\(\int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+C \cos ^2(c+d x))}{(a+b \cos (c+d x))^2} \, dx\) [714]

Optimal result

Integrand size = 35, antiderivative size = 370 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\frac {\left (3 a^2 b^2 (5 A-8 C)+35 a^4 C-2 b^4 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^4 \left (a^2-b^2\right ) d}-\frac {a \left (a^2 b^2 (9 A-20 C)+21 a^4 C-4 b^4 (3 A+C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 b^5 \left (a^2-b^2\right ) d}-\frac {a^2 \left (5 A b^4-3 a^2 b^2 (A-3 C)-7 a^4 C\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{(a-b) b^5 (a+b)^2 d}-\frac {a \left (3 A b^2+7 a^2 C-4 b^2 C\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right ) d}+\frac {\left (5 A b^2+7 a^2 C-2 b^2 C\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}-\frac {\left (A b^2+a^2 C\right ) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \] Output:

1/5*(3*a^2*b^2*(5*A-8*C)+35*a^4*C-2*b^4*(5*A+3*C))*EllipticE(sin(1/2*d*x+1 
/2*c),2^(1/2))/b^4/(a^2-b^2)/d-1/3*a*(a^2*b^2*(9*A-20*C)+21*a^4*C-4*b^4*(3 
*A+C))*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/b^5/(a^2-b^2)/d-a^2*(5*A*b^4 
-3*a^2*b^2*(A-3*C)-7*a^4*C)*EllipticPi(sin(1/2*d*x+1/2*c),2*b/(a+b),2^(1/2 
))/(a-b)/b^5/(a+b)^2/d-1/3*a*(3*A*b^2+7*C*a^2-4*C*b^2)*cos(d*x+c)^(1/2)*si 
n(d*x+c)/b^3/(a^2-b^2)/d+1/5*(5*A*b^2+7*C*a^2-2*C*b^2)*cos(d*x+c)^(3/2)*si 
n(d*x+c)/b^2/(a^2-b^2)/d-(A*b^2+C*a^2)*cos(d*x+c)^(5/2)*sin(d*x+c)/b/(a^2- 
b^2)/d/(a+b*cos(d*x+c))
 

Mathematica [A] (warning: unable to verify)

Time = 5.09 (sec) , antiderivative size = 354, normalized size of antiderivative = 0.96 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\frac {\frac {\frac {2 \left (a^2 b^2 (15 A-32 C)+35 a^4 C-6 b^4 (5 A+3 C)\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {8 a \left (15 A b^2+\left (14 a^2+b^2\right ) C\right ) \left ((a+b) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )\right )}{a+b}+\frac {6 \left (3 a^2 b^2 (5 A-8 C)+35 a^4 C-2 b^4 (5 A+3 C)\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (-2 a^2+b^2\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a b^2 \sqrt {\sin ^2(c+d x)}}}{(a-b) (a+b)}+4 \sqrt {\cos (c+d x)} \left (-20 a C \sin (c+d x)-\frac {15 a^2 \left (A b^2+a^2 C\right ) \sin (c+d x)}{\left (a^2-b^2\right ) (a+b \cos (c+d x))}+3 b C \sin (2 (c+d x))\right )}{60 b^3 d} \] Input:

Integrate[(Cos[c + d*x]^(5/2)*(A + C*Cos[c + d*x]^2))/(a + b*Cos[c + d*x]) 
^2,x]
 

Output:

(((2*(a^2*b^2*(15*A - 32*C) + 35*a^4*C - 6*b^4*(5*A + 3*C))*EllipticPi[(2* 
b)/(a + b), (c + d*x)/2, 2])/(a + b) + (8*a*(15*A*b^2 + (14*a^2 + b^2)*C)* 
((a + b)*EllipticF[(c + d*x)/2, 2] - a*EllipticPi[(2*b)/(a + b), (c + d*x) 
/2, 2]))/(a + b) + (6*(3*a^2*b^2*(5*A - 8*C) + 35*a^4*C - 2*b^4*(5*A + 3*C 
))*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*a*(a + b)*Ellipti 
cF[ArcSin[Sqrt[Cos[c + d*x]]], -1] + (-2*a^2 + b^2)*EllipticPi[-(b/a), Arc 
Sin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(a*b^2*Sqrt[Sin[c + d*x]^2]))/ 
((a - b)*(a + b)) + 4*Sqrt[Cos[c + d*x]]*(-20*a*C*Sin[c + d*x] - (15*a^2*( 
A*b^2 + a^2*C)*Sin[c + d*x])/((a^2 - b^2)*(a + b*Cos[c + d*x])) + 3*b*C*Si 
n[2*(c + d*x)]))/(60*b^3*d)
 

