\(\int \cos ^2(c+d x) \sqrt {a+b \sec (c+d x)} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [940]

Optimal result

Integrand size = 43, antiderivative size = 435 \[ \int \cos ^2(c+d x) \sqrt {a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {(a-b) \sqrt {a+b} (A b+4 a B) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{4 a b d}+\frac {\sqrt {a+b} (A b+2 a (A+2 B+4 C)) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{4 a d}+\frac {\sqrt {a+b} \left (A b^2-4 a b B-4 a^2 (A+2 C)\right ) \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{4 a^2 d}+\frac {(A b+4 a B) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{4 a d}+\frac {A \cos (c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{2 d} \] Output:

1/4*(a-b)*(a+b)^(1/2)*(A*b+4*B*a)*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)/a/b/d+1/4*(a+b)^(1/2)*(A*b+2*a*(A+2*B+4*C))*cot( 
d*x+c)*EllipticF((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)/a/d+1/4*(a+b 
)^(1/2)*(A*b^2-4*B*a*b-4*a^2*(A+2*C))*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^2/d+1/4*(A*b+4*B*a)*(a+b*sec(d*x+ 
c))^(1/2)*sin(d*x+c)/a/d+1/2*A*cos(d*x+c)*(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. \(1371\) vs. \(2(435)=870\).

Time = 19.06 (sec) , antiderivative size = 1371, normalized size of antiderivative = 3.15 \[ \int \cos ^2(c+d x) \sqrt {a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \] Input:

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

Output:

(A*Sqrt[a + b*Sec[c + d*x]]*Sin[2*(c + d*x)])/(4*d) + (Sqrt[a + b*Sec[c + 
d*x]]*(-(a*A*b*Tan[(c + d*x)/2]) - A*b^2*Tan[(c + d*x)/2] - 4*a^2*B*Tan[(c 
 + d*x)/2] - 4*a*b*B*Tan[(c + d*x)/2] + 2*a*A*b*Tan[(c + d*x)/2]^3 + 8*a^2 
*B*Tan[(c + d*x)/2]^3 - a*A*b*Tan[(c + d*x)/2]^5 + A*b^2*Tan[(c + d*x)/2]^ 
5 - 4*a^2*B*Tan[(c + d*x)/2]^5 + 4*a*b*B*Tan[(c + d*x)/2]^5 - 8*a^2*A*Elli 
pticPi[-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^2*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqr 
t[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)] - 8*a*b*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)] - 16*a^2*C*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)] - 8*a^2*A*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)] + 2*A*b^2*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)] - 8*a*b*B*EllipticPi[ 
-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[...
 

Rubi [A] (verified)

Time = 1.86 (sec) , antiderivative size = 433, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.326, Rules used = {3042, 4582, 27, 3042, 4592, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4582

\(\displaystyle \frac {1}{2} \int \frac {\cos (c+d x) \left (b (A+4 C) \sec ^2(c+d x)+2 (2 b B+a (A+2 C)) \sec (c+d x)+A b+4 a B\right )}{2 \sqrt {a+b \sec (c+d x)}}dx+\frac {A \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} \int \frac {\cos (c+d x) \left (b (A+4 C) \sec ^2(c+d x)+2 (2 b B+a (A+2 C)) \sec (c+d x)+A b+4 a B\right )}{\sqrt {a+b \sec (c+d x)}}dx+\frac {A \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \int \frac {b (A+4 C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+2 (2 b B+a (A+2 C)) \csc \left (c+d x+\frac {\pi }{2}\right )+A b+4 a B}{\csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {A \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 4592

\(\displaystyle \frac {1}{4} \left (\frac {(4 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{a d}-\frac {\int \frac {-4 (A+2 C) a^2-4 b B a-2 b (A+4 C) \sec (c+d x) a+A b^2+b (A b+4 a B) \sec ^2(c+d x)}{2 \sqrt {a+b \sec (c+d x)}}dx}{a}\right )+\frac {A \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} \left (\frac {(4 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{a d}-\frac {\int \frac {-4 (A+2 C) a^2-4 b B a-2 b (A+4 C) \sec (c+d x) a+A b^2+b (A b+4 a B) \sec ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}}dx}{2 a}\right )+\frac {A \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {(4 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{a d}-\frac {\int \frac {-4 (A+2 C) a^2-4 b B a-2 b (A+4 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a+A b^2+b (A b+4 a B) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a}\right )+\frac {A \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 4546

\(\displaystyle \frac {1}{4} \left (\frac {(4 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{a d}-\frac {\int \frac {-4 (A+2 C) a^2-4 b B a+A b^2+(-b (A b+4 a B)-2 a b (A+4 C)) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx+b (4 a B+A b) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx}{2 a}\right )+\frac {A \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {(4 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{a d}-\frac {\int \frac {-4 (A+2 C) a^2-4 b B a+A b^2+(-b (A b+4 a B)-2 a b (A+4 C)) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+b (4 a B+A b) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a}\right )+\frac {A \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 4409

