3.10.47 \(\int x^4 (a+b x^2+c x^4)^{3/2} \, dx\) [947]

Optimal. Leaf size=495 \[ \frac {\left (8 b^4-51 a b^2 c+60 a^2 c^2\right ) x \sqrt {a+b x^2+c x^4}}{1155 c^3}-\frac {8 b \left (2 b^2-9 a c\right ) \left (b^2-3 a c\right ) x \sqrt {a+b x^2+c x^4}}{1155 c^{7/2} \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {x^3 \left (b \left (2 b^2+a c\right )+10 c \left (b^2-3 a c\right ) x^2\right ) \sqrt {a+b x^2+c x^4}}{385 c^2}+\frac {x^3 \left (b+3 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{33 c}+\frac {8 \sqrt [4]{a} b \left (2 b^2-9 a c\right ) \left (b^2-3 a c\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{1155 c^{15/4} \sqrt {a+b x^2+c x^4}}-\frac {\sqrt [4]{a} \left (8 b \left (2 b^2-9 a c\right ) \left (b^2-3 a c\right )+\sqrt {a} \sqrt {c} \left (8 b^4-51 a b^2 c+60 a^2 c^2\right )\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{2310 c^{15/4} \sqrt {a+b x^2+c x^4}} \]

[Out]

1/33*x^3*(3*c*x^2+b)*(c*x^4+b*x^2+a)^(3/2)/c+1/1155*(60*a^2*c^2-51*a*b^2*c+8*b^4)*x*(c*x^4+b*x^2+a)^(1/2)/c^3-
1/385*x^3*(b*(a*c+2*b^2)+10*c*(-3*a*c+b^2)*x^2)*(c*x^4+b*x^2+a)^(1/2)/c^2-8/1155*b*(-9*a*c+2*b^2)*(-3*a*c+b^2)
*x*(c*x^4+b*x^2+a)^(1/2)/c^(7/2)/(a^(1/2)+x^2*c^(1/2))+8/1155*a^(1/4)*b*(-9*a*c+2*b^2)*(-3*a*c+b^2)*(cos(2*arc
tan(c^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x/a^(1/4)))*EllipticE(sin(2*arctan(c^(1/4)*x/a^(1/4))),1
/2*(2-b/a^(1/2)/c^(1/2))^(1/2))*(a^(1/2)+x^2*c^(1/2))*((c*x^4+b*x^2+a)/(a^(1/2)+x^2*c^(1/2))^2)^(1/2)/c^(15/4)
/(c*x^4+b*x^2+a)^(1/2)-1/2310*a^(1/4)*(cos(2*arctan(c^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x/a^(1/4
)))*EllipticF(sin(2*arctan(c^(1/4)*x/a^(1/4))),1/2*(2-b/a^(1/2)/c^(1/2))^(1/2))*(a^(1/2)+x^2*c^(1/2))*(8*b*(-9
*a*c+2*b^2)*(-3*a*c+b^2)+(60*a^2*c^2-51*a*b^2*c+8*b^4)*a^(1/2)*c^(1/2))*((c*x^4+b*x^2+a)/(a^(1/2)+x^2*c^(1/2))
^2)^(1/2)/c^(15/4)/(c*x^4+b*x^2+a)^(1/2)

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Rubi [A]
time = 0.28, antiderivative size = 495, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1130, 1287, 1293, 1211, 1117, 1209} \begin {gather*} -\frac {\sqrt [4]{a} \left (\sqrt {a} \sqrt {c} \left (60 a^2 c^2-51 a b^2 c+8 b^4\right )+8 b \left (2 b^2-9 a c\right ) \left (b^2-3 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 \text {ArcTan}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{2310 c^{15/4} \sqrt {a+b x^2+c x^4}}+\frac {x \left (60 a^2 c^2-51 a b^2 c+8 b^4\right ) \sqrt {a+b x^2+c x^4}}{1155 c^3}+\frac {8 \sqrt [4]{a} b \left (2 b^2-9 a c\right ) \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 \text {ArcTan}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{1155 c^{15/4} \sqrt {a+b x^2+c x^4}}-\frac {8 b x \left (2 b^2-9 a c\right ) \left (b^2-3 a c\right ) \sqrt {a+b x^2+c x^4}}{1155 c^{7/2} \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {x^3 \left (10 c x^2 \left (b^2-3 a c\right )+b \left (a c+2 b^2\right )\right ) \sqrt {a+b x^2+c x^4}}{385 c^2}+\frac {x^3 \left (b+3 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{33 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4*(a + b*x^2 + c*x^4)^(3/2),x]

[Out]

