\(\int \sqrt {\cos (c+d x)} (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx\) [486]

Optimal result

Integrand size = 31, antiderivative size = 66 \[ \int \sqrt {\cos (c+d x)} (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {2 a (A-B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 a (A+B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {2 a B \sin (c+d x)}{d \sqrt {\cos (c+d x)}} \] Output:

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

Mathematica [C] (warning: unable to verify)

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

Time = 4.41 (sec) , antiderivative size = 252, normalized size of antiderivative = 3.82 \[ \int \sqrt {\cos (c+d x)} (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {a (1+\cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (\frac {\csc (c) \left (3 (A-B) \cos (c-d x-\arctan (\tan (c))) \sec (c)+(A-B) \cos (c+d x+\arctan (\tan (c))) \sec (c)-2 ((A-2 B) \cos (d x)+A \cos (2 c+d x)) \sqrt {\sec ^2(c)}\right )}{\sqrt {\sec ^2(c)}}-4 (A+B) \cos (c+d x) \sqrt {\cos ^2(d x-\arctan (\cot (c)))} \sqrt {\csc ^2(c)} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec (d x-\arctan (\cot (c))) \sin (c)-\frac {2 (A-B) \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sec (c) \sin (d x+\arctan (\tan (c)))}{\sqrt {\sec ^2(c)} \sqrt {\sin ^2(d x+\arctan (\tan (c)))}}\right )}{4 d \sqrt {\cos (c+d x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.387, Rules used = {3042, 3433, 3042, 3447, 3042, 3500, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3433

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3447

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3500

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

Input:

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

Output:

(2*a*(A - B)*EllipticE[(c + d*x)/2, 2])/d + (2*a*(A + B)*EllipticF[(c + d* 
x)/2, 2])/d + (2*a*B*Sin[c + d*x])/(d*Sqrt[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 3447

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a 
 + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), 
 x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
 

rule 3500

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 
 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)* 
(a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x 
])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A 
*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, 
B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(239\) vs. \(2(67)=134\).

Time = 2.12 (sec) , antiderivative size = 240, normalized size of antiderivative = 3.64

method	result	size
default	\(-\frac {2 a \left (A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}-A \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}-2 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+B \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}+B \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(240\)

Input:

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

Output:

-2*a*(A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(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*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))* 
(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)-2*B*cos(1/2* 
d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(s 
in(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)+B*EllipticE(co 
s(1/2*d*x+1/2*c),2^(1/2))*(sin(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*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 = 173, normalized size of antiderivative = 2.62 \[ \int \sqrt {\cos (c+d x)} (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {-i \, \sqrt {2} {\left (A + B\right )} a \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + i \, \sqrt {2} {\left (A + B\right )} a \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + i \, \sqrt {2} {\left (A - B\right )} a \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - i \, \sqrt {2} {\left (A - B\right )} a \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, B a \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{d \cos \left (d x + c\right )} \] Input:

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

Output:

(-I*sqrt(2)*(A + B)*a*cos(d*x + c)*weierstrassPInverse(-4, 0, cos(d*x + c) 
 + I*sin(d*x + c)) + I*sqrt(2)*(A + B)*a*cos(d*x + c)*weierstrassPInverse( 
-4, 0, cos(d*x + c) - I*sin(d*x + c)) + I*sqrt(2)*(A - B)*a*cos(d*x + c)*w 
eierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x 
+ c))) - I*sqrt(2)*(A - B)*a*cos(d*x + c)*weierstrassZeta(-4, 0, weierstra 
ssPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*B*a*sqrt(cos(d*x + c 
))*sin(d*x + c))/(d*cos(d*x + c))
 

Sympy [F]

\[ \int \sqrt {\cos (c+d x)} (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=a \left (\int A \sqrt {\cos {\left (c + d x \right )}}\, dx + \int A \sqrt {\cos {\left (c + d x \right )}} \sec {\left (c + d x \right )}\, dx + \int B \sqrt {\cos {\left (c + d x \right )}} \sec {\left (c + d x \right )}\, dx + \int B \sqrt {\cos {\left (c + d x \right )}} \sec ^{2}{\left (c + d x \right )}\, dx\right ) \] Input:

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

Output:

a*(Integral(A*sqrt(cos(c + d*x)), x) + Integral(A*sqrt(cos(c + d*x))*sec(c 
 + d*x), x) + Integral(B*sqrt(cos(c + d*x))*sec(c + d*x), x) + Integral(B* 
sqrt(cos(c + d*x))*sec(c + d*x)**2, x))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 12.38 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.45 \[ \int \sqrt {\cos (c+d x)} (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {2\,A\,a\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,A\,a\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,B\,a\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,B\,a\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \] Input:

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

Output:

(2*A*a*ellipticE(c/2 + (d*x)/2, 2))/d + (2*A*a*ellipticF(c/2 + (d*x)/2, 2) 
)/d + (2*B*a*ellipticF(c/2 + (d*x)/2, 2))/d + (2*B*a*sin(c + d*x)*hypergeo 
m([-1/4, 1/2], 3/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(1/2)*(sin(c + d*x)^2 
)^(1/2))
 

Reduce [F]

\[ \int \sqrt {\cos (c+d x)} (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=a \left (\left (\int \sqrt {\cos \left (d x +c \right )}d x \right ) a +\left (\int \sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}d x \right ) b +\left (\int \sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )d x \right ) a +\left (\int \sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )d x \right ) b \right ) \] Input:

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

Output:

a*(int(sqrt(cos(c + d*x)),x)*a + int(sqrt(cos(c + d*x))*sec(c + d*x)**2,x) 
*b + int(sqrt(cos(c + d*x))*sec(c + d*x),x)*a + int(sqrt(cos(c + d*x))*sec 
(c + d*x),x)*b)