\(\int \frac {(a+b \cos (c+d x))^3 (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\) [1083]

Optimal result

Integrand size = 43, antiderivative size = 271 \[ \int \frac {(a+b \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {2 \left (5 a^3 B-15 a b^2 B+15 a^2 b (A-C)-b^3 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (9 a^2 b B+b^3 B+3 a b^2 (3 A+C)+a^3 (A+3 C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}-\frac {2 b \left (6 a^2 B-b^2 B+3 a b (5 A-C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}-\frac {2 b^2 (35 A b+15 a B-3 b C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 (2 A b+a B) (a+b \cos (c+d x))^2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)} \] Output:

-2/5*(5*B*a^3-15*B*a*b^2+15*a^2*b*(A-C)-b^3*(5*A+3*C))*EllipticE(sin(1/2*d 
*x+1/2*c),2^(1/2))/d+2/3*(9*B*a^2*b+B*b^3+3*a*b^2*(3*A+C)+a^3*(A+3*C))*Inv 
erseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/d-2/3*b*(6*B*a^2-B*b^2+3*a*b*(5*A-C))* 
cos(d*x+c)^(1/2)*sin(d*x+c)/d-2/15*b^2*(35*A*b+15*B*a-3*C*b)*cos(d*x+c)^(3 
/2)*sin(d*x+c)/d+2*(2*A*b+B*a)*(a+b*cos(d*x+c))^2*sin(d*x+c)/d/cos(d*x+c)^ 
(1/2)+2/3*A*(a+b*cos(d*x+c))^3*sin(d*x+c)/d/cos(d*x+c)^(3/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 4.98 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.69 \[ \int \frac {(a+b \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 \left (-15 a^3 B+45 a b^2 B-45 a^2 b (A-C)+3 b^3 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+10 \left (9 a^2 b B+b^3 B+3 a b^2 (3 A+C)+a^3 (A+3 C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\frac {6 \left (5 a^2 (3 A b+a B)+b^3 C \cos ^2(c+d x)\right ) \sin (c+d x)+5 \left (b^2 (b B+3 a C) \sin (2 (c+d x))+2 a^3 A \tan (c+d x)\right )}{\sqrt {\cos (c+d x)}}}{15 d} \] Input:

Integrate[((a + b*Cos[c + d*x])^3*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)) 
/Cos[c + d*x]^(5/2),x]
 

Output:

(2*(-15*a^3*B + 45*a*b^2*B - 45*a^2*b*(A - C) + 3*b^3*(5*A + 3*C))*Ellipti 
cE[(c + d*x)/2, 2] + 10*(9*a^2*b*B + b^3*B + 3*a*b^2*(3*A + C) + a^3*(A + 
3*C))*EllipticF[(c + d*x)/2, 2] + (6*(5*a^2*(3*A*b + a*B) + b^3*C*Cos[c + 
d*x]^2)*Sin[c + d*x] + 5*(b^2*(b*B + 3*a*C)*Sin[2*(c + d*x)] + 2*a^3*A*Tan 
[c + d*x]))/Sqrt[Cos[c + d*x]])/(15*d)
 

Rubi [A] (verified)

Time = 1.78 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.01, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.395, Rules used = {3042, 3526, 27, 3042, 3526, 27, 3042, 3512, 27, 3042, 3502, 27, 3042, 3227, 3042, 3119, 3120}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3526

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3526

\(\displaystyle \frac {1}{3} \left (2 \int \frac {(a+b \cos (c+d x)) \left ((A+3 C) a^2+15 b B a+24 A b^2-b (35 A b-3 C b+15 a B) \cos ^2(c+d x)-\left (3 B a^2+10 A b a-6 b C a-3 b^2 B\right ) \cos (c+d x)\right )}{2 \sqrt {\cos (c+d x)}}dx+\frac {6 (a B+2 A b) \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (\int \frac {(a+b \cos (c+d x)) \left ((A+3 C) a^2+15 b B a+24 A b^2-b (35 A b-3 C b+15 a B) \cos ^2(c+d x)-\left (3 B a^2+10 A b a-6 b C a-3 b^2 B\right ) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}}dx+\frac {6 (a B+2 A b) \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left ((A+3 C) a^2+15 b B a+24 A b^2-b (35 A b-3 C b+15 a B) \sin \left (c+d x+\frac {\pi }{2}\right )^2+\left (-3 B a^2-10 A b a+6 b C a+3 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {6 (a B+2 A b) \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 3512

