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

Optimal. Leaf size=478 \[ \frac {\sqrt {\sqrt {b^2-4 a c}+b} \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} \operatorname {EllipticF}\left (\tan ^{-1}\left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {\sqrt {b^2-4 a c}+b}}\right ),-\frac {2 \sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )}{\sqrt {2} \sqrt {c} \sqrt {\frac {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1}{\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1}} \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1}}+\frac {x \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1}}{\sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1}}-\frac {\sqrt {\sqrt {b^2-4 a c}+b} \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} E\left (\tan ^{-1}\left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )|-\frac {2 \sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )}{\sqrt {2} \sqrt {c} \sqrt {\frac {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1}{\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1}} \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1}} \]

[Out]

x*(1+2*c*x^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)-1/2*(1/(1+2*c*x^2/(b+(-4*a
*c+b^2)^(1/2))))^(1/2)*EllipticE(x*2^(1/2)*c^(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),(-2*(-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)*2^(1/2)/c^(1/2)/((1+2*c*x^2/(b-(-4*a*c+b^2)^(1/2)))/(1+2*c*x^2/(b+(-4*a*c+b^2)^(1/2))
))^(1/2)+1/2*(1/(1+2*c*x^2/(b+(-4*a*c+b^2)^(1/2))))^(1/2)*EllipticF(x*2^(1/2)*c^(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),(-2*(-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)*2^(1/2)/c^(1/2)/((1+2*c*x^2/(b-(-4*a*c+b^2)^(1/2))
)/(1+2*c*x^2/(b+(-4*a*c+b^2)^(1/2))))^(1/2)

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Rubi [A]  time = 0.37, antiderivative size = 478, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 59, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.068, Rules used = {422, 418, 492, 411} \[ \frac {x \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1}}{\sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1}}+\frac {\sqrt {\sqrt {b^2-4 a c}+b} \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} F\left (\tan ^{-1}\left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )|-\frac {2 \sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )}{\sqrt {2} \sqrt {c} \sqrt {\frac {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1}{\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1}} \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1}}-\frac {\sqrt {\sqrt {b^2-4 a c}+b} \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} E\left (\tan ^{-1}\left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )|-\frac {2 \sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )}{\sqrt {2} \sqrt {c} \sqrt {\frac {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1}{\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1}} \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1}} \]

Antiderivative was successfully verified.

[In]

Int[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])],x]

[Out]

(x*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])] - (Sqrt[b + Sqrt[b
^2 - 4*a*c]]*Sqrt[1 + (2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*EllipticE[ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b + Sqrt[b^
2 - 4*a*c]]], (-2*Sqrt[b^2 - 4*a*c])/(b - Sqrt[b^2 - 4*a*c])])/(Sqrt[2]*Sqrt[c]*Sqrt[(1 + (2*c*x^2)/(b - Sqrt[
b^2 - 4*a*c]))/(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])]) + (Sqrt[b
+ Sqrt[b^2 - 4*a*c]]*Sqrt[1 + (2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*EllipticF[ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b
+ Sqrt[b^2 - 4*a*c]]], (-2*Sqrt[b^2 - 4*a*c])/(b - Sqrt[b^2 - 4*a*c])])/(Sqrt[2]*Sqrt[c]*Sqrt[(1 + (2*c*x^2)/(
b - Sqrt[b^2 - 4*a*c]))/(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])])

Rule 411

Int[Sqrt[(a_) + (b_.)*(x_)^2]/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticE[ArcTan
[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(c*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]), x] /;
FreeQ[{a, b, c, d}, x] && PosQ[b/a] && PosQ[d/c]

Rule 418

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticF[ArcT
an[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(a*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]), x] /
; FreeQ[{a, b, c, d}, x] && PosQ[d/c] && PosQ[b/a] &&  !SimplerSqrtQ[b/a, d/c]

Rule 422

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Dist[a, Int[1/(Sqrt[a + b*x^2]*Sqrt[c +
d*x^2]), x], x] + Dist[b, Int[x^2/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]), x], x] /; FreeQ[{a, b, c, d}, x] && PosQ[
d/c] && PosQ[b/a]

Rule 492

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(x*Sqrt[a + b*x^2])/(b*Sqr
t[c + d*x^2]), x] - Dist[c/b, Int[Sqrt[a + b*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b
*c - a*d, 0] && PosQ[b/a] && PosQ[d/c] &&  !SimplerSqrtQ[b/a, d/c]

