\(\int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx\) [498]

Optimal result

Integrand size = 33, antiderivative size = 88 \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=\frac {(3 A-B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}-\frac {(A-B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{a d}-\frac {(A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{d (a+a \cos (c+d x))} \] Output:

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

Mathematica [C] (warning: unable to verify)

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

Time = 7.00 (sec) , antiderivative size = 972, normalized size of antiderivative = 11.05 \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

(Cos[c/2 + (d*x)/2]^2*Sqrt[Cos[c + d*x]]*(A + B*Sec[c + d*x])*((-2*(A - B 
+ 2*A*Cos[c])*Csc[c])/d + (2*Sec[c/2]*Sec[c/2 + (d*x)/2]*(-(A*Sin[(d*x)/2] 
) + B*Sin[(d*x)/2]))/d))/((B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])) + (A* 
Cos[c/2 + (d*x)/2]^2*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x 
 - ArcTan[Cot[c]]]^2]*Sec[c/2]*(A + B*Sec[c + d*x])*Sec[d*x - ArcTan[Cot[c 
]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*S 
in[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(d*(B + A* 
Cos[c + d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d*x])) - (B*Cos[c/2 + (d*x 
)/2]^2*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[ 
c]]]^2]*Sec[c/2]*(A + B*Sec[c + d*x])*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - S 
in[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTa 
n[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(d*(B + A*Cos[c + d*x])* 
Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d*x])) - (3*A*Cos[c/2 + (d*x)/2]^2*Csc[c 
/2]*Sec[c/2]*(A + B*Sec[c + d*x])*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, 
 Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - 
Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c] 
*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin 
[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + 
ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos 
[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(2*d*(B + A*Cos[c + d*x])*...
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.06, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.303, Rules used = {3042, 3433, 3042, 3456, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3433

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3456

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

Input:

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

Output:

-1/2*((-2*a*(3*A - B)*EllipticE[(c + d*x)/2, 2])/d + (2*a*(A - B)*Elliptic 
F[(c + d*x)/2, 2])/d)/a^2 - ((A - B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*( 
a + a*Cos[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 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 3433

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]* 
(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Sim 
p[g^(m + n)   Int[(g*Sin[e + f*x])^(p - m - n)*(b + a*Sin[e + f*x])^m*(d + 
c*Sin[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 3456

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/( 
a*f*(2*m + 1))), x] - Simp[1/(a*b*(2*m + 1))   Int[(a + b*Sin[e + f*x])^(m 
+ 1)*(c + d*Sin[e + f*x])^(n - 1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + 
 b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], x], x] /; Fre 
eQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] & 
& NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (In 
tegerQ[2*n] || EqQ[c, 0])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(243\) vs. \(2(89)=178\).

Time = 3.09 (sec) , antiderivative size = 244, normalized size of antiderivative = 2.77

method	result	size
default	\(\frac {\sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \left (A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 A \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-B \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-B \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )+\left (2 A -2 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-A +B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(244\)

Input:

int(cos(d*x+c)^(1/2)*(A+B*sec(d*x+c))/(a+a*sec(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 
)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(A*Ellipti 
cF(cos(1/2*d*x+1/2*c),2^(1/2))+3*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 = 0.09 (sec) , antiderivative size = 241, normalized size of antiderivative = 2.74 \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=-\frac {2 \, {\left (A - B\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - {\left (\sqrt {2} {\left (i \, A - i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A - i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - {\left (\sqrt {2} {\left (-i \, A + i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A + i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - {\left (\sqrt {2} {\left (3 i \, A - i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (3 i \, A - i \, B\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - {\left (\sqrt {2} {\left (-3 i \, A + i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-3 i \, A + i \, B\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{2 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \] Input:

integrate(cos(d*x+c)^(1/2)*(A+B*sec(d*x+c))/(a+a*sec(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)*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))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - (sqrt(2)*( 
3*I*A - I*B)*cos(d*x + c) + sqrt(2)*(3*I*A - I*B))*weierstrassZeta(-4, 0, 
weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - (sqrt(2)*(-3* 
I*A + I*B)*cos(d*x + c) + sqrt(2)*(-3*I*A + I*B))*weierstrassZeta(-4, 0, w 
eierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/(a*d*cos(d*x + 
c) + a*d)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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