\(\displaystyle \frac {1}{3} \left (\frac {2}{5} \int \frac {-15 b \left (6 B a^2+15 A b a-3 b C a-b^2 B\right ) \cos ^2(c+d x)-3 \left (5 B a^3+15 b (A-C) a^2-15 b^2 B a-b^3 (5 A+3 C)\right ) \cos (c+d x)+5 a \left ((A+3 C) a^2+15 b B a+24 A b^2\right )}{2 \sqrt {\cos (c+d x)}}dx-\frac {2 b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (15 a B+35 A b-3 b C)}{5 d}+\frac {6 (a B+2 A b) \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (\frac {1}{5} \int \frac {-15 b \left (6 B a^2+15 A b a-3 b C a-b^2 B\right ) \cos ^2(c+d x)-3 \left (5 B a^3+15 b (A-C) a^2-15 b^2 B a-b^3 (5 A+3 C)\right ) \cos (c+d x)+5 a \left ((A+3 C) a^2+15 b B a+24 A b^2\right )}{\sqrt {\cos (c+d x)}}dx-\frac {2 b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (15 a B+35 A b-3 b C)}{5 d}+\frac {6 (a B+2 A b) \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{5} \int \frac {-15 b \left (6 B a^2+15 A b a-3 b C a-b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-3 \left (5 B a^3+15 b (A-C) a^2-15 b^2 B a-b^3 (5 A+3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+5 a \left ((A+3 C) a^2+15 b B a+24 A b^2\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (15 a B+35 A b-3 b C)}{5 d}+\frac {6 (a B+2 A b) \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {1}{3} \left (\frac {1}{5} \left (\frac {2}{3} \int \frac {3 \left (5 \left ((A+3 C) a^3+9 b B a^2+3 b^2 (3 A+C) a+b^3 B\right )-3 \left (5 B a^3+15 b (A-C) a^2-15 b^2 B a-b^3 (5 A+3 C)\right ) \cos (c+d x)\right )}{2 \sqrt {\cos (c+d x)}}dx-\frac {10 b \sin (c+d x) \sqrt {\cos (c+d x)} \left (6 a^2 B+3 a b (5 A-C)-b^2 B\right )}{d}\right )-\frac {2 b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (15 a B+35 A b-3 b C)}{5 d}+\frac {6 (a B+2 A b) \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (\frac {1}{5} \left (\int \frac {5 \left ((A+3 C) a^3+9 b B a^2+3 b^2 (3 A+C) a+b^3 B\right )-3 \left (5 B a^3+15 b (A-C) a^2-15 b^2 B a-b^3 (5 A+3 C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx-\frac {10 b \sin (c+d x) \sqrt {\cos (c+d x)} \left (6 a^2 B+3 a b (5 A-C)-b^2 B\right )}{d}\right )-\frac {2 b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (15 a B+35 A b-3 b C)}{5 d}+\frac {6 (a B+2 A b) \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{5} \left (\int \frac {5 \left ((A+3 C) a^3+9 b B a^2+3 b^2 (3 A+C) a+b^3 B\right )-3 \left (5 B a^3+15 b (A-C) a^2-15 