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

Optimal. Leaf size=271 \[ \frac {\left (x^2 \sqrt {\frac {c}{a}}+1\right ) \sqrt {\frac {a+b x^2+c x^4}{a \left (x^2 \sqrt {\frac {c}{a}}+1\right )^2}} \left (d \sqrt {\frac {c}{a}}+e\right ) \Pi \left (-\frac {\left (\sqrt {\frac {c}{a}} d-e\right )^2}{4 \sqrt {\frac {c}{a}} d e};2 \tan ^{-1}\left (\sqrt [4]{\frac {c}{a}} x\right )|\frac {1}{4} \left (2-\frac {b \sqrt {\frac {c}{a}}}{c}\right )\right )}{4 d e \sqrt [4]{\frac {c}{a}} \sqrt {a+b x^2+c x^4}}-\frac {\left (d \sqrt {\frac {c}{a}}-e\right ) \tan ^{-1}\left (\frac {x \sqrt {a e^2-b d e+c d^2}}{\sqrt {d} \sqrt {e} \sqrt {a+b x^2+c x^4}}\right )}{2 \sqrt {d} \sqrt {e} \sqrt {a e^2-b d e+c d^2}} \]

[Out]

-1/2*arctan(x*(a*e^2-b*d*e+c*d^2)^(1/2)/d^(1/2)/e^(1/2)/(c*x^4+b*x^2+a)^(1/2))*(-e+d*(c/a)^(1/2))/d^(1/2)/e^(1
/2)/(a*e^2-b*d*e+c*d^2)^(1/2)+1/4*(cos(2*arctan((c/a)^(1/4)*x))^2)^(1/2)/cos(2*arctan((c/a)^(1/4)*x))*Elliptic
Pi(sin(2*arctan((c/a)^(1/4)*x)),-1/4*(-e+d*(c/a)^(1/2))^2/d/e/(c/a)^(1/2),1/2*(2-b*(c/a)^(1/2)/c)^(1/2))*(e+d*
(c/a)^(1/2))*(1+x^2*(c/a)^(1/2))*((c*x^4+b*x^2+a)/a/(1+x^2*(c/a)^(1/2))^2)^(1/2)/(c/a)^(1/4)/d/e/(c*x^4+b*x^2+
a)^(1/2)

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Rubi [A]  time = 0.19, antiderivative size = 271, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.024, Rules used = {1706} \[ \frac {\left (x^2 \sqrt {\frac {c}{a}}+1\right ) \sqrt {\frac {a+b x^2+c x^4}{a \left (x^2 \sqrt {\frac {c}{a}}+1\right )^2}} \left (d \sqrt {\frac {c}{a}}+e\right ) \Pi \left (-\frac {\left (\sqrt {\frac {c}{a}} d-e\right )^2}{4 \sqrt {\frac {c}{a}} d e};2 \tan ^{-1}\left (\sqrt [4]{\frac {c}{a}} x\right )|\frac {1}{4} \left (2-\frac {b \sqrt {\frac {c}{a}}}{c}\right )\right )}{4 d e \sqrt [4]{\frac {c}{a}} \sqrt {a+b x^2+c x^4}}-\frac {\left (d \sqrt {\frac {c}{a}}-e\right ) \tan ^{-1}\left (\frac {x \sqrt {a e^2-b d e+c d^2}}{\sqrt {d} \sqrt {e} \sqrt {a+b x^2+c x^4}}\right )}{2 \sqrt {d} \sqrt {e} \sqrt {a e^2-b d e+c d^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1706

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> With[
{q = Rt[B/A, 2]}, -Simp[((B*d - A*e)*ArcTan[(Rt[-b + (c*d)/e + (a*e)/d, 2]*x)/Sqrt[a + b*x^2 + c*x^4]])/(2*d*e
*Rt[-b + (c*d)/e + (a*e)/d, 2]), x] + Simp[((B*d + A*e)*(A + B*x^2)*Sqrt[(A^2*(a + b*x^2 + c*x^4))/(a*(A + B*x
^2)^2)]*EllipticPi[Cancel[-((B*d - A*e)^2/(4*d*e*A*B))], 2*ArcTan[q*x], 1/2 - (b*A)/(4*a*B)])/(4*d*e*A*q*Sqrt[
a + b*x^2 + c*x^4]), x]] /; FreeQ[{a, b, c, d, e, A, B}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^
2, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[c/a] && EqQ[c*A^2 - a*B^2, 0]

Rubi steps

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

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Mathematica [C]  time = 0.43, size = 312, normalized size = 1.15 \[ -\frac {i \sqrt {\frac {\sqrt {b^2-4 a c}+b+2 c x^2}{\sqrt {b^2-4 a c}+b}} \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} \left (\left (e-d \sqrt {\frac {c}{a}}\right ) \Pi \left (\frac {\left (b+\sqrt {b^2-4 a c}\right ) e}{2 c d};i \sinh ^{-1}\left (\sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} x\right )|\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )+d \sqrt {\frac {c}{a}} F\left (i \sinh ^{-1}\left (\sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} x\right )|\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )\right )}{\sqrt {2} d e \sqrt {\frac {c}{\sqrt {b^2-4 a c}+b}} \sqrt {a+b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])]*Sqrt[1 + (2*c*x^2)/(b - Sqrt[b^2 - 4*a*c
])]*(Sqrt[c/a]*d*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[c/a]*d) + e)*EllipticPi[((b + Sqrt[b^2 - 4*a*c])*e)/(2*c*d), I*ArcSinh[Sqrt[2]*S
qrt[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 + Sqr
t[b^2 - 4*a*c])]*d*e*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+x^2*(c/a)^(1/2))/(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 \[ \int \frac {x^{2} \sqrt {\frac {c}{a}} + 1}{\sqrt {c x^{4} + b x^{2} + a} {\left (e x^{2} + d\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.08, size = 369, normalized size = 1.36 \[ \frac {\sqrt {\frac {c}{a}}\, \sqrt {2}\, \sqrt {-\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}+4}\, \sqrt {\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}+4}\, \EllipticF \left (\frac {\sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, x}{2}, \frac {\sqrt {\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) b}{a c}-4}}{2}\right )}{4 \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, e}+\frac {\left (-\sqrt {\frac {c}{a}}\, d +e \right ) \sqrt {2}\, \sqrt {\frac {b \,x^{2}}{2 a}-\frac {\sqrt {-4 a c +b^{2}}\, x^{2}}{2 a}+1}\, \sqrt {\frac {b \,x^{2}}{2 a}+\frac {\sqrt {-4 a c +b^{2}}\, x^{2}}{2 a}+1}\, \EllipticPi \left (\frac {\sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, x}{2}, -\frac {2 a e}{\left (-b +\sqrt {-4 a c +b^{2}}\right ) d}, \frac {\sqrt {-\frac {b +\sqrt {-4 a c +b^{2}}}{2 a}}\, \sqrt {2}}{\sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}\right )}{\sqrt {-\frac {b}{a}+\frac {\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, d e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((x^2*sqrt(c/a) + 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 = -0.00 \[ \int \frac {x^2\,\sqrt {\frac {c}{a}}+1}{\left (e\,x^2+d\right )\,\sqrt {c\,x^4+b\,x^2+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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