3.2.76 \(\int \frac {x^4 (A+B x^2)}{\sqrt {a+b x^2+c x^4}} \, dx\) [176]

Optimal. Leaf size=403 \[ -\frac {(4 b B-5 A c) x \sqrt {a+b x^2+c x^4}}{15 c^2}+\frac {B x^3 \sqrt {a+b x^2+c x^4}}{5 c}+\frac {\left (8 b^2 B-10 A b c-9 a B c\right ) x \sqrt {a+b x^2+c x^4}}{15 c^{5/2} \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {\sqrt [4]{a} \left (8 b^2 B-10 A b c-9 a B c\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{15 c^{11/4} \sqrt {a+b x^2+c x^4}}+\frac {\sqrt [4]{a} \left (8 b^2 B-10 A b c-9 a B c+\sqrt {a} \sqrt {c} (4 b B-5 A c)\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{30 c^{11/4} \sqrt {a+b x^2+c x^4}} \]

[Out]

-1/15*(-5*A*c+4*B*b)*x*(c*x^4+b*x^2+a)^(1/2)/c^2+1/5*B*x^3*(c*x^4+b*x^2+a)^(1/2)/c+1/15*(-10*A*b*c-9*B*a*c+8*B
*b^2)*x*(c*x^4+b*x^2+a)^(1/2)/c^(5/2)/(a^(1/2)+x^2*c^(1/2))-1/15*a^(1/4)*(-10*A*b*c-9*B*a*c+8*B*b^2)*(cos(2*ar
ctan(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)+1/30*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^2*B-
10*A*b*c-9*a*B*c+(-5*A*c+4*B*b)*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 = 0.18, antiderivative size = 403, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {1293, 1211, 1117, 1209} \begin {gather*} \frac {\sqrt [4]{a} \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}} \left (\sqrt {a} \sqrt {c} (4 b B-5 A c)-9 a B c-10 A b c+8 b^2 B\right ) 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 )}{30 c^{11/4} \sqrt {a+b x^2+c x^4}}-\frac {\sqrt [4]{a} \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}} \left (-9 a B c-10 A b c+8 b^2 B\right ) 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 )}{15 c^{11/4} \sqrt {a+b x^2+c x^4}}+\frac {x \sqrt {a+b x^2+c x^4} \left (-9 a B c-10 A b c+8 b^2 B\right )}{15 c^{5/2} \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {x \sqrt {a+b x^2+c x^4} (4 b B-5 A c)}{15 c^2}+\frac {B x^3 \sqrt {a+b x^2+c x^4}}{5 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/15*((4*b*B - 5*A*c)*x*Sqrt[a + b*x^2 + c*x^4])/c^2 + (B*x^3*Sqrt[a + b*x^2 + c*x^4])/(5*c) + ((8*b^2*B - 10
*A*b*c - 9*a*B*c)*x*Sqrt[a + b*x^2 + c*x^4])/(15*c^(5/2)*(Sqrt[a] + Sqrt[c]*x^2)) - (a^(1/4)*(8*b^2*B - 10*A*b
*c - 9*a*B*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])/(15*c^(11/4)*Sqrt[a + b*x^2 + c*x^4]) + (a^(1/4)*(8*b^2*B -
10*A*b*c - 9*a*B*c + Sqrt[a]*Sqrt[c]*(4*b*B - 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])/(30*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 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 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 \frac {x^4 \left (A+B x^2\right )}{\sqrt {a+b x^2+c x^4}} \, dx &=\frac {B x^3 \sqrt {a+b x^2+c x^4}}{5 c}-\frac {\int \frac {x^2 \left (3 a B+(4 b B-5 A c) x^2\right )}{\sqrt {a+b x^2+c x^4}} \, dx}{5 c}\\ &=-\frac {(4 b B-5 A c) x \sqrt {a+b x^2+c x^4}}{15 c^2}+\frac {B x^3 \sqrt {a+b x^2+c x^4}}{5 c}+\frac {\int \frac {a (4 b B-5 A c)+\left (8 b^2 B-10 A b c-9 a B c\right ) x^2}{\sqrt {a+b x^2+c x^4}} \, dx}{15 c^2}\\ &=-\frac {(4 b B-5 A c) x \sqrt {a+b x^2+c x^4}}{15 c^2}+\frac {B x^3 \sqrt {a+b x^2+c x^4}}{5 c}-\frac {\left (\sqrt {a} \left (8 b^2 B-10 A b c-9 a B c\right )\right ) \int \frac {1-\frac {\sqrt {c} x^2}{\sqrt {a}}}{\sqrt {a+b x^2+c x^4}} \, dx}{15 c^{5/2}}+\frac {\left (\sqrt {a} \left (8 b^2 B-10 A b c-9 a B c+\sqrt {a} \sqrt {c} (4 b B-5 A c)\right )\right ) \int \frac {1}{\sqrt {a+b x^2+c x^4}} \, dx}{15 c^{5/2}}\\ &=-\frac {(4 b B-5 A c) x \sqrt {a+b x^2+c x^4}}{15 c^2}+\frac {B x^3 \sqrt {a+b x^2+c x^4}}{5 c}+\frac {\left (8 b^2 B-10 A b c-9 a B c\right ) x \sqrt {a+b x^2+c x^4}}{15 c^{5/2} \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {\sqrt [4]{a} \left (8 b^2 B-10 A b c-9 a B 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 )}{15 c^{11/4} \sqrt {a+b x^2+c x^4}}+\frac {\sqrt [4]{a} \left (8 b^2 B-10 A b c-9 a B c+\sqrt {a} \sqrt {c} (4 b B-5 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 )}{30 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 = 11.36, size = 532, normalized size = 1.32 \begin {gather*} \frac {4 c \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} x \left (-4 b B+5 A c+3 B c x^2\right ) \left (a+b x^2+c x^4\right )+i \left (8 b^2 B-10 A b c-9 a B c\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^3 B+b c \left (17 a B-10 A \sqrt {b^2-4 a c}\right )+2 b^2 \left (5 A c+4 B \sqrt {b^2-4 a c}\right )-a c \left (10 A c+9 B \sqrt {b^2-4 a c}\right )\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 )}{60 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^4*(A + B*x^2))/Sqrt[a + b*x^2 + c*x^4],x]

