\(\int \frac {(a+b \cos (c+d x))^3 (A+C \cos ^2(c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\) [691]

Optimal result

Integrand size = 35, antiderivative size = 244 \[ \int \frac {(a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {2 a \left (5 a^2 (A-C)-3 b^2 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 b \left (21 a^2 (3 A+C)+b^2 (7 A+5 C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}-\frac {2 b \left (6 a^2 (7 A-3 C)-b^2 (7 A+5 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}-\frac {2 a b^2 (35 A-11 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{35 d}-\frac {2 b (7 A-C) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d}+\frac {2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{d \sqrt {\cos (c+d x)}} \] Output:

-2/5*a*(5*a^2*(A-C)-3*b^2*(5*A+3*C))*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2)) 
/d+2/21*b*(21*a^2*(3*A+C)+b^2*(7*A+5*C))*InverseJacobiAM(1/2*d*x+1/2*c,2^( 
1/2))/d-2/21*b*(6*a^2*(7*A-3*C)-b^2*(7*A+5*C))*cos(d*x+c)^(1/2)*sin(d*x+c) 
/d-2/35*a*b^2*(35*A-11*C)*cos(d*x+c)^(3/2)*sin(d*x+c)/d-2/7*b*(7*A-C)*cos( 
d*x+c)^(1/2)*(a+b*cos(d*x+c))^2*sin(d*x+c)/d+2*A*(a+b*cos(d*x+c))^3*sin(d* 
x+c)/d/cos(d*x+c)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 3.93 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.70 \[ \int \frac {(a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {-84 \left (5 a^3 (A-C)-3 a b^2 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+20 b \left (21 a^2 (3 A+C)+b^2 (7 A+5 C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\frac {\left (5 b \left (28 A b^2+84 a^2 C+29 b^2 C\right ) \cos (c+d x)+3 \left (140 a^3 A+42 a b^2 C+42 a b^2 C \cos (2 (c+d x))+5 b^3 C \cos (3 (c+d x))\right )\right ) \sin (c+d x)}{\sqrt {\cos (c+d x)}}}{210 d} \] Input:

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

Output:

(-84*(5*a^3*(A - C) - 3*a*b^2*(5*A + 3*C))*EllipticE[(c + d*x)/2, 2] + 20* 
b*(21*a^2*(3*A + C) + b^2*(7*A + 5*C))*EllipticF[(c + d*x)/2, 2] + ((5*b*( 
28*A*b^2 + 84*a^2*C + 29*b^2*C)*Cos[c + d*x] + 3*(140*a^3*A + 42*a*b^2*C + 
 42*a*b^2*C*Cos[2*(c + d*x)] + 5*b^3*C*Cos[3*(c + d*x)]))*Sin[c + d*x])/Sq 
rt[Cos[c + d*x]])/(210*d)
 

Rubi [A] (verified)

Time = 1.65 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.05, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.486, Rules used = {3042, 3527, 27, 3042, 3528, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3527

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3528

\(\displaystyle \frac {2}{7} \int \frac {(a+b \cos (c+d x)) \left (-a b (35 A-11 C) \cos ^2(c+d x)-\left (7 a^2 (A-C)-b^2 (7 A+5 C)\right ) \cos (c+d x)+a b (35 A+C)\right )}{2 \sqrt {\cos (c+d x)}}dx-\frac {2 b (7 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}{7 d}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{d \sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{7} \int \frac {(a+b \cos (c+d x)) \left (-a b (35 A-11 C) \cos ^2(c+d x)-\left (7 a^2 (A-C)-b^2 (7 A+5 C)\right ) \cos (c+d x)+a b (35 A+C)\right )}{\sqrt {\cos (c+d x)}}dx-\frac {2 b (7 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}{7 d}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{d \sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (-a b (35 A-11 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+\left (b^2 (7 A+5 C)-7 a^2 (A-C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+a b (35 A+C)\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 b (7 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}{7 d}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{d \sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 3512

