3.930 \(\int x^4 \sqrt{a+b x^2+c x^4} \, dx\)

Optimal. Leaf size=395 \[ \frac{\sqrt [4]{a} \left (2 \sqrt{a} \sqrt{c} \left (2 b^2-5 a c\right )-29 a b c+8 b^3\right ) \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{210 c^{11/4} \sqrt{a+b x^2+c x^4}}-\frac{2 x \left (2 b^2-5 a c\right ) \sqrt{a+b x^2+c x^4}}{105 c^2}+\frac{b x \left (8 b^2-29 a c\right ) \sqrt{a+b x^2+c x^4}}{105 c^{5/2} \left (\sqrt{a}+\sqrt{c} x^2\right )}-\frac{\sqrt [4]{a} b \left (8 b^2-29 a c\right ) \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{105 c^{11/4} \sqrt{a+b x^2+c x^4}}+\frac{x^3 \left (b+5 c x^2\right ) \sqrt{a+b x^2+c x^4}}{35 c} \]

[Out]

(-2*(2*b^2 - 5*a*c)*x*Sqrt[a + b*x^2 + c*x^4])/(105*c^2) + (b*(8*b^2 - 29*a*c)*x*Sqrt[a + b*x^2 + c*x^4])/(105
*c^(5/2)*(Sqrt[a] + Sqrt[c]*x^2)) + (x^3*(b + 5*c*x^2)*Sqrt[a + b*x^2 + c*x^4])/(35*c) - (a^(1/4)*b*(8*b^2 - 2
9*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])/(105*c^(11/4)*Sqrt[a + b*x^2 + c*x^4]) + (a^(1/4)*(8*b^3 - 29*a*b*
c + 2*Sqrt[a]*Sqrt[c]*(2*b^2 - 5*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])/(210*c^(11/4)*Sqrt[a + b*x^2 + c*x
^4])

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Rubi [A]  time = 0.251072, antiderivative size = 395, 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 = {1116, 1279, 1197, 1103, 1195} \[ -\frac{2 x \left (2 b^2-5 a c\right ) \sqrt{a+b x^2+c x^4}}{105 c^2}+\frac{b x \left (8 b^2-29 a c\right ) \sqrt{a+b x^2+c x^4}}{105 c^{5/2} \left (\sqrt{a}+\sqrt{c} x^2\right )}+\frac{\sqrt [4]{a} \left (2 \sqrt{a} \sqrt{c} \left (2 b^2-5 a c\right )-29 a b c+8 b^3\right ) \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{210 c^{11/4} \sqrt{a+b x^2+c x^4}}-\frac{\sqrt [4]{a} b \left (8 b^2-29 a c\right ) \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{105 c^{11/4} \sqrt{a+b x^2+c x^4}}+\frac{x^3 \left (b+5 c x^2\right ) \sqrt{a+b x^2+c x^4}}{35 c} \]

Antiderivative was successfully verified.

[In]

Int[x^4*Sqrt[a + b*x^2 + c*x^4],x]

[Out]

(-2*(2*b^2 - 5*a*c)*x*Sqrt[a + b*x^2 + c*x^4])/(105*c^2) + (b*(8*b^2 - 29*a*c)*x*Sqrt[a + b*x^2 + c*x^4])/(105
*c^(5/2)*(Sqrt[a] + Sqrt[c]*x^2)) + (x^3*(b + 5*c*x^2)*Sqrt[a + b*x^2 + c*x^4])/(35*c) - (a^(1/4)*b*(8*b^2 - 2
9*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])/(105*c^(11/4)*Sqrt[a + b*x^2 + c*x^4]) + (a^(1/4)*(8*b^3 - 29*a*b*
c + 2*Sqrt[a]*Sqrt[c]*(2*b^2 - 5*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])/(210*c^(11/4)*Sqrt[a + b*x^2 + c*x
^4])