Rubi [A] (verified)

Time = 2.62 (sec) , antiderivative size = 356, normalized size of antiderivative = 0.96, number of steps used = 18, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.514, Rules used = {3042, 3527, 27, 3042, 3528, 27, 3042, 3528, 27, 3042, 3538, 27, 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.

\(\displaystyle \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\)

\(\Big \downarrow \) 3527

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3528

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3528

\(\displaystyle -\frac {-\frac {\frac {2 \int -\frac {5 \left (7 C a^2+3 A b^2-4 b^2 C\right ) a^2-2 b \left (15 A b^2+\left (14 a^2+b^2\right ) C\right ) \cos (c+d x) a-3 \left (35 C a^4+3 b^2 (5 A-8 C) a^2-2 b^4 (5 A+3 C)\right ) \cos ^2(c+d x)}{2 \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{3 b}-\frac {10 a \left (7 a^2 C+3 A b^2-4 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}-\frac {2 \left (7 a^2 C+5 A b^2-2 b^2 C\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {-\frac {-\frac {\int \frac {5 \left (7 C a^2+3 A b^2-4 b^2 C\right ) a^2-2 b \left (15 A b^2+\left (14 a^2+b^2\right ) C\right ) \cos (c+d x) a-3 \left (35 C a^4+3 b^2 (5 A-8 C) a^2-2 b^4 (5 A+3 C)\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{3 b}-\frac {10 a \left (7 a^2 C+3 A b^2-4 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}-\frac {2 \left (7 a^2 C+5 A b^2-2 b^2 C\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {-\frac {\int \frac {5 \left (7 C a^2+3 A b^2-4 b^2 C\right ) a^2-2 b \left (15 A b^2+\left (14 a^2+b^2\right ) C\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a-3 \left (35 C a^4+3 b^2 (5 A-8 C) a^2-2 b^4 (5 A+3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{3 b}-\frac {10 a \left (7 a^2 C+3 A b^2-4 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}-\frac {2 \left (7 a^2 C+5 A b^2-2 b^2 C\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3538

\(\displaystyle -\frac {-\frac {-\frac {-\frac {3 \left (35 a^4 C+3 a^2 b^2 (5 A-8 C)-2 b^4 (5 A+3 C)\right ) \int \sqrt {\cos (c+d x)}dx}{b}-\frac {\int -\frac {5 \left (b \left (7 C a^2+3 A b^2-4 b^2 C\right ) a^2+\left (21 C a^4+b^2 (9 A-20 C) a^2-4 b^4 (3 A+C)\right ) \cos (c+d x) a\right )}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{3 b}-\frac {10 a \left (7 a^2 C+3 A b^2-4 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}-\frac {2 \left (7 a^2 C+5 A b^2-2 b^2 C\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {-\frac {-\frac {\frac {5 \int \frac {b \left (7 C a^2+3 A b^2-4 b^2 C\right ) a^2+\left (21 C a^4+b^2 (9 A-20 C) a^2-4 b^4 (3 A+C)\right ) \cos (c+d x) a}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}-\frac {3 \left (35 a^4 C+3 a^2 b^2 (5 A-8 C)-2 b^4 (5 A+3 C)\right ) \int \sqrt {\cos (c+d x)}dx}{b}}{3 b}-\frac {10 a \left (7 a^2 C+3 A b^2-4 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}-\frac {2 \left (7 a^2 C+5 A b^2-2 b^2 C\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {-\frac {\frac {5 \int \frac {b \left (7 C a^2+3 A b^2-4 b^2 C\right ) a^2+\left (21 C a^4+b^2 (9 A-20 C) a^2-4 b^4 (3 A+C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}-\frac {3 \left (35 a^4 C+3 a^2 b^2 (5 A-8 C)-2 b^4 (5 A+3 C)\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}}{3 b}-\frac {10 a \left (7 a^2 C+3 A b^2-4 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}-\frac {2 \left (7 a^2 C+5 A b^2-2 b^2 C\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3119