\(\displaystyle \frac {1}{7} \left (\frac {2}{5} \int \frac {5 b (35 A+C) a^2-7 \left (5 a^2 (A-C)-3 b^2 (5 A+3 C)\right ) \cos (c+d x) a-5 b \left (6 a^2 (7 A-3 C)-b^2 (7 A+5 C)\right ) \cos ^2(c+d x)}{2 \sqrt {\cos (c+d x)}}dx-\frac {2 a b^2 (35 A-11 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )-\frac {2 b (7 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}{7 d}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{d \sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \int \frac {5 b (35 A+C) a^2-7 \left (5 a^2 (A-C)-3 b^2 (5 A+3 C)\right ) \cos (c+d x) a-5 b \left (6 a^2 (7 A-3 C)-b^2 (7 A+5 C)\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)}}dx-\frac {2 a b^2 (35 A-11 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )-\frac {2 b (7 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}{7 d}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{d \sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \int \frac {5 b (35 A+C) a^2-7 \left (5 a^2 (A-C)-3 b^2 (5 A+3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a-5 b \left (6 a^2 (7 A-3 C)-b^2 (7 A+5 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 a b^2 (35 A-11 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )-\frac {2 b (7 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}{7 d}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{d \sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {2}{3} \int \frac {5 b \left (21 (3 A+C) a^2+b^2 (7 A+5 C)\right )-21 a \left (5 a^2 (A-C)-3 b^2 (5 A+3 C)\right ) \cos (c+d x)}{2 \sqrt {\cos (c+d x)}}dx-\frac {10 b \left (6 a^2 (7 A-3 C)-b^2 (7 A+5 C)\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-\frac {2 a b^2 (35 A-11 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )-\frac {2 b (7 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}{7 d}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{d \sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {1}{3} \int \frac {5 b \left (21 (3 A+C) a^2+b^2 (7 A+5 C)\right )-21 a \left (5 a^2 (A-C)-3 b^2 (5 A+3 C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx-\frac {10 b \left (6 a^2 (7 A-3 C)-b^2 (7 A+5 C)\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-\frac {2 a b^2 (35 A-11 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )-\frac {2 b (7 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}{7 d}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{d \sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {1}{3} \int \frac {5 b \left (21 (3 A+C) a^2+b^2 (7 A+5 C)\right )-21 a \left (5 a^2 (A-C)-3 b^2 (5 A+3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {10 b \left (6 a^2 (7 A-3 C)-b^2 (7 A+5 C)\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-\frac {2 a b^2 (35 A-11 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )-\frac {2 b (7 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}{7 d}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{d \sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {1}{3} \left (5 b \left (21 a^2 (3 A+C)+b^2 (7 A+5 C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx-21 a \left (5 a^2 (A-C)-3 b^2 (5 A+3 C)\right ) \int \sqrt {\cos (c+d x)}dx\right )-\frac {10 b \left (6 a^2 (7 A-3 C)-b^2 (7 A+5 C)\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-\frac {2 a b^2 (35 A-11 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )-\frac {2 b (7 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}{7 d}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{d \sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {1}{3} \left (5 b \left (21 a^2 (3 A+C)+b^2 (7 A+5 C)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-21 a \left (5 a^2 (A-C)-3 b^2 (5 A+3 C)\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )-\frac {10 b \left (6 a^2 (7 A-3 C)-b^2 (7 A+5 C)\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-\frac {2 a b^2 (35 A-11 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )-\frac {2 b (7 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}{7 d}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{d \sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {1}{3} \left (5 b \left (21 a^2 (3 A+C)+b^2 (7 A+5 C)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {42 a \left (5 a^2 (A-C)-3 b^2 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-\frac {10 b \left (6 a^2 (7 A-3 C)-b^2 (7 A+5 C)\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-\frac {2 a b^2 (35 A-11 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )-\frac {2 b (7 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}{7 d}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{d \sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {1}{3} \left (\frac {10 b \left (21 a^2 (3 A+C)+b^2 (7 A+5 C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}-\frac {42 a \left (5 a^2 (A-C)-3 b^2 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-\frac {10 b \left (6 a^2 (7 A-3 C)-b^2 (7 A+5 C)\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-\frac {2 a b^2 (35 A-11 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )-\frac {2 b (7 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}{7 d}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{d \sqrt {\cos (c+d x)}}\)

Input:

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

Output:

(-2*b*(7*A - C)*Sqrt[Cos[c + d*x]]*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(7 
*d) + (2*A*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]) + ( 
(-2*a*b^2*(35*A - 11*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*d) + (((-42*a* 
(5*a^2*(A - C) - 3*b^2*(5*A + 3*C))*EllipticE[(c + d*x)/2, 2])/d + (10*b*( 
21*a^2*(3*A + C) + b^2*(7*A + 5*C))*EllipticF[(c + d*x)/2, 2])/d)/3 - (10* 
b*(6*a^2*(7*A - 3*C) - b^2*(7*A + 5*C))*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/( 
3*d))/5)/7
 

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 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])))
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(942\) vs. \(2(229)=458\).