Rule 1116

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

Rule 1279

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(e*f
*(f*x)^(m - 1)*(a + b*x^2 + c*x^4)^(p + 1))/(c*(m + 4*p + 3)), x] - Dist[f^2/(c*(m + 4*p + 3)), Int[(f*x)^(m -
2)*(a + b*x^2 + c*x^4)^p*Simp[a*e*(m - 1) + (b*e*(m + 2*p + 1) - c*d*(m + 4*p + 3))*x^2, x], x], x] /; FreeQ[
{a, b, c, d, e, f, p}, x] && NeQ[b^2 - 4*a*c, 0] && GtQ[m, 1] && NeQ[m + 4*p + 3, 0] && IntegerQ[2*p] && (Inte
gerQ[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 x^4 \sqrt{a+b x^2+c x^4} \, dx &=\frac{x^3 \left (b+5 c x^2\right ) \sqrt{a+b x^2+c x^4}}{35 c}-\frac{\int \frac{x^2 \left (3 a b+2 \left (2 b^2-5 a c\right ) x^2\right )}{\sqrt{a+b x^2+c x^4}} \, dx}{35 c}\\ &=-\frac{2 \left (2 b^2-5 a c\right ) x \sqrt{a+b x^2+c x^4}}{105 c^2}+\frac{x^3 \left (b+5 c x^2\right ) \sqrt{a+b x^2+c x^4}}{35 c}+\frac{\int \frac{2 a \left (2 b^2-5 a c\right )+b \left (8 b^2-29 a c\right ) x^2}{\sqrt{a+b x^2+c x^4}} \, dx}{105 c^2}\\ &=-\frac{2 \left (2 b^2-5 a c\right ) x \sqrt{a+b x^2+c x^4}}{105 c^2}+\frac{x^3 \left (b+5 c x^2\right ) \sqrt{a+b x^2+c x^4}}{35 c}-\frac{\left (\sqrt{a} b \left (8 b^2-29 a c\right )\right ) \int \frac{1-\frac{\sqrt{c} x^2}{\sqrt{a}}}{\sqrt{a+b x^2+c x^4}} \, dx}{105 c^{5/2}}+\frac{\left (\sqrt{a} \left (b \left (8 b^2-29 a c\right )+2 \sqrt{a} \sqrt{c} \left (2 b^2-5 a c\right )\right )\right ) \int \frac{1}{\sqrt{a+b x^2+c x^4}} \, dx}{105 c^{5/2}}\\ &=-\frac{2 \left (2 b^2-5 a c\right ) x \sqrt{a+b x^2+c x^4}}{105 c^2}+\frac{b \left (8 b^2-29 a c\right ) x \sqrt{a+b x^2+c x^4}}{105 c^{5/2} \left (\sqrt{a}+\sqrt{c} x^2\right )}+\frac{x^3 \left (b+5 c x^2\right ) \sqrt{a+b x^2+c x^4}}{35 c}-\frac{\sqrt [4]{a} b \left (8 b^2-29 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 )}{105 c^{11/4} \sqrt{a+b x^2+c x^4}}+\frac{\sqrt [4]{a} \left (8 b^3-29 a b c+2 \sqrt{a} \sqrt{c} \left (2 b^2-5 a c\right )\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 )}{210 c^{11/4} \sqrt{a+b x^2+c x^4}}\\ \end{align*}

Mathematica [C]  time = 1.55288, size = 538, normalized size = 1.36 \[ \frac{-i \left (-20 a^2 c^2+8 b^3 \sqrt{b^2-4 a c}+37 a b^2 c-29 a b c \sqrt{b^2-4 a c}-8 b^4\right ) \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x^2}{\sqrt{b^2-4 a c}+b}} \sqrt{\frac{-2 \sqrt{b^2-4 a c}+2 b+4 c x^2}{b-\sqrt{b^2-4 a c}}} \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{2} x \sqrt{\frac{c}{\sqrt{b^2-4 a c}+b}}\right ),\frac{\sqrt{b^2-4 a c}+b}{b-\sqrt{b^2-4 a c}}\right )+4 c x \sqrt{\frac{c}{\sqrt{b^2-4 a c}+b}} \left (10 a^2 c+a \left (-4 b^2+13 b c x^2+25 c^2 x^4\right )-b^2 c x^4-4 b^3 x^2+18 b c^2 x^6+15 c^3 x^8\right )+i b \left (8 b^2-29 a c\right ) \left (\sqrt{b^2-4 a c}-b\right ) \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x^2}{\sqrt{b^2-4 a c}+b}} \sqrt{\frac{-2 \sqrt{b^2-4 a c}+2 b+4 c x^2}{b-\sqrt{b^2-4 a c}}} E\left (i \sinh ^{-1}\left (\sqrt{2} \sqrt{\frac{c}{b+\sqrt{b^2-4 a c}}} x\right )|\frac{b+\sqrt{b^2-4 a c}}{b-\sqrt{b^2-4 a c}}\right )}{420 c^3 \sqrt{\frac{c}{\sqrt{b^2-4 a c}+b}} \sqrt{a+b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*Sqrt[a + b*x^2 + c*x^4],x]

[Out]

(4*c*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*x*(10*a^2*c - 4*b^3*x^2 - b^2*c*x^4 + 18*b*c^2*x^6 + 15*c^3*x^8 + a*(-4*b
^2 + 13*b*c*x^2 + 25*c^2*x^4)) + I*b*(8*b^2 - 29*a*c)*(-b + Sqrt[b^2 - 4*a*c])*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*(-
8*b^4 + 37*a*b^2*c - 20*a^2*c^2 + 8*b^3*Sqrt[b^2 - 4*a*c] - 29*a*b*c*Sqrt[b^2 - 4*a*c])*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]
)])/(420*c^3*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*Sqrt[a + b*x^2 + c*x^4])

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Maple [A]  time = 0.273, size = 476, normalized size = 1.2 \begin{align*}{\frac{{x}^{5}}{7}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{b{x}^{3}}{35\,c}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{x}{3\,c} \left ({\frac{2\,a}{7}}-{\frac{4\,{b}^{2}}{35\,c}} \right ) \sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{a\sqrt{2}}{12\,c} \left ({\frac{2\,a}{7}}-{\frac{4\,{b}^{2}}{35\,c}} \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{3\,ab}{35\,c}}-{\frac{2\,b}{3\,c} \left ({\frac{2\,a}{7}}-{\frac{4\,{b}^{2}}{35\,c}} \right ) } \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(x^4*(c*x^4+b*x^2+a)^(1/2),x)

[Out]

1/7*x^5*(c*x^4+b*x^2+a)^(1/2)+1/35*b/c*x^3*(c*x^4+b*x^2+a)^(1/2)+1/3*(2/7*a-4/35*b^2/c)/c*x*(c*x^4+b*x^2+a)^(1
/2)-1/12*(2/7*a-4/35*b^2/c)/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)*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))-1/2*(-3/35*b/c*a-2/3*(2/7*a-4/35*b^2/c)/c*b)*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 \sqrt{c x^{4} + b x^{2} + a} x^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c*x^4 + b*x^2 + a)*x^4, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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