3.4.61 \(\int \cos ^2(c+d x) (a+b \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx\) [361]

3.4.61.1 Optimal result

Integrand size = 33, antiderivative size = 428 \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\frac {(a-b) \sqrt {a+b} (5 A b+4 a B) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{4 b d}+\frac {\sqrt {a+b} (2 a A+5 A b+4 a B+8 b B) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{4 d}-\frac {\sqrt {a+b} \left (4 a^2 A+3 A b^2+12 a b B\right ) \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{4 a d}+\frac {(5 A b+4 a B) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{4 d}+\frac {a A \cos (c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{2 d} \]

output

1/4*(a-b)*(5*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))*(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+1/4*(2*A*a+5*A*b+4*B*a+8*B*b)*cot(d*x+c)*E 
llipticF((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(a+b)^(1/ 
2)*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/d-1/4*(4 
*A*a^2+3*A*b^2+12*B*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))*(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*(5*A*b+4*B*a)*sin(d*x+c)*(a+ 
b*sec(d*x+c))^(1/2)/d+1/2*a*A*cos(d*x+c)*sin(d*x+c)*(a+b*sec(d*x+c))^(1/2) 
/d
 

3.4.61.2 Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(1197\) vs. \(2(428)=856\).

Time = 20.07 (sec) , antiderivative size = 1197, normalized size of antiderivative = 2.80 \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx =\text {Too large to display} \]

input

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

output

(a*A*Cos[c + d*x]*(a + b*Sec[c + d*x])^(3/2)*Sin[2*(c + d*x)])/(4*d*(b + a 
*Cos[c + d*x])) - ((a + b*Sec[c + d*x])^(3/2)*(5*a*A*b*Tan[(c + d*x)/2] + 
5*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] - 10*a*A*b*Tan[(c + d*x)/2]^3 - 8*a^2*B*Tan[(c + d*x)/2]^3 + 5*a*A*b* 
Tan[(c + d*x)/2]^5 - 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*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)] + 6*A*b^2*EllipticPi[-1, 
ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sq 
rt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + 24*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)] + 8*a^2*A*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)] + 6*A*b^2*EllipticPi[-1, Arc 
Sin[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)] + 24*a*b*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)] + (a + b)*(5*A*b + 4*a*B)*...
 

3.4.61.3 Rubi [A] (verified)

Time = 1.86 (sec) , antiderivative size = 431, normalized size of antiderivative = 1.01, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.424, Rules used = {3042, 4513, 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.

\(\displaystyle \int \cos ^2(c+d x) (a+b \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4513

\(\displaystyle \frac {a A \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}-\frac {1}{2} \int -\frac {\cos (c+d x) \left (b (a A+4 b B) \sec ^2(c+d x)+2 \left (A a^2+4 b B a+2 A b^2\right ) \sec (c+d x)+a (5 A b+4 a B)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} \int \frac {\cos (c+d x) \left (b (a A+4 b B) \sec ^2(c+d x)+2 \left (A a^2+4 b B a+2 A b^2\right ) \sec (c+d x)+a (5 A b+4 a B)\right )}{\sqrt {a+b \sec (c+d x)}}dx+\frac {a A \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \int \frac {b (a A+4 b B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+2 \left (A a^2+4 b B a+2 A b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+a (5 A b+4 a B)}{\csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {a A \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 4592

\(\displaystyle \frac {1}{4} \left (\frac {(4 a B+5 A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}-\frac {\int -\frac {-a b (5 A b+4 a B) \sec ^2(c+d x)+2 a b (a A+4 b B) \sec (c+d x)+a \left (4 A a^2+12 b B a+3 A b^2\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{a}\right )+\frac {a A \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} \left (\frac {\int \frac {-a b (5 A b+4 a B) \sec ^2(c+d x)+2 a b (a A+4 b B) \sec (c+d x)+a \left (4 A a^2+12 b B a+3 A b^2\right )}{\sqrt {a+b \sec (c+d x)}}dx}{2 a}+\frac {(4 a B+5 A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\right )+\frac {a A \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {\int \frac {-a b (5 A b+4 a B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+2 a b (a A+4 b B) \csc \left (c+d x+\frac {\pi }{2}\right )+a \left (4 A a^2+12 b B a+3 A b^2\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a}+\frac {(4 a B+5 A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\right )+\frac {a A \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 4546

\(\displaystyle \frac {1}{4} \left (\frac {\int \frac {a \left (4 A a^2+12 b B a+3 A b^2\right )+(a b (5 A b+4 a B)+2 a b (a A+4 b B)) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx-a b (4 a B+5 A b) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx}{2 a}+\frac {(4 a B+5 A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\right )+\frac {a A \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {\int \frac {a \left (4 A a^2+12 b B a+3 A b^2\right )+(a b (5 A b+4 a B)+2 a b (a A+4 b B)) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-a b (4 a B+5 A b) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a}+\frac {(4 a B+5 A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\right )+\frac {a A \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 4409

