\(\int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}} \, dx\) [1444]

Optimal result
Mathematica [B] (warning: unable to verify)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [F(-1)]
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 37, antiderivative size = 563 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}} \, dx=-\frac {\left (2 A b^2+3 a^2 C-b^2 C\right ) \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{a b^2 \sqrt {a+b} d \sqrt {\sec (c+d x)}}+\frac {\left (2 A b^2+a (3 a+b) C\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{a b^2 \sqrt {a+b} d \sqrt {\sec (c+d x)}}+\frac {3 a \sqrt {a+b} C \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{b^3 d \sqrt {\sec (c+d x)}}-\frac {2 \left (A b^2+a^2 C\right ) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)}}+\frac {\left (2 A b^2+3 a^2 C-b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{b^2 \left (a^2-b^2\right ) d} \] Output:

-(2*A*b^2+3*C*a^2-C*b^2)*cos(d*x+c)^(1/2)*csc(d*x+c)*EllipticE((a+b*cos(d* 
x+c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),(-(a+b)/(a-b))^(1/2))*(a*(1-sec(d 
*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/a/b^2/(a+b)^(1/2)/d/sec 
(d*x+c)^(1/2)+(2*A*b^2+a*(3*a+b)*C)*cos(d*x+c)^(1/2)*csc(d*x+c)*EllipticF( 
(a+b*cos(d*x+c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),(-(a+b)/(a-b))^(1/2))* 
(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/a/b^2/(a+b)^ 
(1/2)/d/sec(d*x+c)^(1/2)+3*a*(a+b)^(1/2)*C*cos(d*x+c)^(1/2)*csc(d*x+c)*Ell 
ipticPi((a+b*cos(d*x+c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),(a+b)/b,(-(a+b 
)/(a-b))^(1/2))*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1 
/2)/b^3/d/sec(d*x+c)^(1/2)-2*(A*b^2+C*a^2)*sin(d*x+c)/b/(a^2-b^2)/d/(a+b*c 
os(d*x+c))^(1/2)/sec(d*x+c)^(1/2)+(2*A*b^2+3*C*a^2-C*b^2)*(a+b*cos(d*x+c)) 
^(1/2)*sec(d*x+c)^(1/2)*sin(d*x+c)/b^2/(a^2-b^2)/d
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(1155\) vs. \(2(563)=1126\).

Time = 15.22 (sec) , antiderivative size = 1155, normalized size of antiderivative = 2.05 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}} \, dx =\text {Too large to display} \] Input:

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

Output:

(Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*(A*b^2 + a^2*C)*Sin[c + d 
*x])/(b^2*(-a^2 + b^2)) - (2*(a*A*b^2*Sin[c + d*x] + a^3*C*Sin[c + d*x]))/ 
(b^2*(-a^2 + b^2)*(a + b*Cos[c + d*x]))))/d + (Sqrt[(1 - Tan[(c + d*x)/2]^ 
2)^(-1)]*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(1 + T 
an[(c + d*x)/2]^2)]*(2*a*A*b^2*Tan[(c + d*x)/2] + 2*A*b^3*Tan[(c + d*x)/2] 
 + 3*a^3*C*Tan[(c + d*x)/2] + 3*a^2*b*C*Tan[(c + d*x)/2] - a*b^2*C*Tan[(c 
+ d*x)/2] - b^3*C*Tan[(c + d*x)/2] - 4*A*b^3*Tan[(c + d*x)/2]^3 - 6*a^2*b* 
C*Tan[(c + d*x)/2]^3 + 2*b^3*C*Tan[(c + d*x)/2]^3 - 2*a*A*b^2*Tan[(c + d*x 
)/2]^5 + 2*A*b^3*Tan[(c + d*x)/2]^5 - 3*a^3*C*Tan[(c + d*x)/2]^5 + 3*a^2*b 
*C*Tan[(c + d*x)/2]^5 + a*b^2*C*Tan[(c + d*x)/2]^5 - b^3*C*Tan[(c + d*x)/2 
]^5 - 6*a^3*C*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*S 
qrt[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*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)] - 6*a^3*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)] + 6*a*b^2*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)*(2*A*b^2 + 3*a^2*...
 

Rubi [A] (verified)

Time = 2.68 (sec) , antiderivative size = 553, normalized size of antiderivative = 0.98, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.405, Rules used = {3042, 4709, 3042, 3527, 27, 3042, 3540, 3042, 3532, 3042, 3288, 3477, 3042, 3295, 3473}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4709

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

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+A\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx\)

\(\Big \downarrow \) 3527

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\int \frac {C a^2-b (A+C) \sin \left (c+d x+\frac {\pi }{2}\right ) a+A b^2+\left (-3 C a^2-2 A b^2+b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\right )\)

\(\Big \downarrow \) 3540

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

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {\int \frac {3 a \left (a^2-b^2\right ) C \sin \left (c+d x+\frac {\pi }{2}\right )^2+2 b \left (C a^2+A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+a \left (3 C a^2+2 A b^2-b^2 C\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 b}-\frac {\left (3 a^2 C+2 A b^2-b^2 C\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\right )\)

