\(\int (a+a \cos (c+d x))^{5/2} (A+C \cos ^2(c+d x)) \sec ^7(c+d x) \, dx\) [101]

Optimal result

Integrand size = 35, antiderivative size = 290 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^7(c+d x) \, dx=\frac {a^{5/2} (1015 A+1304 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{512 d}+\frac {a^3 (1015 A+1304 C) \tan (c+d x)}{512 d \sqrt {a+a \cos (c+d x)}}+\frac {a^3 (1015 A+1304 C) \sec (c+d x) \tan (c+d x)}{768 d \sqrt {a+a \cos (c+d x)}}+\frac {a^3 (109 A+136 C) \sec ^2(c+d x) \tan (c+d x)}{192 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (23 A+24 C) \sqrt {a+a \cos (c+d x)} \sec ^3(c+d x) \tan (c+d x)}{96 d}+\frac {a A (a+a \cos (c+d x))^{3/2} \sec ^4(c+d x) \tan (c+d x)}{12 d}+\frac {A (a+a \cos (c+d x))^{5/2} \sec ^5(c+d x) \tan (c+d x)}{6 d} \] Output:

1/512*a^(5/2)*(1015*A+1304*C)*arctanh(a^(1/2)*sin(d*x+c)/(a+a*cos(d*x+c))^ 
(1/2))/d+1/512*a^3*(1015*A+1304*C)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/7 
68*a^3*(1015*A+1304*C)*sec(d*x+c)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/19 
2*a^3*(109*A+136*C)*sec(d*x+c)^2*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/96* 
a^2*(23*A+24*C)*(a+a*cos(d*x+c))^(1/2)*sec(d*x+c)^3*tan(d*x+c)/d+1/12*a*A* 
(a+a*cos(d*x+c))^(3/2)*sec(d*x+c)^4*tan(d*x+c)/d+1/6*A*(a+a*cos(d*x+c))^(5 
/2)*sec(d*x+c)^5*tan(d*x+c)/d
 

Mathematica [A] (verified)

Time = 3.07 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.68 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^7(c+d x) \, dx=\frac {a^2 \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sec ^6(c+d x) \left (24 \sqrt {2} (1015 A+1304 C) \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^6(c+d x)+(27412 A+18720 C+14 (4591 A+4056 C) \cos (c+d x)+16 (1711 A+1496 C) \cos (2 (c+d x))+21721 A \cos (3 (c+d x))+25448 C \cos (3 (c+d x))+4060 A \cos (4 (c+d x))+5216 C \cos (4 (c+d x))+3045 A \cos (5 (c+d x))+3912 C \cos (5 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{24576 d} \] Input:

Integrate[(a + a*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^7 
,x]
 

Output:

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sec[c + d*x]^6*(24*Sqrt[2 
]*(1015*A + 1304*C)*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x]^6 + (27 
412*A + 18720*C + 14*(4591*A + 4056*C)*Cos[c + d*x] + 16*(1711*A + 1496*C) 
*Cos[2*(c + d*x)] + 21721*A*Cos[3*(c + d*x)] + 25448*C*Cos[3*(c + d*x)] + 
4060*A*Cos[4*(c + d*x)] + 5216*C*Cos[4*(c + d*x)] + 3045*A*Cos[5*(c + d*x) 
] + 3912*C*Cos[5*(c + d*x)])*Sin[(c + d*x)/2]))/(24576*d)
 

Rubi [A] (verified)

Time = 1.80 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.01, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.514, Rules used = {3042, 3523, 27, 3042, 3454, 27, 3042, 3454, 27, 3042, 3459, 3042, 3251, 3042, 3251, 3042, 3252, 219}

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.

\(\displaystyle \int \sec ^7(c+d x) (a \cos (c+d x)+a)^{5/2} \left (A+C \cos ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3523

\(\displaystyle \frac {\int \frac {1}{2} (\cos (c+d x) a+a)^{5/2} (5 a A+a (5 A+12 C) \cos (c+d x)) \sec ^6(c+d x)dx}{6 a}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^{5/2}}{6 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int (\cos (c+d x) a+a)^{5/2} (5 a A+a (5 A+12 C) \cos (c+d x)) \sec ^6(c+d x)dx}{12 a}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^{5/2}}{6 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (5 a A+a (5 A+12 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^6}dx}{12 a}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^{5/2}}{6 d}\)