b^2 B a-b^3 (5 A+3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {10 b \sin (c+d x) \sqrt {\cos (c+d x)} \left (6 a^2 B+3 a b (5 A-C)-b^2 B\right )}{d}\right )-\frac {2 b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (15 a B+35 A b-3 b C)}{5 d}+\frac {6 (a B+2 A b) \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {1}{3} \left (\frac {1}{5} \left (5 \left (a^3 (A+3 C)+9 a^2 b B+3 a b^2 (3 A+C)+b^3 B\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx-3 \left (5 a^3 B+15 a^2 b (A-C)-15 a b^2 B-b^3 (5 A+3 C)\right ) \int \sqrt {\cos (c+d x)}dx-\frac {10 b \sin (c+d x) \sqrt {\cos (c+d x)} \left (6 a^2 B+3 a b (5 A-C)-b^2 B\right )}{d}\right )-\frac {2 b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (15 a B+35 A b-3 b C)}{5 d}+\frac {6 (a B+2 A b) \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{5} \left (5 \left (a^3 (A+3 C)+9 a^2 b B+3 a b^2 (3 A+C)+b^3 B\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-3 \left (5 a^3 B+15 a^2 b (A-C)-15 a b^2 B-b^3 (5 A+3 C)\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {10 b \sin (c+d x) \sqrt {\cos (c+d x)} \left (6 a^2 B+3 a b (5 A-C)-b^2 B\right )}{d}\right )-\frac {2 b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (15 a B+35 A b-3 b C)}{5 d}+\frac {6 (a B+2 A b) \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {1}{3} \left (\frac {1}{5} \left (5 \left (a^3 (A+3 C)+9 a^2 b B+3 a b^2 (3 A+C)+b^3 B\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {10 b \sin (c+d x) \sqrt {\cos (c+d x)} \left (6 a^2 B+3 a b (5 A-C)-b^2 B\right )}{d}-\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^3 B+15 a^2 b (A-C)-15 a b^2 B-b^3 (5 A+3 C)\right )}{d}\right )-\frac {2 b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (15 a B+35 A b-3 b C)}{5 d}+\frac {6 (a B+2 A b) \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {1}{3} \left (\frac {1}{5} \left (-\frac {10 b \sin (c+d x) \sqrt {\cos (c+d x)} \left (6 a^2 B+3 a b (5 A-C)-b^2 B\right )}{d}+\frac {10 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (a^3 (A+3 C)+9 a^2 b B+3 a b^2 (3 A+C)+b^3 B\right )}{d}-\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^3 B+15 a^2 b (A-C)-15 a b^2 B-b^3 (5 A+3 C)\right )}{d}\right )-\frac {2 b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (15 a B+35 A b-3 b C)}{5 d}+\frac {6 (a B+2 A b) \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\)