\(\Big \downarrow \) 3532

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

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {\int \frac {a \left (3 C a^2+2 A b^2-b^2 C\right )+2 b \left (C a^2+A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+3 a C \left (a^2-b^2\right ) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 b}-\frac {\left (3 a^2 C+2 A b^2-b^2 C\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\right )\)

\(\Big \downarrow \) 3288

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {\int \frac {a \left (3 C a^2+2 A b^2-b^2 C\right )+2 b \left (C a^2+A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 a C \sqrt {a+b} \left (a^2-b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}}{2 b}-\frac {\left (3 a^2 C+2 A b^2-b^2 C\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\right )\)

\(\Big \downarrow \) 3477

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {a \left (3 a^2 C+2 A b^2-b^2 C\right ) \int \frac {\cos (c+d x)+1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx-(a-b) \left (a C (3 a+b)+2 A b^2\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx-\frac {6 a C \sqrt {a+b} \left (a^2-b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}}{2 b}-\frac {\left (3 a^2 C+2 A b^2-b^2 C\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {a \left (3 a^2 C+2 A b^2-b^2 C\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-(a-b) \left (a C (3 a+b)+2 A b^2\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 a C \sqrt {a+b} \left (a^2-b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}}{2 b}-\frac {\left (3 a^2 C+2 A b^2-b^2 C\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\right )\)

\(\Big \downarrow \) 3295

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {a \left (3 a^2 C+2 A b^2-b^2 C\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 a C \sqrt {a+b} \left (a^2-b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}-\frac {2 (a-b) \sqrt {a+b} \left (a C (3 a+b)+2 A b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}}{2 b}-\frac {\left (3 a^2 C+2 A b^2-b^2 C\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\right )\)

\(\Big \downarrow \) 3473

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {\frac {2 (a-b) \sqrt {a+b} \left (3 a^2 C+2 A b^2-b^2 C\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a d}-\frac {6 a C \sqrt {a+b} \left (a^2-b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}-\frac {2 (a-b) \sqrt {a+b} \left (a C (3 a+b)+2 A b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}}{2 b}-\frac {\left (3 a^2 C+2 A b^2-b^2 C\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\right )\)

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((-2*(A*b^2 + a^2*C)*Sqrt[Cos[c + d* 
x]]*Sin[c + d*x])/(b*(a^2 - b^2)*d*Sqrt[a + b*Cos[c + d*x]]) - (((2*(a - b 
)*Sqrt[a + b]*(2*A*b^2 + 3*a^2*C - b^2*C)*Cot[c + d*x]*EllipticE[ArcSin[Sq 
rt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b 
))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - 
b)])/(a*d) - (2*(a - b)*Sqrt[a + b]*(2*A*b^2 + a*(3*a + b)*C)*Cot[c + d*x] 
*EllipticF[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]] 
)], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + 
Sec[c + d*x]))/(a - b)])/(a*d) - (6*a*Sqrt[a + b]*(a^2 - b^2)*C*Cot[c + d* 
x]*EllipticPi[(a + b)/b, ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt 
[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)] 
*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(b*d))/(2*b) - ((2*A*b^2 + 3*a^2*C 
- b^2*C)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(b*d*Sqrt[Cos[c + d*x]]))/ 
(b*(a^2 - b^2)))
 

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 3288
Int[Sqrt[(b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.) 
*(x_)]], x_Symbol] :> Simp[2*b*(Tan[e + f*x]/(d*f))*Rt[(c + d)/b, 2]*Sqrt[c 
*((1 + Csc[e + f*x])/(c - d))]*Sqrt[c*((1 - Csc[e + f*x])/(c + d))]*Ellipti 
cPi[(c + d)/d, ArcSin[Sqrt[c + d*Sin[e + f*x]]/Sqrt[b*Sin[e + f*x]]/Rt[(c + 
 d)/b, 2]], -(c + d)/(c - d)], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - 
 d^2, 0] && PosQ[(c + d)/b]
 

rule 3295
Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f 
_.)*(x_)]]), x_Symbol] :> Simp[-2*(Tan[e + f*x]/(a*f))*Rt[(a + b)/d, 2]*Sqr 
t[a*((1 - Csc[e + f*x])/(a + b))]*Sqrt[a*((1 + Csc[e + f*x])/(a - b))]*Elli 
pticF[ArcSin[Sqrt[a + b*Sin[e + f*x]]/Sqrt[d*Sin[e + f*x]]/Rt[(a + b)/d, 2] 
], -(a + b)/(a - b)], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] 
&& PosQ[(a + b)/d]
 

rule 3473
Int[((A_) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((b_.)*sin[(e_.) + (f_.)*(x_)]) 
^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*A* 
(c - d)*(Tan[e + f*x]/(f*b*c^2))*Rt[(c + d)/b, 2]*Sqrt[c*((1 + Csc[e + f*x] 
)/(c - d))]*Sqrt[c*((1 - Csc[e + f*x])/(c + d))]*EllipticE[ArcSin[Sqrt[c + 
d*Sin[e + f*x]]/Sqrt[b*Sin[e + f*x]]/Rt[(c + d)/b, 2]], -(c + d)/(c - d)], 
x] /; FreeQ[{b, c, d, e, f, A, B}, x] && NeQ[c^2 - d^2, 0] && EqQ[A, B] && 
PosQ[(c + d)/b]
 