\(\Big \downarrow \) 3454

\(\displaystyle \frac {\frac {1}{5} \int \frac {5}{2} (\cos (c+d x) a+a)^{3/2} \left ((23 A+24 C) a^2+3 (5 A+8 C) \cos (c+d x) a^2\right ) \sec ^5(c+d x)dx+\frac {a^2 A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{d}}{12 a}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^{5/2}}{6 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{2} \int (\cos (c+d x) a+a)^{3/2} \left ((23 A+24 C) a^2+3 (5 A+8 C) \cos (c+d x) a^2\right ) \sec ^5(c+d x)dx+\frac {a^2 A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{d}}{12 a}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^{5/2}}{6 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{2} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left ((23 A+24 C) a^2+3 (5 A+8 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx+\frac {a^2 A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{d}}{12 a}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^{5/2}}{6 d}\)

\(\Big \downarrow \) 3454

\(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{4} \int \frac {1}{2} \sqrt {\cos (c+d x) a+a} \left (3 (109 A+136 C) a^3+(235 A+312 C) \cos (c+d x) a^3\right ) \sec ^4(c+d x)dx+\frac {a^3 (23 A+24 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\right )+\frac {a^2 A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{d}}{12 a}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^{5/2}}{6 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{8} \int \sqrt {\cos (c+d x) a+a} \left (3 (109 A+136 C) a^3+(235 A+312 C) \cos (c+d x) a^3\right ) \sec ^4(c+d x)dx+\frac {a^3 (23 A+24 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\right )+\frac {a^2 A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{d}}{12 a}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^{5/2}}{6 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{8} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (3 (109 A+136 C) a^3+(235 A+312 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx+\frac {a^3 (23 A+24 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\right )+\frac {a^2 A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{d}}{12 a}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^{5/2}}{6 d}\)

\(\Big \downarrow \) 3459

\(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{8} \left (\frac {1}{2} a^3 (1015 A+1304 C) \int \sqrt {\cos (c+d x) a+a} \sec ^3(c+d x)dx+\frac {a^4 (109 A+136 C) \tan (c+d x) \sec ^2(c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^3 (23 A+24 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\right )+\frac {a^2 A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{d}}{12 a}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^{5/2}}{6 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{8} \left (\frac {1}{2} a^3 (1015 A+1304 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {a^4 (109 A+136 C) \tan (c+d x) \sec ^2(c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^3 (23 A+24 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\right )+\frac {a^2 A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{d}}{12 a}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^{5/2}}{6 d}\)

\(\Big \downarrow \) 3251

\(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{8} \left (\frac {1}{2} a^3 (1015 A+1304 C) \left (\frac {3}{4} \int \sqrt {\cos (c+d x) a+a} \sec ^2(c+d x)dx+\frac {a \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^4 (109 A+136 C) \tan (c+d x) \sec ^2(c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^3 (23 A+24 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\right )+\frac {a^2 A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{d}}{12 a}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^{5/2}}{6 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{8} \left (\frac {1}{2} a^3 (1015 A+1304 C) \left (\frac {3}{4} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^4 (109 A+136 C) \tan (c+d x) \sec ^2(c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^3 (23 A+24 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\right )+\frac {a^2 A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{d}}{12 a}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^{5/2}}{6 d}\)

\(\Big \downarrow \) 3251

\(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{8} \left (\frac {1}{2} a^3 (1015 A+1304 C) \left (\frac {3}{4} \left (\frac {1}{2} \int \sqrt {\cos (c+d x) a+a} \sec (c+d x)dx+\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^4 (109 A+136 C) \tan (c+d x) \sec ^2(c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^3 (23 A+24 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\right )+\frac {a^2 A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{d}}{12 a}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^{5/2}}{6 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{8} \left (\frac {1}{2} a^3 (1015 A+1304 C) \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^4 (109 A+136 C) \tan (c+d x) \sec ^2(c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^3 (23 A+24 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\right )+\frac {a^2 A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{d}}{12 a}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^{5/2}}{6 d}\)

\(\Big \downarrow \) 3252

\(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{8} \left (\frac {1}{2} a^3 (1015 A+1304 C) \left (\frac {3}{4} \left (\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {a \int \frac {1}{a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}\right )+\frac {a \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^4 (109 A+136 C) \tan (c+d x) \sec ^2(c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^3 (23 A+24 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\right )+\frac {a^2 A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{d}}{12 a}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^{5/2}}{6 d}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {a^2 A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{d}+\frac {1}{2} \left (\frac {a^3 (23 A+24 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}+\frac {1}{8} \left (\frac {a^4 (109 A+136 C) \tan (c+d x) \sec ^2(c+d x)}{d \sqrt {a \cos (c+d x)+a}}+\frac {1}{2} a^3 (1015 A+1304 C) \left (\frac {3}{4} \left (\frac {\sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )\right )\right )}{12 a}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^{5/2}}{6 d}\)