Input:

Int[((a + b*Cos[c + d*x])^3*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Cos[c 
 + d*x]^(5/2),x]
 

Output:

(2*A*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + ((-2* 
b^2*(35*A*b + 15*a*B - 3*b*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*d) + (6* 
(2*A*b + a*B)*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]) 
+ ((-6*(5*a^3*B - 15*a*b^2*B + 15*a^2*b*(A - C) - b^3*(5*A + 3*C))*Ellipti 
cE[(c + d*x)/2, 2])/d + (10*(9*a^2*b*B + b^3*B + 3*a*b^2*(3*A + C) + a^3*( 
A + 3*C))*EllipticF[(c + d*x)/2, 2])/d - (10*b*(6*a^2*B - b^2*B + 3*a*b*(5 
*A - C))*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/d)/5)/3
 

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 3227

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

rule 3502

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m 
+ 2))   Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m 
 + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] 
 &&  !LtQ[m, -1]
 

rule 3512

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

rule 3526

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^2*C - B*c*d + 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* 
d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 
1) - a*c*(n + 2))) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x 
] + b*(d*(B*c - A*d)*(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, B, 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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(996\) vs. \(2(256)=512\).

Time = 5.69 (sec) , antiderivative size = 997, normalized size of antiderivative = 3.68

method	result	size
parts	\(\text {Expression too large to display}\)	\(997\)
default	\(\text {Expression too large to display}\)	\(1837\)

Input:

int((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2),x, 
method=_RETURNVERBOSE)
 

Output:

-2*(3*A*a^2*b+B*a^3)*(-2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1 
/2)*cos(1/2*d*x+1/2*c)*sin(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)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c 
)^2)^(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)/sin(1/2*d*x+1/2*c)/(-1+2*cos(1/2*d*x+1/2*c)^2 
)^(1/2)/d-2/3*(B*b^3+3*C*a*b^2)*((-1+2*cos(1/2*d*x+1/2*c)^2)*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)^4-2*cos(1/2*d*x+1/ 
2*c)*sin(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)))/(-2*sin(1/2*d*x+1/2*c 
)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(-1+2*cos(1/2*d*x+1/2*c 
)^2)^(1/2)/d+2*(A*b^3+3*B*a*b^2+3*C*a^2*b)*((-1+2*cos(1/2*d*x+1/2*c)^2)*si 
n(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(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)/sin(1/2*d*x+1/2*c)/(-1+2*cos(1/2*d*x+1/2*c 
)^2)^(1/2)/d+2*(3*A*a*b^2+3*B*a^2*b+C*a^3)/d*InverseJacobiAM(1/2*d*x+1/2*c 
,2^(1/2))-2/3*A*a^3*(-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))*sin(1/2*d*x+1/2*c)^2-2*c 
os(1/2*d*x+1/2*c)*sin(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)))*((-1+2*c 
os(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)...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.13 (sec) , antiderivative size = 369, normalized size of antiderivative = 1.36 \[ \int \frac {(a+b \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {5 \, \sqrt {2} {\left (i \, {\left (A + 3 \, C\right )} a^{3} + 9 i \, B a^{2} b + 3 i \, {\left (3 \, A + C\right )} a b^{2} + i \, B b^{3}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-i \, {\left (A + 3 \, C\right )} a^{3} - 9 i \, B a^{2} b - 3 i \, {\left (3 \, A + C\right )} a b^{2} - i \, B b^{3}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, \sqrt {2} {\left (5 i \, B a^{3} + 15 i \, {\left (A - C\right )} a^{2} b - 15 i \, B a b^{2} - i \, {\left (5 \, A + 3 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {2} {\left (-5 i \, B a^{3} - 15 i \, {\left (A - C\right )} a^{2} b + 15 i \, B a b^{2} + i \, {\left (5 \, A + 3 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, {\left (3 \, C b^{3} \cos \left (d x + c\right )^{3} + 5 \, A a^{3} + 5 \, {\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{15 \, d \cos \left (d x + c\right )^{2}} \] Input:

integrate((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5 
/2),x, algorithm="fricas")
 

Output:

-1/15*(5*sqrt(2)*(I*(A + 3*C)*a^3 + 9*I*B*a^2*b + 3*I*(3*A + C)*a*b^2 + I* 
B*b^3)*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x 
+ c)) + 5*sqrt(2)*(-I*(A + 3*C)*a^3 - 9*I*B*a^2*b - 3*I*(3*A + C)*a*b^2 - 
I*B*b^3)*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d* 
x + c)) + 3*sqrt(2)*(5*I*B*a^3 + 15*I*(A - C)*a^2*b - 15*I*B*a*b^2 - I*(5* 
A + 3*C)*b^3)*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4 
, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*sqrt(2)*(-5*I*B*a^3 - 15*I*(A - C 
)*a^2*b + 15*I*B*a*b^2 + I*(5*A + 3*C)*b^3)*cos(d*x + c)^2*weierstrassZeta 
(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(3* 
C*b^3*cos(d*x + c)^3 + 5*A*a^3 + 5*(3*C*a*b^2 + B*b^3)*cos(d*x + c)^2 + 15 
*(B*a^3 + 3*A*a^2*b)*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos 
(d*x + c)^2)
 

Sympy [F(-1)]

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

integrate((a+b*cos(d*x+c))**3*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)* 
*(5/2),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5 
/2),x, algorithm="maxima")
 

Output:

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^3/c 
os(d*x + c)^(5/2), x)
 

