3.392 \(\int \frac {d+e x^2}{\sqrt {-a+b x^2-c x^4}} \, dx\)

Optimal. Leaf size=293 \[ \frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a-b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (\frac {\sqrt {c} d}{\sqrt {a}}+e\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (\frac {b}{\sqrt {a} \sqrt {c}}+2\right )\right )}{2 c^{3/4} \sqrt {-a+b x^2-c x^4}}-\frac {\sqrt [4]{a} e \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a-b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (\frac {b}{\sqrt {a} \sqrt {c}}+2\right )\right )}{c^{3/4} \sqrt {-a+b x^2-c x^4}}-\frac {e x \sqrt {-a+b x^2-c x^4}}{\sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 293, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {1197, 1103, 1195} \[ \frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a-b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (\frac {\sqrt {c} d}{\sqrt {a}}+e\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (\frac {b}{\sqrt {a} \sqrt {c}}+2\right )\right )}{2 c^{3/4} \sqrt {-a+b x^2-c x^4}}-\frac {\sqrt [4]{a} e \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a-b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (\frac {b}{\sqrt {a} \sqrt {c}}+2\right )\right )}{c^{3/4} \sqrt {-a+b x^2-c x^4}}-\frac {e x \sqrt {-a+b x^2-c x^4}}{\sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x^2)/Sqrt[-a + b*x^2 - c*x^4],x]

[Out]

-((e*x*Sqrt[-a + b*x^2 - c*x^4])/(Sqrt[c]*(Sqrt[a] + Sqrt[c]*x^2))) - (a^(1/4)*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])/(c^(3/4)*Sqrt[-a + b*x^2 - c*x^4]) + (a^(1/4)*((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*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]

Rubi steps

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

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Mathematica [C]  time = 0.31, size = 295, normalized size = 1.01 \[ -\frac {i \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}-b}+1} \sqrt {1-\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}} \left (\left (e \left (b-\sqrt {b^2-4 a c}\right )+2 c d\right ) F\left (i \sinh ^{-1}\left (\sqrt {2} \sqrt {-\frac {c}{b+\sqrt {b^2-4 a c}}} x\right )|\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )+e \left (\sqrt {b^2-4 a c}-b\right ) E\left (i \sinh ^{-1}\left (\sqrt {2} \sqrt {-\frac {c}{b+\sqrt {b^2-4 a c}}} x\right )|\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )\right )}{2 \sqrt {2} c \sqrt {-\frac {c}{\sqrt {b^2-4 a c}+b}} \sqrt {-a+b x^2-c x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x^2)/Sqrt[-a + b*x^2 - c*x^4],x]

[Out]

((-1/2*I)*Sqrt[1 + (2*c*x^2)/(-b + Sqrt[b^2 - 4*a*c])]*Sqrt[1 - (2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])]*((-b + Sqrt
[b^2 - 4*a*c])*e*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])] + (2*c*d + (b - Sqrt[b^2 - 4*a*c])*e)*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])]))/(Sqrt[2]*c*Sqrt[-(c/(b + Sqrt[b^2 - 4*a*c
]))]*Sqrt[-a + b*x^2 - c*x^4])

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fricas [F]  time = 0.63, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-c x^{4} + b x^{2} - a} {\left (e x^{2} + d\right )}}{c x^{4} - b x^{2} + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(-c*x^4 + b*x^2 - a)*(e*x^2 + d)/(c*x^4 - b*x^2 + a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {e x^{2} + d}{\sqrt {-c x^{4} + b x^{2} - a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x^2 + d)/sqrt(-c*x^4 + b*x^2 - a), x)

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maple [A]  time = 0.04, size = 357, normalized size = 1.22 \[ \frac {\sqrt {\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}+4}\, \sqrt {-\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}+4}\, \left (-\EllipticE \left (\frac {\sqrt {-\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right )}{a}}\, x}{2}, \frac {\sqrt {\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) b}{a c}-4}}{2}\right )+\EllipticF \left (\frac {\sqrt {-\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right )}{a}}\, x}{2}, \frac {\sqrt {\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) b}{a c}-4}}{2}\right )\right ) a e}{\sqrt {-\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right )}{a}}\, \sqrt {-c \,x^{4}+b \,x^{2}-a}\, \left (b +\sqrt {-4 a c +b^{2}}\right )}+\frac {\sqrt {\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}+4}\, \sqrt {-\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}+4}\, d \EllipticF \left (\frac {\sqrt {-\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right )}{a}}\, x}{2}, \frac {\sqrt {\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) b}{a c}-4}}{2}\right )}{2 \sqrt {-\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right )}{a}}\, \sqrt {-c \,x^{4}+b \,x^{2}-a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)/(-c*x^4+b*x^2-a)^(1/2),x)

[Out]

e*a/(-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*(-b+(-4*a*c+b^2)^(1/2))/a)^(1
/2),1/2*(2*(b+(-4*a*c+b^2)^(1/2))/a*b/c-4)^(1/2))-EllipticE(1/2*x*(-2*(-b+(-4*a*c+b^2)^(1/2))/a)^(1/2),1/2*(2*
(b+(-4*a*c+b^2)^(1/2))/a*b/c-4)^(1/2)))+1/2*d/(-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*(-b+(-4*a*c
+b^2)^(1/2))/a)^(1/2),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 \[ \int \frac {e x^{2} + d}{\sqrt {-c x^{4} + b x^{2} - a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x^2 + d)/sqrt(-c*x^4 + b*x^2 - a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {e\,x^2+d}{\sqrt {-c\,x^4+b\,x^2-a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x^2)/(b*x^2 - a - c*x^4)^(1/2),x)

[Out]

int((d + e*x^2)/(b*x^2 - a - c*x^4)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {d + e x^{2}}{\sqrt {- a + b x^{2} - c x^{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((d + e*x**2)/sqrt(-a + b*x**2 - c*x**4), x)

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