3.993 \(\int \frac{1}{x^2 (a+b x^2+c x^4)^{3/2}} \, dx\)

Optimal. Leaf size=428 \[ \frac{\sqrt [4]{c} \left (\sqrt{a} b \sqrt{c}-6 a c+2 b^2\right ) \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{2 a^{7/4} \left (b^2-4 a c\right ) \sqrt{a+b x^2+c x^4}}-\frac{2 \left (b^2-3 a c\right ) \sqrt{a+b x^2+c x^4}}{a^2 x \left (b^2-4 a c\right )}+\frac{2 \sqrt{c} x \left (b^2-3 a c\right ) \sqrt{a+b x^2+c x^4}}{a^2 \left (b^2-4 a c\right ) \left (\sqrt{a}+\sqrt{c} x^2\right )}-\frac{2 \sqrt [4]{c} \left (b^2-3 a c\right ) \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{a^{7/4} \left (b^2-4 a c\right ) \sqrt{a+b x^2+c x^4}}+\frac{-2 a c+b^2+b c x^2}{a x \left (b^2-4 a c\right ) \sqrt{a+b x^2+c x^4}} \]

[Out]

(b^2 - 2*a*c + b*c*x^2)/(a*(b^2 - 4*a*c)*x*Sqrt[a + b*x^2 + c*x^4]) - (2*(b^2 - 3*a*c)*Sqrt[a + b*x^2 + c*x^4]
)/(a^2*(b^2 - 4*a*c)*x) + (2*Sqrt[c]*(b^2 - 3*a*c)*x*Sqrt[a + b*x^2 + c*x^4])/(a^2*(b^2 - 4*a*c)*(Sqrt[a] + Sq
rt[c]*x^2)) - (2*c^(1/4)*(b^2 - 3*a*c)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + b*x^2 + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2
)^2]*EllipticE[2*ArcTan[(c^(1/4)*x)/a^(1/4)], (2 - b/(Sqrt[a]*Sqrt[c]))/4])/(a^(7/4)*(b^2 - 4*a*c)*Sqrt[a + b*
x^2 + c*x^4]) + (c^(1/4)*(2*b^2 + Sqrt[a]*b*Sqrt[c] - 6*a*c)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + b*x^2 + c*x^4)/
(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)], (2 - b/(Sqrt[a]*Sqrt[c]))/4])/(2*a^(7/4)*(
b^2 - 4*a*c)*Sqrt[a + b*x^2 + c*x^4])

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Rubi [A]  time = 0.216941, antiderivative size = 428, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {1121, 1281, 1197, 1103, 1195} \[ -\frac{2 \left (b^2-3 a c\right ) \sqrt{a+b x^2+c x^4}}{a^2 x \left (b^2-4 a c\right )}+\frac{2 \sqrt{c} x \left (b^2-3 a c\right ) \sqrt{a+b x^2+c x^4}}{a^2 \left (b^2-4 a c\right ) \left (\sqrt{a}+\sqrt{c} x^2\right )}+\frac{\sqrt [4]{c} \left (\sqrt{a} b \sqrt{c}-6 a c+2 b^2\right ) \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{2 a^{7/4} \left (b^2-4 a c\right ) \sqrt{a+b x^2+c x^4}}-\frac{2 \sqrt [4]{c} \left (b^2-3 a c\right ) \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{a^{7/4} \left (b^2-4 a c\right ) \sqrt{a+b x^2+c x^4}}+\frac{-2 a c+b^2+b c x^2}{a x \left (b^2-4 a c\right ) \sqrt{a+b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a + b*x^2 + c*x^4)^(3/2)),x]

[Out]

