\(\int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x)+C \sec ^2(c+d x))}{a+a \sec (c+d x)} \, dx\) [1220]

Optimal result

Integrand size = 43, antiderivative size = 93 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {(3 A-B+C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}-\frac {(A-B-C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{a d}-\frac {(A-B+C) \sqrt {\cos (c+d x)} \sin (c+d x)}{d (a+a \cos (c+d x))} \] Output:

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

Mathematica [C] (warning: unable to verify)

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

Time = 7.67 (sec) , antiderivative size = 1612, normalized size of antiderivative = 17.33 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

(Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d 
*x]^2)*((-4*(A - B + C + 2*A*Cos[c])*Csc[c])/d - (4*Sec[c/2]*Sec[c/2 + (d* 
x)/2]*(A*Sin[(d*x)/2] - B*Sin[(d*x)/2] + C*Sin[(d*x)/2]))/d))/((A + 2*C + 
2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])) + (2*A*Cos[c/ 
2 + (d*x)/2]^2*Cos[c + d*x]*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, 
Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*(A + B*Sec[c + d*x] + C*Sec[c + d*x] 
^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(S 
qrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - Ar 
cTan[Cot[c]]]])/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[ 
1 + Cot[c]^2]*(a + a*Sec[c + d*x])) - (2*B*Cos[c/2 + (d*x)/2]^2*Cos[c + d* 
x]*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]] 
^2]*Sec[c/2]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[ 
c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]* 
Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(d*(A + 2 
*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[ 
c + d*x])) - (2*C*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*Csc[c/2]*Hypergeometri 
cPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*(A + B*Sec[c 
 + d*x] + C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - A 
rcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]] 
)]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(d*(A + 2*C + 2*B*Cos[c + d*x] ...
 

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.05, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {3042, 4600, 3042, 3520, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4600

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3520

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

\(\displaystyle -\frac {a (A-B-C) \int \frac {1}{\sqrt {\cos (c+d x)}}dx-a (3 A-B+C) \int \sqrt {\cos (c+d x)}dx}{2 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {a (A-B-C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-a (3 A-B+C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}\)

\(\Big \downarrow \) 3119

\(\displaystyle -\frac {a (A-B-C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 a (3 A-B+C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{2 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}\)

\(\Big \downarrow \) 3120

\(\displaystyle -\frac {\frac {2 a (A-B-C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}-\frac {2 a (3 A-B+C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{2 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}\)

Input:

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

Output:

-1/2*((-2*a*(3*A - B + C)*EllipticE[(c + d*x)/2, 2])/d + (2*a*(A - B - C)* 
EllipticF[(c + d*x)/2, 2])/d)/a^2 - ((A - B + C)*Sqrt[Cos[c + d*x]]*Sin[c 
+ d*x])/(d*(a + a*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 3520

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

rule 4600

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(280\) vs. \(2(94)=188\).

Time = 3.30 (sec) , antiderivative size = 281, normalized size of antiderivative = 3.02

method	result	size
default	\(\frac {\sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \left (A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 A \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-B \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-B \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-C \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+C \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )+\left (2 A -2 B +2 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-A +B -C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(281\)

Input:

int(cos(d*x+c)^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x,me 
thod=_RETURNVERBOSE)
 

Output:

((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(cos(1/2*d*x+1/2*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)*(A*Ellipti 
cF(cos(1/2*d*x+1/2*c),2^(1/2))+3*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-B 
*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-B*EllipticE(cos(1/2*d*x+1/2*c),2^(1 
/2))-C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+C*EllipticE(cos(1/2*d*x+1/2*c 
),2^(1/2)))+(2*A-2*B+2*C)*sin(1/2*d*x+1/2*c)^4+(-A+B-C)*sin(1/2*d*x+1/2*c) 
^2)/a/cos(1/2*d*x+1/2*c)/(-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 [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 266, normalized size of antiderivative = 2.86 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=-\frac {2 \, {\left (A - B + C\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - {\left (\sqrt {2} {\left (i \, A - i \, B - i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A - i \, B - i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - {\left (\sqrt {2} {\left (-i \, A + i \, B + i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A + i \, B + i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - {\left (\sqrt {2} {\left (3 i \, A - i \, B + i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (3 i \, A - i \, B + i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - {\left (\sqrt {2} {\left (-3 i \, A + i \, B - i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-3 i \, A + i \, B - i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{2 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \] Input:

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

Output:

-1/2*(2*(A - B + C)*sqrt(cos(d*x + c))*sin(d*x + c) - (sqrt(2)*(I*A - I*B 
- I*C)*cos(d*x + c) + sqrt(2)*(I*A - I*B - I*C))*weierstrassPInverse(-4, 0 
, cos(d*x + c) + I*sin(d*x + c)) - (sqrt(2)*(-I*A + I*B + I*C)*cos(d*x + c 
) + sqrt(2)*(-I*A + I*B + I*C))*weierstrassPInverse(-4, 0, cos(d*x + c) - 
I*sin(d*x + c)) - (sqrt(2)*(3*I*A - I*B + I*C)*cos(d*x + c) + sqrt(2)*(3*I 
*A - I*B + I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x 
 + c) + I*sin(d*x + c))) - (sqrt(2)*(-3*I*A + I*B - I*C)*cos(d*x + c) + sq 
rt(2)*(-3*I*A + I*B - I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 
 0, cos(d*x + c) - I*sin(d*x + c))))/(a*d*cos(d*x + c) + a*d)
 

Sympy [F]

\[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {\int \frac {A \sqrt {\cos {\left (c + d x \right )}}}{\sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sqrt {\cos {\left (c + d x \right )}} \sec {\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sqrt {\cos {\left (c + d x \right )}} \sec ^{2}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \] Input:

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

Output:

(Integral(A*sqrt(cos(c + d*x))/(sec(c + d*x) + 1), x) + Integral(B*sqrt(co 
s(c + d*x))*sec(c + d*x)/(sec(c + d*x) + 1), x) + Integral(C*sqrt(cos(c + 
d*x))*sec(c + d*x)**2/(sec(c + d*x) + 1), x))/a
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\sec \left (d x +c \right )+1}d x \right ) a +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}}{\sec \left (d x +c \right )+1}d x \right ) c +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )}{\sec \left (d x +c \right )+1}d x \right ) b}{a} \] Input:

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

Output:

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