Giac [F]

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

integrate((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5 
/2),x, algorithm="giac")
 

Output:

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^3/c 
os(d*x + c)^(5/2), x)
 

Mupad [B] (verification not implemented)

Time = 2.03 (sec) , antiderivative size = 379, normalized size of antiderivative = 1.40 \[ \int \frac {(a+b \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2\,\left (A\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\,b^3+3\,A\,a\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\,b^2\right )}{d}+\frac {B\,b^3\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}+\frac {2\,C\,a^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {6\,B\,a\,b^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {6\,B\,a^2\,b\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {6\,C\,a^2\,b\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {3\,C\,a\,b^2\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}+\frac {2\,A\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,B\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,C\,b^3\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {6\,A\,a^2\,b\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \] Input:

int(((a + b*cos(c + d*x))^3*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c 
 + d*x)^(5/2),x)
 

Output:

(2*(A*b^3*ellipticE(c/2 + (d*x)/2, 2) + 3*A*a*b^2*ellipticF(c/2 + (d*x)/2, 
 2)))/d + (B*b^3*((2*cos(c + d*x)^(1/2)*sin(c + d*x))/3 + (2*ellipticF(c/2 
 + (d*x)/2, 2))/3))/d + (2*C*a^3*ellipticF(c/2 + (d*x)/2, 2))/d + (6*B*a*b 
^2*ellipticE(c/2 + (d*x)/2, 2))/d + (6*B*a^2*b*ellipticF(c/2 + (d*x)/2, 2) 
)/d + (6*C*a^2*b*ellipticE(c/2 + (d*x)/2, 2))/d + (3*C*a*b^2*((2*cos(c + d 
*x)^(1/2)*sin(c + d*x))/3 + (2*ellipticF(c/2 + (d*x)/2, 2))/3))/d + (2*A*a 
^3*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(3*d*cos(c + 
d*x)^(3/2)*(sin(c + d*x)^2)^(1/2)) + (2*B*a^3*sin(c + d*x)*hypergeom([-1/4 
, 1/2], 3/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2) 
) - (2*C*b^3*cos(c + d*x)^(7/2)*sin(c + d*x)*hypergeom([1/2, 7/4], 11/4, c 
os(c + d*x)^2))/(7*d*(sin(c + d*x)^2)^(1/2)) + (6*A*a^2*b*sin(c + d*x)*hyp 
ergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(1/2)*(sin(c + d 
*x)^2)^(1/2))
 

Reduce [F]

\[ \int \frac {(a+b \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )}d x \right ) a^{3} c +6 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )}d x \right ) a^{2} b^{2}+\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{3}}d x \right ) a^{4}+4 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2}}d x \right ) a^{3} b +3 \left (\int \sqrt {\cos \left (d x +c \right )}d x \right ) a^{2} b c +4 \left (\int \sqrt {\cos \left (d x +c \right )}d x \right ) a \,b^{3}+3 \left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )d x \right ) a \,b^{2} c +\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )d x \right ) b^{4}+\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2}d x \right ) b^{3} c \] Input:

int((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2),x)
 

Output:

int(sqrt(cos(c + d*x))/cos(c + d*x),x)*a**3*c + 6*int(sqrt(cos(c + d*x))/c 
os(c + d*x),x)*a**2*b**2 + int(sqrt(cos(c + d*x))/cos(c + d*x)**3,x)*a**4 
+ 4*int(sqrt(cos(c + d*x))/cos(c + d*x)**2,x)*a**3*b + 3*int(sqrt(cos(c + 
d*x)),x)*a**2*b*c + 4*int(sqrt(cos(c + d*x)),x)*a*b**3 + 3*int(sqrt(cos(c 
+ d*x))*cos(c + d*x),x)*a*b**2*c + int(sqrt(cos(c + d*x))*cos(c + d*x),x)* 
b**4 + int(sqrt(cos(c + d*x))*cos(c + d*x)**2,x)*b**3*c