3.382 \(\int \frac{x^2 \sqrt{1-x^2}}{a+b x^2+c x^4} \, dx\)

Optimal. Leaf size=263 \[ \frac{\left (-\frac{-2 a c+b^2+b c}{\sqrt{b^2-4 a c}}+b+c\right ) \tan ^{-1}\left (\frac{x \sqrt{-\sqrt{b^2-4 a c}+b+2 c}}{\sqrt{1-x^2} \sqrt{b-\sqrt{b^2-4 a c}}}\right )}{c \sqrt{b-\sqrt{b^2-4 a c}} \sqrt{-\sqrt{b^2-4 a c}+b+2 c}}+\frac{\left (\frac{-2 a c+b^2+b c}{\sqrt{b^2-4 a c}}+b+c\right ) \tan ^{-1}\left (\frac{x \sqrt{\sqrt{b^2-4 a c}+b+2 c}}{\sqrt{1-x^2} \sqrt{\sqrt{b^2-4 a c}+b}}\right )}{c \sqrt{\sqrt{b^2-4 a c}+b} \sqrt{\sqrt{b^2-4 a c}+b+2 c}}-\frac{\sin ^{-1}(x)}{c} \]

[Out]

-(ArcSin[x]/c) + ((b + c - (b^2 - 2*a*c + b*c)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[b + 2*c - Sqrt[b^2 - 4*a*c]]*x)
/(Sqrt[b - Sqrt[b^2 - 4*a*c]]*Sqrt[1 - x^2])])/(c*Sqrt[b - Sqrt[b^2 - 4*a*c]]*Sqrt[b + 2*c - Sqrt[b^2 - 4*a*c]
]) + ((b + c + (b^2 - 2*a*c + b*c)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[b + 2*c + Sqrt[b^2 - 4*a*c]]*x)/(Sqrt[b + S
qrt[b^2 - 4*a*c]]*Sqrt[1 - x^2])])/(c*Sqrt[b + Sqrt[b^2 - 4*a*c]]*Sqrt[b + 2*c + Sqrt[b^2 - 4*a*c]])

________________________________________________________________________________________

Rubi [A]  time = 2.13482, antiderivative size = 263, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {1293, 216, 1692, 377, 205} \[ \frac{\left (-\frac{-2 a c+b^2+b c}{\sqrt{b^2-4 a c}}+b+c\right ) \tan ^{-1}\left (\frac{x \sqrt{-\sqrt{b^2-4 a c}+b+2 c}}{\sqrt{1-x^2} \sqrt{b-\sqrt{b^2-4 a c}}}\right )}{c \sqrt{b-\sqrt{b^2-4 a c}} \sqrt{-\sqrt{b^2-4 a c}+b+2 c}}+\frac{\left (\frac{-2 a c+b^2+b c}{\sqrt{b^2-4 a c}}+b+c\right ) \tan ^{-1}\left (\frac{x \sqrt{\sqrt{b^2-4 a c}+b+2 c}}{\sqrt{1-x^2} \sqrt{\sqrt{b^2-4 a c}+b}}\right )}{c \sqrt{\sqrt{b^2-4 a c}+b} \sqrt{\sqrt{b^2-4 a c}+b+2 c}}-\frac{\sin ^{-1}(x)}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcSin[x]/c) + ((b + c - (b^2 - 2*a*c + b*c)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[b + 2*c - Sqrt[b^2 - 4*a*c]]*x)
/(Sqrt[b - Sqrt[b^2 - 4*a*c]]*Sqrt[1 - x^2])])/(c*Sqrt[b - Sqrt[b^2 - 4*a*c]]*Sqrt[b + 2*c - Sqrt[b^2 - 4*a*c]
]) + ((b + c + (b^2 - 2*a*c + b*c)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[b + 2*c + Sqrt[b^2 - 4*a*c]]*x)/(Sqrt[b + S
qrt[b^2 - 4*a*c]]*Sqrt[1 - x^2])])/(c*Sqrt[b + Sqrt[b^2 - 4*a*c]]*Sqrt[b + 2*c + Sqrt[b^2 - 4*a*c]])

