\(\int \frac {\cos (c+d x) (A+C \cos ^2(c+d x))}{(a+b \cos (c+d x))^{5/2}} \, dx\) [665]

Optimal result

Integrand size = 33, antiderivative size = 314 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{5/2}} \, dx=-\frac {2 \left (3 b^4 (A-C)-8 a^4 C+a^2 b^2 (A+15 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 a \left (A b^2-8 a^2 C+9 b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{3 b^3 \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}+\frac {2 a \left (A b^2+a^2 C\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac {2 \left (3 A b^4-5 a^4 C+a^2 b^2 (A+9 C)\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}} \] Output:

-2/3*(3*b^4*(A-C)-8*a^4*C+a^2*b^2*(A+15*C))*(a+b*cos(d*x+c))^(1/2)*Ellipti 
cE(sin(1/2*d*x+1/2*c),2^(1/2)*(b/(a+b))^(1/2))/b^3/(a^2-b^2)^2/d/((a+b*cos 
(d*x+c))/(a+b))^(1/2)+2/3*a*(A*b^2-8*C*a^2+9*C*b^2)*((a+b*cos(d*x+c))/(a+b 
))^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2)*(b/(a+b))^(1/2))/b^3/(a^2-b 
^2)/d/(a+b*cos(d*x+c))^(1/2)+2/3*a*(A*b^2+C*a^2)*sin(d*x+c)/b^2/(a^2-b^2)/ 
d/(a+b*cos(d*x+c))^(3/2)+2/3*(3*A*b^4-5*a^4*C+a^2*b^2*(A+9*C))*sin(d*x+c)/ 
b^2/(a^2-b^2)^2/d/(a+b*cos(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 3.14 (sec) , antiderivative size = 227, normalized size of antiderivative = 0.72 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{5/2}} \, dx=\frac {2 \left (\frac {\left (\frac {a+b \cos (c+d x)}{a+b}\right )^{3/2} \left (\left (8 a^4 C+3 b^4 (-A+C)-a^2 b^2 (A+15 C)\right ) E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-a (a-b) \left (-A b^2+8 a^2 C-9 b^2 C\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )\right )}{(a-b)^2}+\frac {b \left (2 a A b^4-4 a^5 C+2 a^3 b^2 (A+4 C)+\left (3 A b^5-5 a^4 b C+a^2 b^3 (A+9 C)\right ) \cos (c+d x)\right ) \sin (c+d x)}{\left (a^2-b^2\right )^2}\right )}{3 b^3 d (a+b \cos (c+d x))^{3/2}} \] Input:

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

Output:

(2*((((a + b*Cos[c + d*x])/(a + b))^(3/2)*((8*a^4*C + 3*b^4*(-A + C) - a^2 
*b^2*(A + 15*C))*EllipticE[(c + d*x)/2, (2*b)/(a + b)] - a*(a - b)*(-(A*b^ 
2) + 8*a^2*C - 9*b^2*C)*EllipticF[(c + d*x)/2, (2*b)/(a + b)]))/(a - b)^2 
+ (b*(2*a*A*b^4 - 4*a^5*C + 2*a^3*b^2*(A + 4*C) + (3*A*b^5 - 5*a^4*b*C + a 
^2*b^3*(A + 9*C))*Cos[c + d*x])*Sin[c + d*x])/(a^2 - b^2)^2))/(3*b^3*d*(a 
+ b*Cos[c + d*x])^(3/2))
 

Rubi [A] (verified)

Time = 1.62 (sec) , antiderivative size = 322, normalized size of antiderivative = 1.03, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.485, Rules used = {3042, 3511, 27, 3042, 3500, 27, 25, 3042, 3231, 3042, 3134, 3042, 3132, 3142, 3042, 3140}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3511

