3.5.27 \(\int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx\) [427]

3.5.27.1 Optimal result

Integrand size = 31, antiderivative size = 65 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx=\frac {2 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 (3 A+C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 C \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d} \]

output

2*B*(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))/d+2/3*(3*A+C)*(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))/d+2/3*C*sin(d*x+c)*cos(d*x+c)^(1 
/2)/d
 

3.5.27.2 Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.88 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx=\frac {2 \left (3 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+(3 A+C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+C \sqrt {\cos (c+d x)} \sin (c+d x)\right )}{3 d} \]

input

Integrate[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/Sqrt[Cos[c + d*x]],x]
 

output

(2*(3*B*EllipticE[(c + d*x)/2, 2] + (3*A + C)*EllipticF[(c + d*x)/2, 2] + 
C*Sqrt[Cos[c + d*x]]*Sin[c + d*x]))/(3*d)
 

3.5.27.3 Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {3042, 3502, 27, 3042, 3227, 3042, 3119, 3120}

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.

\(\displaystyle \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {2}{3} \int \frac {3 A+C+3 B \cos (c+d x)}{2 \sqrt {\cos (c+d x)}}dx+\frac {2 C \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \int \frac {3 A+C+3 B \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx+\frac {2 C \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \int \frac {3 A+C+3 B \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 C \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {1}{3} \left ((3 A+C) \int \frac {1}{\sqrt {\cos (c+d x)}}dx+3 B \int \sqrt {\cos (c+d x)}dx\right )+\frac {2 C \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left ((3 A+C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+3 B \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )+\frac {2 C \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {1}{3} \left ((3 A+C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {6 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {2 C \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {1}{3} \left (\frac {2 (3 A+C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {6 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {2 C \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\)

input

Int[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/Sqrt[Cos[c + d*x]],x]
 

output

((6*B*EllipticE[(c + d*x)/2, 2])/d + (2*(3*A + C)*EllipticF[(c + d*x)/2, 2 
])/d)/3 + (2*C*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d)
 

3.5.27.3.1 Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3119

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]
 

rule 3120

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]
 

rule 3227

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

rule 3502

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m 
+ 2))   Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m 
 + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] 
 &&  !LtQ[m, -1]
 

3.5.27.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(273\) vs. \(2(111)=222\).

Time = 10.34 (sec) , antiderivative size = 274, normalized size of antiderivative = 4.22

method	result	size
default	\(-\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (4 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-2 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{3 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(274\)
parts	\(\frac {2 A \,\operatorname {am}^{-1}\left (\frac {d x}{2}+\frac {c}{2}| \sqrt {2}\right )}{d}+\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{\sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}-\frac {2 C \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (4 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{3 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(332\)

input

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2),x,method=_RETURNVERBO 
SE)
 

output

-2/3*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4*C*cos(1/2* 
d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+3*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/ 
2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*B*(sin(1/2 
*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d* 
x+1/2*c),2^(1/2))-2*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+C*(sin(1/2*d 
*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+ 
1/2*c),2^(1/2)))/(-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
 

3.5.27.5 Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.09 (sec) , antiderivative size = 135, normalized size of antiderivative = 2.08 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx=\frac {2 \, C \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + \sqrt {2} {\left (-3 i \, A - i \, C\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {2} {\left (3 i \, A + i \, C\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 i \, \sqrt {2} B {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 i \, \sqrt {2} B {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{3 \, d} \]

input

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2),x, algorithm="f 
ricas")
 

output

1/3*(2*C*sqrt(cos(d*x + c))*sin(d*x + c) + sqrt(2)*(-3*I*A - I*C)*weierstr 
assPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + sqrt(2)*(3*I*A + I*C)* 
weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*I*sqrt(2)*B* 
weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x 
 + c))) - 3*I*sqrt(2)*B*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, 
cos(d*x + c) - I*sin(d*x + c))))/d
 

3.5.27.6 Sympy [F(-1)]

Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx=\text {Timed out} \]

input

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)**(1/2),x)
 

output

Timed out
 

3.5.27.7 Maxima [F]

\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{\sqrt {\cos \left (d x + c\right )}} \,d x } \]

input

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2),x, algorithm="m 
axima")
 

output

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)/sqrt(cos(d*x + c)), x)
 

3.5.27.8 Giac [F]

\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{\sqrt {\cos \left (d x + c\right )}} \,d x } \]

input

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2),x, algorithm="g 
iac")
 

output

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)/sqrt(cos(d*x + c)), x)
 

3.5.27.9 Mupad [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.06 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx=\frac {2\,A\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,B\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,C\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3\,d}+\frac {2\,C\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3\,d} \]

input

int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/cos(c + d*x)^(1/2),x)
 

output

(2*A*ellipticF(c/2 + (d*x)/2, 2))/d + (2*B*ellipticE(c/2 + (d*x)/2, 2))/d 
+ (2*C*ellipticF(c/2 + (d*x)/2, 2))/(3*d) + (2*C*cos(c + d*x)^(1/2)*sin(c 
+ d*x))/(3*d)