\(\int \frac {A+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3} \, dx\) [724]

Optimal result

Integrand size = 35, antiderivative size = 345 \[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3} \, dx=\frac {\left (3 A b^4-a^4 C-a^2 b^2 (9 A+5 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^2 b \left (a^2-b^2\right )^2 d}+\frac {\left (A b^4+a^4 C-7 a^2 b^2 (A+C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{4 a b^2 \left (a^2-b^2\right )^2 d}+\frac {\left (3 A b^6-3 a^2 b^4 (2 A-C)-a^6 C+5 a^4 b^2 (3 A+2 C)\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{4 a^2 (a-b)^2 b^2 (a+b)^3 d}+\frac {\left (A b^2+a^2 C\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac {\left (3 A b^4-a^4 C-a^2 b^2 (9 A+5 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))} \] Output:

1/4*(3*A*b^4-a^4*C-a^2*b^2*(9*A+5*C))*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2) 
)/a^2/b/(a^2-b^2)^2/d+1/4*(A*b^4+a^4*C-7*a^2*b^2*(A+C))*InverseJacobiAM(1/ 
2*d*x+1/2*c,2^(1/2))/a/b^2/(a^2-b^2)^2/d+1/4*(3*A*b^6-3*a^2*b^4*(2*A-C)-a^ 
6*C+5*a^4*b^2*(3*A+2*C))*EllipticPi(sin(1/2*d*x+1/2*c),2*b/(a+b),2^(1/2))/ 
a^2/(a-b)^2/b^2/(a+b)^3/d+1/2*(A*b^2+C*a^2)*cos(d*x+c)^(1/2)*sin(d*x+c)/a/ 
(a^2-b^2)/d/(a+b*cos(d*x+c))^2-1/4*(3*A*b^4-a^4*C-a^2*b^2*(9*A+5*C))*cos(d 
*x+c)^(1/2)*sin(d*x+c)/a^2/(a^2-b^2)^2/d/(a+b*cos(d*x+c))
 

Mathematica [A] (warning: unable to verify)

Time = 5.56 (sec) , antiderivative size = 366, normalized size of antiderivative = 1.06 \[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3} \, dx=\frac {\frac {4 \sqrt {\cos (c+d x)} \left (-5 a A b^4+3 a^5 C+a^3 b^2 (11 A+3 C)+b \left (-3 A b^4+a^4 C+a^2 b^2 (9 A+5 C)\right ) \cos (c+d x)\right ) \sin (c+d x)}{\left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}+\frac {\frac {2 \left (9 A b^4+a^2 b^2 (-19 A+C)+a^4 (16 A+5 C)\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {16 \left (a A b^2-a^3 (4 A+3 C)\right ) \left ((a+b) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )\right )}{a+b}-\frac {2 \left (-3 A b^4+a^4 C+a^2 b^2 (9 A+5 C)\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (-2 a^2+b^2\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a b^2 \sqrt {\sin ^2(c+d x)}}}{(a-b)^2 (a+b)^2}}{16 a^2 d} \] Input:

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

Output:

((4*Sqrt[Cos[c + d*x]]*(-5*a*A*b^4 + 3*a^5*C + a^3*b^2*(11*A + 3*C) + b*(- 
3*A*b^4 + a^4*C + a^2*b^2*(9*A + 5*C))*Cos[c + d*x])*Sin[c + d*x])/((a^2 - 
 b^2)^2*(a + b*Cos[c + d*x])^2) + ((2*(9*A*b^4 + a^2*b^2*(-19*A + C) + a^4 
*(16*A + 5*C))*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b) + (16*(a 
*A*b^2 - a^3*(4*A + 3*C))*((a + b)*EllipticF[(c + d*x)/2, 2] - a*EllipticP 
i[(2*b)/(a + b), (c + d*x)/2, 2]))/(a + b) - (2*(-3*A*b^4 + a^4*C + a^2*b^ 
2*(9*A + 5*C))*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*a*(a 
+ b)*EllipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] + (-2*a^2 + b^2)*EllipticPi 
[-(b/a), ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(a*b^2*Sqrt[Sin[c 
+ d*x]^2]))/((a - b)^2*(a + b)^2))/(16*a^2*d)
 

Rubi [A] (verified)

Time = 2.25 (sec) , antiderivative size = 342, normalized size of antiderivative = 0.99, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3535, 27, 3042, 3534, 27, 3042, 3538, 25, 3042, 3119, 3481, 3042, 3120, 3284}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3535

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\int \frac {-\left ((4 A+C) a^2\right )+4 b (A+C) \sin \left (c+d x+\frac {\pi }{2}\right ) a+3 A b^2+\left (-C a^2-A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx}{4 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3534