(b^2 - 2*a*c + b*c*x^2)/(a*(b^2 - 4*a*c)*x*Sqrt[a + b*x^2 + c*x^4]) - (2*(b^2 - 3*a*c)*Sqrt[a + b*x^2 + c*x^4]
)/(a^2*(b^2 - 4*a*c)*x) + (2*Sqrt[c]*(b^2 - 3*a*c)*x*Sqrt[a + b*x^2 + c*x^4])/(a^2*(b^2 - 4*a*c)*(Sqrt[a] + Sq
rt[c]*x^2)) - (2*c^(1/4)*(b^2 - 3*a*c)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + b*x^2 + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2
)^2]*EllipticE[2*ArcTan[(c^(1/4)*x)/a^(1/4)], (2 - b/(Sqrt[a]*Sqrt[c]))/4])/(a^(7/4)*(b^2 - 4*a*c)*Sqrt[a + b*
x^2 + c*x^4]) + (c^(1/4)*(2*b^2 + Sqrt[a]*b*Sqrt[c] - 6*a*c)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + b*x^2 + c*x^4)/
(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)], (2 - b/(Sqrt[a]*Sqrt[c]))/4])/(2*a^(7/4)*(
b^2 - 4*a*c)*Sqrt[a + b*x^2 + c*x^4])

Rule 1121

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> -Simp[((d*x)^(m + 1)*(b^2 - 2*a
*c + b*c*x^2)*(a + b*x^2 + c*x^4)^(p + 1))/(2*a*d*(p + 1)*(b^2 - 4*a*c)), x] + Dist[1/(2*a*(p + 1)*(b^2 - 4*a*
c)), Int[(d*x)^m*(a + b*x^2 + c*x^4)^(p + 1)*Simp[b^2*(m + 2*p + 3) - 2*a*c*(m + 4*p + 5) + b*c*(m + 4*p + 7)*
x^2, x], x], x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && IntegerQ[2*p] && (Integer
Q[p] || IntegerQ[m])

Rule 1281

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(d*(
f*x)^(m + 1)*(a + b*x^2 + c*x^4)^(p + 1))/(a*f*(m + 1)), x] + Dist[1/(a*f^2*(m + 1)), Int[(f*x)^(m + 2)*(a + b
*x^2 + c*x^4)^p*Simp[a*e*(m + 1) - b*d*(m + 2*p + 3) - c*d*(m + 4*p + 5)*x^2, x], x], x] /; FreeQ[{a, b, c, d,
e, f, p}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[m, -1] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])

Rule 1197

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 2]}, Dist[(
e + d*q)/q, Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] - Dist[e/q, Int[(1 - q*x^2)/Sqrt[a + b*x^2 + c*x^4], x], x]
/; NeQ[e + d*q, 0]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1103

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(
a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2 - (b*q^2)/(4*c)])/(2*q*Sqrt[a + b*x^2 + c
*x^4]), x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1195

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[
(d*x*Sqrt[a + b*x^2 + c*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + b*x^2 + c*x^4)/(a*(1 + q
^2*x^2)^2)]*EllipticE[2*ArcTan[q*x], 1/2 - (b*q^2)/(4*c)])/(q*Sqrt[a + b*x^2 + c*x^4]), x] /; EqQ[e + d*q^2, 0
]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rubi steps