[Out]

(4*c*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*x*(-4*b*B + 5*A*c + 3*B*c*x^2)*(a + b*x^2 + c*x^4) + I*(8*b^2*B - 10*A*b*
c - 9*a*B*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^3*B + b*c*(17*a*B - 10*A*Sqrt[b^2 - 4
*a*c]) + 2*b^2*(5*A*c + 4*B*Sqrt[b^2 - 4*a*c]) - a*c*(10*A*c + 9*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])]*E
llipticF[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])
])/(60*c^3*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*Sqrt[a + b*x^2 + c*x^4])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(814\) vs. \(2(391)=782\).
time = 0.04, size = 815, normalized size = 2.02 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(B*x^2+A)/(c*x^4+b*x^2+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

B*(1/5/c*x^3*(c*x^4+b*x^2+a)^(1/2)-4/15*b/c^2*x*(c*x^4+b*x^2+a)^(1/2)+1/15*a*b/c^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)*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*(-3/5*a/c+8/15*b^2/c^2)*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))))+A*(1/3*x*(c*x^4+b*x^2+
a)^(1/2)/c-1/12*a/c*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))+1/3*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)/(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*(B*x^2+A)/(c*x^4+b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)*x^4/sqrt(c*x^4 + b*x^2 + a), 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*(B*x^2+A)/(c*x^4+b*x^2+a)^(1/2),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 \frac {x^{4} \left (A + B x^{2}\right )}{\sqrt {a + b x^{2} + c x^{4}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**4*(A + B*x**2)/sqrt(a + b*x**2 + c*x**4), 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*(B*x^2+A)/(c*x^4+b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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