3.4.96 \(\int \frac {(a+b \sec (c+d x)) (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx\) [396]

3.4.96.1 Optimal result
3.4.96.2 Mathematica [A] (verified)
3.4.96.3 Rubi [A] (verified)
3.4.96.4 Maple [A] (verified)
3.4.96.5 Fricas [C] (verification not implemented)
3.4.96.6 Sympy [F]
3.4.96.7 Maxima [F]
3.4.96.8 Giac [F]
3.4.96.9 Mupad [F(-1)]

3.4.96.1 Optimal result

Integrand size = 31, antiderivative size = 111 \[ \int \frac {(a+b \sec (c+d x)) (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\frac {2 (a A-b B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{d}+\frac {2 (A b+a B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{d}+\frac {2 b B \sqrt {\sec (c+d x)} \sin (c+d x)}{d} \]

output
2*b*B*sin(d*x+c)*sec(d*x+c)^(1/2)/d+2*(A*a-B*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))*cos(d*x+c)^(1/ 
2)*sec(d*x+c)^(1/2)/d+2*(A*b+B*a)*(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))*cos(d*x+c)^(1/2)*sec(d*x+c)^ 
(1/2)/d
 
3.4.96.2 Mathematica [A] (verified)

Time = 2.26 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.76 \[ \int \frac {(a+b \sec (c+d x)) (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\frac {2 \sqrt {\sec (c+d x)} \left ((a A-b B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+(A b+a B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+b B \sin (c+d x)\right )}{d} \]

input
Integrate[((a + b*Sec[c + d*x])*(A + B*Sec[c + d*x]))/Sqrt[Sec[c + d*x]],x 
]
 
output
(2*Sqrt[Sec[c + d*x]]*((a*A - b*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/ 
2, 2] + (A*b + a*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + b*B*Sin 
[c + d*x]))/d
 
3.4.96.3 Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.323, Rules used = {3042, 4485, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4485

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4274

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

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

3.4.96.3.1 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 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 4485
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B*Cot[ 
e + f*x]*((d*Csc[e + f*x])^n/(f*(n + 1))), x] + Simp[1/(n + 1)   Int[(d*Csc 
[e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*a)*(n + 1)*Csc[e + f*x], x 
], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] &&  !LeQ[ 
n, -1]
 
3.4.96.4 Maple [A] (verified)

Time = 19.48 (sec) , antiderivative size = 244, normalized size of antiderivative = 2.20

method result size
default \(-\frac {2 \left (A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, b -A \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, a -2 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +B \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, a +B \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, b \right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) \(244\)
parts \(-\frac {2 \left (A b +B a \right ) \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}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {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}-\frac {2 B b \left (-2 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\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 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\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}+\frac {2 a A \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}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {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}\) \(458\)

input
int((a+b*sec(d*x+c))*(A+B*sec(d*x+c))/sec(d*x+c)^(1/2),x,method=_RETURNVER 
BOSE)
 
output
-2*(A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/ 
2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*b-A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))* 
(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-2*B*cos(1/ 
2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*b+B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2) 
)*(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+B*Ellipt 
icE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2* 
d*x+1/2*c)^2)^(1/2)*b)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2) 
/d
 
3.4.96.5 Fricas [C] (verification not implemented)

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

Time = 0.11 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.38 \[ \int \frac {(a+b \sec (c+d x)) (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\frac {\frac {2 \, B b \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}} + \sqrt {2} {\left (-i \, B a - i \, A b\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {2} {\left (i \, B a + i \, A b\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + \sqrt {2} {\left (i \, A a - i \, B b\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + \sqrt {2} {\left (-i \, A a + i \, B b\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{d} \]

input
integrate((a+b*sec(d*x+c))*(A+B*sec(d*x+c))/sec(d*x+c)^(1/2),x, algorithm= 
"fricas")
 
output
(2*B*b*sin(d*x + c)/sqrt(cos(d*x + c)) + sqrt(2)*(-I*B*a - I*A*b)*weierstr 
assPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + sqrt(2)*(I*B*a + I*A*b 
)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + sqrt(2)*(I*A 
*a - I*B*b)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) 
 + I*sin(d*x + c))) + sqrt(2)*(-I*A*a + I*B*b)*weierstrassZeta(-4, 0, weie 
rstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/d
 
3.4.96.6 Sympy [F]

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

input
integrate((a+b*sec(d*x+c))*(A+B*sec(d*x+c))/sec(d*x+c)**(1/2),x)
 
output
Integral((A + B*sec(c + d*x))*(a + b*sec(c + d*x))/sqrt(sec(c + d*x)), x)
 
3.4.96.7 Maxima [F]

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

input
integrate((a+b*sec(d*x+c))*(A+B*sec(d*x+c))/sec(d*x+c)^(1/2),x, algorithm= 
"maxima")
 
output
integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)/sqrt(sec(d*x + c)), x)
 
3.4.96.8 Giac [F]

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

input
integrate((a+b*sec(d*x+c))*(A+B*sec(d*x+c))/sec(d*x+c)^(1/2),x, algorithm= 
"giac")
 
output
integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)/sqrt(sec(d*x + c)), x)
 
3.4.96.9 Mupad [F(-1)]

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

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)