\(\int \frac {(A+C \cos ^2(c+d x)) \sec ^{\frac {7}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx\) [1240]

Optimal result

Integrand size = 37, antiderivative size = 268 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=-\frac {(15 A+7 C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{2 \sqrt {2} a^{3/2} d}+\frac {(49 A+25 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{10 a d \sqrt {a+a \cos (c+d x)}}-\frac {(13 A+5 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A+C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(9 A+5 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt {a+a \cos (c+d x)}} \] Output:

-1/4*(15*A+7*C)*arctan(1/2*a^(1/2)*sin(d*x+c)*2^(1/2)/cos(d*x+c)^(1/2)/(a+ 
a*cos(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)*2^(1/2)/a^(3/2)/d+1 
/10*(49*A+25*C)*sec(d*x+c)^(1/2)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(1/2)-1/1 
0*(13*A+5*C)*sec(d*x+c)^(3/2)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(1/2)-1/2*(A 
+C)*sec(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(3/2)+1/10*(9*A+5*C)*se 
c(d*x+c)^(5/2)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(1/2)
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 8.08 (sec) , antiderivative size = 2281, normalized size of antiderivative = 8.51 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\text {Result too large to show} \] Input:

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

Output:

(2*Cos[c/2 + (d*x)/2]^3*Sqrt[(1 - 2*Sin[c/2 + (d*x)/2]^2)^(-1)]*Sqrt[1 - 2 
*Sin[c/2 + (d*x)/2]^2]*((4*C*Sin[c/2 + (d*x)/2])/(5*(1 - 2*Sin[c/2 + (d*x) 
/2]^2)^(5/2)) - ((A + C)*(1 - 2*Sin[c/2 + (d*x)/2]))/(20*(1 + Sin[c/2 + (d 
*x)/2])*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(5/2)) + ((A + C)*(1 + 2*Sin[c/2 + (d 
*x)/2]))/(20*(1 - Sin[c/2 + (d*x)/2])*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(5/2)) 
+ (16*C*Sin[c/2 + (d*x)/2])/(15*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(3/2)) + (32* 
C*Sin[c/2 + (d*x)/2])/(15*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]) - ((A + C)*(-1 
05*ArcTan[(1 - 2*Sin[c/2 + (d*x)/2])/Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]] + ( 
4 + 3*Sin[c/2 + (d*x)/2])/((1 - Sin[c/2 + (d*x)/2])*(1 - 2*Sin[c/2 + (d*x) 
/2]^2)^(3/2)) - (19 + 29*Sin[c/2 + (d*x)/2])/((1 - Sin[c/2 + (d*x)/2])*Sqr 
t[1 - 2*Sin[c/2 + (d*x)/2]^2]) - (67*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2])/(1 
- Sin[c/2 + (d*x)/2])))/30 + ((A + C)*(-105*ArcTan[(1 + 2*Sin[c/2 + (d*x)/ 
2])/Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]] + (4 - 3*Sin[c/2 + (d*x)/2])/((1 + S 
in[c/2 + (d*x)/2])*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(3/2)) - (19 - 29*Sin[c/2 
+ (d*x)/2])/((1 + Sin[c/2 + (d*x)/2])*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]) - 
(67*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2])/(1 + Sin[c/2 + (d*x)/2])))/30 + ((-A 
 + 7*C)*Csc[c/2 + (d*x)/2]^7*(4725*Sin[c/2 + (d*x)/2]^2 - 48825*Sin[c/2 + 
(d*x)/2]^4 + 210105*Sin[c/2 + (d*x)/2]^6 - 486630*Sin[c/2 + (d*x)/2]^8 + 6 
55812*Sin[c/2 + (d*x)/2]^10 - 710*Hypergeometric2F1[2, 9/2, 11/2, Sin[c/2 
+ (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^10 - 40*...
 

Rubi [A] (verified)

Time = 1.63 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.06, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.459, Rules used = {3042, 4709, 3042, 3521, 27, 3042, 3463, 27, 3042, 3463, 27, 3042, 3463, 27, 3042, 3261, 218}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4709

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {C \cos ^2(c+d x)+A}{\cos ^{\frac {7}{2}}(c+d x) (\cos (c+d x) a+a)^{3/2}}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3521

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {a (9 A+5 C)-2 a (3 A+C) \cos (c+d x)}{2 \cos ^{\frac {7}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}-\frac {(A+C) \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {a (9 A+5 C)-2 a (3 A+C) \cos (c+d x)}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{4 a^2}-\frac {(A+C) \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {a (9 A+5 C)-2 a (3 A+C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {(A+C) \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 3463

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 \int -\frac {3 a^2 (13 A+5 C)-4 a^2 (9 A+5 C) \cos (c+d x)}{2 \cos ^{\frac {5}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{5 a}+\frac {2 a (9 A+5 C) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}}{4 a^2}-\frac {(A+C) \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (9 A+5 C) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {3 a^2 (13 A+5 C)-4 a^2 (9 A+5 C) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{5 a}}{4 a^2}-\frac {(A+C) \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (9 A+5 C) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {3 a^2 (13 A+5 C)-4 a^2 (9 A+5 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{5 a}}{4 a^2}-\frac {(A+C) \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 3463

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (9 A+5 C) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 \int -\frac {3 \left (a^3 (49 A+25 C)-2 a^3 (13 A+5 C) \cos (c+d x)\right )}{2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{3 a}+\frac {2 a^2 (13 A+5 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}}{5 a}}{4 a^2}-\frac {(A+C) \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (9 A+5 C) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^2 (13 A+5 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {a^3 (49 A+25 C)-2 a^3 (13 A+5 C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{a}}{5 a}}{4 a^2}-\frac {(A+C) \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (9 A+5 C) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^2 (13 A+5 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {a^3 (49 A+25 C)-2 a^3 (13 A+5 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{5 a}}{4 a^2}-\frac {(A+C) \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 3463