\(\displaystyle \frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\frac {\int -\frac {(8 A+3 C) a^4-b^2 (5 A-3 C) a^2+4 b \left (A b^2-a^2 (4 A+3 C)\right ) \cos (c+d x) a+3 A b^4+\left (-C a^4-b^2 (9 A+5 C) a^2+3 A b^4\right ) \cos ^2(c+d x)}{2 \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{a \left (a^2-b^2\right )}+\frac {\left (a^4 (-C)-a^2 b^2 (9 A+5 C)+3 A b^4\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\frac {\left (a^4 (-C)-a^2 b^2 (9 A+5 C)+3 A b^4\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\int \frac {(8 A+3 C) a^4-b^2 (5 A-3 C) a^2+4 b \left (A b^2-a^2 (4 A+3 C)\right ) \cos (c+d x) a+3 A b^4+\left (-C a^4-b^2 (9 A+5 C) a^2+3 A b^4\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{2 a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\frac {\left (a^4 (-C)-a^2 b^2 (9 A+5 C)+3 A b^4\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\int \frac {(8 A+3 C) a^4-b^2 (5 A-3 C) a^2+4 b \left (A b^2-a^2 (4 A+3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a+3 A b^4+\left (-C a^4-b^2 (9 A+5 C) a^2+3 A b^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3538

\(\displaystyle \frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\frac {\left (a^4 (-C)-a^2 b^2 (9 A+5 C)+3 A b^4\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\left (a^4 (-C)-a^2 b^2 (9 A+5 C)+3 A b^4\right ) \int \sqrt {\cos (c+d x)}dx}{b}-\frac {\int -\frac {b \left ((8 A+3 C) a^4-b^2 (5 A-3 C) a^2+3 A b^4\right )+a \left (C a^4-7 b^2 (A+C) a^2+A b^4\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{2 a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\frac {\left (a^4 (-C)-a^2 b^2 (9 A+5 C)+3 A b^4\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\left (a^4 (-C)-a^2 b^2 (9 A+5 C)+3 A b^4\right ) \int \sqrt {\cos (c+d x)}dx}{b}+\frac {\int \frac {b \left ((8 A+3 C) a^4-b^2 (5 A-3 C) a^2+3 A b^4\right )+a \left (C a^4-7 b^2 (A+C) a^2+A b^4\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{2 a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\frac {\left (a^4 (-C)-a^2 b^2 (9 A+5 C)+3 A b^4\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\left (a^4 (-C)-a^2 b^2 (9 A+5 C)+3 A b^4\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}+\frac {\int \frac {b \left ((8 A+3 C) a^4-b^2 (5 A-3 C) a^2+3 A b^4\right )+a \left (C a^4-7 b^2 (A+C) a^2+A b^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}}{2 a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\frac {\left (a^4 (-C)-a^2 b^2 (9 A+5 C)+3 A b^4\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\int \frac {b \left ((8 A+3 C) a^4-b^2 (5 A-3 C) a^2+3 A b^4\right )+a \left (C a^4-7 b^2 (A+C) a^2+A b^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}+\frac {2 \left (a^4 (-C)-a^2 b^2 (9 A+5 C)+3 A b^4\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{2 a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3481

\(\displaystyle \frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\frac {\left (a^4 (-C)-a^2 b^2 (9 A+5 C)+3 A b^4\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\frac {a \left (a^4 C-7 a^2 b^2 (A+C)+A b^4\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{b}+\frac {\left (a^6 (-C)+5 a^4 b^2 (3 A+2 C)-3 a^2 b^4 (2 A-C)+3 A b^6\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{b}+\frac {2 \left (a^4 (-C)-a^2 b^2 (9 A+5 C)+3 A b^4\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{2 a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\frac {\left (a^4 (-C)-a^2 b^2 (9 A+5 C)+3 A b^4\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\frac {a \left (a^4 C-7 a^2 b^2 (A+C)+A b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}+\frac {\left (a^6 (-C)+5 a^4 b^2 (3 A+2 C)-3 a^2 b^4 (2 A-C)+3 A b^6\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}}{b}+\frac {2 \left (a^4 (-C)-a^2 b^2 (9 A+5 C)+3 A b^4\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{2 a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\frac {\left (a^4 (-C)-a^2 b^2 (9 A+5 C)+3 A b^4\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\frac {\left (a^6 (-C)+5 a^4 b^2 (3 A+2 C)-3 a^2 b^4 (2 A-C)+3 A b^6\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}+\frac {2 a \left (a^4 C-7 a^2 b^2 (A+C)+A b^4\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d}}{b}+\frac {2 \left (a^4 (-C)-a^2 b^2 (9 A+5 C)+3 A b^4\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{2 a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3284

\(\displaystyle \frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\frac {\left (a^4 (-C)-a^2 b^2 (9 A+5 C)+3 A b^4\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {2 \left (a^4 (-C)-a^2 b^2 (9 A+5 C)+3 A b^4\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}+\frac {\frac {2 a \left (a^4 C-7 a^2 b^2 (A+C)+A b^4\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d}+\frac {2 \left (a^6 (-C)+5 a^4 b^2 (3 A+2 C)-3 a^2 b^4 (2 A-C)+3 A b^6\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b d (a+b)}}{b}}{2 a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\)