Rule 1293

Int[(((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_)^2)^(q_))/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Dist[(
e*f^2)/c, Int[(f*x)^(m - 2)*(d + e*x^2)^(q - 1), x], x] - Dist[f^2/c, Int[((f*x)^(m - 2)*(d + e*x^2)^(q - 1)*S
imp[a*e - (c*d - b*e)*x^2, x])/(a + b*x^2 + c*x^4), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b^2 - 4*a*c,
0] &&  !IntegerQ[q] && GtQ[q, 0] && GtQ[m, 1] && LeQ[m, 3]

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 1692

Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandInteg
rand[Px*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, q}, x] && PolyQ[Px, x^2] && NeQ[b
^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{x^2 \sqrt{1-x^2}}{a+b x^2+c x^4} \, dx &=-\frac{\int \frac{1}{\sqrt{1-x^2}} \, dx}{c}-\frac{\int \frac{-a-(b+c) x^2}{\sqrt{1-x^2} \left (a+b x^2+c x^4\right )} \, dx}{c}\\ &=-\frac{\sin ^{-1}(x)}{c}-\frac{\int \left (\frac{-b-c+\frac{b^2-2 a c+b c}{\sqrt{b^2-4 a c}}}{\sqrt{1-x^2} \left (b-\sqrt{b^2-4 a c}+2 c x^2\right )}+\frac{-b-c-\frac{b^2-2 a c+b c}{\sqrt{b^2-4 a c}}}{\sqrt{1-x^2} \left (b+\sqrt{b^2-4 a c}+2 c x^2\right )}\right ) \, dx}{c}\\ &=-\frac{\sin ^{-1}(x)}{c}+\frac{\left (b+c-\frac{b^2-2 a c+b c}{\sqrt{b^2-4 a c}}\right ) \int \frac{1}{\sqrt{1-x^2} \left (b-\sqrt{b^2-4 a c}+2 c x^2\right )} \, dx}{c}+\frac{\left (b+c+\frac{b^2-2 a c+b c}{\sqrt{b^2-4 a c}}\right ) \int \frac{1}{\sqrt{1-x^2} \left (b+\sqrt{b^2-4 a c}+2 c x^2\right )} \, dx}{c}\\ &=-\frac{\sin ^{-1}(x)}{c}+\frac{\left (b+c-\frac{b^2-2 a c+b c}{\sqrt{b^2-4 a c}}\right ) \operatorname{Subst}\left (\int \frac{1}{b-\sqrt{b^2-4 a c}-\left (-b-2 c+\sqrt{b^2-4 a c}\right ) x^2} \, dx,x,\frac{x}{\sqrt{1-x^2}}\right )}{c}+\frac{\left (b+c+\frac{b^2-2 a c+b c}{\sqrt{b^2-4 a c}}\right ) \operatorname{Subst}\left (\int \frac{1}{b+\sqrt{b^2-4 a c}-\left (-b-2 c-\sqrt{b^2-4 a c}\right ) x^2} \, dx,x,\frac{x}{\sqrt{1-x^2}}\right )}{c}\\ &=-\frac{\sin ^{-1}(x)}{c}+\frac{\left (b+c-\frac{b^2-2 a c+b c}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\sqrt{b+2 c-\sqrt{b^2-4 a c}} x}{\sqrt{b-\sqrt{b^2-4 a c}} \sqrt{1-x^2}}\right )}{c \sqrt{b-\sqrt{b^2-4 a c}} \sqrt{b+2 c-\sqrt{b^2-4 a c}}}+\frac{\left (b+c+\frac{b^2-2 a c+b c}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\sqrt{b+2 c+\sqrt{b^2-4 a c}} x}{\sqrt{b+\sqrt{b^2-4 a c}} \sqrt{1-x^2}}\right )}{c \sqrt{b+\sqrt{b^2-4 a c}} \sqrt{b+2 c+\sqrt{b^2-4 a c}}}\\ \end{align*}

Mathematica [B]  time = 6.14809, size = 7543, normalized size = 28.68 \[ \text{Result too large to show} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Result too large to show