rule 3477
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ 
.)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> S 
imp[(A - B)/(a - b)   Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f* 
x]]), x], x] - Simp[(A*b - a*B)/(a - b)   Int[(1 + Sin[e + f*x])/((a + b*Si 
n[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e 
, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 
0] && NeQ[A, B]
 

rule 3527
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> 
Simp[(-(c^2*C + A*d^2))*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + 
 f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^ 
2))   Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A* 
d*(b*d*m + a*c*(n + 1)) + c*C*(b*c*m + a*d*(n + 1)) - (A*d*(a*d*(n + 2) - b 
*c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*( 
A*d^2*(m + n + 2) + C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x]^2, x], x], 
x] /; FreeQ[{a, b, c, d, e, f, A, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - 
b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]
 

rule 3532
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 
2)/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)*Sqrt[(c_.) + (d_.)*sin[(e 
_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[C/b^2   Int[Sqrt[a + b*Sin[e + f*x]] 
/Sqrt[c + d*Sin[e + f*x]], x], x] + Simp[1/b^2   Int[(A*b^2 - a^2*C + b*(b* 
B - 2*a*C)*Sin[e + f*x])/((a + b*Sin[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x 
]]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] & 
& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
 

rule 3540
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 
2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_) + (d_.)*sin[(e_.) 
 + (f_.)*(x_)]]), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(Sqrt[c + d*Sin[e + f 
*x]]/(d*f*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[1/(2*d)   Int[(1/((a + b*Si 
n[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x]]))*Simp[2*a*A*d - C*(b*c - a*d) - 
 2*(a*c*C - d*(A*b + a*B))*Sin[e + f*x] + (2*b*B*d - C*(b*c + a*d))*Sin[e + 
 f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a 
*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
 

rule 4709
Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Simp[(c*Sec[a 
+ b*x])^m*(c*Cos[a + b*x])^m   Int[ActivateTrig[u]/(c*Cos[a + b*x])^m, x], 
x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u, x]
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1359\) vs. \(2(510)=1020\).

Time = 15.86 (sec) , antiderivative size = 1360, normalized size of antiderivative = 2.42

method result size
default \(\text {Expression too large to display}\) \(1360\)
parts \(\text {Expression too large to display}\) \(1398\)

Input:

int((A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(3/2)/sec(d*x+c)^(1/2),x,method=_R 
ETURNVERBOSE)
 

Output:

1/d/b^2/(a-b)/(a+b)*(a+b*cos(d*x+c))^(1/2)/(b*cos(d*x+c)^2+a*cos(d*x+c)+b* 
cos(d*x+c)+a)/sec(d*x+c)^(1/2)*(C*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)) 
)^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*a^3*EllipticPi(-csc(d*x+c)+cot(d 
*x+c),-1,(-(a-b)/(a+b))^(1/2))*(6*cos(d*x+c)+12+6*sec(d*x+c))+C*(1/(a+b)*( 
a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*a* 
b^2*EllipticPi(-csc(d*x+c)+cot(d*x+c),-1,(-(a-b)/(a+b))^(1/2))*(-6*cos(d*x 
+c)-12-6*sec(d*x+c))+A*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(co 
s(d*x+c)/(1+cos(d*x+c)))^(1/2)*a*b^2*EllipticE(-csc(d*x+c)+cot(d*x+c),(-(a 
-b)/(a+b))^(1/2))*(-2*cos(d*x+c)-4-2*sec(d*x+c))+A*(1/(a+b)*(a+b*cos(d*x+c 
))/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*b^3*EllipticE(- 
csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))*(-2*cos(d*x+c)-4-2*sec(d*x+c)) 
+C*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+ 
c)))^(1/2)*a^3*EllipticE(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))*(-3* 
cos(d*x+c)-6-3*sec(d*x+c))+C*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/ 
2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*a^2*b*EllipticE(-csc(d*x+c)+cot(d*x+c 
),(-(a-b)/(a+b))^(1/2))*(-3*cos(d*x+c)-6-3*sec(d*x+c))+C*(1/(a+b)*(a+b*cos 
(d*x+c))/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*a*b^2*Ell 
ipticE(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))*(cos(d*x+c)+2+sec(d*x+ 
c))+C*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d 
*x+c)))^(1/2)*b^3*EllipticE(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2)...
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

Integral((A + C*cos(c + d*x)**2)/((a + b*cos(c + d*x))**(3/2)*sqrt(sec(c + 
 d*x))), x)
 

Maxima [F]

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

integrate((A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(3/2)/sec(d*x+c)^(1/2),x, al 
gorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(3/2)/sec(d*x+c)^(1/2),x, al 
gorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+A}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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