\(\int \sqrt {\cos (c+d x)} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [1183]

Optimal result

Integrand size = 31, antiderivative size = 61 \[ \int \sqrt {\cos (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 (A-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 B \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {2 C \sin (c+d x)}{d \sqrt {\cos (c+d x)}} \] Output:

2*(A-C)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+2*B*InverseJacobiAM(1/2*d* 
x+1/2*c,2^(1/2))/d+2*C*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 = 4.49 (sec) , antiderivative size = 272, normalized size of antiderivative = 4.46 \[ \int \sqrt {\cos (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac {\csc (c) \left (3 (A-C) \cos (c-d x-\arctan (\tan (c))) \sec (c)+(A-C) \cos (c+d x+\arctan (\tan (c))) \sec (c)-2 ((A-2 C) \cos (d x)+A \cos (2 c+d x)) \sqrt {\sec ^2(c)}\right )}{\sqrt {\sec ^2(c)}}-4 B \cos (c+d x) \sqrt {\cos ^2(d x-\arctan (\cot (c)))} \sqrt {\csc ^2(c)} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec (d x-\arctan (\cot (c))) \sin (c)-\frac {2 (A-C) \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sec (c) \sin (d x+\arctan (\tan (c)))}{\sqrt {\sec ^2(c)} \sqrt {\sin ^2(d x+\arctan (\tan (c)))}}\right )}{d (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x)))} \] Input:

Integrate[Sqrt[Cos[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 61, 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, 4552, 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.

\(\displaystyle \int \sqrt {\cos (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sqrt {\cos (c+d x)} \left (A+B \sec (c+d x)+C \sec (c+d x)^2\right )dx\)

\(\Big \downarrow \) 4552

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3500

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

Input:

Int[Sqrt[Cos[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
 

Output:

(2*(A - C)*EllipticE[(c + d*x)/2, 2])/d + (2*B*EllipticF[(c + d*x)/2, 2])/ 
d + (2*C*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 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]
 

rule 4552

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(m_)*((A_.) + (B_.)*sec[(e_.) + (f_.)*( 
x_)] + (C_.)*sec[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[b^2   Int[(b*Cos 
[e + f*x])^(m - 2)*(C + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; FreeQ 
[{b, e, f, A, B, C, m}, x] &&  !IntegerQ[m]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(194\) vs. \(2(62)=124\).

Time = 3.63 (sec) , antiderivative size = 195, normalized size of antiderivative = 3.20

method	result	size
default	\(\frac {2 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-2 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) C -2 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(195\)

Input:

int(cos(d*x+c)^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x,method=_RETURNVERBO 
SE)
 

Output:

2*(A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ellipti 
cE(cos(1/2*d*x+1/2*c),2^(1/2))-B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d 
*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+2*sin(1/2*d*x+1 
/2*c)^2*cos(1/2*d*x+1/2*c)*C-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)*EllipticE(cos(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 = 0.08 (sec) , antiderivative size = 166, normalized size of antiderivative = 2.72 \[ \int \sqrt {\cos (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {-i \, \sqrt {2} B \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + i \, \sqrt {2} B \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + \sqrt {2} {\left (i \, A - i \, C\right )} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + \sqrt {2} {\left (-i \, A + i \, C\right )} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, C \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{d \cos \left (d x + c\right )} \] Input:

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

Output:

(-I*sqrt(2)*B*cos(d*x + c)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin 
(d*x + c)) + I*sqrt(2)*B*cos(d*x + c)*weierstrassPInverse(-4, 0, cos(d*x + 
 c) - I*sin(d*x + c)) + sqrt(2)*(I*A - I*C)*cos(d*x + c)*weierstrassZeta(- 
4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + sqrt(2) 
*(-I*A + I*C)*cos(d*x + c)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 
0, cos(d*x + c) - I*sin(d*x + c))) + 2*C*sqrt(cos(d*x + c))*sin(d*x + c))/ 
(d*cos(d*x + c))
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sqrt(cos(d*x + c)), x)
 

Giac [F]

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

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

Output:

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sqrt(cos(d*x + c)), x)
 

Mupad [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.25 \[ \int \sqrt {\cos (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2\,A\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,B\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,C\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \] Input:

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

Output:

(2*A*ellipticE(c/2 + (d*x)/2, 2))/d + (2*B*ellipticF(c/2 + (d*x)/2, 2))/d 
+ (2*C*sin(c + d*x)*hypergeom([-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]

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

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

Output:

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