\(\displaystyle \frac {1}{4} \left (\frac {(4 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{a d}-\frac {\left (-4 a^2 (A+2 C)-4 a b B+A b^2\right ) \int \frac {1}{\sqrt {a+b \sec (c+d x)}}dx-b (2 a (A+2 B+4 C)+A b) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx+b (4 a B+A b) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a}\right )+\frac {A \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {(4 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{a d}-\frac {\left (-4 a^2 (A+2 C)-4 a b B+A b^2\right ) \int \frac {1}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b (2 a (A+2 B+4 C)+A b) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+b (4 a B+A b) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a}\right )+\frac {A \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 4271

\(\displaystyle \frac {1}{4} \left (\frac {(4 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{a d}-\frac {-b (2 a (A+2 B+4 C)+A b) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+b (4 a B+A b) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {a+b} \cot (c+d x) \left (-4 a^2 (A+2 C)-4 a b B+A b^2\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}}{2 a}\right )+\frac {A \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 4319

\(\displaystyle \frac {1}{4} \left (\frac {(4 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{a d}-\frac {b (4 a B+A b) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {a+b} \cot (c+d x) \left (-4 a^2 (A+2 C)-4 a b B+A b^2\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}-\frac {2 \sqrt {a+b} \cot (c+d x) (2 a (A+2 B+4 C)+A b) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}}{2 a}\right )+\frac {A \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 4492

\(\displaystyle \frac {1}{4} \left (\frac {(4 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{a d}-\frac {-\frac {2 \sqrt {a+b} \cot (c+d x) \left (-4 a^2 (A+2 C)-4 a b B+A b^2\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}-\frac {2 \sqrt {a+b} \cot (c+d x) (2 a (A+2 B+4 C)+A b) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}-\frac {2 (a-b) \sqrt {a+b} (4 a B+A b) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b d}}{2 a}\right )+\frac {A \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

Input:

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

Output:

(A*Cos[c + d*x]*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(2*d) + (-1/2*((-2* 
(a - b)*Sqrt[a + b]*(A*b + 4*a*B)*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*a*(A + 2*B + 4*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))])/d - (2*Sqrt[a + b]*(A*b^2 - 4*a 
*b*B - 4*a^2*(A + 2*C))*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))/a + ((A*b + 4*a 
*B)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(a*d))/4
 

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]
 

rule 4592

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 + 1)*((d 
*Csc[e + f*x])^n/(a*f*n)), x] + Simp[1/(a*d*n)   Int[(a + b*Csc[e + f*x])^m 
*(d*Csc[e + f*x])^(n + 1)*Simp[a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)* 
Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d 
, e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1468\) vs. \(2(394)=788\).

Time = 12.44 (sec) , antiderivative size = 1469, normalized size of antiderivative = 3.38

method	result	size
default	\(\text {Expression too large to display}\)	\(1469\)

Input:

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

Output:

-1/4/d/a*((8*cos(d*x+c)^2+16*cos(d*x+c)+8)*A*(1/(a+b)*(b+a*cos(d*x+c))/(co 
s(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^2*EllipticPi(-csc(d 
*x+c)+cot(d*x+c),-1,((a-b)/(a+b))^(1/2))+(-2*cos(d*x+c)^2-4*cos(d*x+c)-2)* 
A*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+ 
1))^(1/2)*b^2*EllipticPi(-csc(d*x+c)+cot(d*x+c),-1,((a-b)/(a+b))^(1/2))+(8 
*cos(d*x+c)^2+16*cos(d*x+c)+8)*B*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1)) 
^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a*b*EllipticPi(-csc(d*x+c)+cot(d* 
x+c),-1,((a-b)/(a+b))^(1/2))+(16*cos(d*x+c)^2+32*cos(d*x+c)+16)*C*(1/(a+b) 
*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)* 
a^2*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*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d 
*x+c)/(cos(d*x+c)+1))^(1/2)*a*b*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*(1/(a+b)*(b+a*cos(d*x+c))/(cos 
(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*b^2*EllipticE(-csc(d*x 
+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+(4*cos(d*x+c)^2+8*cos(d*x+c)+4)*B*(1/( 
a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1 
/2)*a^2*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+(4*cos(d*x+c 
)^2+8*cos(d*x+c)+4)*B*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos 
(d*x+c)/(cos(d*x+c)+1))^(1/2)*a*b*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/ 
(a+b))^(1/2))+(-4*cos(d*x+c)^2-8*cos(d*x+c)-4)*A*(1/(a+b)*(b+a*cos(d*x+...
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

integrate(cos(d*x+c)**2*(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)**2, x)
 

Maxima [F]

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

integrate(cos(d*x+c)^2*(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)^2, x)
 

Giac [F]

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

integrate(cos(d*x+c)^2*(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)^2, x)
 

Mupad [F(-1)]

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

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

Reduce [F]

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

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