Input:

Int[(a + a*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^7,x]
 

Output:

(A*(a + a*Cos[c + d*x])^(5/2)*Sec[c + d*x]^5*Tan[c + d*x])/(6*d) + ((a^2*A 
*(a + a*Cos[c + d*x])^(3/2)*Sec[c + d*x]^4*Tan[c + d*x])/d + ((a^3*(23*A + 
 24*C)*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + ((a^4 
*(109*A + 136*C)*Sec[c + d*x]^2*Tan[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]]) 
 + (a^3*(1015*A + 1304*C)*((a*Sec[c + d*x]*Tan[c + d*x])/(2*d*Sqrt[a + a*C 
os[c + d*x]]) + (3*((Sqrt[a]*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos 
[c + d*x]]])/d + (a*Tan[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]])))/4))/2)/8) 
/2)/(12*a)
 

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 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3251

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)*Cos[e + f*x]*((c + d*Sin[e 
+ f*x])^(n + 1)/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]])), x] + Sim 
p[(2*n + 3)*((b*c - a*d)/(2*b*(n + 1)*(c^2 - d^2)))   Int[Sqrt[a + b*Sin[e 
+ f*x]]*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x 
] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, 
-1] && NeQ[2*n + 3, 0] && IntegerQ[2*n]
 

rule 3252

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

rule 3454

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(-b^2)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[ 
e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a*d))), x] - Simp[b/(d*(n + 1)*(b*c + 
 a*d))   Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp 
[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n + 1) - B 
*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f 
, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] 
&& GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0 
])
 

rule 3459

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

rule 3523

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/(b*d*(n + 1)*(c^2 - 
d^2))   Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a 
*d*m + b*c*(n + 1)) + c*C*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(m + n + 2) + C* 
(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, 
 e, f, A, C, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - 
d^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2298\) vs. \(2(258)=516\).

Time = 0.20 (sec) , antiderivative size = 2299, normalized size of antiderivative = 7.93

\[\text {Expression too large to display}\]

Input:

int((a+a*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^7,x)
 

Output:

1/48*a^(3/2)*cos(1/2*d*x+1/2*c)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*(192*a*(101 
5*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1 
/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+2*a))+1015*A*ln(-4/(2*cos(1/2*d 
*x+1/2*c)-2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2)*(sin(1/2* 
d*x+1/2*c)^2*a)^(1/2)-2*a))+1304*C*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a* 
2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+ 
2*a))+1304*C*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1 
/2*c)-a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-2*a)))*sin(1/2*d*x+1/ 
2*c)^12-192*(1015*A*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*a^(1/2)+1304*C* 
2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*a^(1/2)+3045*A*ln(4/(2*cos(1/2*d*x+ 
1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(sin(1/2*d*x 
+1/2*c)^2*a)^(1/2)+2*a))*a+3045*A*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(a* 
2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)- 
2*a))*a+3912*C*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+ 
1/2*c)+a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+2*a))*a+3912*C*ln(-4 
/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1 
/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-2*a))*a)*sin(1/2*d*x+1/2*c)^10+16*(3451 
0*A*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*a^(1/2)+44336*C*2^(1/2)*(sin(1/ 
2*d*x+1/2*c)^2*a)^(1/2)*a^(1/2)+45675*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2) 
)*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a...
 

Fricas [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 266, normalized size of antiderivative = 0.92 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^7(c+d x) \, dx=\frac {3 \, {\left ({\left (1015 \, A + 1304 \, C\right )} a^{2} \cos \left (d x + c\right )^{7} + {\left (1015 \, A + 1304 \, C\right )} a^{2} \cos \left (d x + c\right )^{6}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left (3 \, {\left (1015 \, A + 1304 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} + 2 \, {\left (1015 \, A + 1304 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 8 \, {\left (203 \, A + 184 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 48 \, {\left (29 \, A + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 896 \, A a^{2} \cos \left (d x + c\right ) + 256 \, A a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{6144 \, {\left (d \cos \left (d x + c\right )^{7} + d \cos \left (d x + c\right )^{6}\right )}} \] Input:

integrate((a+a*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^7,x, algori 
thm="fricas")
 

Output:

1/6144*(3*((1015*A + 1304*C)*a^2*cos(d*x + c)^7 + (1015*A + 1304*C)*a^2*co 
s(d*x + c)^6)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*sqrt( 
a*cos(d*x + c) + a)*sqrt(a)*(cos(d*x + c) - 2)*sin(d*x + c) + 8*a)/(cos(d* 
x + c)^3 + cos(d*x + c)^2)) + 4*(3*(1015*A + 1304*C)*a^2*cos(d*x + c)^5 + 
2*(1015*A + 1304*C)*a^2*cos(d*x + c)^4 + 8*(203*A + 184*C)*a^2*cos(d*x + c 
)^3 + 48*(29*A + 8*C)*a^2*cos(d*x + c)^2 + 896*A*a^2*cos(d*x + c) + 256*A* 
a^2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(d*cos(d*x + c)^7 + d*cos(d*x 
+ c)^6)
 

Sympy [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^7(c+d x) \, dx=\text {Timed out} \] Input:

integrate((a+a*cos(d*x+c))**(5/2)*(A+C*cos(d*x+c)**2)*sec(d*x+c)**7,x)
 

Output:

Timed out
 

Maxima [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^7(c+d x) \, dx=\text {Timed out} \] Input:

integrate((a+a*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^7,x, algori 
thm="maxima")
 

Output:

Timed out
 

Giac [A] (verification not implemented)

Time = 1.49 (sec) , antiderivative size = 430, normalized size of antiderivative = 1.48 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^7(c+d x) \, dx =\text {Too large to display} \] Input:

integrate((a+a*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^7,x, algori 
thm="giac")
 

Output:

-1/6144*sqrt(2)*(3*sqrt(2)*(1015*A*a^2*sgn(cos(1/2*d*x + 1/2*c)) + 1304*C* 
a^2*sgn(cos(1/2*d*x + 1/2*c)))*log(abs(-2*sqrt(2) + 4*sin(1/2*d*x + 1/2*c) 
)/abs(2*sqrt(2) + 4*sin(1/2*d*x + 1/2*c))) + 4*(97440*A*a^2*sgn(cos(1/2*d* 
x + 1/2*c))*sin(1/2*d*x + 1/2*c)^11 + 125184*C*a^2*sgn(cos(1/2*d*x + 1/2*c 
))*sin(1/2*d*x + 1/2*c)^11 - 276080*A*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/ 
2*d*x + 1/2*c)^9 - 354688*C*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/ 
2*c)^9 + 321552*A*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^7 + 4 
08192*C*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^7 - 195576*A*a^ 
2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^5 - 238272*C*a^2*sgn(cos( 
1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^5 + 63794*A*a^2*sgn(cos(1/2*d*x + 1 
/2*c))*sin(1/2*d*x + 1/2*c)^3 + 70352*C*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin( 
1/2*d*x + 1/2*c)^3 - 9243*A*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/ 
2*c) - 8376*C*a^2*sgn(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)^6)*sqrt(a)/d
 

Mupad [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^7(c+d x) \, dx=\int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^7} \,d x \] Input:

int(((A + C*cos(c + d*x)^2)*(a + a*cos(c + d*x))^(5/2))/cos(c + d*x)^7,x)
 

Output:

int(((A + C*cos(c + d*x)^2)*(a + a*cos(c + d*x))^(5/2))/cos(c + d*x)^7, x)
 

Reduce [F]

\[ \int (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^7(c+d x) \, dx=\sqrt {a}\, a^{2} \left (2 \left (\int \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{7}d x \right ) a +\left (\int \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{4} \sec \left (d x +c \right )^{7}d x \right ) c +2 \left (\int \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )^{7}d x \right ) c +\left (\int \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{7}d x \right ) a +\left (\int \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{7}d x \right ) c +\left (\int \sqrt {\cos \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{7}d x \right ) a \right ) \] Input:

int((a+a*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^7,x)
 

Output:

sqrt(a)*a**2*(2*int(sqrt(cos(c + d*x) + 1)*cos(c + d*x)*sec(c + d*x)**7,x) 
*a + int(sqrt(cos(c + d*x) + 1)*cos(c + d*x)**4*sec(c + d*x)**7,x)*c + 2*i 
nt(sqrt(cos(c + d*x) + 1)*cos(c + d*x)**3*sec(c + d*x)**7,x)*c + int(sqrt( 
cos(c + d*x) + 1)*cos(c + d*x)**2*sec(c + d*x)**7,x)*a + int(sqrt(cos(c + 
d*x) + 1)*cos(c + d*x)**2*sec(c + d*x)**7,x)*c + int(sqrt(cos(c + d*x) + 1 
)*sec(c + d*x)**7,x)*a)