________________________________________________________________________________________

Maple [C]  time = 0.02, size = 175, normalized size = 0.7 \begin{align*} -{\frac{1}{4\,c}\sum _{{\it \_R}={\it RootOf} \left ( a{{\it \_Z}}^{8}+ \left ( 4\,a+4\,b \right ){{\it \_Z}}^{6}+ \left ( 6\,a+8\,b+16\,c \right ){{\it \_Z}}^{4}+ \left ( 4\,a+4\,b \right ){{\it \_Z}}^{2}+a \right ) }{\frac{{{\it \_R}}^{6}a+ \left ( 4\,c+3\,a+4\,b \right ){{\it \_R}}^{4}+ \left ( 4\,c+3\,a+4\,b \right ){{\it \_R}}^{2}+a}{{{\it \_R}}^{7}a+3\,{{\it \_R}}^{5}a+3\,{{\it \_R}}^{5}b+3\,{{\it \_R}}^{3}a+4\,{{\it \_R}}^{3}b+8\,{{\it \_R}}^{3}c+{\it \_R}\,a+{\it \_R}\,b}\ln \left ({\frac{1}{x} \left ( \sqrt{-{x}^{2}+1}-1 \right ) }-{\it \_R} \right ) }}+2\,{\frac{1}{c}\arctan \left ({\frac{\sqrt{-{x}^{2}+1}-1}{x}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(-x^2+1)^(1/2)/(c*x^4+b*x^2+a),x)

[Out]

-1/4/c*sum((_R^6*a+(4*c+3*a+4*b)*_R^4+(4*c+3*a+4*b)*_R^2+a)/(_R^7*a+3*_R^5*a+3*_R^5*b+3*_R^3*a+4*_R^3*b+8*_R^3
*c+_R*a+_R*b)*ln(((-x^2+1)^(1/2)-1)/x-_R),_R=RootOf(a*_Z^8+(4*a+4*b)*_Z^6+(6*a+8*b+16*c)*_Z^4+(4*a+4*b)*_Z^2+a
))+2/c*arctan(((-x^2+1)^(1/2)-1)/x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{-x^{2} + 1} x^{2}}{c x^{4} + b x^{2} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(-x^2+1)^(1/2)/(c*x^4+b*x^2+a),x, algorithm="maxima")

[Out]

integrate(sqrt(-x^2 + 1)*x^2/(c*x^4 + b*x^2 + a), x)

________________________________________________________________________________________

Fricas [B]  time = 5.62704, size = 2969, normalized size = 11.29 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(-x^2+1)^(1/2)/(c*x^4+b*x^2+a),x, algorithm="fricas")

[Out]

