\(\int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))} \, dx\) [419]

Optimal result

Integrand size = 33, antiderivative size = 149 \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))} \, dx=\frac {2 A \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a d}-\frac {2 (A b-a B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{a^2 d}+\frac {2 b (A b-a B) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{a^2 (a+b) d} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 20.04 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.48 \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))} \, dx=\frac {\cot (c+d x) \left (-a A \sec ^{\frac {3}{2}}(c+d x)-a A \cos (2 (c+d x)) \sec ^{\frac {3}{2}}(c+d x)+a A \sec ^{\frac {7}{2}}(c+d x)+a A \cos (2 (c+d x)) \sec ^{\frac {7}{2}}(c+d x)-2 a A E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {-\tan ^2(c+d x)}+2 a A \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {-\tan ^2(c+d x)}-2 A b \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {-\tan ^2(c+d x)}+2 a B \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {-\tan ^2(c+d x)}\right )}{a^2 d} \] Input:

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

Output:

(Cot[c + d*x]*(-(a*A*Sec[c + d*x]^(3/2)) - a*A*Cos[2*(c + d*x)]*Sec[c + d* 
x]^(3/2) + a*A*Sec[c + d*x]^(7/2) + a*A*Cos[2*(c + d*x)]*Sec[c + d*x]^(7/2 
) - 2*a*A*EllipticE[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[-Tan[c + d*x]^2] 
+ 2*a*A*EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[-Tan[c + d*x]^2] - 
2*A*b*EllipticPi[-(b/a), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[-Tan[c + d*x 
]^2] + 2*a*B*EllipticPi[-(b/a), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[-Tan[ 
c + d*x]^2]))/(a^2*d)
 

Rubi [A] (verified)

Time = 0.88 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.85, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3042, 4526, 3042, 4258, 3042, 3119, 4335, 3042, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4526

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 4335

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3282

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

\(\Big \downarrow \) 3284

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

Input:

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

Output:

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

Defintions of rubi rules used

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 3282

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]
 

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 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 4335

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a 
_)), x_Symbol] :> Simp[Sqrt[d*Sin[e + f*x]]*(Sqrt[d*Csc[e + f*x]]/d)   Int[ 
Sqrt[d*Sin[e + f*x]]/(b + a*Sin[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, 
 x] && NeQ[a^2 - b^2, 0]
 

rule 4526

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(294\) vs. \(2(142)=284\).

Time = 3.94 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.98

method	result	size
default	\(\frac {2 \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \left (A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a b -A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b^{2}+A \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{2}-A \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a b +A \operatorname {EllipticPi}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \frac {2 a}{a -b}, \sqrt {2}\right ) b^{2}-B \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{2}+B \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a b -B \operatorname {EllipticPi}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \frac {2 a}{a -b}, \sqrt {2}\right ) a b \right )}{a^{2} \left (a -b \right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(295\)

Input:

int((A+B*sec(d*x+c))/sec(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*b-A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*b^2+A*EllipticE( 
cos(1/2*d*x+1/2*c),2^(1/2))*a^2-A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a* 
b+A*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))*b^2-B*EllipticF(cos(1 
/2*d*x+1/2*c),2^(1/2))*a^2+B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a*b-B*E 
llipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))*a*b)/a^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
 

Fricas [F]

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

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

Output:

integral((B*sec(d*x + c) + A)*sqrt(sec(d*x + c))/(b*sec(d*x + c)^2 + a*sec 
(d*x + c)), x)
 

Sympy [F]

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

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

Maxima [F]

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

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

Giac [F]

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

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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