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3500

\(\displaystyle \frac {2 a \left (a^2 C+A b^2\right ) \sin (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}-\frac {-\frac {2 \int -\frac {2 a \left (2 A b^2-\left (a^2-3 b^2\right ) C\right ) b^2+\left (-8 C a^4+b^2 (A+15 C) a^2+3 b^4 (A-C)\right ) \cos (c+d x) b}{2 \sqrt {a+b \cos (c+d x)}}dx}{b \left (a^2-b^2\right )}-\frac {2 \left (-5 a^4 C+a^2 b^2 (A+9 C)+3 A b^4\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 b^2 \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 a \left (a^2 C+A b^2\right ) \sin (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}-\frac {\frac {\int -\frac {2 a b^2 \left (a^2 C-b^2 (2 A+3 C)\right )-b \left (-8 C a^4+b^2 (A+15 C) a^2+3 b^4 (A-C)\right ) \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b \left (a^2-b^2\right )}-\frac {2 \left (-5 a^4 C+a^2 b^2 (A+9 C)+3 A b^4\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 b^2 \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 a \left (a^2 C+A b^2\right ) \sin (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}-\frac {-\frac {\int \frac {2 a b^2 \left (a^2 C-b^2 (2 A+3 C)\right )-b \left (-8 C a^4+b^2 (A+15 C) a^2+3 b^4 (A-C)\right ) \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b \left (a^2-b^2\right )}-\frac {2 \left (-5 a^4 C+a^2 b^2 (A+9 C)+3 A b^4\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 b^2 \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 a \left (a^2 C+A b^2\right ) \sin (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}-\frac {-\frac {\int \frac {2 a b^2 \left (a^2 C-b^2 (2 A+3 C)\right )-b \left (-8 C a^4+b^2 (A+15 C) a^2+3 b^4 (A-C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b \left (a^2-b^2\right )}-\frac {2 \left (-5 a^4 C+a^2 b^2 (A+9 C)+3 A b^4\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 b^2 \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3231

\(\displaystyle \frac {2 a \left (a^2 C+A b^2\right ) \sin (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}-\frac {-\frac {a \left (a^2-b^2\right ) \left (-8 a^2 C+A b^2+9 b^2 C\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}}dx-\left (-8 a^4 C+a^2 b^2 (A+15 C)+3 b^4 (A-C)\right ) \int \sqrt {a+b \cos (c+d x)}dx}{b \left (a^2-b^2\right )}-\frac {2 \left (-5 a^4 C+a^2 b^2 (A+9 C)+3 A b^4\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 b^2 \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 a \left (a^2 C+A b^2\right ) \sin (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}-\frac {-\frac {a \left (a^2-b^2\right ) \left (-8 a^2 C+A b^2+9 b^2 C\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\left (-8 a^4 C+a^2 b^2 (A+15 C)+3 b^4 (A-C)\right ) \int \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b \left (a^2-b^2\right )}-\frac {2 \left (-5 a^4 C+a^2 b^2 (A+9 C)+3 A b^4\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 b^2 \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3134

\(\displaystyle \frac {2 a \left (a^2 C+A b^2\right ) \sin (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}-\frac {-\frac {a \left (a^2-b^2\right ) \left (-8 a^2 C+A b^2+9 b^2 C\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {\left (-8 a^4 C+a^2 b^2 (A+15 C)+3 b^4 (A-C)\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}dx}{\sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{b \left (a^2-b^2\right )}-\frac {2 \left (-5 a^4 C+a^2 b^2 (A+9 C)+3 A b^4\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 b^2 \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 a \left (a^2 C+A b^2\right ) \sin (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}-\frac {-\frac {a \left (a^2-b^2\right ) \left (-8 a^2 C+A b^2+9 b^2 C\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {\left (-8 a^4 C+a^2 b^2 (A+15 C)+3 b^4 (A-C)\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}dx}{\sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{b \left (a^2-b^2\right )}-\frac {2 \left (-5 a^4 C+a^2 b^2 (A+9 C)+3 A b^4\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 b^2 \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3132

\(\displaystyle \frac {2 a \left (a^2 C+A b^2\right ) \sin (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}-\frac {-\frac {a \left (a^2-b^2\right ) \left (-8 a^2 C+A b^2+9 b^2 C\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \left (-8 a^4 C+a^2 b^2 (A+15 C)+3 b^4 (A-C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{b \left (a^2-b^2\right )}-\frac {2 \left (-5 a^4 C+a^2 b^2 (A+9 C)+3 A b^4\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 b^2 \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3142

