\(\int \frac {(a+b \cos (c+d x))^3 (A+C \cos ^2(c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\) [693]

Optimal result

Integrand size = 35, antiderivative size = 229 \[ \int \frac {(a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {2 a \left (15 b^2 (A-C)+a^2 (3 A+5 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 b \left (b^2 (3 A+C)+3 a^2 (A+3 C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 a \left (8 A b^2+a^2 (3 A+5 C)\right ) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}-\frac {2 b^3 (9 A-5 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d}+\frac {4 A b (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)} \] Output:

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

Mathematica [A] (verified)

Time = 3.89 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.86 \[ \int \frac {(a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {-6 a \left (15 b^2 (A-C)+a^2 (3 A+5 C)\right ) \cos ^{\frac {3}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+10 b \left (b^2 (3 A+C)+3 a^2 (A+3 C)\right ) \cos ^{\frac {3}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+30 a^2 A b \sin (c+d x)+10 b^3 C \cos ^2(c+d x) \sin (c+d x)+9 a^3 A \sin (2 (c+d x))+45 a A b^2 \sin (2 (c+d x))+15 a^3 C \sin (2 (c+d x))+6 a^3 A \tan (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)} \] Input:

Integrate[((a + b*Cos[c + d*x])^3*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(7/ 
2),x]
 

Output:

(-6*a*(15*b^2*(A - C) + a^2*(3*A + 5*C))*Cos[c + d*x]^(3/2)*EllipticE[(c + 
 d*x)/2, 2] + 10*b*(b^2*(3*A + C) + 3*a^2*(A + 3*C))*Cos[c + d*x]^(3/2)*El 
lipticF[(c + d*x)/2, 2] + 30*a^2*A*b*Sin[c + d*x] + 10*b^3*C*Cos[c + d*x]^ 
2*Sin[c + d*x] + 9*a^3*A*Sin[2*(c + d*x)] + 45*a*A*b^2*Sin[2*(c + d*x)] + 
15*a^3*C*Sin[2*(c + d*x)] + 6*a^3*A*Tan[c + d*x])/(15*d*Cos[c + d*x]^(3/2) 
)
 

Rubi [A] (verified)

Time = 1.53 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.01, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.486, Rules used = {3042, 3527, 27, 3042, 3526, 27, 3042, 3510, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3527

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3526

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3510

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (5 b \left (3 a^2 (A+3 C)+b^2 (3 A+C)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 a \left (a^2 (3 A+5 C)+15 b^2 (A-C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {2 a \left (a^2 (3 A+5 C)+8 A b^2\right ) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {4 A b \sin (c+d x) (a+b \cos (c+d x))^2}{d \cos ^{\frac {3}{2}}(c+d x)}-\frac {2 b^3 (9 A-5 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {1}{5} \left (\frac {2 a \left (a^2 (3 A+5 C)+8 A b^2\right ) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {1}{3} \left (\frac {10 b \left (3 a^2 (A+3 C)+b^2 (3 A+C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}-\frac {6 a \left (a^2 (3 A+5 C)+15 b^2 (A-C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {4 A b \sin (c+d x) (a+b \cos (c+d x))^2}{d \cos ^{\frac {3}{2}}(c+d x)}-\frac {2 b^3 (9 A-5 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

Input:

Int[((a + b*Cos[c + d*x])^3*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(7/2),x]
 

Output:

(2*A*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + (((-6 
*a*(15*b^2*(A - C) + a^2*(3*A + 5*C))*EllipticE[(c + d*x)/2, 2])/d + (10*b 
*(b^2*(3*A + C) + 3*a^2*(A + 3*C))*EllipticF[(c + d*x)/2, 2])/d)/3 + (2*a* 
(8*A*b^2 + a^2*(3*A + 5*C))*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]) - (2*b^3* 
(9*A - 5*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d) + (4*A*b*(a + b*Cos[c + 
 d*x])^2*Sin[c + d*x])/(d*Cos[c + d*x]^(3/2)))/5
 

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 3510

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[(-(b*c - a*d))*(A*b^2 - a*b*B + 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] - S 
imp[1/(b^2*(m + 1)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*( 
m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b^2*d*(m 
 + 1) - a*b*c*(m + 2)) + (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] /; F 
reeQ[{a, b, c, d, e, f, A, B, C}, 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]
 

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1122\) vs. \(2(212)=424\).