\(\displaystyle \frac {1}{4} \left (\frac {a \left (4 a^2 A+12 a b B+3 A b^2\right ) \int \frac {1}{\sqrt {a+b \sec (c+d x)}}dx-a b (4 a B+5 A b) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+a b (2 a A+4 a B+5 A b+8 b B) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx}{2 a}+\frac {(4 a B+5 A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\right )+\frac {a A \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {a \left (4 a^2 A+12 a b B+3 A b^2\right ) \int \frac {1}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+a b (2 a A+4 a B+5 A b+8 b B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-a b (4 a B+5 A b) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a}+\frac {(4 a B+5 A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\right )+\frac {a A \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 4271

\(\displaystyle \frac {1}{4} \left (\frac {a b (2 a A+4 a B+5 A b+8 b B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-a b (4 a B+5 A b) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {a+b} \left (4 a^2 A+12 a b B+3 A b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}}{2 a}+\frac {(4 a B+5 A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\right )+\frac {a A \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 4319

\(\displaystyle \frac {1}{4} \left (\frac {-a b (4 a B+5 A b) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {a+b} \left (4 a^2 A+12 a b B+3 A b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}+\frac {2 a \sqrt {a+b} (2 a A+4 a B+5 A b+8 b B) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}}{2 a}+\frac {(4 a B+5 A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\right )+\frac {a A \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 4492

\(\displaystyle \frac {1}{4} \left (\frac {-\frac {2 \sqrt {a+b} \left (4 a^2 A+12 a b B+3 A b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}+\frac {2 a \sqrt {a+b} (2 a A+4 a B+5 A b+8 b B) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}+\frac {2 a (a-b) \sqrt {a+b} (4 a B+5 A b) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b d}}{2 a}+\frac {(4 a B+5 A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\right )+\frac {a A \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

input

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

output

(a*A*Cos[c + d*x]*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(2*d) + (((2*a*(a 
 - b)*Sqrt[a + b]*(5*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*a*Sqrt[a + b] 
*(2*a*A + 5*A*b + 4*a*B + 8*b*B)*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 
*A + 3*A*b^2 + 12*a*b*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))])/d)/(2*a) + ((5*A*b + 
4*a*B)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/d)/4
 

3.4.61.3.1 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 4513

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[a*A*Cot 
[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*n)), x] + Sim 
p[1/(d*n)   Int[(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^(n + 1)*Simp[ 
a*(a*B*n - A*b*(m - n - 1)) + (2*a*b*B*n + A*(b^2*n + a^2*(1 + n)))*Csc[e + 
 f*x] + b*(b*B*n + a*A*(m + n))*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] && GtQ[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 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]
 

3.4.61.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(3285\) vs. \(2(387)=774\).

Time = 9.98 (sec) , antiderivative size = 3286, normalized size of antiderivative = 7.68

method	result	size
default	\(\text {Expression too large to display}\)	\(3286\)

input

int(cos(d*x+c)^2*(a+b*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)),x,method=_RETURNV 
ERBOSE)
 

output

1/4/d*(8*A*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(co 
s(d*x+c)+1))^(1/2)*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*b^ 
2*cos(d*x+c)^2-16*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)*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^ 
(1/2))*b^2*cos(d*x+c)-8*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)*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/( 
a+b))^(1/2))*b^2*cos(d*x+c)^2-6*A*EllipticPi(cot(d*x+c)-csc(d*x+c),-1,((a- 
b)/(a+b))^(1/2))*(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^2*cos(d*x+c)^2+4*A*EllipticF(cot(d*x+c)-csc(d*x 
+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*c 
os(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^2*cos(d*x+c)^2-5*A*(1/(a+b)*(b+a*cos(d* 
x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^ 
(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a*b-2*A*(1/(a+b)*(b+a*cos(d*x+c)) 
/(cos(d*x+c)+1))^(1/2)*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2) 
)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a*b+16*B*(1/(a+b)*(b+a*cos(d*x+c))/(co 
s(d*x+c)+1))^(1/2)*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(c 
os(d*x+c)/(cos(d*x+c)+1))^(1/2)*a*b-16*A*EllipticPi(cot(d*x+c)-csc(d*x+c), 
-1,((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*co 
s(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^2*cos(d*x+c)-8*A*EllipticPi(cot(d*x+c)-c 
sc(d*x+c),-1,((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/...
 

3.4.61.5 Fricas [F]

\[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{2} \,d x } \]

input

integrate(cos(d*x+c)^2*(a+b*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)),x, algorith 
m="fricas")
 

output

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

3.4.61.6 Sympy [F(-1)]

Timed out. \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\text {Timed out} \]

input

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

output

Timed out
 

3.4.61.7 Maxima [F]

\[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{2} \,d x } \]

input

integrate(cos(d*x+c)^2*(a+b*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)),x, algorith 
m="maxima")
 

output

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

3.4.61.8 Giac [F]

\[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{2} \,d x } \]

input

integrate(cos(d*x+c)^2*(a+b*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)),x, algorith 
m="giac")
 

output

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

3.4.61.9 Mupad [F(-1)]

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

input

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

output

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