\(\displaystyle \frac {2 a \left (a^2 C+A b^2\right ) \sin (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}-\frac {-\frac {\frac {a \left (a^2-b^2\right ) \left (-8 a^2 C+A b^2+9 b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}-\frac {2 \left (-8 a^4 C+a^2 b^2 (A+15 C)+3 b^4 (A-C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{b \left (a^2-b^2\right )}-\frac {2 \left (-5 a^4 C+a^2 b^2 (A+9 C)+3 A b^4\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 b^2 \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 a \left (a^2 C+A b^2\right ) \sin (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}-\frac {-\frac {\frac {a \left (a^2-b^2\right ) \left (-8 a^2 C+A b^2+9 b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}-\frac {2 \left (-8 a^4 C+a^2 b^2 (A+15 C)+3 b^4 (A-C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{b \left (a^2-b^2\right )}-\frac {2 \left (-5 a^4 C+a^2 b^2 (A+9 C)+3 A b^4\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 b^2 \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3140

\(\displaystyle \frac {2 a \left (a^2 C+A b^2\right ) \sin (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}-\frac {-\frac {2 \left (-5 a^4 C+a^2 b^2 (A+9 C)+3 A b^4\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {2 a \left (a^2-b^2\right ) \left (-8 a^2 C+A b^2+9 b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (-8 a^4 C+a^2 b^2 (A+15 C)+3 b^4 (A-C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{b \left (a^2-b^2\right )}}{3 b^2 \left (a^2-b^2\right )}\)

Input:

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

Output:

(2*a*(A*b^2 + a^2*C)*Sin[c + d*x])/(3*b^2*(a^2 - b^2)*d*(a + b*Cos[c + d*x 
])^(3/2)) - (-(((-2*(3*b^4*(A - C) - 8*a^4*C + a^2*b^2*(A + 15*C))*Sqrt[a 
+ b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(d*Sqrt[(a + b*Co 
s[c + d*x])/(a + b)]) + (2*a*(a^2 - b^2)*(A*b^2 - 8*a^2*C + 9*b^2*C)*Sqrt[ 
(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(d*Sq 
rt[a + b*Cos[c + d*x]]))/(b*(a^2 - b^2))) - (2*(3*A*b^4 - 5*a^4*C + a^2*b^ 
2*(A + 9*C))*Sin[c + d*x])/((a^2 - b^2)*d*Sqrt[a + b*Cos[c + d*x]]))/(3*b^ 
2*(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 3132

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

rule 3134

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

rule 3140

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

rule 3142

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

rule 3231

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + ( 
f_.)*(x_)]], x_Symbol] :> Simp[(b*c - a*d)/b   Int[1/Sqrt[a + b*Sin[e + f*x 
]], x], x] + Simp[d/b   Int[Sqrt[a + b*Sin[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 3500

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((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)/(b*f*(m + 1)* 
(a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x 
])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A 
*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, 
B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
 

rule 3511

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[ 
(-(b*c - a*d))*(A*b^2 + a^2*C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/( 
b^2*f*(m + 1)*(a^2 - b^2))), x] + Simp[1/(b^2*(m + 1)*(a^2 - b^2))   Int[(a 
 + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1)*(a*C*(b*c - a*d) + A*b*(a*c - b*d 
)) - ((b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Sin[e + f*x] + b 
*C*d*(m + 1)*(a^2 - b^2)*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] && LtQ[m, -1]
 

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(929\) vs. \(2(303)=606\).

Time = 17.95 (sec) , antiderivative size = 930, normalized size of antiderivative = 2.96

method	result	size
default	\(\text {Expression too large to display}\)	\(930\)
parts	\(\text {Expression too large to display}\)	\(1659\)

Input:

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

Output:

-(-(-2*b*cos(1/2*d*x+1/2*c)^2-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*C/b^3/( 
-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1 
/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*(3*a*El 
lipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-a*EllipticE(cos(1/2*d*x+1/2 
*c),(-2*b/(a-b))^(1/2))+b*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2)) 
)-2/b^3*(A*b^2+3*C*a^2)/sin(1/2*d*x+1/2*c)^2/(2*b*sin(1/2*d*x+1/2*c)^2-a-b 
)/(a^2-b^2)*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*( 
2*b*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+(-2*b/(a-b)*sin(1/2*d*x+1/2*c) 
^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*(si 
n(1/2*d*x+1/2*c)^2)^(1/2)*a-(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^ 
(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*(sin(1/2*d*x+1/2*c) 
^2)^(1/2)*b)-2*a*(A*b^2+C*a^2)/b^3*(1/6/b/(a-b)/(a+b)*cos(1/2*d*x+1/2*c)*( 
-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1 
/2*c)^2+1/2/b*(a-b))^2+8/3*b*sin(1/2*d*x+1/2*c)^2/(a-b)^2/(a+b)^2*cos(1/2* 
d*x+1/2*c)*a/(-(-2*b*cos(1/2*d*x+1/2*c)^2-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2) 
+(3*a-b)/(3*a^3+3*a^2*b-3*a*b^2-3*b^3)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b* 
cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*si 
n(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2)) 
-4/3*a/(a-b)/(a+b)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c) 
^2+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.15 (sec) , antiderivative size = 992, normalized size of antiderivative = 3.16 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

-2/9*(2*sqrt(1/2)*(-8*I*C*a^7 + I*(A + 18*C)*a^5*b^2 - 3*I*(A + 4*C)*a^3*b 
^4 + (-8*I*C*a^5*b^2 + I*(A + 18*C)*a^3*b^4 - 3*I*(A + 4*C)*a*b^6)*cos(d*x 
 + c)^2 + 2*(-8*I*C*a^6*b + I*(A + 18*C)*a^4*b^3 - 3*I*(A + 4*C)*a^2*b^5)* 
cos(d*x + c))*sqrt(b)*weierstrassPInverse(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*( 
8*a^3 - 9*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) + 3*I*b*sin(d*x + c) + 2*a)/b) 
 + 2*sqrt(1/2)*(8*I*C*a^7 - I*(A + 18*C)*a^5*b^2 + 3*I*(A + 4*C)*a^3*b^4 + 
 (8*I*C*a^5*b^2 - I*(A + 18*C)*a^3*b^4 + 3*I*(A + 4*C)*a*b^6)*cos(d*x + c) 
^2 + 2*(8*I*C*a^6*b - I*(A + 18*C)*a^4*b^3 + 3*I*(A + 4*C)*a^2*b^5)*cos(d* 
x + c))*sqrt(b)*weierstrassPInverse(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 
- 9*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) - 3*I*b*sin(d*x + c) + 2*a)/b) + 3*s 
qrt(1/2)*(-8*I*C*a^6*b + I*(A + 15*C)*a^4*b^3 + 3*I*(A - C)*a^2*b^5 + (-8* 
I*C*a^4*b^3 + I*(A + 15*C)*a^2*b^5 + 3*I*(A - C)*b^7)*cos(d*x + c)^2 + 2*( 
-8*I*C*a^5*b^2 + I*(A + 15*C)*a^3*b^4 + 3*I*(A - C)*a*b^6)*cos(d*x + c))*s 
qrt(b)*weierstrassZeta(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^ 
3, weierstrassPInverse(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^ 
3, 1/3*(3*b*cos(d*x + c) + 3*I*b*sin(d*x + c) + 2*a)/b)) + 3*sqrt(1/2)*(8* 
I*C*a^6*b - I*(A + 15*C)*a^4*b^3 - 3*I*(A - C)*a^2*b^5 + (8*I*C*a^4*b^3 - 
I*(A + 15*C)*a^2*b^5 - 3*I*(A - C)*b^7)*cos(d*x + c)^2 + 2*(8*I*C*a^5*b^2 
- I*(A + 15*C)*a^3*b^4 - 3*I*(A - C)*a*b^6)*cos(d*x + c))*sqrt(b)*weierstr 
assZeta(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, weierstra...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{5/2}} \, dx=\left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )}{\cos \left (d x +c \right )^{3} b^{3}+3 \cos \left (d x +c \right )^{2} a \,b^{2}+3 \cos \left (d x +c \right ) a^{2} b +a^{3}}d x \right ) a +\left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )^{3}}{\cos \left (d x +c \right )^{3} b^{3}+3 \cos \left (d x +c \right )^{2} a \,b^{2}+3 \cos \left (d x +c \right ) a^{2} b +a^{3}}d x \right ) c \] Input:

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

Output:

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