Integrand size = 42, antiderivative size = 89 \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx=\frac {2 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}+\frac {2 (b B-a C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b^2 d}-\frac {2 a (b B-a C) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b^2 (a+b) d} \]
2*C*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+ 1/2*c),2^(1/2))/b/d+2*(B*b-C*a)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1 /2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))/b^2/d-2*a*(B*b-C*a)*(cos(1/2*d *x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2*b/(a +b),2^(1/2))/b^2/(a+b)/d
Time = 1.16 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.44 \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx=\frac {b B \left (2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-\frac {2 a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}\right )-\frac {2 C \left (b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )-(a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+a \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{\sqrt {\sin ^2(c+d x)}}}{b^2 d} \]
(b*B*(2*EllipticF[(c + d*x)/2, 2] - (2*a*EllipticPi[(2*b)/(a + b), (c + d* x)/2, 2])/(a + b)) - (2*C*(b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] - ( a + b)*EllipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] + a*EllipticPi[-(b/a), Ar cSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/Sqrt[Sin[c + d*x]^2])/(b^2*d)
Time = 0.70 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.97, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 3508, 3042, 3481, 3042, 3119, 3282, 3042, 3120, 3284}
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.
| \(\displaystyle \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx\) |
| \(\Big \downarrow \) 3508 |
| \(\displaystyle \int \frac {\sqrt {\cos (c+d x)} (B+C \cos (c+d x))}{a+b \cos (c+d x)}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (B+C \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 3481 |
| \(\displaystyle \frac {(b B-a C) \int \frac {\sqrt {\cos (c+d x)}}{a+b \cos (c+d x)}dx}{b}+\frac {C \int \sqrt {\cos (c+d x)}dx}{b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(b B-a C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}+\frac {C \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {(b B-a C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}+\frac {2 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}\) |
| \(\Big \downarrow \) 3282 |
| \(\displaystyle \frac {(b B-a C) \left (\frac {\int \frac {1}{\sqrt {\cos (c+d x)}}dx}{b}-\frac {a \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}\right )}{b}+\frac {2 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(b B-a C) \left (\frac {\int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {a \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}\right )}{b}+\frac {2 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {(b B-a C) \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d}-\frac {a \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}\right )}{b}+\frac {2 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}\) |
| \(\Big \downarrow \) 3284 |
| \(\displaystyle \frac {(b B-a C) \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d}-\frac {2 a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b d (a+b)}\right )}{b}+\frac {2 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}\) |
(2*C*EllipticE[(c + d*x)/2, 2])/(b*d) + ((b*B - a*C)*((2*EllipticF[(c + d* x)/2, 2])/(b*d) - (2*a*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(b*(a + b)*d)))/b
3.9.79.3.1 Defintions of rubi rules used
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
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]
Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[ B/d Int[(a + b*Sin[e + f*x])^m, x], x] - Simp[(B*c - A*d)/d Int[(a + b* Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_ .) + (f_.)*(x_)]^2), x_Symbol] :> Simp[1/b^2 Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Time = 4.61 (sec) , antiderivative size = 295, normalized size of antiderivative = 3.31
| method | result | size |
| default | \(-\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \left (B F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a b -B F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b^{2}-B \Pi \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), -\frac {2 b}{a -b}, \sqrt {2}\right ) a b -C F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{2}+C F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a b -C E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a b +C E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b^{2}+C \Pi \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), -\frac {2 b}{a -b}, \sqrt {2}\right ) a^{2}\right )}{b^{2} \left (a -b \right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\) | \(295\) |
-2*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/ 2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*(B*EllipticF(cos(1/2*d*x+1 /2*c),2^(1/2))*a*b-B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*b^2-B*EllipticP i(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))*a*b-C*EllipticF(cos(1/2*d*x+1/2*c ),2^(1/2))*a^2+C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a*b-C*EllipticE(cos (1/2*d*x+1/2*c),2^(1/2))*a*b+C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*b^2+C *EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))*a^2)/b^2/(a-b)/(-2*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)/(2*cos(1/2 *d*x+1/2*c)^2-1)^(1/2)/d
Timed out. \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx=\text {Timed out} \]
integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2)/(a+b*cos(d*x+c)), x, algorithm="fricas")
Timed out. \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx=\text {Timed out} \]
\[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )}{{\left (b \cos \left (d x + c\right ) + a\right )} \sqrt {\cos \left (d x + c\right )}} \,d x } \]
integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2)/(a+b*cos(d*x+c)), x, algorithm="maxima")
\[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )}{{\left (b \cos \left (d x + c\right ) + a\right )} \sqrt {\cos \left (d x + c\right )}} \,d x } \]
integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2)/(a+b*cos(d*x+c)), x, algorithm="giac")
Timed out. \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )}{\sqrt {\cos \left (c+d\,x\right )}\,\left (a+b\,\cos \left (c+d\,x\right )\right )} \,d x \]