^2} \, dx=\frac {{\left (2 \, a x^{3} + 3 \, x^{2}\right )} \sqrt {-a^{2} c}}{6 \, a} \]

[In]

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

[Out]

1/6*(2*a*x^3 + 3*x^2)*sqrt(-a^2*c)/a

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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