\(\int \frac {(A+B \cos (c+d x)) \sqrt {\sec (c+d x)}}{a+a \cos (c+d x)} \, dx\) [478]

Optimal result

Integrand size = 33, antiderivative size = 123 \[ \int \frac {(A+B \cos (c+d x)) \sqrt {\sec (c+d x)}}{a+a \cos (c+d x)} \, dx=\frac {(A-B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a d}+\frac {(A+B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{a d}-\frac {(A-B) \sqrt {\sec (c+d x)} \sin (c+d x)}{d (a+a \sec (c+d x))} \] Output:

(A-B)*cos(d*x+c)^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*sec(d*x+c)^(1 
/2)/a/d+(A+B)*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))*sec( 
d*x+c)^(1/2)/a/d-(A-B)*sec(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 3.02 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.63 \[ \int \frac {(A+B \cos (c+d x)) \sqrt {\sec (c+d x)}}{a+a \cos (c+d x)} \, dx=-\frac {e^{-\frac {1}{2} i (4 c+d x)} \left (-1+e^{2 i c}\right ) \left (3 i (A+B) \left (1+e^{i (c+d x)}\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+(A-B) \left (-3 \left (1+e^{2 i (c+d x)}\right )+e^{i (c+d x)} \left (1+e^{i (c+d x)}\right ) \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )\right ) \left (\csc \left (\frac {c}{2}\right )+i \sec \left (\frac {c}{2}\right )\right ) \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)}}{24 a d} \] Input:

Integrate[((A + B*Cos[c + d*x])*Sqrt[Sec[c + d*x]])/(a + a*Cos[c + d*x]),x 
]
 

Output:

-1/24*((-1 + E^((2*I)*c))*((3*I)*(A + B)*(1 + E^(I*(c + d*x)))*Sqrt[Cos[c 
+ d*x]]*EllipticF[(c + d*x)/2, 2] + (A - B)*(-3*(1 + E^((2*I)*(c + d*x))) 
+ E^(I*(c + d*x))*(1 + E^(I*(c + d*x)))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hype 
rgeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]))*(Csc[c/2] + I*Sec[c/2 
])*Sec[(c + d*x)/2]*Sqrt[Sec[c + d*x]])/(a*d*E^((I/2)*(4*c + d*x)))
 

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.05, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3042, 3439, 3042, 4507, 27, 3042, 4274, 3042, 4258, 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.

\(\displaystyle \int \frac {\sqrt {\sec (c+d x)} (A+B \cos (c+d x))}{a \cos (c+d x)+a} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3439

\(\displaystyle \int \frac {\sqrt {\sec (c+d x)} (A \sec (c+d x)+B)}{a \sec (c+d x)+a}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4507

\(\displaystyle \frac {\int \frac {a (A-B)+a (A+B) \sec (c+d x)}{2 \sqrt {\sec (c+d x)}}dx}{a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (a \sec (c+d x)+a)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {a (A-B)+a (A+B) \sec (c+d x)}{\sqrt {\sec (c+d x)}}dx}{2 a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (a \sec (c+d x)+a)}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4274

\(\displaystyle \frac {a (A-B) \int \frac {1}{\sqrt {\sec (c+d x)}}dx+a (A+B) \int \sqrt {\sec (c+d x)}dx}{2 a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (a \sec (c+d x)+a)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a (A-B) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+a (A+B) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx}{2 a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (a \sec (c+d x)+a)}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {a (A+B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+a (A-B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx}{2 a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (a \sec (c+d x)+a)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a (A+B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+a (A-B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (a \sec (c+d x)+a)}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {a (A+B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a (A-B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{2 a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (a \sec (c+d x)+a)}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {2 a (A+B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {2 a (A-B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{2 a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (a \sec (c+d x)+a)}\)

Input:

Int[((A + B*Cos[c + d*x])*Sqrt[Sec[c + d*x]])/(a + a*Cos[c + d*x]),x]
 

Output:

((2*a*(A - B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d* 
x]])/d + (2*a*(A + B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Se 
c[c + d*x]])/d)/(2*a^2) - ((A - B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*(a 
+ a*Sec[c + d*x]))
 

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 3439

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)* 
(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Sim 
p[g^(m + n)   Int[(g*Csc[e + f*x])^(p - m - n)*(b + a*Csc[e + f*x])^m*(d + 
c*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c 
- a*d, 0] &&  !IntegerQ[p] && IntegerQ[m] && IntegerQ[n]
 

rule 4258

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] 
)^n*Sin[c + d*x]^n   Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && 
 EqQ[n^2, 1/4]
 

rule 4274

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_)), x_Symbol] :> Simp[a   Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d   In 
t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
 

rule 4507

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[d*(A*b 
- a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^(n - 1)/(a*f*( 
2*m + 1))), x] - Simp[1/(a*b*(2*m + 1))   Int[(a + b*Csc[e + f*x])^(m + 1)* 
(d*Csc[e + f*x])^(n - 1)*Simp[A*(a*d*(n - 1)) - B*(b*d*(n - 1)) - d*(a*B*(m 
 - n + 1) + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, 
A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && G 
tQ[n, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(242\) vs. \(2(116)=232\).

