\(\int \cos ^3(c+d x) \sqrt {a+b \sec (c+d x)} (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [820]

Optimal result

Integrand size = 42, antiderivative size = 429 \[ \int \cos ^3(c+d x) \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {(a-b) \sqrt {a+b} (b B+4 a C) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{4 a b d}+\frac {\sqrt {a+b} (b B+2 a (B+2 C)) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{4 a d}-\frac {\sqrt {a+b} \left (4 a^2 B-b^2 B+4 a b C\right ) \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{4 a^2 d}+\frac {(b B+4 a C) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{4 a d}+\frac {B \cos (c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{2 d} \] Output:

1/4*(a-b)*(a+b)^(1/2)*(B*b+4*C*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)*(B*b+2*a*(B+2*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)*(4*B*a^2-B*b^2+4*C*a*b)*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*(B*b+4*C*a)*(a+b*sec(d*x+c))^(1/2)* 
sin(d*x+c)/a/d+1/2*B*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. \(1149\) vs. \(2(429)=858\).

Time = 17.15 (sec) , antiderivative size = 1149, normalized size of antiderivative = 2.68 \[ \int \cos ^3(c+d x) \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \] Input:

Integrate[Cos[c + d*x]^3*Sqrt[a + b*Sec[c + d*x]]*(B*Sec[c + d*x] + C*Sec[ 
c + d*x]^2),x]
 

Output:

(B*Sqrt[a + b*Sec[c + d*x]]*Sin[2*(c + d*x)])/(4*d) + (Sqrt[a + b*Sec[c + 
d*x]]*Sqrt[(1 - Tan[(c + d*x)/2]^2)^(-1)]*(a*b*B*Tan[(c + d*x)/2] + b^2*B* 
Tan[(c + d*x)/2] + 4*a^2*C*Tan[(c + d*x)/2] + 4*a*b*C*Tan[(c + d*x)/2] - 2 
*a*b*B*Tan[(c + d*x)/2]^3 - 8*a^2*C*Tan[(c + d*x)/2]^3 + a*b*B*Tan[(c + d* 
x)/2]^5 - b^2*B*Tan[(c + d*x)/2]^5 + 4*a^2*C*Tan[(c + d*x)/2]^5 - 4*a*b*C* 
Tan[(c + d*x)/2]^5 + 8*a^2*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*b^2*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)] + 8*a*b*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*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)] - 2*b^2*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)] + 8*a* 
b*C*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)] + (a + b)*(b*B + 4*a*C)*EllipticE[ArcSin[...
 

Rubi [A] (verified)

Time = 1.98 (sec) , antiderivative size = 427, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 4560, 3042, 4520, 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.

\(\displaystyle \int \cos ^3(c+d x) \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )} \left (B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx\)

\(\Big \downarrow \) 4560

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4520

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4592

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4546

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4409

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4271

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

\(\Big \downarrow \) 4319

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

\(\Big \downarrow \) 4492

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

Input:

Int[Cos[c + d*x]^3*Sqrt[a + b*Sec[c + d*x]]*(B*Sec[c + d*x] + C*Sec[c + d* 
x]^2),x]
 

Output:

(B*Cos[c + d*x]*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(2*d) + (((2*(a - b 
)*Sqrt[a + b]*(b*B + 4*a*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]*(b*B + 
2*a*(B + 2*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]*(4*a^2*B - b^2*B + 4*a*b 
*C)*Cot[c + d*x]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqr 
t[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) + ((b*B + 4*a*C)*Sqrt[a + b*S 
ec[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 4520

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), 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*A*(n + 1))*Csc[e + f*x] - A*b*(m + n + 1)*Csc[e + f*x]^2 
, x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && NeQ 
[a^2 - b^2, 0] && LtQ[0, m, 1] && LeQ[n, -1]
 

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 4560

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

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. \(1168\) vs. \(2(388)=776\).

Time = 8.62 (sec) , antiderivative size = 1169, normalized size of antiderivative = 2.72

method	result	size
default	\(\text {Expression too large to display}\)	\(1169\)

Input:

int(cos(d*x+c)^3*(a+b*sec(d*x+c))^(1/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x,me 
thod=_RETURNVERBOSE)
                                                                                    
                                                                                    
 

Output:

1/4/d/a*((-8*cos(d*x+c)^2-16*cos(d*x+c)-8)*B*(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)*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)*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)*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*b*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)*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))+(-cos(d*x+c)^2-2*cos 
(d*x+c)-1)*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)*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)*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*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)*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*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)*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*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*(1/(a+b)*(b+a*cos(d*x+c))...
 

Fricas [F]

\[ \int \cos ^3(c+d x) \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \sqrt {b \sec \left (d x + c\right ) + a} \cos \left (d x + c\right )^{3} \,d x } \] Input:

integrate(cos(d*x+c)^3*(a+b*sec(d*x+c))^(1/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2 
),x, algorithm="fricas")
 

Output:

integral((C*cos(d*x + c)^3*sec(d*x + c)^2 + B*cos(d*x + c)^3*sec(d*x + c)) 
*sqrt(b*sec(d*x + c) + a), x)
 

Sympy [F(-1)]

Timed out. \[ \int \cos ^3(c+d x) \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**3*(a+b*sec(d*x+c))**(1/2)*(B*sec(d*x+c)+C*sec(d*x+c) 
**2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \cos ^3(c+d x) \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \sqrt {b \sec \left (d x + c\right ) + a} \cos \left (d x + c\right )^{3} \,d x } \] Input:

integrate(cos(d*x+c)^3*(a+b*sec(d*x+c))^(1/2)*(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))*sqrt(b*sec(d*x + c) + a)*cos 
(d*x + c)^3, x)
 

Giac [F]

\[ \int \cos ^3(c+d x) \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \sqrt {b \sec \left (d x + c\right ) + a} \cos \left (d x + c\right )^{3} \,d x } \] Input:

integrate(cos(d*x+c)^3*(a+b*sec(d*x+c))^(1/2)*(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))*sqrt(b*sec(d*x + c) + a)*cos 
(d*x + c)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \cos ^3(c+d x) \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int {\cos \left (c+d\,x\right )}^3\,\left (\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}} \,d x \] Input:

int(cos(c + d*x)^3*(B/cos(c + d*x) + C/cos(c + d*x)^2)*(a + b/cos(c + d*x) 
)^(1/2),x)
 

Output:

int(cos(c + d*x)^3*(B/cos(c + d*x) + C/cos(c + d*x)^2)*(a + b/cos(c + d*x) 
)^(1/2), x)
 

Reduce [F]

\[ \int \cos ^3(c+d x) \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\left (\int \sqrt {\sec \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )^{2}d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )d x \right ) b \] Input:

int(cos(d*x+c)^3*(a+b*sec(d*x+c))^(1/2)*(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)**3*sec(c + d*x)**2,x)*c + int(sq 
rt(sec(c + d*x)*b + a)*cos(c + d*x)**3*sec(c + d*x),x)*b