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

Optimal. Leaf size=76 \[ \frac {x^3 \sqrt {c-a^2 c x^2}}{4 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {x^2 \sqrt {c-a^2 c x^2}}{3 a \sqrt {1-\frac {1}{a^2 x^2}}} \]

[Out]

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

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Rubi [A]  time = 0.22, antiderivative size = 76, 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 = {6192, 6193, 43} \[ \frac {x^3 \sqrt {c-a^2 c x^2}}{4 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {x^2 \sqrt {c-a^2 c x^2}}{3 a \sqrt {1-\frac {1}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 45, normalized size = 0.59 \[ \frac {x^2 (3 a x-4) \sqrt {c-a^2 c x^2}}{12 a \sqrt {1-\frac {1}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.69, size = 25, normalized size = 0.33 \[ \frac {{\left (3 \, a x^{4} - 4 \, x^{3}\right )} \sqrt {-a^{2} c}}{12 \, a} \]

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*x+1))^(1/2),x, algorithm="fricas")

[Out]

1/12*(3*a*x^4 - 4*x^3)*sqrt(-a^2*c)/a

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

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*x+1))^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.04, size = 47, normalized size = 0.62 \[ \frac {x^{3} \left (3 a x -4\right ) \sqrt {-a^{2} c \,x^{2}+c}\, \sqrt {\frac {a x -1}{a x +1}}}{12 a x -12} \]

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*x+1))^(1/2),x)

[Out]

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

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

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*x+1))^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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*x+1))**(1/2),x)

[Out]

Timed out

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