Time = 20.16 (sec) , antiderivative size = 1123, normalized size of antiderivative = 4.90

method	result	size
parts	\(\text {Expression too large to display}\)	\(1123\)
default	\(\text {Expression too large to display}\)	\(1306\)

Input:

int((a+b*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x,method=_RETUR 
NVERBOSE)
 

Output:

2*(A*b^3+3*C*a^2*b)/d*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))-2*(3*A*a*b^2+ 
C*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)*Ell 
ipticE(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)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d-2/5*A 
*a^3*(-(1-2*cos(1/2*d*x+1/2*c)^2)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(8*sin(1/2*d 
*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1/2*d*x+ 
1/2*c)^3*(24*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-12*(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))*(sin(1/2*d*x+1/2*c)^2 
)^(1/2)*sin(1/2*d*x+1/2*c)^4-24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+12 
*(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))*(s 
in(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2+8*sin(1/2*d*x+1/2*c)^2*cos 
(1/2*d*x+1/2*c)-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*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d-2/3*C*b^3*((2*c 
os(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4*sin(1/2*d*x+1/2*c)^4 
*cos(1/2*d*x+1/2*c)-2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*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)/sin...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.12 (sec) , antiderivative size = 315, normalized size of antiderivative = 1.38 \[ \int \frac {(a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {5 \, \sqrt {2} {\left (3 i \, {\left (A + 3 \, C\right )} a^{2} b + i \, {\left (3 \, A + C\right )} b^{3}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-3 i \, {\left (A + 3 \, C\right )} a^{2} b - i \, {\left (3 \, A + C\right )} b^{3}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, \sqrt {2} {\left (i \, {\left (3 \, A + 5 \, C\right )} a^{3} + 15 i \, {\left (A - C\right )} a b^{2}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {2} {\left (-i \, {\left (3 \, A + 5 \, C\right )} a^{3} - 15 i \, {\left (A - C\right )} a b^{2}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, {\left (5 \, C b^{3} \cos \left (d x + c\right )^{3} + 15 \, A a^{2} b \cos \left (d x + c\right ) + 3 \, A a^{3} + 3 \, {\left ({\left (3 \, A + 5 \, C\right )} a^{3} + 15 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{15 \, d \cos \left (d x + c\right )^{3}} \] Input:

integrate((a+b*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x, algori 
thm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:

integrate((a+b*cos(d*x+c))**3*(A+C*cos(d*x+c)**2)/cos(d*x+c)**(7/2),x)
 

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [F]

\[ \int \frac {(a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \] Input:

integrate((a+b*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x, algori 
thm="maxima")
 

Output:

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)^3/cos(d*x + c)^(7/2) 
, x)
 

Giac [F]

\[ \int \frac {(a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \] Input:

integrate((a+b*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x, algori 
thm="giac")
 

Output:

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)^3/cos(d*x + c)^(7/2) 
, x)
 

Mupad [B] (verification not implemented)

Time = 1.83 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.24 \[ \int \frac {(a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {C\,b^3\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}+\frac {2\,A\,b^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {6\,C\,a\,b^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {6\,C\,a^2\,b\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,A\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{5\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,C\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {6\,A\,a\,b^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,A\,a^2\,b\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \] Input:

int(((A + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^3)/cos(c + d*x)^(7/2),x)
 

Output:

(C*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*A*b^3*ellipticF(c/2 + (d*x)/2, 2))/d + (6*C*a*b^2*ellipti 
cE(c/2 + (d*x)/2, 2))/d + (6*C*a^2*b*ellipticF(c/2 + (d*x)/2, 2))/d + (2*A 
*a^3*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2))/(5*d*cos(c 
 + d*x)^(5/2)*(sin(c + d*x)^2)^(1/2)) + (2*C*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)) + (6*A*a*b^2*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*A*a^2*b*sin(c + d*x)* 
hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(3/2)*(sin(c 
+ d*x)^2)^(1/2))
 

Reduce [F]

\[ \int \frac {(a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=3 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )}d x \right ) a^{2} b c +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )}d x \right ) a \,b^{3}+\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{4}}d x \right ) a^{4}+3 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{3}}d x \right ) a^{3} b +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2}}d x \right ) a^{3} c +3 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2}}d x \right ) a^{2} b^{2}+3 \left (\int \sqrt {\cos \left (d x +c \right )}d x \right ) a \,b^{2} c +\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )d x \right ) b^{3} c \] Input:

int((a+b*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x)
 

Output:

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