\begin{align*} \int \frac{1}{x^2 \left (a+b x^2+c x^4\right )^{3/2}} \, dx &=\frac{b^2-2 a c+b c x^2}{a \left (b^2-4 a c\right ) x \sqrt{a+b x^2+c x^4}}-\frac{\int \frac{-2 \left (b^2-3 a c\right )-b c x^2}{x^2 \sqrt{a+b x^2+c x^4}} \, dx}{a \left (b^2-4 a c\right )}\\ &=\frac{b^2-2 a c+b c x^2}{a \left (b^2-4 a c\right ) x \sqrt{a+b x^2+c x^4}}-\frac{2 \left (b^2-3 a c\right ) \sqrt{a+b x^2+c x^4}}{a^2 \left (b^2-4 a c\right ) x}+\frac{\int \frac{a b c+2 c \left (b^2-3 a c\right ) x^2}{\sqrt{a+b x^2+c x^4}} \, dx}{a^2 \left (b^2-4 a c\right )}\\ &=\frac{b^2-2 a c+b c x^2}{a \left (b^2-4 a c\right ) x \sqrt{a+b x^2+c x^4}}-\frac{2 \left (b^2-3 a c\right ) \sqrt{a+b x^2+c x^4}}{a^2 \left (b^2-4 a c\right ) x}-\frac{\left (2 \sqrt{c} \left (b^2-3 a c\right )\right ) \int \frac{1-\frac{\sqrt{c} x^2}{\sqrt{a}}}{\sqrt{a+b x^2+c x^4}} \, dx}{a^{3/2} \left (b^2-4 a c\right )}+\frac{\left (\sqrt{a} b c^{3/2}+2 c \left (b^2-3 a c\right )\right ) \int \frac{1}{\sqrt{a+b x^2+c x^4}} \, dx}{a^{3/2} \sqrt{c} \left (b^2-4 a c\right )}\\ &=\frac{b^2-2 a c+b c x^2}{a \left (b^2-4 a c\right ) x \sqrt{a+b x^2+c x^4}}-\frac{2 \left (b^2-3 a c\right ) \sqrt{a+b x^2+c x^4}}{a^2 \left (b^2-4 a c\right ) x}+\frac{2 \sqrt{c} \left (b^2-3 a c\right ) x \sqrt{a+b x^2+c x^4}}{a^2 \left (b^2-4 a c\right ) \left (\sqrt{a}+\sqrt{c} x^2\right )}-\frac{2 \sqrt [4]{c} \left (b^2-3 a c\right ) \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{a^{7/4} \left (b^2-4 a c\right ) \sqrt{a+b x^2+c x^4}}+\frac{\sqrt [4]{c} \left (2 b^2+\sqrt{a} b \sqrt{c}-6 a c\right ) \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{2 a^{7/4} \left (b^2-4 a c\right ) \sqrt{a+b x^2+c x^4}}\\ \end{align*}

Mathematica [C]  time = 1.32956, size = 515, normalized size = 1.2 \[ -\frac{i x \left (b^2 \sqrt{b^2-4 a c}-3 a c \sqrt{b^2-4 a c}+4 a b c-b^3\right ) \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x^2}{\sqrt{b^2-4 a c}+b}} \sqrt{\frac{-2 \sqrt{b^2-4 a c}+2 b+4 c x^2}{b-\sqrt{b^2-4 a c}}} \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{2} x \sqrt{\frac{c}{\sqrt{b^2-4 a c}+b}}\right ),\frac{\sqrt{b^2-4 a c}+b}{b-\sqrt{b^2-4 a c}}\right )+2 \sqrt{\frac{c}{\sqrt{b^2-4 a c}+b}} \left (-4 a^2 c+a \left (b^2-7 b c x^2-6 c^2 x^4\right )+2 b^2 x^2 \left (b+c x^2\right )\right )-i x \left (b^2-3 a c\right ) \left (\sqrt{b^2-4 a c}-b\right ) \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x^2}{\sqrt{b^2-4 a c}+b}} \sqrt{\frac{-2 \sqrt{b^2-4 a c}+2 b+4 c x^2}{b-\sqrt{b^2-4 a c}}} E\left (i \sinh ^{-1}\left (\sqrt{2} \sqrt{\frac{c}{b+\sqrt{b^2-4 a c}}} x\right )|\frac{b+\sqrt{b^2-4 a c}}{b-\sqrt{b^2-4 a c}}\right )}{2 a^2 x \left (b^2-4 a c\right ) \sqrt{\frac{c}{\sqrt{b^2-4 a c}+b}} \sqrt{a+b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(a + b*x^2 + c*x^4)^(3/2)),x]

[Out]

