\(\int \cos ^2(c+d x) \sqrt {a+b \sec (c+d x)} (A+C \sec ^2(c+d x)) \, dx\) [714]

Optimal result

Integrand size = 35, antiderivative size = 411 \[ \int \cos ^2(c+d x) \sqrt {a+b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {A (a-b) \sqrt {a+b} \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{4 a d}+\frac {\sqrt {a+b} (A b+2 a (A+4 C)) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{4 a d}+\frac {\sqrt {a+b} \left (A b^2-4 a^2 (A+2 C)\right ) \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{4 a^2 d}+\frac {A b \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{4 a d}+\frac {A \cos (c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{2 d} \] Output:

1/4*A*(a-b)*(a+b)^(1/2)*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/d+1/4*(a+b)^(1/2)*(A*b+2*a*(A+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 
*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*(a+b*sec(d*x+c))^(1/2)*sin(d*x+c)/a/d+1/2*A*c 
os(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. \(1084\) vs. \(2(411)=822\).

Time = 18.64 (sec) , antiderivative size = 1084, normalized size of antiderivative = 2.64 \[ \int \cos ^2(c+d x) \sqrt {a+b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \] Input:

Integrate[Cos[c + d*x]^2*Sqrt[a + b*Sec[c + d*x]]*(A + 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] + 2*a*A*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 - 8*a^2 
*A*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^2*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*Ta 
n[(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*EllipticPi[-1, ArcS 
in[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)] - 16*a^2*C*EllipticPi[-1, ArcSi 
n[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 + b)*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*S 
qrt[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*(2*a*(A + 2*C) - b...
 

Rubi [A] (verified)

Time = 1.71 (sec) , antiderivative size = 409, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4583, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4583

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4592

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4546

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4409

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4271

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

\(\Big \downarrow \) 4319

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

\(\Big \downarrow \) 4492

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

Input:

Int[Cos[c + d*x]^2*Sqrt[a + b*Sec[c + d*x]]*(A + 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*(a - b)*Sqrt[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)]*Sqr 
t[-((b*(1 + Sec[c + d*x]))/(a - b))])/d - (2*Sqrt[a + b]*(A*b + 2*a*(A + 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^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*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 4583

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_. 
))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[A*Co 
t[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*(C*n + A*(n + 1))*Csc[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, C}, x] && NeQ[a^2 - b^2, 0] && Gt 
Q[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. \(1032\) vs. \(2(370)=740\).

Time = 11.38 (sec) , antiderivative size = 1033, normalized size of antiderivative = 2.51

method	result	size
default	\(\text {Expression too large to display}\)	\(1033\)

Input:

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

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))+(-1 
6*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*c 
os(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)*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^2*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)*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*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+(8*cos(d*x+c)^ 
2+16*cos(d*x+c)+8)*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*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/( 
a+b))^(1/2))+(-8*cos(d*x+c)^2-16*cos(d*x+c)-8)*C*(1/(a+b)*(b+a*cos(d*x+...
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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