Integrand size = 12, antiderivative size = 43 \[ \int \frac {1}{\sqrt {a+b \cos ^2(x)}} \, dx=\frac {\sqrt {\frac {a+b \cos ^2(x)}{a}} \operatorname {EllipticF}\left (\frac {\pi }{2}+x,-\frac {b}{a}\right )}{\sqrt {a+b \cos ^2(x)}} \] Output:
((a+b*cos(x)^2)/a)^(1/2)*InverseJacobiAM(1/2*Pi+x,(-b/a)^(1/2))/(a+b*cos(x )^2)^(1/2)
Time = 0.08 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.07 \[ \int \frac {1}{\sqrt {a+b \cos ^2(x)}} \, dx=\frac {\sqrt {\frac {2 a+b+b \cos (2 x)}{a+b}} \operatorname {EllipticF}\left (x,\frac {b}{a+b}\right )}{\sqrt {2 a+b+b \cos (2 x)}} \] Input:
Integrate[1/Sqrt[a + b*Cos[x]^2],x]
Output:
(Sqrt[(2*a + b + b*Cos[2*x])/(a + b)]*EllipticF[x, b/(a + b)])/Sqrt[2*a + b + b*Cos[2*x]]
Time = 0.27 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.98, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3662, 3042, 3661}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{\sqrt {a+b \cos ^2(x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sqrt {a+b \sin \left (x+\frac {\pi }{2}\right )^2}}dx\) |
\(\Big \downarrow \) 3662 |
\(\displaystyle \frac {\sqrt {\frac {b \cos ^2(x)}{a}+1} \int \frac {1}{\sqrt {\frac {b \cos ^2(x)}{a}+1}}dx}{\sqrt {a+b \cos ^2(x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\frac {b \cos ^2(x)}{a}+1} \int \frac {1}{\sqrt {\frac {b \sin \left (x+\frac {\pi }{2}\right )^2}{a}+1}}dx}{\sqrt {a+b \cos ^2(x)}}\) |
\(\Big \downarrow \) 3661 |
\(\displaystyle \frac {\sqrt {\frac {b \cos ^2(x)}{a}+1} \operatorname {EllipticF}\left (x+\frac {\pi }{2},-\frac {b}{a}\right )}{\sqrt {a+b \cos ^2(x)}}\) |
Input:
Int[1/Sqrt[a + b*Cos[x]^2],x]
Output:
(Sqrt[1 + (b*Cos[x]^2)/a]*EllipticF[Pi/2 + x, -(b/a)])/Sqrt[a + b*Cos[x]^2 ]
Int[1/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Simp[(1/(S qrt[a]*f))*EllipticF[e + f*x, -b/a], x] /; FreeQ[{a, b, e, f}, x] && GtQ[a, 0]
Int[1/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Simp[Sqrt[ 1 + b*(Sin[e + f*x]^2/a)]/Sqrt[a + b*Sin[e + f*x]^2] Int[1/Sqrt[1 + (b*Si n[e + f*x]^2)/a], x], x] /; FreeQ[{a, b, e, f}, x] && !GtQ[a, 0]
Time = 0.08 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.12
method | result | size |
default | \(-\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (2 x \right )}{2}}\, \sqrt {\frac {a +b \cos \left (x \right )^{2}}{a}}\, \operatorname {EllipticF}\left (\cos \left (x \right ), \sqrt {-\frac {b}{a}}\right )}{\sin \left (x \right ) \sqrt {a +b \cos \left (x \right )^{2}}}\) | \(48\) |
Input:
int(1/(a+b*cos(x)^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
-(sin(x)^2)^(1/2)*((a+b*cos(x)^2)/a)^(1/2)*EllipticF(cos(x),(-b/a)^(1/2))/ sin(x)/(a+b*cos(x)^2)^(1/2)