-(2*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*(-4*a^2*c + 2*b^2*x^2*(b + c*x^2) + a*(b^2 - 7*b*c*x^2 - 6*c^2*x^4)) - I*(
b^2 - 3*a*c)*(-b + Sqrt[b^2 - 4*a*c])*x*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])]*Sqrt[(
2*b - 2*Sqrt[b^2 - 4*a*c] + 4*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*EllipticE[I*ArcSinh[Sqrt[2]*Sqrt[c/(b + Sqrt[b^2
- 4*a*c])]*x], (b + Sqrt[b^2 - 4*a*c])/(b - Sqrt[b^2 - 4*a*c])] + I*(-b^3 + 4*a*b*c + b^2*Sqrt[b^2 - 4*a*c] -
3*a*c*Sqrt[b^2 - 4*a*c])*x*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])]*Sqrt[(2*b - 2*Sqrt
[b^2 - 4*a*c] + 4*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*EllipticF[I*ArcSinh[Sqrt[2]*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*
x], (b + Sqrt[b^2 - 4*a*c])/(b - Sqrt[b^2 - 4*a*c])])/(2*a^2*(b^2 - 4*a*c)*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*x*S
qrt[a + b*x^2 + c*x^4])

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Maple [A]  time = 0.229, size = 536, normalized size = 1.3 \begin{align*} -2\,{c \left ( 1/2\,{\frac{ \left ( 2\,ac-{b}^{2} \right ){x}^{3}}{ \left ( 4\,ac-{b}^{2} \right ){a}^{2}}}+1/2\,{\frac{b \left ( 3\,ac-{b}^{2} \right ) x}{{a}^{2} \left ( 4\,ac-{b}^{2} \right ) c}} \right ){\frac{1}{\sqrt{ \left ({x}^{4}+{\frac{b{x}^{2}}{c}}+{\frac{a}{c}} \right ) c}}}}-{\frac{1}{{a}^{2}x}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{\sqrt{2}}{4} \left ( -{\frac{b}{{a}^{2}}}+{\frac{b \left ( 3\,ac-{b}^{2} \right ) }{ \left ( 4\,ac-{b}^{2} \right ){a}^{2}}} \right ) \sqrt{4-2\,{\frac{ \left ( -b+\sqrt{-4\,ac+{b}^{2}} \right ){x}^{2}}{a}}}\sqrt{4+2\,{\frac{ \left ( b+\sqrt{-4\,ac+{b}^{2}} \right ){x}^{2}}{a}}}{\it EllipticF} \left ({\frac{x\sqrt{2}}{2}\sqrt{{\frac{1}{a} \left ( -b+\sqrt{-4\,ac+{b}^{2}} \right ) }}},{\frac{1}{2}\sqrt{-4+2\,{\frac{b \left ( b+\sqrt{-4\,ac+{b}^{2}} \right ) }{ac}}}} \right ){\frac{1}{\sqrt{{\frac{1}{a} \left ( -b+\sqrt{-4\,ac+{b}^{2}} \right ) }}}}{\frac{1}{\sqrt{c{x}^{4}+b{x}^{2}+a}}}}-{\frac{a\sqrt{2}}{2} \left ({\frac{c \left ( 2\,ac-{b}^{2} \right ) }{ \left ( 4\,ac-{b}^{2} \right ){a}^{2}}}+{\frac{c}{{a}^{2}}} \right ) \sqrt{4-2\,{\frac{ \left ( -b+\sqrt{-4\,ac+{b}^{2}} \right ){x}^{2}}{a}}}\sqrt{4+2\,{\frac{ \left ( b+\sqrt{-4\,ac+{b}^{2}} \right ){x}^{2}}{a}}} \left ({\it EllipticF} \left ({\frac{x\sqrt{2}}{2}\sqrt{{\frac{1}{a} \left ( -b+\sqrt{-4\,ac+{b}^{2}} \right ) }}},{\frac{1}{2}\sqrt{-4+2\,{\frac{b \left ( b+\sqrt{-4\,ac+{b}^{2}} \right ) }{ac}}}} \right ) -{\it EllipticE} \left ({\frac{x\sqrt{2}}{2}\sqrt{{\frac{1}{a} \left ( -b+\sqrt{-4\,ac+{b}^{2}} \right ) }}},{\frac{1}{2}\sqrt{-4+2\,{\frac{b \left ( b+\sqrt{-4\,ac+{b}^{2}} \right ) }{ac}}}} \right ) \right ){\frac{1}{\sqrt{{\frac{1}{a} \left ( -b+\sqrt{-4\,ac+{b}^{2}} \right ) }}}}{\frac{1}{\sqrt{c{x}^{4}+b{x}^{2}+a}}} \left ( b+\sqrt{-4\,ac+{b}^{2}} \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^2/(c*x^4+b*x^2+a)^(3/2),x)