Time = 3.04 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.98

method	result	size
default	\(\frac {\sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (-\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-A \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+B \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+B \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )+\left (2 A -2 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-A +B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(243\)

Input:

int((A+B*cos(d*x+c))*sec(d*x+c)^(1/2)/(a+a*cos(d*x+c)),x,method=_RETURNVER 
BOSE)
 

Output:

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.96 \[ \int \frac {(A+B \cos (c+d x)) \sqrt {\sec (c+d x)}}{a+a \cos (c+d x)} \, dx=-\frac {2 \, {\left (A - B\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - {\left (\sqrt {2} {\left (-i \, A - i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A - i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - {\left (\sqrt {2} {\left (i \, A + i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A + i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - {\left (\sqrt {2} {\left (i \, A - i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A - i \, B\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - {\left (\sqrt {2} {\left (-i \, A + i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A + i \, B\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{2 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \] Input:

integrate((A+B*cos(d*x+c))*sec(d*x+c)^(1/2)/(a+a*cos(d*x+c)),x, algorithm= 
"fricas")
 

Output:

-1/2*(2*(A - B)*sqrt(cos(d*x + c))*sin(d*x + c) - (sqrt(2)*(-I*A - I*B)*co 
s(d*x + c) + sqrt(2)*(-I*A - I*B))*weierstrassPInverse(-4, 0, cos(d*x + c) 
 + I*sin(d*x + c)) - (sqrt(2)*(I*A + I*B)*cos(d*x + c) + sqrt(2)*(I*A + I* 
B))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - (sqrt(2)*( 
I*A - I*B)*cos(d*x + c) + sqrt(2)*(I*A - I*B))*weierstrassZeta(-4, 0, weie 
rstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - (sqrt(2)*(-I*A + 
I*B)*cos(d*x + c) + sqrt(2)*(-I*A + I*B))*weierstrassZeta(-4, 0, weierstra 
ssPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/(a*d*cos(d*x + c) + a*d 
)
 

Sympy [F]

\[ \int \frac {(A+B \cos (c+d x)) \sqrt {\sec (c+d x)}}{a+a \cos (c+d x)} \, dx=\frac {\int \frac {A \sqrt {\sec {\left (c + d x \right )}}}{\cos {\left (c + d x \right )} + 1}\, dx + \int \frac {B \cos {\left (c + d x \right )} \sqrt {\sec {\left (c + d x \right )}}}{\cos {\left (c + d x \right )} + 1}\, dx}{a} \] Input:

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

Output:

(Integral(A*sqrt(sec(c + d*x))/(cos(c + d*x) + 1), x) + Integral(B*cos(c + 
 d*x)*sqrt(sec(c + d*x))/(cos(c + d*x) + 1), x))/a
 

Maxima [F]

\[ \int \frac {(A+B \cos (c+d x)) \sqrt {\sec (c+d x)}}{a+a \cos (c+d x)} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \sqrt {\sec \left (d x + c\right )}}{a \cos \left (d x + c\right ) + a} \,d x } \] Input:

integrate((A+B*cos(d*x+c))*sec(d*x+c)^(1/2)/(a+a*cos(d*x+c)),x, algorithm= 
"maxima")
 

Output:

integrate((B*cos(d*x + c) + A)*sqrt(sec(d*x + c))/(a*cos(d*x + c) + a), x)
 

Giac [F]

\[ \int \frac {(A+B \cos (c+d x)) \sqrt {\sec (c+d x)}}{a+a \cos (c+d x)} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \sqrt {\sec \left (d x + c\right )}}{a \cos \left (d x + c\right ) + a} \,d x } \] Input:

integrate((A+B*cos(d*x+c))*sec(d*x+c)^(1/2)/(a+a*cos(d*x+c)),x, algorithm= 
"giac")
 

Output:

integrate((B*cos(d*x + c) + A)*sqrt(sec(d*x + c))/(a*cos(d*x + c) + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B \cos (c+d x)) \sqrt {\sec (c+d x)}}{a+a \cos (c+d x)} \, dx=\int \frac {\left (A+B\,\cos \left (c+d\,x\right )\right )\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}}{a+a\,\cos \left (c+d\,x\right )} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {(A+B \cos (c+d x)) \sqrt {\sec (c+d x)}}{a+a \cos (c+d x)} \, dx=\frac {\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\cos \left (d x +c \right )+1}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}d x \right ) b}{a} \] Input:

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

Output:

(int(sqrt(sec(c + d*x))/(cos(c + d*x) + 1),x)*a + int((sqrt(sec(c + d*x))* 
cos(c + d*x))/(cos(c + d*x) + 1),x)*b)/a