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 276, normalized size of antiderivative = 6.42 \[ \int \frac {1}{\sqrt {a+b \cos ^2(x)}} \, dx=-\frac {{\left (-2 i \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} - 2 i \, a - i \, b\right )} \sqrt {b} \sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} - 2 \, a - b}{b}} F(\arcsin \left (\sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} - 2 \, a - b}{b}} {\left (\cos \left (x\right ) + i \, \sin \left (x\right )\right )}\right )\,|\,\frac {8 \, a^{2} + 8 \, a b + b^{2} + 4 \, {\left (2 \, a b + b^{2}\right )} \sqrt {\frac {a^{2} + a b}{b^{2}}}}{b^{2}}) + {\left (2 i \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 i \, a + i \, b\right )} \sqrt {b} \sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} - 2 \, a - b}{b}} F(\arcsin \left (\sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} - 2 \, a - b}{b}} {\left (\cos \left (x\right ) - i \, \sin \left (x\right )\right )}\right )\,|\,\frac {8 \, a^{2} + 8 \, a b + b^{2} + 4 \, {\left (2 \, a b + b^{2}\right )} \sqrt {\frac {a^{2} + a b}{b^{2}}}}{b^{2}})}{b^{2}} \] Input:
integrate(1/(a+b*cos(x)^2)^(1/2),x, algorithm="fricas")
Output:
-((-2*I*b*sqrt((a^2 + a*b)/b^2) - 2*I*a - I*b)*sqrt(b)*sqrt((2*b*sqrt((a^2 + a*b)/b^2) - 2*a - b)/b)*elliptic_f(arcsin(sqrt((2*b*sqrt((a^2 + a*b)/b^ 2) - 2*a - b)/b)*(cos(x) + I*sin(x))), (8*a^2 + 8*a*b + b^2 + 4*(2*a*b + b ^2)*sqrt((a^2 + a*b)/b^2))/b^2) + (2*I*b*sqrt((a^2 + a*b)/b^2) + 2*I*a + I *b)*sqrt(b)*sqrt((2*b*sqrt((a^2 + a*b)/b^2) - 2*a - b)/b)*elliptic_f(arcsi n(sqrt((2*b*sqrt((a^2 + a*b)/b^2) - 2*a - b)/b)*(cos(x) - I*sin(x))), (8*a ^2 + 8*a*b + b^2 + 4*(2*a*b + b^2)*sqrt((a^2 + a*b)/b^2))/b^2))/b^2
\[ \int \frac {1}{\sqrt {a+b \cos ^2(x)}} \, dx=\int \frac {1}{\sqrt {a + b \cos ^{2}{\left (x \right )}}}\, dx \] Input:
integrate(1/(a+b*cos(x)**2)**(1/2),x)
Output:
Integral(1/sqrt(a + b*cos(x)**2), x)
\[ \int \frac {1}{\sqrt {a+b \cos ^2(x)}} \, dx=\int { \frac {1}{\sqrt {b \cos \left (x\right )^{2} + a}} \,d x } \] Input:
integrate(1/(a+b*cos(x)^2)^(1/2),x, algorithm="maxima")
Output:
integrate(1/sqrt(b*cos(x)^2 + a), x)
\[ \int \frac {1}{\sqrt {a+b \cos ^2(x)}} \, dx=\int { \frac {1}{\sqrt {b \cos \left (x\right )^{2} + a}} \,d x } \] Input:
integrate(1/(a+b*cos(x)^2)^(1/2),x, algorithm="giac")
Output:
integrate(1/sqrt(b*cos(x)^2 + a), x)
Timed out. \[ \int \frac {1}{\sqrt {a+b \cos ^2(x)}} \, dx=\int \frac {1}{\sqrt {b\,{\cos \left (x\right )}^2+a}} \,d x \] Input:
int(1/(a + b*cos(x)^2)^(1/2),x)
Output:
int(1/(a + b*cos(x)^2)^(1/2), x)
\[ \int \frac {1}{\sqrt {a+b \cos ^2(x)}} \, dx=\int \frac {\sqrt {\cos \left (x \right )^{2} b +a}}{\cos \left (x \right )^{2} b +a}d x \] Input:
int(1/(a+b*cos(x)^2)^(1/2),x)
Output:
int(sqrt(cos(x)**2*b + a)/(cos(x)**2*b + a),x)