\(\int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+b \sec (c+d x))} \, dx\) [577]

Optimal result

Integrand size = 33, antiderivative size = 61 \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+b \sec (c+d x))} \, dx=\frac {2 A \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{a d}-\frac {2 (A b-a B) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a (a+b) d} \] Output:

2*A*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/a/d-2*(A*b-B*a)*EllipticPi(sin( 
1/2*d*x+1/2*c),2*a/(a+b),2^(1/2))/a/(a+b)/d
 

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.95 \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+b \sec (c+d x))} \, dx=\frac {2 \left (A (a+b) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+(-A b+a B) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )\right )}{a (a+b) d} \] Input:

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

Output:

(2*(A*(a + b)*EllipticF[(c + d*x)/2, 2] + (-(A*b) + a*B)*EllipticPi[(2*a)/ 
(a + b), (c + d*x)/2, 2]))/(a*(a + b)*d)
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3042, 3433, 3042, 3481, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3433

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3481

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

\(\displaystyle \frac {2 A \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{a d}-\frac {(A b-a B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}\)

\(\Big \downarrow \) 3284

\(\displaystyle \frac {2 A \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{a d}-\frac {2 (A b-a B) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a d (a+b)}\)

Input:

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

Output:

(2*A*EllipticF[(c + d*x)/2, 2])/(a*d) - (2*(A*b - a*B)*EllipticPi[(2*a)/(a 
 + b), (c + d*x)/2, 2])/(a*(a + b)*d)
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, 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 3284

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]
 

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 3481

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(216\) vs. \(2(64)=128\).

Time = 2.10 (sec) , antiderivative size = 217, normalized size of antiderivative = 3.56

method	result	size
default	\(-\frac {2 \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \left (A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a -A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b +A \operatorname {EllipticPi}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \frac {2 a}{a -b}, \sqrt {2}\right ) b -B \operatorname {EllipticPi}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \frac {2 a}{a -b}, \sqrt {2}\right ) a \right )}{a \left (a -b \right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(217\)

Input:

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

Output:

-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)*(A*EllipticF(cos(1/2*d*x+1 
/2*c),2^(1/2))*a-A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*b+A*EllipticPi(co 
s(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))*b-B*EllipticPi(cos(1/2*d*x+1/2*c),2*a/ 
(a-b),2^(1/2))*a)/a/(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
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(sqrt(cos(c + d*x))/cos(c + d*x),x)