\(\displaystyle -\frac {-\frac {-\frac {\frac {5 \int \frac {b \left (7 C a^2+3 A b^2-4 b^2 C\right ) a^2+\left (21 C a^4+b^2 (9 A-20 C) a^2-4 b^4 (3 A+C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}-\frac {6 \left (35 a^4 C+3 a^2 b^2 (5 A-8 C)-2 b^4 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{3 b}-\frac {10 a \left (7 a^2 C+3 A b^2-4 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}-\frac {2 \left (7 a^2 C+5 A b^2-2 b^2 C\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3481

\(\displaystyle -\frac {-\frac {-\frac {\frac {5 \left (\frac {3 a^2 \left (-7 a^4 C-3 a^2 b^2 (A-3 C)+5 A b^4\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}+\frac {a \left (21 a^4 C+a^2 b^2 (9 A-20 C)-4 b^4 (3 A+C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{b}\right )}{b}-\frac {6 \left (35 a^4 C+3 a^2 b^2 (5 A-8 C)-2 b^4 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{3 b}-\frac {10 a \left (7 a^2 C+3 A b^2-4 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}-\frac {2 \left (7 a^2 C+5 A b^2-2 b^2 C\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {-\frac {\frac {5 \left (\frac {3 a^2 \left (-7 a^4 C-3 a^2 b^2 (A-3 C)+5 A b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}+\frac {a \left (21 a^4 C+a^2 b^2 (9 A-20 C)-4 b^4 (3 A+C)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}\right )}{b}-\frac {6 \left (35 a^4 C+3 a^2 b^2 (5 A-8 C)-2 b^4 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{3 b}-\frac {10 a \left (7 a^2 C+3 A b^2-4 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}-\frac {2 \left (7 a^2 C+5 A b^2-2 b^2 C\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3120

\(\displaystyle -\frac {-\frac {-\frac {\frac {5 \left (\frac {3 a^2 \left (-7 a^4 C-3 a^2 b^2 (A-3 C)+5 A b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}+\frac {2 a \left (21 a^4 C+a^2 b^2 (9 A-20 C)-4 b^4 (3 A+C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d}\right )}{b}-\frac {6 \left (35 a^4 C+3 a^2 b^2 (5 A-8 C)-2 b^4 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{3 b}-\frac {10 a \left (7 a^2 C+3 A b^2-4 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}-\frac {2 \left (7 a^2 C+5 A b^2-2 b^2 C\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3284

\(\displaystyle -\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {-\frac {2 \left (7 a^2 C+5 A b^2-2 b^2 C\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}-\frac {-\frac {10 a \left (7 a^2 C+3 A b^2-4 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}-\frac {\frac {5 \left (\frac {2 a \left (21 a^4 C+a^2 b^2 (9 A-20 C)-4 b^4 (3 A+C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d}+\frac {6 a^2 \left (-7 a^4 C-3 a^2 b^2 (A-3 C)+5 A b^4\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b d (a+b)}\right )}{b}-\frac {6 \left (35 a^4 C+3 a^2 b^2 (5 A-8 C)-2 b^4 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{3 b}}{5 b}}{2 b \left (a^2-b^2\right )}\)

Input:

Int[(Cos[c + d*x]^(5/2)*(A + C*Cos[c + d*x]^2))/(a + b*Cos[c + d*x])^2,x]
 

Output:

