\(\int \cos (c+d x) \sqrt {a+b \sec (c+d x)} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [939]

Optimal result

Integrand size = 41, antiderivative size = 362 \[ \int \cos (c+d x) \sqrt {a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {(a-b) \sqrt {a+b} (A-2 C) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{b d}+\frac {\sqrt {a+b} (A b+2 b B+2 a C-2 b C) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{b d}-\frac {\sqrt {a+b} (A b+2 a B) \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a d}+\frac {A \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{d} \] Output:

(a-b)*(a+b)^(1/2)*(A-2*C)*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1/2)/(a+b 
)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x 
+c))/(a-b))^(1/2)/b/d+(a+b)^(1/2)*(A*b+2*B*b+2*C*a-2*C*b)*cot(d*x+c)*Ellip 
ticF((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x 
+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/b/d-(a+b)^(1/2)*(A*b+2*B 
*a)*cot(d*x+c)*EllipticPi((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),(a+b)/a,((a+b 
)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^( 
1/2)/a/d+A*(a+b*sec(d*x+c))^(1/2)*sin(d*x+c)/d
                                                                                    
                                                                                    
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(922\) vs. \(2(362)=724\).

Time = 15.43 (sec) , antiderivative size = 922, normalized size of antiderivative = 2.55 \[ \int \cos (c+d x) \sqrt {a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \] Input:

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

Output:

(2*C*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/d + (Sqrt[a + b*Sec[c + d*x]]* 
Sqrt[(1 - Tan[(c + d*x)/2]^2)^(-1)]*(a*A*Tan[(c + d*x)/2] + A*b*Tan[(c + d 
*x)/2] - 2*a*C*Tan[(c + d*x)/2] - 2*b*C*Tan[(c + d*x)/2] - 2*a*A*Tan[(c + 
d*x)/2]^3 + 4*a*C*Tan[(c + d*x)/2]^3 + a*A*Tan[(c + d*x)/2]^5 - A*b*Tan[(c 
 + d*x)/2]^5 - 2*a*C*Tan[(c + d*x)/2]^5 + 2*b*C*Tan[(c + d*x)/2]^5 + 2*A*b 
*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c 
 + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/( 
a + b)] + 4*a*B*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]* 
Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c 
 + d*x)/2]^2)/(a + b)] + 2*A*b*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a 
 - b)/(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b 
 - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + 4*a*B*EllipticP 
i[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[1 
 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x 
)/2]^2)/(a + b)] + (a + b)*(A - 2*C)*EllipticE[ArcSin[Tan[(c + d*x)/2]], ( 
a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*(1 + Tan[(c + d*x)/2]^2)*Sqrt 
[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] - 2*(A*b + 
 a*(B - C) - b*(B + C))*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b 
)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*(1 + Tan[(c + d*x)/2]^2)*Sqrt[(a + b - a*T 
an[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)]))/(d*Sqrt[b + a*Cos[...
 

Rubi [A] (verified)

Time = 1.35 (sec) , antiderivative size = 369, normalized size of antiderivative = 1.02, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.268, Rules used = {3042, 4582, 27, 3042, 4546, 3042, 4409, 3042, 4271, 4319, 4492}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4582

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4546

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4409

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4271

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

\(\Big \downarrow \) 4319

\(\displaystyle \frac {1}{2} \left (-b (A-2 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sqrt {a+b} \cot (c+d x) (2 a C+A b+2 b B-2 b C) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\frac {2 \sqrt {a+b} (2 a B+A b) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}\right )+\frac {A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\)

\(\Big \downarrow \) 4492

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

Input:

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

Output:

((2*(a - b)*Sqrt[a + b]*(A - 2*C)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b 
*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/ 
(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b*d) + (2*Sqrt[a + b]*( 
A*b + 2*b*B + 2*a*C - 2*b*C)*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[ 
c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + 
b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b*d) - (2*Sqrt[a + b]*(A*b + 
 2*a*B)*Cot[c + d*x]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Sec[c + d*x]] 
/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[ 
-((b*(1 + Sec[c + d*x]))/(a - b))])/(a*d))/2 + (A*Sqrt[a + b*Sec[c + d*x]] 
*Sin[c + d*x])/d
 

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 4271

Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[2*(Rt[a 
 + b, 2]/(a*d*Cot[c + d*x]))*Sqrt[b*((1 - Csc[c + d*x])/(a + b))]*Sqrt[(-b) 
*((1 + Csc[c + d*x])/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Csc[ 
c + d*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, c, d}, x] && 
NeQ[a^2 - b^2, 0]
 

rule 4319

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S 
ymbol] :> Simp[-2*(Rt[a + b, 2]/(b*f*Cot[e + f*x]))*Sqrt[(b*(1 - Csc[e + f* 
x]))/(a + b)]*Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]*EllipticF[ArcSin[Sqrt 
[a + b*Csc[e + f*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, e, 
 f}, x] && NeQ[a^2 - b^2, 0]
 

rule 4409

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_ 
.) + (a_)], x_Symbol] :> Simp[c   Int[1/Sqrt[a + b*Csc[e + f*x]], x], x] + 
Simp[d   Int[Csc[e + f*x]/Sqrt[a + b*Csc[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]
 

rule 4492

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[c 
sc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(A*b - a*B)*Rt[a 
 + b*(B/A), 2]*Sqrt[b*((1 - Csc[e + f*x])/(a + b))]*(Sqrt[(-b)*((1 + Csc[e 
+ f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + 
 f*x]]/Rt[a + b*(B/A), 2]], (a*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, 
 f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]
 

rule 4546

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

rule 4582

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1195\) vs. \(2(333)=666\).

Time = 13.72 (sec) , antiderivative size = 1196, normalized size of antiderivative = 3.30

method	result	size
default	\(\text {Expression too large to display}\)	\(1196\)

Input:

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

Output:

-1/d*((2*cos(d*x+c)^2+4*cos(d*x+c)+2)*A*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)* 
(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*b*EllipticPi(-csc(d*x+c)+c 
ot(d*x+c),-1,((a-b)/(a+b))^(1/2))+(4*cos(d*x+c)^2+8*cos(d*x+c)+4)*B*(cos(d 
*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2 
)*a*EllipticPi(-csc(d*x+c)+cot(d*x+c),-1,((a-b)/(a+b))^(1/2))+(cos(d*x+c)^ 
2+2*cos(d*x+c)+1)*A*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d* 
x+c))/(cos(d*x+c)+1))^(1/2)*a*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b 
))^(1/2))+(cos(d*x+c)^2+2*cos(d*x+c)+1)*A*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2 
)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*b*EllipticE(-csc(d*x+c)+ 
cot(d*x+c),((a-b)/(a+b))^(1/2))+(-2*cos(d*x+c)^2-4*cos(d*x+c)-2)*C*(cos(d* 
x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2) 
*a*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+(-2*cos(d*x+c)^2- 
4*cos(d*x+c)-2)*C*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+ 
c))/(cos(d*x+c)+1))^(1/2)*b*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b)) 
^(1/2))+(-2*cos(d*x+c)^2-4*cos(d*x+c)-2)*A*(cos(d*x+c)/(cos(d*x+c)+1))^(1/ 
2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*b*EllipticF(-csc(d*x+c) 
+cot(d*x+c),((a-b)/(a+b))^(1/2))+(-2*cos(d*x+c)^2-4*cos(d*x+c)-2)*B*(cos(d 
*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2 
)*a*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+(2*cos(d*x+c)^2+ 
4*cos(d*x+c)+2)*B*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d...
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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