3.26 \(\int \frac {(A+B x^2) (d+e x^2)}{(a+b x^2+c x^4)^{3/2}} \, dx\)

Optimal. Leaf size=481 \[ \frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right ) (A c (b d-2 a e)-a B (2 c d-b e))}{a^{3/4} c^{3/4} \left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}}+\frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \left (\sqrt {a} B-A \sqrt {c}\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (\sqrt {c} d-\sqrt {a} e\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{2 a^{3/4} c^{3/4} \left (b-2 \sqrt {a} \sqrt {c}\right ) \sqrt {a+b x^2+c x^4}}-\frac {x \sqrt {a+b x^2+c x^4} (A c (b d-2 a e)-a B (2 c d-b e))}{a \sqrt {c} \left (b^2-4 a c\right ) \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {x \left (-A \left (-a b e-2 a c d+b^2 d\right )-\left (x^2 (A c (b d-2 a e)-a B (2 c d-b e))\right )+a B (b d-2 a e)\right )}{a \left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}} \]

[Out]

-x*(a*B*(-2*a*e+b*d)-A*(-a*b*e-2*a*c*d+b^2*d)-(A*c*(-2*a*e+b*d)-a*B*(-b*e+2*c*d))*x^2)/a/(-4*a*c+b^2)/(c*x^4+b
*x^2+a)^(1/2)-(A*c*(-2*a*e+b*d)-a*B*(-b*e+2*c*d))*x*(c*x^4+b*x^2+a)^(1/2)/a/(-4*a*c+b^2)/c^(1/2)/(a^(1/2)+x^2*
c^(1/2))+(A*c*(-2*a*e+b*d)-a*B*(-b*e+2*c*d))*(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)/a^(3/4)/c^(3/4)/(-4*a*c+b^2)/(c*x^4+b*x^2+a)^(1/2)+1/2*(cos(2*a
rctan(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))*(B*a^(1/2)-A*c^(1/2))*(-e*a^(1/2)+d*c^(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)/a^(3/4)/c^(3/4)/(-2*a^(1/2)*c^(1/2)+b)/(c*x^4+b*x^2+a)^(1/2)

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Rubi [A]  time = 0.39, antiderivative size = 481, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {1678, 1197, 1103, 1195} \[ \frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right ) (A c (b d-2 a e)-a B (2 c d-b e))}{a^{3/4} c^{3/4} \left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}}+\frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \left (\sqrt {a} B-A \sqrt {c}\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (\sqrt {c} d-\sqrt {a} e\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{2 a^{3/4} c^{3/4} \left (b-2 \sqrt {a} \sqrt {c}\right ) \sqrt {a+b x^2+c x^4}}-\frac {x \sqrt {a+b x^2+c x^4} (A c (b d-2 a e)-a B (2 c d-b e))}{a \sqrt {c} \left (b^2-4 a c\right ) \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {x \left (-A \left (-a b e-2 a c d+b^2 d\right )+x^2 (-(A c (b d-2 a e)-a B (2 c d-b e)))+a B (b d-2 a e)\right )}{a \left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(d + e*x^2))/(a + b*x^2 + c*x^4)^(3/2),x]

[Out]

-((x*(a*B*(b*d - 2*a*e) - A*(b^2*d - 2*a*c*d - a*b*e) - (A*c*(b*d - 2*a*e) - a*B*(2*c*d - b*e))*x^2))/(a*(b^2
- 4*a*c)*Sqrt[a + b*x^2 + c*x^4])) - ((A*c*(b*d - 2*a*e) - a*B*(2*c*d - b*e))*x*Sqrt[a + b*x^2 + c*x^4])/(a*Sq
rt[c]*(b^2 - 4*a*c)*(Sqrt[a] + Sqrt[c]*x^2)) + ((A*c*(b*d - 2*a*e) - a*B*(2*c*d - b*e))*(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^(3/4)*c^(3/4)*(b^2 - 4*a*c)*Sqrt[a + b*x^2 + c*x^4]) + ((Sqrt[a]*B - A*Sqrt[c])*(Sqrt[c]*d -
Sqrt[a]*e)*(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^(3/4)*(b - 2*Sqrt[a]*Sqrt[c])*c^(3/4)*Sqrt[a + b*x^2 + c
*x^4])

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]

