Integrand size = 21, antiderivative size = 50 \[ \int \sqrt {3-4 \cos (c+d x)} \sec (c+d x) \, dx=-\frac {8 \operatorname {EllipticF}\left (\frac {1}{2} (c+\pi +d x),\frac {8}{7}\right )}{\sqrt {7} d}-\frac {6 \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+\pi +d x),\frac {8}{7}\right )}{\sqrt {7} d} \]
8/7*(sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)*EllipticF(cos(1/2*d*x+ 1/2*c),2/7*14^(1/2))/d*7^(1/2)+6/7*(sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d* x+1/2*c)*EllipticPi(cos(1/2*d*x+1/2*c),2,2/7*14^(1/2))/d*7^(1/2)
Time = 0.11 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.22 \[ \int \sqrt {3-4 \cos (c+d x)} \sec (c+d x) \, dx=\frac {2 \sqrt {-3+4 \cos (c+d x)} \left (-4 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),8\right )+3 \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),8\right )\right )}{d \sqrt {3-4 \cos (c+d x)}} \]
(2*Sqrt[-3 + 4*Cos[c + d*x]]*(-4*EllipticF[(c + d*x)/2, 8] + 3*EllipticPi[ 2, (c + d*x)/2, 8]))/(d*Sqrt[3 - 4*Cos[c + d*x]])
Time = 0.35 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 3282, 3042, 3141, 3285}
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.
\(\displaystyle \int \sqrt {3-4 \cos (c+d x)} \sec (c+d x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {3-4 \sin \left (c+d x+\frac {\pi }{2}\right )}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 3282 |
\(\displaystyle 3 \int \frac {\sec (c+d x)}{\sqrt {3-4 \cos (c+d x)}}dx-4 \int \frac {1}{\sqrt {3-4 \cos (c+d x)}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 3 \int \frac {1}{\sqrt {3-4 \sin \left (c+d x+\frac {\pi }{2}\right )} \sin \left (c+d x+\frac {\pi }{2}\right )}dx-4 \int \frac {1}{\sqrt {3-4 \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 3141 |
\(\displaystyle 3 \int \frac {1}{\sqrt {3-4 \sin \left (c+d x+\frac {\pi }{2}\right )} \sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {8 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x+\pi ),\frac {8}{7}\right )}{\sqrt {7} d}\) |
\(\Big \downarrow \) 3285 |
\(\displaystyle -\frac {8 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x+\pi ),\frac {8}{7}\right )}{\sqrt {7} d}-\frac {6 \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x+\pi ),\frac {8}{7}\right )}{\sqrt {7} d}\) |
(-8*EllipticF[(c + Pi + d*x)/2, 8/7])/(Sqrt[7]*d) - (6*EllipticPi[2, (c + Pi + d*x)/2, 8/7])/(Sqrt[7]*d)
3.6.20.3.1 Defintions of rubi rules used
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*S qrt[a - b]))*EllipticF[(1/2)*(c + Pi/2 + d*x), -2*(b/(a - b))], x] /; FreeQ [{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a - b, 0]
Int[Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]/((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d/b Int[1/Sqrt[c + d*Sin[e + f*x]], x], x ] + Simp[(b*c - a*d)/b Int[1/((a + b*Sin[e + f*x])*Sqrt[c + d*Sin[e + f*x ]]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a - b)*Sqrt[c - d]))*EllipticPi[ -2*(b/(a - b)), (1/2)*(e + Pi/2 + f*x), -2*(d/(c - d))], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2 , 0] && GtQ[c - d, 0]
Time = 3.01 (sec) , antiderivative size = 159, normalized size of antiderivative = 3.18
method | result | size |
default | \(\frac {2 \sqrt {-\left (8 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-7\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {56 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-7}\, \left (4 F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \frac {2 \sqrt {14}}{7}\right )+3 \Pi \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), 2, \frac {2 \sqrt {14}}{7}\right )\right )}{7 \sqrt {8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-8 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+7}\, d}\) | \(159\) |
2/7*(-(8*cos(1/2*d*x+1/2*c)^2-7)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+ 1/2*c)^2)^(1/2)*(56*sin(1/2*d*x+1/2*c)^2-7)^(1/2)*(4*EllipticF(cos(1/2*d*x +1/2*c),2/7*14^(1/2))+3*EllipticPi(cos(1/2*d*x+1/2*c),2,2/7*14^(1/2)))/(8* sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(-8*co s(1/2*d*x+1/2*c)^2+7)^(1/2)/d
\[ \int \sqrt {3-4 \cos (c+d x)} \sec (c+d x) \, dx=\int { \sqrt {-4 \, \cos \left (d x + c\right ) + 3} \sec \left (d x + c\right ) \,d x } \]
\[ \int \sqrt {3-4 \cos (c+d x)} \sec (c+d x) \, dx=\int \sqrt {3 - 4 \cos {\left (c + d x \right )}} \sec {\left (c + d x \right )}\, dx \]
\[ \int \sqrt {3-4 \cos (c+d x)} \sec (c+d x) \, dx=\int { \sqrt {-4 \, \cos \left (d x + c\right ) + 3} \sec \left (d x + c\right ) \,d x } \]
\[ \int \sqrt {3-4 \cos (c+d x)} \sec (c+d x) \, dx=\int { \sqrt {-4 \, \cos \left (d x + c\right ) + 3} \sec \left (d x + c\right ) \,d x } \]
Timed out. \[ \int \sqrt {3-4 \cos (c+d x)} \sec (c+d x) \, dx=\int \frac {\sqrt {3-4\,\cos \left (c+d\,x\right )}}{\cos \left (c+d\,x\right )} \,d x \]