Rubi steps

\begin {align*} \int \frac {\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}}}} \, dx &=\frac {(2 c) \int \frac {x^2}{\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}}}} \, dx}{b-\sqrt {b^2-4 a c}}+\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}}}} \, dx\\ &=\frac {x \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}}}}+\frac {\sqrt {b+\sqrt {b^2-4 a c}} \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} F\left (\tan ^{-1}\left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )|-\frac {2 \sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )}{\sqrt {2} \sqrt {c} \sqrt {\frac {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}}{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}}}}-\int \frac {\sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}}}{\left (1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )^{3/2}} \, dx\\ &=\frac {x \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}}}}-\frac {\sqrt {b+\sqrt {b^2-4 a c}} \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} E\left (\tan ^{-1}\left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )|-\frac {2 \sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )}{\sqrt {2} \sqrt {c} \sqrt {\frac {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}}{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}}}}+\frac {\sqrt {b+\sqrt {b^2-4 a c}} \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} F\left (\tan ^{-1}\left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )|-\frac {2 \sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )}{\sqrt {2} \sqrt {c} \sqrt {\frac {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}}{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}}}}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 102, normalized size = 0.21 \[ \frac {\sqrt {-\sqrt {b^2-4 a c}-b} E\left (\sin ^{-1}\left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {-b-\sqrt {b^2-4 a c}}}\right )|\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )}{\sqrt {2} \sqrt {c}} \]

Antiderivative was successfully verified.

[In]

Integrate[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])],x]

[Out]

(Sqrt[-b - Sqrt[b^2 - 4*a*c]]*EllipticE[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])

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fricas [F]  time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{2} + \sqrt {b^{2} - 4 \, a c} x^{2} + 2 \, a\right )} \sqrt {\frac {b x^{2} + \sqrt {b^{2} - 4 \, a c} x^{2} + 2 \, a}{a}} \sqrt {\frac {b x^{2} - \sqrt {b^{2} - 4 \, a c} x^{2} + 2 \, a}{a}}}{4 \, {\left (c x^{4} + b x^{2} + a\right )}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*c*x^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),x, algorithm="fric
as")

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*c*x^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),x, algorithm="giac
")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Valuesym2poly/r2sym(const ge
n & e,const index_m & i,const vecteur & l) Error: Bad Argument ValueWarning, choosing root of [1,0,%%%{8,[1,0,
1]%%%}+%%%{-2,[1,0,0]%%%}+%%%{-2,[0,2,0]%%%},0,%%%{16,[2,0,2]%%%}+%%%{8,[2,0,1]%%%}+%%%{1,[2,0,0]%%%}+%%%{-8,[
1,2,1]%%%}+%%%{-2,[1,2,0]%%%}+%%%{1,[0,4,0]%%%}] at parameters values [71.707969239,70,22]Warning, choosing ro
ot of [1,0,%%%{8,[1,0,1]%%%}+%%%{-2,[1,0,0]%%%}+%%%{-2,[0,2,0]%%%},0,%%%{16,[2,0,2]%%%}+%%%{8,[2,0,1]%%%}+%%%{
1,[2,0,0]%%%}+%%%{-8,[1,2,1]%%%}+%%%{-2,[1,2,0]%%%}+%%%{1,[0,4,0]%%%}] at parameters values [78.6493344628,0,0
]Warning, choosing root of [1,0,%%%{8,[1,0,1]%%%}+%%%{-2,[1,0,0]%%%}+%%%{-2,[0,2,0]%%%},0,%%%{16,[2,0,2]%%%}+%
%%{8,[2,0,1]%%%}+%%%{1,[2,0,0]%%%}+%%%{-8,[1,2,1]%%%}+%%%{-2,[1,2,0]%%%}+%%%{1,[0,4,0]%%%}] at parameters valu
es [50.5901726987,49,-6]Warning, choosing root of [1,0,%%%{8,[1,0,1]%%%}+%%%{-2,[1,0,0]%%%}+%%%{-2,[0,2,0]%%%}
,0,%%%{16,[2,0,2]%%%}+%%%{8,[2,0,1]%%%}+%%%{1,[2,0,0]%%%}+%%%{-8,[1,2,1]%%%}+%%%{-2,[1,2,0]%%%}+%%%{1,[0,4,0]%
%%}] at parameters values [91.0141688026,0,0]Warning, choosing root of [1,0,%%%{8,[1,0,1]%%%}+%%%{-2,[1,0,0]%%
%}+%%%{-2,[0,2,0]%%%},0,%%%{16,[2,0,2]%%%}+%%%{8,[2,0,1]%%%}+%%%{1,[2,0,0]%%%}+%%%{-8,[1,2,1]%%%}+%%%{-2,[1,2,
0]%%%}+%%%{1,[0,4,0]%%%}] at parameters values [94.3029591851,0,0]Warning, choosing root of [1,0,%%%{8,[1,1,0]
%%%}+%%%{-4,[1,0,0]%%%}+%%%{-2,[0,0,2]%%%},0,%%%{16,[2,2,0]%%%}+%%%{16,[2,1,0]%%%}+%%%{4,[2,0,0]%%%}+%%%{-8,[1
,1,2]%%%}+%%%{-4,[1,0,2]%%%}+%%%{1,[0,0,4]%%%}] at parameters values [-64,2,62]Evaluation time: 21.17

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maple [F]  time = 0.33, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\frac {2 c \,x^{2}}{b -\sqrt {-4 a c +b^{2}}}+1}}{\sqrt {\frac {2 c \,x^{2}}{b +\sqrt {-4 a c +b^{2}}}+1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*c*x^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),x, algorithm="maxi
ma")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*c*x**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),x)

[Out]

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

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