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 1678

Int[(Pq_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{d = Coeff[PolynomialRemainder[Pq, a +
b*x^2 + c*x^4, x], x, 0], e = Coeff[PolynomialRemainder[Pq, a + b*x^2 + c*x^4, x], x, 2]}, Simp[(x*(a + b*x^2
+ c*x^4)^(p + 1)*(a*b*e - d*(b^2 - 2*a*c) - c*(b*d - 2*a*e)*x^2))/(2*a*(p + 1)*(b^2 - 4*a*c)), x] + Dist[1/(2*
a*(p + 1)*(b^2 - 4*a*c)), Int[(a + b*x^2 + c*x^4)^(p + 1)*ExpandToSum[2*a*(p + 1)*(b^2 - 4*a*c)*PolynomialQuot
ient[Pq, a + b*x^2 + c*x^4, x] + b^2*d*(2*p + 3) - 2*a*c*d*(4*p + 5) - a*b*e + c*(4*p + 7)*(b*d - 2*a*e)*x^2,
x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x^2] && Expon[Pq, x^2] > 1 && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1
]

Rubi steps

\begin {align*} \int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )}{\left (a+b x^2+c x^4\right )^{3/2}} \, dx &=-\frac {x \left (a B (b d-2 a e)-A \left (b^2 d-2 a c d-a b e\right )-(A c (b d-2 a e)-a B (2 c d-b e)) x^2\right )}{a \left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}}-\frac {\int \frac {-a (b B d-2 A c d+A b e-2 a B e)+(A c (b d-2 a e)-a B (2 c d-b e)) x^2}{\sqrt {a+b x^2+c x^4}} \, dx}{a \left (b^2-4 a c\right )}\\ &=-\frac {x \left (a B (b d-2 a e)-A \left (b^2 d-2 a c d-a b e\right )-(A c (b d-2 a e)-a B (2 c d-b e)) x^2\right )}{a \left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}}+\frac {\left (\left (\sqrt {a} B-A \sqrt {c}\right ) \left (\sqrt {c} d-\sqrt {a} e\right )\right ) \int \frac {1}{\sqrt {a+b x^2+c x^4}} \, dx}{\sqrt {a} \left (b-2 \sqrt {a} \sqrt {c}\right ) \sqrt {c}}+\frac {(A c (b d-2 a e)-a B (2 c d-b e)) \int \frac {1-\frac {\sqrt {c} x^2}{\sqrt {a}}}{\sqrt {a+b x^2+c x^4}} \, dx}{\sqrt {a} \sqrt {c} \left (b^2-4 a c\right )}\\ &=-\frac {x \left (a B (b d-2 a e)-A \left (b^2 d-2 a c d-a b e\right )-(A c (b d-2 a e)-a B (2 c d-b e)) x^2\right )}{a \left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}}-\frac {(A c (b d-2 a e)-a B (2 c d-b e)) x \sqrt {a+b x^2+c x^4}}{a \sqrt {c} \left (b^2-4 a c\right ) \left (\sqrt {a}+\sqrt {c} x^2\right )}+\frac {(A c (b d-2 a e)-a B (2 c d-b e)) \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^{3/4} c^{3/4} \left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}}+\frac {\left (\sqrt {a} B-A \sqrt {c}\right ) \left (\sqrt {c} d-\sqrt {a} e\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^{3/4} \left (b-2 \sqrt {a} \sqrt {c}\right ) c^{3/4} \sqrt {a+b x^2+c x^4}}\\ \end {align*}

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Mathematica [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {\$Aborted} \]

Verification is Not applicable to the result.

[In]

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

[Out]

$Aborted

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fricas [F]  time = 1.28, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B e x^{4} + {\left (B d + A e\right )} x^{2} + A d\right )} \sqrt {c x^{4} + b x^{2} + a}}{c^{2} x^{8} + 2 \, b c x^{6} + {\left (b^{2} + 2 \, a c\right )} x^{4} + 2 \, a b x^{2} + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*e*x^4 + (B*d + A*e)*x^2 + A*d)*sqrt(c*x^4 + b*x^2 + a)/(c^2*x^8 + 2*b*c*x^6 + (b^2 + 2*a*c)*x^4 +
2*a*b*x^2 + a^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B x^{2} + A\right )} {\left (e x^{2} + d\right )}}{{\left (c x^{4} + b x^{2} + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*x^2 + A)*(e*x^2 + d)/(c*x^4 + b*x^2 + a)^(3/2), x)

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maple [B]  time = 0.01, size = 1390, normalized size = 2.89 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)*(e*x^2+d)/(c*x^4+b*x^2+a)^(3/2),x)

[Out]