-(((A*b^2 + a^2*C)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(b*(a^2 - b^2)*d*(a + 
b*Cos[c + d*x]))) - ((-2*(5*A*b^2 + 7*a^2*C - 2*b^2*C)*Cos[c + d*x]^(3/2)* 
Sin[c + d*x])/(5*b*d) - (-1/3*((-6*(3*a^2*b^2*(5*A - 8*C) + 35*a^4*C - 2*b 
^4*(5*A + 3*C))*EllipticE[(c + d*x)/2, 2])/(b*d) + (5*((2*a*(a^2*b^2*(9*A 
- 20*C) + 21*a^4*C - 4*b^4*(3*A + C))*EllipticF[(c + d*x)/2, 2])/(b*d) + ( 
6*a^2*(5*A*b^4 - 3*a^2*b^2*(A - 3*C) - 7*a^4*C)*EllipticPi[(2*b)/(a + b), 
(c + d*x)/2, 2])/(b*(a + b)*d)))/b)/b - (10*a*(3*A*b^2 + 7*a^2*C - 4*b^2*C 
)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*b*d))/(5*b))/(2*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 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 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 3528

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[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x 
])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Simp[1/(d*(m + 
n + 2))   Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A* 
d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2) - C*(a 
*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + 
 n + 2))*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] && GtQ[ 
m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 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. \(1336\) vs. \(2(361)=722\).

Time = 28.78 (sec) , antiderivative size = 1337, normalized size of antiderivative = 3.61

method	result	size
default	\(\text {Expression too large to display}\)	\(1337\)

Input:

int(cos(d*x+c)^(5/2)*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2,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*a^2/b^4*(3*A 
*b^2+5*C*a^2)/(-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*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*Elli 
pticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))-2*(2*A*a*b^2+A*b^3+4*C*a^3+3 
*C*a^2*b+2*C*a*b^2+C*b^3)/b^5*(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)*Ell 
ipticF(cos(1/2*d*x+1/2*c),2^(1/2))-4/5*C/b^2/(-2*sin(1/2*d*x+1/2*c)^4+sin( 
1/2*d*x+1/2*c)^2)^(1/2)*(4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-14*sin( 
1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+6*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2 
*c)-5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ellipt 
icF(cos(1/2*d*x+1/2*c),2^(1/2))+9*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2* 
d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+2/b^4*(A*b^2+ 
3*C*a^2+4*C*a*b+3*C*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)*(Ellipti 
cF(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))-4/3* 
C/b^3*(2*a+3*b)*(2*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-sin(1/2*d*x+1/2 
*c)^2*cos(1/2*d*x+1/2*c)+2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2 
*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*(sin(1/2*d*x+1/2*c) 
^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(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)-2*a^3*(A*b...
 

Fricas [F]

\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}} \,d x } \] Input:

integrate(cos(d*x+c)^(5/2)*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2,x, algori 
thm="fricas")
 

Output:

integral((C*cos(d*x + c)^4 + A*cos(d*x + c)^2)*sqrt(cos(d*x + c))/(b^2*cos 
(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**(5/2)*(A+C*cos(d*x+c)**2)/(a+b*cos(d*x+c))**2,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}} \,d x } \] Input:

integrate(cos(d*x+c)^(5/2)*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2,x, algori 
thm="maxima")
 

Output:

integrate((C*cos(d*x + c)^2 + A)*cos(d*x + c)^(5/2)/(b*cos(d*x + c) + a)^2 
, x)
 

Giac [F]

\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}} \,d x } \] Input:

integrate(cos(d*x+c)^(5/2)*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2,x, algori 
thm="giac")
 

Output:

integrate((C*cos(d*x + c)^2 + A)*cos(d*x + c)^(5/2)/(b*cos(d*x + c) + a)^2 
, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^{5/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2} \,d x \] Input:

int((cos(c + d*x)^(5/2)*(A + C*cos(c + d*x)^2))/(a + b*cos(c + d*x))^2,x)
 

Output:

int((cos(c + d*x)^(5/2)*(A + C*cos(c + d*x)^2))/(a + b*cos(c + d*x))^2, x)
 

Reduce [F]

\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{4}}{\cos \left (d x +c \right )^{2} b^{2}+2 \cos \left (d x +c \right ) a b +a^{2}}d x \right ) c +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2}}{\cos \left (d x +c \right )^{2} b^{2}+2 \cos \left (d x +c \right ) a b +a^{2}}d x \right ) a \] Input:

int(cos(d*x+c)^(5/2)*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2,x)
 

Output:

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