Time = 29.62 (sec) , antiderivative size = 943, normalized size of antiderivative = 3.86

method	result	size
default	\(\text {Expression too large to display}\)	\(943\)
parts	\(\text {Expression too large to display}\)	\(953\)

Input:

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

Output:

-2/105*(240*C*b^3*(-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)^8-72*C*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2 
*d*x+1/2*c)^2)^(1/2)*b^2*(7*a+5*b)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c) 
+28*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*b*(5*A*b^2+15*C*a 
^2+18*C*a*b+10*C*b^2)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-2*(-2*sin(1/ 
2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(105*A*a^3+35*A*b^3+105*C*a^2*b 
+63*C*a*b^2+40*C*b^3)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+315*A*a^2*b* 
(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(co 
s(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)+35*A*b^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 
)*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)+105*A*(-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)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Elliptic 
E(cos(1/2*d*x+1/2*c),2^(1/2))*a^3-315*A*(-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)^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))*a*b^2+105*a^2*b*C*(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)+25*C*b 
^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)*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...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.12 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.27 \[ \int \frac {(a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {5 \, \sqrt {2} {\left (21 i \, {\left (3 \, A + C\right )} a^{2} b + i \, {\left (7 \, A + 5 \, C\right )} b^{3}\right )} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-21 i \, {\left (3 \, A + C\right )} a^{2} b - i \, {\left (7 \, A + 5 \, C\right )} b^{3}\right )} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 21 \, \sqrt {2} {\left (5 i \, {\left (A - C\right )} a^{3} - 3 i \, {\left (5 \, A + 3 \, C\right )} a b^{2}\right )} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 21 \, \sqrt {2} {\left (-5 i \, {\left (A - C\right )} a^{3} + 3 i \, {\left (5 \, A + 3 \, C\right )} a b^{2}\right )} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, {\left (15 \, C b^{3} \cos \left (d x + c\right )^{3} + 63 \, C a b^{2} \cos \left (d x + c\right )^{2} + 105 \, A a^{3} + 5 \, {\left (21 \, C a^{2} b + {\left (7 \, A + 5 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{105 \, d \cos \left (d x + c\right )} \] Input:

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

Output:

-1/105*(5*sqrt(2)*(21*I*(3*A + C)*a^2*b + I*(7*A + 5*C)*b^3)*cos(d*x + c)* 
weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*sqrt(2)*(-21 
*I*(3*A + C)*a^2*b - I*(7*A + 5*C)*b^3)*cos(d*x + c)*weierstrassPInverse(- 
4, 0, cos(d*x + c) - I*sin(d*x + c)) + 21*sqrt(2)*(5*I*(A - C)*a^3 - 3*I*( 
5*A + 3*C)*a*b^2)*cos(d*x + c)*weierstrassZeta(-4, 0, weierstrassPInverse( 
-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 21*sqrt(2)*(-5*I*(A - C)*a^3 + 3* 
I*(5*A + 3*C)*a*b^2)*cos(d*x + c)*weierstrassZeta(-4, 0, weierstrassPInver 
se(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(15*C*b^3*cos(d*x + c)^3 + 6 
3*C*a*b^2*cos(d*x + c)^2 + 105*A*a^3 + 5*(21*C*a^2*b + (7*A + 5*C)*b^3)*co 
s(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c))
 

Sympy [F(-1)]

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

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

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 1.15 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.12 \[ \int \frac {(a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2\,\left (C\,a^3\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+C\,a^2\,b\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+C\,a^2\,b\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )\right )}{d}+\frac {A\,b^3\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}+\frac {6\,A\,a\,b^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {6\,A\,a^2\,b\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,A\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,C\,b^3\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{9\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {6\,C\,a\,b^2\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \] Input:

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

Output:

(2*(C*a^3*ellipticE(c/2 + (d*x)/2, 2) + C*a^2*b*ellipticF(c/2 + (d*x)/2, 2 
) + C*a^2*b*cos(c + d*x)^(1/2)*sin(c + d*x)))/d + (A*b^3*((2*cos(c + d*x)^ 
(1/2)*sin(c + d*x))/3 + (2*ellipticF(c/2 + (d*x)/2, 2))/3))/d + (6*A*a*b^2 
*ellipticE(c/2 + (d*x)/2, 2))/d + (6*A*a^2*b*ellipticF(c/2 + (d*x)/2, 2))/ 
d + (2*A*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)^(9/2)*s 
in(c + d*x)*hypergeom([1/2, 9/4], 13/4, cos(c + d*x)^2))/(9*d*(sin(c + d*x 
)^2)^(1/2)) - (6*C*a*b^2*cos(c + d*x)^(7/2)*sin(c + d*x)*hypergeom([1/2, 7 
/4], 11/4, cos(c + d*x)^2))/(7*d*(sin(c + d*x)^2)^(1/2))
 

Reduce [F]

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

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

Output:

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