3.1202 \(\int e^{-\tanh ^{-1}(a x)} x^2 \sqrt {c-a^2 c x^2} \, dx\)

Optimal. Leaf size=74 \[ \frac {x^3 \sqrt {c-a^2 c x^2}}{3 \sqrt {1-a^2 x^2}}-\frac {a x^4 \sqrt {c-a^2 c x^2}}{4 \sqrt {1-a^2 x^2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.19, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6153, 6150, 43} \[ \frac {x^3 \sqrt {c-a^2 c x^2}}{3 \sqrt {1-a^2 x^2}}-\frac {a x^4 \sqrt {c-a^2 c x^2}}{4 \sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 43

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 6150

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[x^m*(1 -
a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p
] || GtQ[c, 0])

Rule 6153

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(c^IntPart[p]*(c +
d*x^2)^FracPart[p])/(1 - a^2*x^2)^FracPart[p], Int[x^m*(1 - a^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a,
c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] &&  !(IntegerQ[p] || GtQ[c, 0]) &&  !IntegerQ[n/2]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 42, normalized size = 0.57 \[ -\frac {x^3 (3 a x-4) \sqrt {c-a^2 c x^2}}{12 \sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*(x^3*(-4 + 3*a*x)*Sqrt[c - a^2*c*x^2])/Sqrt[1 - a^2*x^2]

________________________________________________________________________________________

fricas [A]  time = 0.50, size = 50, normalized size = 0.68 \[ \frac {\sqrt {-a^{2} c x^{2} + c} {\left (3 \, a x^{4} - 4 \, x^{3}\right )} \sqrt {-a^{2} x^{2} + 1}}{12 \, {\left (a^{2} x^{2} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*sqrt(-a^2*c*x^2 + c)*(3*a*x^4 - 4*x^3)*sqrt(-a^2*x^2 + 1)/(a^2*x^2 - 1)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} x^{2}}{a x + 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.03, size = 51, normalized size = 0.69 \[ \frac {x^{3} \left (3 a x -4\right ) \sqrt {-a^{2} c \,x^{2}+c}\, \sqrt {-a^{2} x^{2}+1}}{12 \left (a x -1\right ) \left (a x +1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*x^3*(3*a*x-4)*(-a^2*c*x^2+c)^(1/2)*(-a^2*x^2+1)^(1/2)/(a*x-1)/(a*x+1)

________________________________________________________________________________________

maxima [A]  time = 0.35, size = 41, normalized size = 0.55 \[ -\frac {{\left (3 \, a \sqrt {c} x^{4} - 4 \, \sqrt {c} x^{3}\right )} {\left (a x + 1\right )} {\left (a x - 1\right )}}{12 \, {\left (a^{2} x^{2} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*a*sqrt(c)*x^4 - 4*sqrt(c)*x^3)*(a*x + 1)*(a*x - 1)/(a^2*x^2 - 1)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^2\,\sqrt {c-a^2\,c\,x^2}\,\sqrt {1-a^2\,x^2}}{a\,x+1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \sqrt {- c \left (a x - 1\right ) \left (a x + 1\right )}}{a x + 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________