((8*b^4 - 51*a*b^2*c + 60*a^2*c^2)*x*Sqrt[a + b*x^2 + c*x^4])/(1155*c^3) - (8*b*(2*b^2 - 9*a*c)*(b^2 - 3*a*c)*
x*Sqrt[a + b*x^2 + c*x^4])/(1155*c^(7/2)*(Sqrt[a] + Sqrt[c]*x^2)) - (x^3*(b*(2*b^2 + a*c) + 10*c*(b^2 - 3*a*c)
*x^2)*Sqrt[a + b*x^2 + c*x^4])/(385*c^2) + (x^3*(b + 3*c*x^2)*(a + b*x^2 + c*x^4)^(3/2))/(33*c) + (8*a^(1/4)*b
*(2*b^2 - 9*a*c)*(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]*Ell
ipticE[2*ArcTan[(c^(1/4)*x)/a^(1/4)], (2 - b/(Sqrt[a]*Sqrt[c]))/4])/(1155*c^(15/4)*Sqrt[a + b*x^2 + c*x^4]) -
(a^(1/4)*(8*b*(2*b^2 - 9*a*c)*(b^2 - 3*a*c) + Sqrt[a]*Sqrt[c]*(8*b^4 - 51*a*b^2*c + 60*a^2*c^2))*(Sqrt[a] + Sq
rt[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])/(2310*c^(15/4)*Sqrt[a + b*x^2 + c*x^4])

Rule 1117

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)]/(2*q*Sqrt[a + b*x^2 + c*x^4]))*EllipticF[2*ArcTan[q*x], 1/2 - b*(q^2/(
4*c))], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1130

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 1209

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)]/(q*Sqrt[a + b*x^2 + c*x^4]))*EllipticE[2*ArcTan[q*x], 1/2 - b*(q^2/(4*c))], x] /; EqQ[e + d*q^2,
0]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1211

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 1287

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

Rule 1293

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])

Rubi steps

\begin {align*} \int x^4 \left (a+b x^2+c x^4\right )^{3/2} \, dx &=\frac {x^3 \left (b+3 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{33 c}-\frac {\int x^2 \left (3 a b+6 \left (b^2-3 a c\right ) x^2\right ) \sqrt {a+b x^2+c x^4} \, dx}{33 c}\\ &=-\frac {x^3 \left (b \left (2 b^2+a c\right )+10 c \left (b^2-3 a c\right ) x^2\right ) \sqrt {a+b x^2+c x^4}}{385 c^2}+\frac {x^3 \left (b+3 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{33 c}-\frac {\int \frac {x^2 \left (-6 a b \left (3 b^2-16 a c\right )-3 \left (8 b^4-51 a b^2 c+60 a^2 c^2\right ) x^2\right )}{\sqrt {a+b x^2+c x^4}} \, dx}{1155 c^2}\\ &=\frac {\left (8 b^4-51 a b^2 c+60 a^2 c^2\right ) x \sqrt {a+b x^2+c x^4}}{1155 c^3}-\frac {x^3 \left (b \left (2 b^2+a c\right )+10 c \left (b^2-3 a c\right ) x^2\right ) \sqrt {a+b x^2+c x^4}}{385 c^2}+\frac {x^3 \left (b+3 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{33 c}+\frac {\int \frac {-3 a \left (8 b^4-51 a b^2 c+60 a^2 c^2\right )-24 b \left (2 b^2-9 a c\right ) \left (b^2-3 a c\right ) x^2}{\sqrt {a+b x^2+c x^4}} \, dx}{3465 c^3}\\ &=\frac {\left (8 b^4-51 a b^2 c+60 a^2 c^2\right ) x \sqrt {a+b x^2+c x^4}}{1155 c^3}-\frac {x^3 \left (b \left (2 b^2+a c\right )+10 c \left (b^2-3 a c\right ) x^2\right ) \sqrt {a+b x^2+c x^4}}{385 c^2}+\frac {x^3 \left (b+3 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{33 c}+\frac {\left (8 \sqrt {a} b \left (2 b^2-9 a c\right ) \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}{1155 c^{7/2}}-\frac {\left (\sqrt {a} \left (8 b \left (2 b^2-9 a c\right ) \left (b^2-3 a c\right )+\sqrt {a} \sqrt {c} \left (8 b^4-51 a b^2 c+60 a^2 c^2\right )\right )\right ) \int \frac {1}{\sqrt {a+b x^2+c x^4}} \, dx}{1155 c^{7/2}}\\ &=\frac {\left (8 b^4-51 a b^2 c+60 a^2 c^2\right ) x \sqrt {a+b x^2+c x^4}}{1155 c^3}-\frac {8 b \left (2 b^2-9 a c\right ) \left (b^2-3 a c\right ) x \sqrt {a+b x^2+c x^4}}{1155 c^{7/2} \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {x^3 \left (b \left (2 b^2+a c\right )+10 c \left (b^2-3 a c\right ) x^2\right ) \sqrt {a+b x^2+c x^4}}{385 c^2}+\frac {x^3 \left (b+3 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{33 c}+\frac {8 \sqrt [4]{a} b \left (2 b^2-9 a c\right ) \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 )}{1155 c^{15/4} \sqrt {a+b x^2+c x^4}}-\frac {\sqrt [4]{a} \left (8 b \left (2 b^2-9 a c\right ) \left (b^2-3 a c\right )+\sqrt {a} \sqrt {c} \left (8 b^4-51 a b^2 c+60 a^2 c^2\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 )}{2310 c^{15/4} \sqrt {a+b x^2+c x^4}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 11.46, size = 657, normalized size = 1.33 \begin {gather*} \frac {2 c \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} x \left (60 a^3 c^2+a^2 c \left (-51 b^2+92 b c x^2+255 c^2 x^4\right )+a \left (8 b^4-57 b^3 c x^2-14 b^2 c^2 x^4+367 b c^3 x^6+300 c^4 x^8\right )+x^2 \left (8 b^5+2 b^4 c x^2-b^3 c^2 x^4+145 b^2 c^3 x^6+245 b c^4 x^8+105 c^5 x^{10}\right )\right )-4 i b \left (2 b^4-15 a b^2 c+27 a^2 c^2\right ) \left (-b+\sqrt {b^2-4 a c}\right ) \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}} \sqrt {\frac {2 b-2 \sqrt {b^2-4 a c}+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 )+i \left (-8 b^6+68 a b^4 c-159 a^2 b^2 c^2+60 a^3 c^3+8 b^5 \sqrt {b^2-4 a c}-60 a b^3 c \sqrt {b^2-4 a c}+108 a^2 b c^2 \sqrt {b^2-4 a c}\right ) \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}} \sqrt {\frac {2 b-2 \sqrt {b^2-4 a c}+4 c x^2}{b-\sqrt {b^2-4 a c}}} F\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 )}{2310 c^4 \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} \sqrt {a+b x^2+c x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*x*(60*a^3*c^2 + a^2*c*(-51*b^2 + 92*b*c*x^2 + 255*c^2*x^4) + a*(8*b^4 - 5
7*b^3*c*x^2 - 14*b^2*c^2*x^4 + 367*b*c^3*x^6 + 300*c^4*x^8) + x^2*(8*b^5 + 2*b^4*c*x^2 - b^3*c^2*x^4 + 145*b^2
*c^3*x^6 + 245*b*c^4*x^8 + 105*c^5*x^10)) - (4*I)*b*(2*b^4 - 15*a*b^2*c + 27*a^2*c^2)*(-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^6 + 68*a*b^4*c - 159*a^2*b^2*c^2 + 60*a^3*c^3 + 8*b^5*Sqrt[b^2 - 4*a*c] -
60*a*b^3*c*Sqrt[b^2 - 4*a*c] + 108*a^2*b*c^2*Sqrt[b^2 - 4*a*c])*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b + Sq
rt[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])])/(2310*c^4*Sqrt[c/(b
+ Sqrt[b^2 - 4*a*c])]*Sqrt[a + b*x^2 + c*x^4])

