3.10.48 \(\int x^2 (a+b x^2+c x^4)^{3/2} \, dx\) [948]

Optimal. Leaf size=443 \[ \frac {\left (8 b^4-57 a b^2 c+84 a^2 c^2\right ) x \sqrt {a+b x^2+c x^4}}{315 c^{5/2} \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {x \left (b \left (4 b^2-9 a c\right )+6 c \left (2 b^2-7 a c\right ) x^2\right ) \sqrt {a+b x^2+c x^4}}{315 c^2}+\frac {x \left (3 b+7 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{63 c}-\frac {\sqrt [4]{a} \left (8 b^4-57 a b^2 c+84 a^2 c^2\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{315 c^{11/4} \sqrt {a+b x^2+c x^4}}+\frac {\sqrt [4]{a} \left (8 b^4-57 a b^2 c+84 a^2 c^2+4 \sqrt {a} b \sqrt {c} \left (b^2-6 a c\right )\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{630 c^{11/4} \sqrt {a+b x^2+c x^4}} \]

[Out]

1/63*x*(7*c*x^2+3*b)*(c*x^4+b*x^2+a)^(3/2)/c-1/315*x*(b*(-9*a*c+4*b^2)+6*c*(-7*a*c+2*b^2)*x^2)*(c*x^4+b*x^2+a)
^(1/2)/c^2+1/315*(84*a^2*c^2-57*a*b^2*c+8*b^4)*x*(c*x^4+b*x^2+a)^(1/2)/c^(5/2)/(a^(1/2)+x^2*c^(1/2))-1/315*a^(
1/4)*(84*a^2*c^2-57*a*b^2*c+8*b^4)*(cos(2*arctan(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^(11/4)/(c*x^4+b*x^2+a)^(1/2)+1/630*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^4-57*a*b^2*c+84*a^2*c^2+4*b*(-6*a*c+b^2)*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^(11/4)/(c*x^4+b*x^2+a)^(1/2)

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Rubi [A]
time = 0.17, antiderivative size = 443, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1130, 1190, 1211, 1117, 1209} \begin {gather*} \frac {\sqrt [4]{a} \left (84 a^2 c^2-57 a b^2 c+4 \sqrt {a} b \sqrt {c} \left (b^2-6 a c\right )+8 b^4\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 )}{630 c^{11/4} \sqrt {a+b x^2+c x^4}}-\frac {\sqrt [4]{a} \left (84 a^2 c^2-57 a b^2 c+8 b^4\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 )}{315 c^{11/4} \sqrt {a+b x^2+c x^4}}+\frac {x \left (84 a^2 c^2-57 a b^2 c+8 b^4\right ) \sqrt {a+b x^2+c x^4}}{315 c^{5/2} \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {x \left (6 c x^2 \left (2 b^2-7 a c\right )+b \left (4 b^2-9 a c\right )\right ) \sqrt {a+b x^2+c x^4}}{315 c^2}+\frac {x \left (3 b+7 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{63 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*x^2 + c*x^4)^(3/2),x]

[Out]

((8*b^4 - 57*a*b^2*c + 84*a^2*c^2)*x*Sqrt[a + b*x^2 + c*x^4])/(315*c^(5/2)*(Sqrt[a] + Sqrt[c]*x^2)) - (x*(b*(4
*b^2 - 9*a*c) + 6*c*(2*b^2 - 7*a*c)*x^2)*Sqrt[a + b*x^2 + c*x^4])/(315*c^2) + (x*(3*b + 7*c*x^2)*(a + b*x^2 +
c*x^4)^(3/2))/(63*c) - (a^(1/4)*(8*b^4 - 57*a*b^2*c + 84*a^2*c^2)*(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])/(315*c^
(11/4)*Sqrt[a + b*x^2 + c*x^4]) + (a^(1/4)*(8*b^4 - 57*a*b^2*c + 84*a^2*c^2 + 4*Sqrt[a]*b*Sqrt[c]*(b^2 - 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])/(630*c^(11/4)*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 1190

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

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]