-1/2*(sqrt(1/2)*c*sqrt(-(b^2 - (2*a - b)*c + (b^2*c^2 - 4*a*c^3)*sqrt((b^2 + 2*b*c + c^2)/(b^2*c^4 - 4*a*c^5))
)/(b^2*c^2 - 4*a*c^3))*log(-(2*(a*b + a*c)*x^2 - 2*a*b - 2*a*c + sqrt(1/2)*((b^3 - 4*a*c^2 - (4*a*b - b^2)*c)*
sqrt(-x^2 + 1)*x - (b^3 - 4*a*c^2 - (4*a*b - b^2)*c)*x - ((b^3*c^2 - 4*a*b*c^3)*sqrt(-x^2 + 1)*x - (b^3*c^2 -
4*a*b*c^3)*x)*sqrt((b^2 + 2*b*c + c^2)/(b^2*c^4 - 4*a*c^5)))*sqrt(-(b^2 - (2*a - b)*c + (b^2*c^2 - 4*a*c^3)*sq
rt((b^2 + 2*b*c + c^2)/(b^2*c^4 - 4*a*c^5)))/(b^2*c^2 - 4*a*c^3)) + 2*(a*b + a*c)*sqrt(-x^2 + 1))/x^2) - sqrt(
1/2)*c*sqrt(-(b^2 - (2*a - b)*c + (b^2*c^2 - 4*a*c^3)*sqrt((b^2 + 2*b*c + c^2)/(b^2*c^4 - 4*a*c^5)))/(b^2*c^2
- 4*a*c^3))*log(-(2*(a*b + a*c)*x^2 - 2*a*b - 2*a*c - sqrt(1/2)*((b^3 - 4*a*c^2 - (4*a*b - b^2)*c)*sqrt(-x^2 +
1)*x - (b^3 - 4*a*c^2 - (4*a*b - b^2)*c)*x - ((b^3*c^2 - 4*a*b*c^3)*sqrt(-x^2 + 1)*x - (b^3*c^2 - 4*a*b*c^3)*
x)*sqrt((b^2 + 2*b*c + c^2)/(b^2*c^4 - 4*a*c^5)))*sqrt(-(b^2 - (2*a - b)*c + (b^2*c^2 - 4*a*c^3)*sqrt((b^2 + 2
*b*c + c^2)/(b^2*c^4 - 4*a*c^5)))/(b^2*c^2 - 4*a*c^3)) + 2*(a*b + a*c)*sqrt(-x^2 + 1))/x^2) + sqrt(1/2)*c*sqrt
(-(b^2 - (2*a - b)*c - (b^2*c^2 - 4*a*c^3)*sqrt((b^2 + 2*b*c + c^2)/(b^2*c^4 - 4*a*c^5)))/(b^2*c^2 - 4*a*c^3))
*log(-(2*(a*b + a*c)*x^2 - 2*a*b - 2*a*c + sqrt(1/2)*((b^3 - 4*a*c^2 - (4*a*b - b^2)*c)*sqrt(-x^2 + 1)*x - (b^
3 - 4*a*c^2 - (4*a*b - b^2)*c)*x + ((b^3*c^2 - 4*a*b*c^3)*sqrt(-x^2 + 1)*x - (b^3*c^2 - 4*a*b*c^3)*x)*sqrt((b^
2 + 2*b*c + c^2)/(b^2*c^4 - 4*a*c^5)))*sqrt(-(b^2 - (2*a - b)*c - (b^2*c^2 - 4*a*c^3)*sqrt((b^2 + 2*b*c + c^2)
/(b^2*c^4 - 4*a*c^5)))/(b^2*c^2 - 4*a*c^3)) + 2*(a*b + a*c)*sqrt(-x^2 + 1))/x^2) - sqrt(1/2)*c*sqrt(-(b^2 - (2
*a - b)*c - (b^2*c^2 - 4*a*c^3)*sqrt((b^2 + 2*b*c + c^2)/(b^2*c^4 - 4*a*c^5)))/(b^2*c^2 - 4*a*c^3))*log(-(2*(a
*b + a*c)*x^2 - 2*a*b - 2*a*c - sqrt(1/2)*((b^3 - 4*a*c^2 - (4*a*b - b^2)*c)*sqrt(-x^2 + 1)*x - (b^3 - 4*a*c^2
- (4*a*b - b^2)*c)*x + ((b^3*c^2 - 4*a*b*c^3)*sqrt(-x^2 + 1)*x - (b^3*c^2 - 4*a*b*c^3)*x)*sqrt((b^2 + 2*b*c +
c^2)/(b^2*c^4 - 4*a*c^5)))*sqrt(-(b^2 - (2*a - b)*c - (b^2*c^2 - 4*a*c^3)*sqrt((b^2 + 2*b*c + c^2)/(b^2*c^4 -
4*a*c^5)))/(b^2*c^2 - 4*a*c^3)) + 2*(a*b + a*c)*sqrt(-x^2 + 1))/x^2) - 4*arctan((sqrt(-x^2 + 1) - 1)/x))/c

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2} \sqrt{- \left (x - 1\right ) \left (x + 1\right )}}{a + b x^{2} + c x^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(-x**2+1)**(1/2)/(c*x**4+b*x**2+a),x)

[Out]

Integral(x**2*sqrt(-(x - 1)*(x + 1))/(a + b*x**2 + c*x**4), x)

________________________________________________________________________________________

Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(-x^2+1)^(1/2)/(c*x^4+b*x^2+a),x, algorithm="giac")

[Out]

Timed out