[Out]

-2*c*(1/2*(2*a*c-b^2)/(4*a*c-b^2)/a^2*x^3+1/2*b*(3*a*c-b^2)/a^2/(4*a*c-b^2)/c*x)/((x^4+b/c*x^2+a/c)*c)^(1/2)-1
/a^2*(c*x^4+b*x^2+a)^(1/2)/x+1/4*(-b/a^2+b*(3*a*c-b^2)/a^2/(4*a*c-b^2))*2^(1/2)/((-b+(-4*a*c+b^2)^(1/2))/a)^(1
/2)*(4-2*(-b+(-4*a*c+b^2)^(1/2))/a*x^2)^(1/2)*(4+2*(b+(-4*a*c+b^2)^(1/2))/a*x^2)^(1/2)/(c*x^4+b*x^2+a)^(1/2)*E
llipticF(1/2*x*2^(1/2)*((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2),1/2*(-4+2*b*(b+(-4*a*c+b^2)^(1/2))/a/c)^(1/2))-1/2*(c
*(2*a*c-b^2)/(4*a*c-b^2)/a^2+1/a^2*c)*a*2^(1/2)/((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*(4-2*(-b+(-4*a*c+b^2)^(1/2))
/a*x^2)^(1/2)*(4+2*(b+(-4*a*c+b^2)^(1/2))/a*x^2)^(1/2)/(c*x^4+b*x^2+a)^(1/2)/(b+(-4*a*c+b^2)^(1/2))*(EllipticF
(1/2*x*2^(1/2)*((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2),1/2*(-4+2*b*(b+(-4*a*c+b^2)^(1/2))/a/c)^(1/2))-EllipticE(1/2*
x*2^(1/2)*((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2),1/2*(-4+2*b*(b+(-4*a*c+b^2)^(1/2))/a/c)^(1/2)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (c x^{4} + b x^{2} + a\right )}^{\frac{3}{2}} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(c*x^4+b*x^2+a)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((c*x^4 + b*x^2 + a)^(3/2)*x^2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{c x^{4} + b x^{2} + a}}{c^{2} x^{10} + 2 \, b c x^{8} +{\left (b^{2} + 2 \, a c\right )} x^{6} + 2 \, a b x^{4} + a^{2} x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(c*x^4+b*x^2+a)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*x^4 + b*x^2 + a)/(c^2*x^10 + 2*b*c*x^8 + (b^2 + 2*a*c)*x^6 + 2*a*b*x^4 + a^2*x^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{2} \left (a + b x^{2} + c x^{4}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**2/(c*x**4+b*x**2+a)**(3/2),x)

[Out]

Integral(1/(x**2*(a + b*x**2 + c*x**4)**(3/2)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (c x^{4} + b x^{2} + a\right )}^{\frac{3}{2}} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(c*x^4+b*x^2+a)^(3/2),x, algorithm="giac")

[Out]

integrate(1/((c*x^4 + b*x^2 + a)^(3/2)*x^2), x)