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Maple [A]
time = 0.05, size = 674, normalized size = 1.36 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(c*x^4+b*x^2+a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/11*c*x^9*(c*x^4+b*x^2+a)^(1/2)+4/33*b*x^7*(c*x^4+b*x^2+a)^(1/2)+1/7*(13/11*a*c+1/33*b^2)/c*x^5*(c*x^4+b*x^2+
a)^(1/2)+1/5*(38/33*a*b-6/7*(13/11*a*c+1/33*b^2)/c*b)/c*x^3*(c*x^4+b*x^2+a)^(1/2)+1/3*(a^2-5/7*(13/11*a*c+1/33
*b^2)/c*a-4/5*(38/33*a*b-6/7*(13/11*a*c+1/33*b^2)/c*b)/c*b)/c*x*(c*x^4+b*x^2+a)^(1/2)-1/12*(a^2-5/7*(13/11*a*c
+1/33*b^2)/c*a-4/5*(38/33*a*b-6/7*(13/11*a*c+1/33*b^2)/c*b)/c*b)/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)*Elli
pticF(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*(-3/5
*(38/33*a*b-6/7*(13/11*a*c+1/33*b^2)/c*b)/c*a-2/3*(a^2-5/7*(13/11*a*c+1/33*b^2)/c*a-4/5*(38/33*a*b-6/7*(13/11*
a*c+1/33*b^2)/c*b)/c*b)/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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> Symbolic function elliptic_ec takes exactly 1 arguments (2 given)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{4} \left (a + b x^{2} + c x^{4}\right )^{\frac {3}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x^4\,{\left (c\,x^4+b\,x^2+a\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a + b*x^2 + c*x^4)^(3/2),x)

[Out]

int(x^4*(a + b*x^2 + c*x^4)^(3/2), x)

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