\(\int \sqrt {a+b \cos (c+d x)} (A+C \cos ^2(c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx\) [1413]

Optimal result

Integrand size = 37, antiderivative size = 385 \[ \int \sqrt {a+b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=-\frac {2 (a-b) \sqrt {a+b} \left (2 A b^2-3 a^2 (3 A+5 C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{15 a^3 d \sqrt {\sec (c+d x)}}-\frac {2 (a-b) \sqrt {a+b} (9 a A+2 A b+15 a C) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{15 a^2 d \sqrt {\sec (c+d x)}}+\frac {2 A b \sqrt {a+b \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 a d}+\frac {2 A \sqrt {a+b \cos (c+d x)} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 14.81 (sec) , antiderivative size = 429, normalized size of antiderivative = 1.11 \[ \int \sqrt {a+b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {2 \sqrt {2} \sqrt {\frac {\cos (c+d x)}{(1+\cos (c+d x))^2}} \sqrt {\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )} \left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^{3/2} \left (-\left ((a+b) \left (\left (-2 A b^2+3 a^2 (3 A+5 C)\right ) E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {-a+b}{a+b}\right )+a (2 A b-3 a (3 A+5 C)) \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right )\right ) \left (\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )\right )^{3/2} \sqrt {\frac {(a+b \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}} \sec (c+d x)\right )+\left (2 A b^2-3 a^2 (3 A+5 C)\right ) \cos (c+d x) (a+b \cos (c+d x)) \sec ^4\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{15 a^2 d \sqrt {\frac {1}{1+\cos (c+d x)}} \sqrt {a+b \cos (c+d x)} \sec ^2\left (\frac {1}{2} (c+d x)\right )^{3/2}}+\frac {\sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \left (9 a^2 A-2 A b^2+15 a^2 C\right ) \sin (c+d x)}{15 a^2}+\frac {2 A b \tan (c+d x)}{15 a}+\frac {2}{5} A \sec (c+d x) \tan (c+d x)\right )}{d} \] Input:

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

Output:

(2*Sqrt[2]*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])^2]*Sqrt[Cos[c + d*x]*Sec[( 
c + d*x)/2]^2]*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(3/2)*(-((a + b)*((-2*A*b 
^2 + 3*a^2*(3*A + 5*C))*EllipticE[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + 
b)] + a*(2*A*b - 3*a*(3*A + 5*C))*EllipticF[ArcSin[Tan[(c + d*x)/2]], (-a 
+ b)/(a + b)])*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(3/2)*Sqrt[((a + b*Cos[c 
+ d*x])*Sec[(c + d*x)/2]^2)/(a + b)]*Sec[c + d*x]) + (2*A*b^2 - 3*a^2*(3*A 
 + 5*C))*Cos[c + d*x]*(a + b*Cos[c + d*x])*Sec[(c + d*x)/2]^4*Tan[(c + d*x 
)/2]))/(15*a^2*d*Sqrt[(1 + Cos[c + d*x])^(-1)]*Sqrt[a + b*Cos[c + d*x]]*(S 
ec[(c + d*x)/2]^2)^(3/2)) + (Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*( 
(2*(9*a^2*A - 2*A*b^2 + 15*a^2*C)*Sin[c + d*x])/(15*a^2) + (2*A*b*Tan[c + 
d*x])/(15*a) + (2*A*Sec[c + d*x]*Tan[c + d*x])/5))/d
 

Rubi [A] (verified)

Time = 1.67 (sec) , antiderivative size = 375, normalized size of antiderivative = 0.97, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.351, Rules used = {3042, 4709, 3042, 3527, 27, 3042, 3534, 27, 3042, 3477, 3042, 3295, 3473}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4709

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3527

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {2 \int -\frac {-3 (3 A+5 C) a^2-b (7 A+15 C) \cos (c+d x) a+2 A b^2}{2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{3 a}+\frac {2 A b \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {2 A b \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\int \frac {-3 (3 A+5 C) a^2-b (7 A+15 C) \cos (c+d x) a+2 A b^2}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {2 A b \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\int \frac {-3 (3 A+5 C) a^2-b (7 A+15 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a+2 A b^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3477

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3295

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

\(\Big \downarrow \) 3473

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

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*A*Sqrt[a + b*Cos[c + d*x]]*Sin[c 
 + d*x])/(5*d*Cos[c + d*x]^(5/2)) + (-1/3*((2*(a - b)*Sqrt[a + b]*(2*A*b^2 
 - 3*a^2*(3*A + 5*C))*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Cos[c + d*x 
]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec 
[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(a^2*d) + (2*(a 
 - b)*Sqrt[a + b]*(9*a*A + 2*A*b + 15*a*C)*Cot[c + d*x]*EllipticF[ArcSin[S 
qrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - 
b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - 
 b)])/(a*d))/a + (2*A*b*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(3*a*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 3295

Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f 
_.)*(x_)]]), x_Symbol] :> Simp[-2*(Tan[e + f*x]/(a*f))*Rt[(a + b)/d, 2]*Sqr 
t[a*((1 - Csc[e + f*x])/(a + b))]*Sqrt[a*((1 + Csc[e + f*x])/(a - b))]*Elli 
pticF[ArcSin[Sqrt[a + b*Sin[e + f*x]]/Sqrt[d*Sin[e + f*x]]/Rt[(a + b)/d, 2] 
], -(a + b)/(a - b)], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] 
&& PosQ[(a + b)/d]
 

rule 3473

Int[((A_) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((b_.)*sin[(e_.) + (f_.)*(x_)]) 
^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*A* 
(c - d)*(Tan[e + f*x]/(f*b*c^2))*Rt[(c + d)/b, 2]*Sqrt[c*((1 + Csc[e + f*x] 
)/(c - d))]*Sqrt[c*((1 - Csc[e + f*x])/(c + d))]*EllipticE[ArcSin[Sqrt[c + 
d*Sin[e + f*x]]/Sqrt[b*Sin[e + f*x]]/Rt[(c + d)/b, 2]], -(c + d)/(c - d)], 
x] /; FreeQ[{b, c, d, e, f, A, B}, x] && NeQ[c^2 - d^2, 0] && EqQ[A, B] && 
PosQ[(c + d)/b]
 

rule 3477

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

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 3534

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[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x 
]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b* 
c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2))   Int 
[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b*c - a* 
d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - 
a*b*B + a^2*C) + (m + 1)*(b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A 
*b^2 - a*b*B + a^2*C)*(m + n + 3)*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] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ 
[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) | 
| EqQ[a, 0])))
 

rule 4709

Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Simp[(c*Sec[a 
+ b*x])^m*(c*Cos[a + b*x])^m   Int[ActivateTrig[u]/(c*Cos[a + b*x])^m, x], 
x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u, x]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1449\) vs. \(2(339)=678\).

Time = 26.68 (sec) , antiderivative size = 1450, normalized size of antiderivative = 3.77

method	result	size
parts	\(\text {Expression too large to display}\)	\(1450\)
default	\(\text {Expression too large to display}\)	\(1477\)

Input:

int((a+b*cos(d*x+c))^(1/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x,method=_R 
ETURNVERBOSE)
 

Output:

2/15*A/d/a^2*(a+b*cos(d*x+c))^(1/2)*sec(d*x+c)^(7/2)/(b*cos(d*x+c)^2+a*cos 
(d*x+c)+b*cos(d*x+c)+a)*((cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*c 
os(d*x+c))/(1+cos(d*x+c)))^(1/2)*a^3*EllipticE(-csc(d*x+c)+cot(d*x+c),(-(a 
-b)/(a+b))^(1/2))*(9*cos(d*x+c)^5+18*cos(d*x+c)^4+9*cos(d*x+c)^3)+(cos(d*x 
+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)* 
a^2*b*EllipticE(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))*(9*cos(d*x+c) 
^5+18*cos(d*x+c)^4+9*cos(d*x+c)^3)+(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a 
+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a*b^2*EllipticE(-csc(d*x+c)+cot 
(d*x+c),(-(a-b)/(a+b))^(1/2))*(-2*cos(d*x+c)^5-4*cos(d*x+c)^4-2*cos(d*x+c) 
^3)+(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x 
+c)))^(1/2)*b^3*EllipticE(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))*(-2 
*cos(d*x+c)^5-4*cos(d*x+c)^4-2*cos(d*x+c)^3)+(cos(d*x+c)/(1+cos(d*x+c)))^( 
1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a^3*EllipticF(-csc(d* 
x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))*(-9*cos(d*x+c)^5-18*cos(d*x+c)^4-9*c 
os(d*x+c)^3)+(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/( 
1+cos(d*x+c)))^(1/2)*a^2*b*EllipticF(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b)) 
^(1/2))*(-7*cos(d*x+c)^5-14*cos(d*x+c)^4-7*cos(d*x+c)^3)+(cos(d*x+c)/(1+co 
s(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a*b^2*Ell 
ipticF(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))*(2*cos(d*x+c)^5+4*cos( 
d*x+c)^4+2*cos(d*x+c)^3)+(9*cos(d*x+c)^2+3*cos(d*x+c)+3)*sin(d*x+c)*a^3...
 

Fricas [F]

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

integrate((a+b*cos(d*x+c))^(1/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x, al 
gorithm="fricas")
 

Output:

integral((C*cos(d*x + c)^2 + A)*sqrt(b*cos(d*x + c) + a)*sec(d*x + c)^(7/2 
), x)
 

Sympy [F(-1)]

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

integrate((a+b*cos(d*x+c))**(1/2)*(A+C*cos(d*x+c)**2)*sec(d*x+c)**(7/2),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((a+b*cos(d*x+c))^(1/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x, al 
gorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((a+b*cos(d*x+c))^(1/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x, al 
gorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \sqrt {a+b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\int \left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}\,\sqrt {a+b\,\cos \left (c+d\,x\right )} \,d x \] Input:

int((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(7/2)*(a + b*cos(c + d*x))^(1/ 
2),x)
 

Output:

int((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(7/2)*(a + b*cos(c + d*x))^(1/ 
2), x)
 

Reduce [F]

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

int((a+b*cos(d*x+c))^(1/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x)
 

Output:

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