Input:

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

Output:

((A*b^2 + a^2*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(2*a*(a^2 - b^2)*d*(a + 
b*Cos[c + d*x])^2) - (-1/2*((2*(3*A*b^4 - a^4*C - a^2*b^2*(9*A + 5*C))*Ell 
ipticE[(c + d*x)/2, 2])/(b*d) + ((2*a*(A*b^4 + a^4*C - 7*a^2*b^2*(A + C))* 
EllipticF[(c + d*x)/2, 2])/(b*d) + (2*(3*A*b^6 - 3*a^2*b^4*(2*A - C) - a^6 
*C + 5*a^4*b^2*(3*A + 2*C))*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(b* 
(a + b)*d))/b)/(a*(a^2 - b^2)) + ((3*A*b^4 - a^4*C - a^2*b^2*(9*A + 5*C))* 
Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])))/( 
4*a*(a^2 - b^2))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 3119

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3120

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 
)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3284

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) 
 + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 
2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a, b, c 
, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 
0] && GtQ[c + d, 0]
 

rule 3481

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

rule 3534

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

rule 3535

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[(-(A*b^2 + a^2*C))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*S 
in[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + Simp[1/((m 
+ 1)*(b*c - a*d)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin 
[e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*(m + n 
+ 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d 
*(A*b^2 + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, 
d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 
- d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) || 
 !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 
 0])))
 

rule 3538

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 
2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])), x_Symbol] :> Simp[C/(b*d)   Int[Sqrt[a + b*Sin[e + f*x]], x] 
, x] - Simp[1/(b*d)   Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[ 
e + f*x], x]/(Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])), x], x] /; Fre 
eQ[{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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1845\) vs. \(2(336)=672\).

Time = 7.54 (sec) , antiderivative size = 1846, normalized size of antiderivative = 5.35

method	result	size
default	\(\text {Expression too large to display}\)	\(1846\)

Input:

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

Output:

-(-(1-2*cos(1/2*d*x+1/2*c)^2)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-4*C/b/(-2*a*b+ 
2*b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(1-2*cos(1/2*d*x+1/2*c)^2)^(1/2)/(-2*s 
in(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2 
*c),-2*b/(a-b),2^(1/2))+2*(A*b^2+C*a^2)/b^2*(-1/2*b^2/a/(a^2-b^2)*cos(1/2* 
d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*b*cos(1 
/2*d*x+1/2*c)^2+a-b)^2-3/4*b^2*(3*a^2-b^2)/a^2/(a^2-b^2)^2*cos(1/2*d*x+1/2 
*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*b*cos(1/2*d*x+ 
1/2*c)^2+a-b)-7/8/(a+b)/(a^2-b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(1-2*cos(1/ 
2*d*x+1/2*c)^2)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2) 
*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+1/4/(a+b)/(a^2-b^2)/a*(sin(1/2*d*x+ 
1/2*c)^2)^(1/2)*(1-2*cos(1/2*d*x+1/2*c)^2)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+ 
sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*b+3/8/(a 
+b)/(a^2-b^2)/a^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(1-2*cos(1/2*d*x+1/2*c)^2)^ 
(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1 
/2*d*x+1/2*c),2^(1/2))*b^2-9/8*b/(a^2-b^2)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)* 
(1-2*cos(1/2*d*x+1/2*c)^2)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2* 
c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3/8*b^3/a^2/(a^2-b^2)^2* 
(sin(1/2*d*x+1/2*c)^2)^(1/2)*(1-2*cos(1/2*d*x+1/2*c)^2)^(1/2)/(-2*sin(1/2* 
d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1 
/2))+9/8*b/(a^2-b^2)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(1-2*cos(1/2*d*x+1/...
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3} \, dx=\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{4} b^{3}+3 \cos \left (d x +c \right )^{3} a \,b^{2}+3 \cos \left (d x +c \right )^{2} a^{2} b +\cos \left (d x +c \right ) a^{3}}d x \right ) a +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )}{\cos \left (d x +c \right )^{3} b^{3}+3 \cos \left (d x +c \right )^{2} a \,b^{2}+3 \cos \left (d x +c \right ) a^{2} b +a^{3}}d x \right ) c \] Input:

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

Output:

int(sqrt(cos(c + d*x))/(cos(c + d*x)**4*b**3 + 3*cos(c + d*x)**3*a*b**2 + 
3*cos(c + d*x)**2*a**2*b + cos(c + d*x)*a**3),x)*a + int((sqrt(cos(c + d*x 
))*cos(c + d*x))/(cos(c + d*x)**3*b**3 + 3*cos(c + d*x)**2*a*b**2 + 3*cos( 
c + d*x)*a**2*b + a**3),x)*c