3.388 \(\int \frac {1}{(d+e x^2) \sqrt {a+b x^2-c x^4}} \, dx\)

Optimal. Leaf size=197 \[ \frac {\sqrt {\sqrt {4 a c+b^2}+b} \sqrt {1-\frac {2 c x^2}{b-\sqrt {4 a c+b^2}}} \sqrt {1-\frac {2 c x^2}{\sqrt {4 a c+b^2}+b}} \Pi \left (-\frac {\left (b+\sqrt {b^2+4 a c}\right ) e}{2 c d};\sin ^{-1}\left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2+4 a c}}}\right )|\frac {b+\sqrt {b^2+4 a c}}{b-\sqrt {b^2+4 a c}}\right )}{\sqrt {2} \sqrt {c} d \sqrt {a+b x^2-c x^4}} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 197, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {1220, 537} \[ \frac {\sqrt {\sqrt {4 a c+b^2}+b} \sqrt {1-\frac {2 c x^2}{b-\sqrt {4 a c+b^2}}} \sqrt {1-\frac {2 c x^2}{\sqrt {4 a c+b^2}+b}} \Pi \left (-\frac {\left (b+\sqrt {b^2+4 a c}\right ) e}{2 c d};\sin ^{-1}\left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2+4 a c}}}\right )|\frac {b+\sqrt {b^2+4 a c}}{b-\sqrt {b^2+4 a c}}\right )}{\sqrt {2} \sqrt {c} d \sqrt {a+b x^2-c x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 537

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(1*Ellipt
icPi[(b*c)/(a*d), ArcSin[Rt[-(d/c), 2]*x], (c*f)/(d*e)])/(a*Sqrt[c]*Sqrt[e]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b,
c, d, e, f}, x] &&  !GtQ[d/c, 0] && GtQ[c, 0] && GtQ[e, 0] &&  !( !GtQ[f/e, 0] && SimplerSqrtQ[-(f/e), -(d/c)
])

Rule 1220

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c,
2]}, Dist[(Sqrt[1 + (2*c*x^2)/(b - q)]*Sqrt[1 + (2*c*x^2)/(b + q)])/Sqrt[a + b*x^2 + c*x^4], Int[1/((d + e*x^
2)*Sqrt[1 + (2*c*x^2)/(b - q)]*Sqrt[1 + (2*c*x^2)/(b + q)]), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 -
4*a*c, 0] && NegQ[c/a]

Rubi steps

\begin {align*} \int \frac {1}{\left (d+e x^2\right ) \sqrt {a+b x^2-c x^4}} \, dx &=\frac {\left (\sqrt {1-\frac {2 c x^2}{b-\sqrt {b^2+4 a c}}} \sqrt {1-\frac {2 c x^2}{b+\sqrt {b^2+4 a c}}}\right ) \int \frac {1}{\sqrt {1-\frac {2 c x^2}{b-\sqrt {b^2+4 a c}}} \sqrt {1-\frac {2 c x^2}{b+\sqrt {b^2+4 a c}}} \left (d+e x^2\right )} \, dx}{\sqrt {a+b x^2-c x^4}}\\ &=\frac {\sqrt {b+\sqrt {b^2+4 a c}} \sqrt {1-\frac {2 c x^2}{b-\sqrt {b^2+4 a c}}} \sqrt {1-\frac {2 c x^2}{b+\sqrt {b^2+4 a c}}} \Pi \left (-\frac {\left (b+\sqrt {b^2+4 a c}\right ) e}{2 c d};\sin ^{-1}\left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2+4 a c}}}\right )|\frac {b+\sqrt {b^2+4 a c}}{b-\sqrt {b^2+4 a c}}\right )}{\sqrt {2} \sqrt {c} d \sqrt {a+b x^2-c x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.23, size = 205, normalized size = 1.04 \[ -\frac {i \sqrt {\frac {2 c x^2}{\sqrt {4 a c+b^2}-b}+1} \sqrt {1-\frac {2 c x^2}{\sqrt {4 a c+b^2}+b}} \Pi \left (-\frac {\left (b+\sqrt {b^2+4 a c}\right ) e}{2 c d};i \sinh ^{-1}\left (\sqrt {2} \sqrt {-\frac {c}{b+\sqrt {b^2+4 a c}}} x\right )|-\frac {b+\sqrt {b^2+4 a c}}{\sqrt {b^2+4 a c}-b}\right )}{\sqrt {2} d \sqrt {-\frac {c}{\sqrt {4 a c+b^2}+b}} \sqrt {a+b x^2-c x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {-c x^{4} + b x^{2} + a} {\left (e x^{2} + d\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 201, normalized size = 1.02 \[ \frac {\sqrt {2}\, \sqrt {\frac {b \,x^{2}}{2 a}-\frac {\sqrt {4 a c +b^{2}}\, x^{2}}{2 a}+1}\, \sqrt {\frac {b \,x^{2}}{2 a}+\frac {\sqrt {4 a c +b^{2}}\, x^{2}}{2 a}+1}\, \EllipticPi \left (\frac {\sqrt {2}\, \sqrt {\frac {-b +\sqrt {4 a c +b^{2}}}{a}}\, x}{2}, -\frac {2 a e}{\left (-b +\sqrt {4 a c +b^{2}}\right ) d}, \frac {\sqrt {-\frac {b +\sqrt {4 a c +b^{2}}}{2 a}}\, \sqrt {2}}{\sqrt {\frac {-b +\sqrt {4 a c +b^{2}}}{a}}}\right )}{\sqrt {-\frac {b}{a}+\frac {\sqrt {4 a c +b^{2}}}{a}}\, \sqrt {-c \,x^{4}+b \,x^{2}+a}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {-c x^{4} + b x^{2} + a} {\left (e x^{2} + d\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\left (e\,x^2+d\right )\,\sqrt {-c\,x^4+b\,x^2+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (d + e x^{2}\right ) \sqrt {a + b x^{2} - c x^{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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