Rubi steps

\begin {align*} \int x^2 \left (a+b x^2+c x^4\right )^{3/2} \, dx &=\frac {x \left (3 b+7 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{63 c}-\frac {\int \left (a b+2 \left (2 b^2-7 a c\right ) x^2\right ) \sqrt {a+b x^2+c x^4} \, dx}{21 c}\\ &=-\frac {x \left (b \left (4 b^2-9 a c\right )+6 c \left (2 b^2-7 a c\right ) x^2\right ) \sqrt {a+b x^2+c x^4}}{315 c^2}+\frac {x \left (3 b+7 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{63 c}-\frac {\int \frac {-4 a b \left (b^2-6 a c\right )+\left (-8 b^4+57 a b^2 c-84 a^2 c^2\right ) x^2}{\sqrt {a+b x^2+c x^4}} \, dx}{315 c^2}\\ &=-\frac {x \left (b \left (4 b^2-9 a c\right )+6 c \left (2 b^2-7 a c\right ) x^2\right ) \sqrt {a+b x^2+c x^4}}{315 c^2}+\frac {x \left (3 b+7 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{63 c}-\frac {\left (\sqrt {a} \left (8 b^4-57 a b^2 c+84 a^2 c^2\right )\right ) \int \frac {1-\frac {\sqrt {c} x^2}{\sqrt {a}}}{\sqrt {a+b x^2+c x^4}} \, dx}{315 c^{5/2}}+\frac {\left (\sqrt {a} \left (8 b^4-57 a b^2 c+84 a^2 c^2+4 \sqrt {a} b \sqrt {c} \left (b^2-6 a c\right )\right )\right ) \int \frac {1}{\sqrt {a+b x^2+c x^4}} \, dx}{315 c^{5/2}}\\ &=\frac {\left (8 b^4-57 a b^2 c+84 a^2 c^2\right ) x \sqrt {a+b x^2+c x^4}}{315 c^{5/2} \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {x \left (b \left (4 b^2-9 a c\right )+6 c \left (2 b^2-7 a c\right ) x^2\right ) \sqrt {a+b x^2+c x^4}}{315 c^2}+\frac {x \left (3 b+7 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{63 c}-\frac {\sqrt [4]{a} \left (8 b^4-57 a b^2 c+84 a^2 c^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}} 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 )}{315 c^{11/4} \sqrt {a+b x^2+c x^4}}+\frac {\sqrt [4]{a} \left (8 b^4-57 a b^2 c+84 a^2 c^2+4 \sqrt {a} b \sqrt {c} \left (b^2-6 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 )}{630 c^{11/4} \sqrt {a+b x^2+c x^4}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 11.23, size = 602, normalized size = 1.36 \begin {gather*} \frac {4 c \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} x \left (-4 b^4 x^2-b^3 c x^4+53 b^2 c^2 x^6+85 b c^3 x^8+35 c^4 x^{10}+a^2 c \left (24 b+77 c x^2\right )+a \left (-4 b^3+27 b^2 c x^2+151 b c^2 x^4+112 c^3 x^6\right )\right )+i \left (8 b^4-57 a b^2 c+84 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^5+65 a b^3 c-132 a^2 b c^2+8 b^4 \sqrt {b^2-4 a c}-57 a b^2 c \sqrt {b^2-4 a c}+84 a^2 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 )}{1260 c^3 \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^2*(a + b*x^2 + c*x^4)^(3/2),x]

[Out]

(4*c*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*x*(-4*b^4*x^2 - b^3*c*x^4 + 53*b^2*c^2*x^6 + 85*b*c^3*x^8 + 35*c^4*x^10 +
a^2*c*(24*b + 77*c*x^2) + a*(-4*b^3 + 27*b^2*c*x^2 + 151*b*c^2*x^4 + 112*c^3*x^6)) + I*(8*b^4 - 57*a*b^2*c +
84*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^5 + 65*a*b^3*c - 132*a^2*b*c^2 + 8*b^4
*Sqrt[b^2 - 4*a*c] - 57*a*b^2*c*Sqrt[b^2 - 4*a*c] + 84*a^2*c^2*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])]*Ellipt
icF[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])])/(1
260*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 = 0.04, size = 545, normalized size = 1.23 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*c*x^7*(c*x^4+b*x^2+a)^(1/2)+10/63*b*x^5*(c*x^4+b*x^2+a)^(1/2)+1/5*(11/9*a*c+1/21*b^2)/c*x^3*(c*x^4+b*x^2+a
)^(1/2)+1/3*(76/63*a*b-4/5*(11/9*a*c+1/21*b^2)/c*b)/c*x*(c*x^4+b*x^2+a)^(1/2)-1/12*(76/63*a*b-4/5*(11/9*a*c+1/
21*b^2)/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)*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*(a^2-3/5*(11/9*a*c+1/21*b^2)/c*a-2/3*(76/63*a*b-4/5*(11/9
*a*c+1/21*b^2)/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^2*(c*x^4+b*x^2+a)^(3/2),x, algorithm="maxima")

[Out]

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

[Out]

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

[Out]

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

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

[Out]

int(x^2*(a + b*x^2 + c*x^4)^(3/2), x)

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