B*e*(-2*(1/2/(4*a*c-b^2)*b/c*x^3+1/(4*a*c-b^2)*a/c*x)/((x^4+b/c*x^2+a/c)*c)^(1/2)*c+1/2*a/(4*a*c-b^2)*2^(1/2)/
((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*(-2*(-b+(-4*a*c+b^2)^(1/2))/a*x^2+4)^(1/2)*(2*(b+(-4*a*c+b^2)^(1/2))/a*x^2+4
)^(1/2)/(c*x^4+b*x^2+a)^(1/2)*EllipticF(1/2*2^(1/2)*((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*x,1/2*(2*(b+(-4*a*c+b^2)
^(1/2))/a*b/c-4)^(1/2))-1/2/(4*a*c-b^2)*b*a*2^(1/2)/((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*(-2*(-b+(-4*a*c+b^2)^(1/
2))/a*x^2+4)^(1/2)*(2*(b+(-4*a*c+b^2)^(1/2))/a*x^2+4)^(1/2)/(c*x^4+b*x^2+a)^(1/2)/(b+(-4*a*c+b^2)^(1/2))*(Elli
pticF(1/2*2^(1/2)*((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*x,1/2*(2*(b+(-4*a*c+b^2)^(1/2))/a*b/c-4)^(1/2))-EllipticE(
1/2*2^(1/2)*((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*x,1/2*(2*(b+(-4*a*c+b^2)^(1/2))/a*b/c-4)^(1/2))))+(A*e+B*d)*(-2*
(-1/(4*a*c-b^2)*x^3-1/2/(4*a*c-b^2)*b/c*x)/((x^4+b/c*x^2+a/c)*c)^(1/2)*c-1/4/(4*a*c-b^2)*b*2^(1/2)/((-b+(-4*a*
c+b^2)^(1/2))/a)^(1/2)*(-2*(-b+(-4*a*c+b^2)^(1/2))/a*x^2+4)^(1/2)*(2*(b+(-4*a*c+b^2)^(1/2))/a*x^2+4)^(1/2)/(c*
x^4+b*x^2+a)^(1/2)*EllipticF(1/2*2^(1/2)*((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*x,1/2*(2*(b+(-4*a*c+b^2)^(1/2))/a*b
/c-4)^(1/2))+c/(4*a*c-b^2)*a*2^(1/2)/((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*(-2*(-b+(-4*a*c+b^2)^(1/2))/a*x^2+4)^(1
/2)*(2*(b+(-4*a*c+b^2)^(1/2))/a*x^2+4)^(1/2)/(c*x^4+b*x^2+a)^(1/2)/(b+(-4*a*c+b^2)^(1/2))*(EllipticF(1/2*2^(1/
2)*((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*x,1/2*(2*(b+(-4*a*c+b^2)^(1/2))/a*b/c-4)^(1/2))-EllipticE(1/2*2^(1/2)*((-
b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*x,1/2*(2*(b+(-4*a*c+b^2)^(1/2))/a*b/c-4)^(1/2))))+A*d*(-2*(1/2/(4*a*c-b^2)/a*b*
x^3-1/2*(2*a*c-b^2)/(4*a*c-b^2)/a/c*x)/((x^4+b/c*x^2+a/c)*c)^(1/2)*c+1/4*(1/a-(2*a*c-b^2)/(4*a*c-b^2)/a)*2^(1/
2)/((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*(-2*(-b+(-4*a*c+b^2)^(1/2))/a*x^2+4)^(1/2)*(2*(b+(-4*a*c+b^2)^(1/2))/a*x^
2+4)^(1/2)/(c*x^4+b*x^2+a)^(1/2)*EllipticF(1/2*2^(1/2)*((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*x,1/2*(2*(b+(-4*a*c+b
^2)^(1/2))/a*b/c-4)^(1/2))-1/2*b/(4*a*c-b^2)*c*2^(1/2)/((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*(-2*(-b+(-4*a*c+b^2)^
(1/2))/a*x^2+4)^(1/2)*(2*(b+(-4*a*c+b^2)^(1/2))/a*x^2+4)^(1/2)/(c*x^4+b*x^2+a)^(1/2)/(b+(-4*a*c+b^2)^(1/2))*(E
llipticF(1/2*2^(1/2)*((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*x,1/2*(2*(b+(-4*a*c+b^2)^(1/2))/a*b/c-4)^(1/2))-Ellipti
cE(1/2*2^(1/2)*((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*x,1/2*(2*(b+(-4*a*c+b^2)^(1/2))/a*b/c-4)^(1/2))))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B x^{2} + A\right )} {\left (e x^{2} + d\right )}}{{\left (c x^{4} + b x^{2} + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*x^2 + A)*(e*x^2 + d)/(c*x^4 + b*x^2 + a)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (B\,x^2+A\right )\,\left (e\,x^2+d\right )}{{\left (c\,x^4+b\,x^2+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^2)*(d + e*x^2))/(a + b*x^2 + c*x^4)^(3/2),x)

[Out]

int(((A + B*x^2)*(d + e*x^2))/(a + b*x^2 + c*x^4)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B x^{2}\right ) \left (d + e x^{2}\right )}{\left (a + b x^{2} + c x^{4}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)*(e*x**2+d)/(c*x**4+b*x**2+a)**(3/2),x)

[Out]

Integral((A + B*x**2)*(d + e*x**2)/(a + b*x**2 + c*x**4)**(3/2), x)

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