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (9 A+5 C) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^2 (13 A+5 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 \int -\frac {5 a^4 (15 A+7 C)}{2 \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{a}+\frac {2 a^3 (49 A+25 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}}{a}}{5 a}}{4 a^2}-\frac {(A+C) \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (9 A+5 C) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^2 (13 A+5 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^3 (49 A+25 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-5 a^3 (15 A+7 C) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{a}}{5 a}}{4 a^2}-\frac {(A+C) \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (9 A+5 C) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^2 (13 A+5 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^3 (49 A+25 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-5 a^3 (15 A+7 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{5 a}}{4 a^2}-\frac {(A+C) \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 3261

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (9 A+5 C) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^2 (13 A+5 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {10 a^4 (15 A+7 C) \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x) a^3}{\cos (c+d x) a+a}+2 a^2}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}\right )}{d}+\frac {2 a^3 (49 A+25 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}}{a}}{5 a}}{4 a^2}-\frac {(A+C) \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 218

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (9 A+5 C) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^2 (13 A+5 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^3 (49 A+25 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-\frac {5 \sqrt {2} a^{5/2} (15 A+7 C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{a}}{5 a}}{4 a^2}-\frac {(A+C) \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\)

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-1/2*((A + C)*Sin[c + d*x])/(d*Cos[ 
c + d*x]^(5/2)*(a + a*Cos[c + d*x])^(3/2)) + ((2*a*(9*A + 5*C)*Sin[c + d*x 
])/(5*d*Cos[c + d*x]^(5/2)*Sqrt[a + a*Cos[c + d*x]]) - ((2*a^2*(13*A + 5*C 
)*Sin[c + d*x])/(d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) - ((-5*Sqr 
t[2]*a^(5/2)*(15*A + 7*C)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[ 
c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/d + (2*a^3*(49*A + 25*C)*Sin[c + d*x 
])/(d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]]))/a)/(5*a))/(4*a^2))
 

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 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 3042

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

rule 3261

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e 
_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*(a/f)   Subst[Int[1/(2*b^2 - (a*c 
 - b*d)*x^2), x], x, b*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*S 
in[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 3463

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

rule 3521

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

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 [A] (verified)

Time = 3.26 (sec) , antiderivative size = 255, normalized size of antiderivative = 0.95

method	result	size
default	\(\frac {\sqrt {2}\, \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \sec \left (d x +c \right )^{\frac {7}{2}} \left (A \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (75 \cos \left (d x +c \right )^{5}+150 \cos \left (d x +c \right )^{4}+75 \cos \left (d x +c \right )^{3}\right )+C \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (35 \cos \left (d x +c \right )^{5}+70 \cos \left (d x +c \right )^{4}+35 \cos \left (d x +c \right )^{3}\right )+\sin \left (d x +c \right ) \left (49 \cos \left (d x +c \right )^{3}+36 \cos \left (d x +c \right )^{2}-4 \cos \left (d x +c \right )+4\right ) \sqrt {2}\, A \cos \left (d x +c \right )+\sin \left (d x +c \right ) \cos \left (d x +c \right )^{3} \left (25 \cos \left (d x +c \right )+20\right ) \sqrt {2}\, C \right )}{20 d \,a^{2} \left (1+\cos \left (d x +c \right )\right )^{2}}\)	\(255\)
parts	\(\frac {A \sqrt {2}\, \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \sec \left (d x +c \right )^{\frac {7}{2}} \left (\sin \left (d x +c \right ) \left (49 \cos \left (d x +c \right )^{3}+36 \cos \left (d x +c \right )^{2}-4 \cos \left (d x +c \right )+4\right ) \sqrt {2}\, \cos \left (d x +c \right )+\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) \left (75 \cos \left (d x +c \right )^{5}+150 \cos \left (d x +c \right )^{4}+75 \cos \left (d x +c \right )^{3}\right )\right )}{20 d \,a^{2} \left (1+\cos \left (d x +c \right )\right )^{2}}+\frac {C \cos \left (d x +c \right )^{3} \sqrt {2}\, \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \left (\sin \left (d x +c \right ) \left (5 \cos \left (d x +c \right )+4\right ) \sqrt {2}+\frac {7 \left (\cos \left (2 d x +2 c \right )+3+4 \cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2}\right ) \sec \left (d x +c \right )^{\frac {7}{2}}}{4 d \,a^{2} \left (1+\cos \left (d x +c \right )\right )^{2}}\)	\(285\)

Input:

int((A+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2)/(a+a*cos(d*x+c))^(3/2),x,method=_R 
ETURNVERBOSE)
 

Output:

1/20/d/a^2*2^(1/2)*((1+cos(d*x+c))*a)^(1/2)*sec(d*x+c)^(7/2)/(1+cos(d*x+c) 
)^2*(A*arcsin(-csc(d*x+c)+cot(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(7 
5*cos(d*x+c)^5+150*cos(d*x+c)^4+75*cos(d*x+c)^3)+C*arcsin(-csc(d*x+c)+cot( 
d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(35*cos(d*x+c)^5+70*cos(d*x+c)^4 
+35*cos(d*x+c)^3)+sin(d*x+c)*(49*cos(d*x+c)^3+36*cos(d*x+c)^2-4*cos(d*x+c) 
+4)*2^(1/2)*A*cos(d*x+c)+sin(d*x+c)*cos(d*x+c)^3*(25*cos(d*x+c)+20)*2^(1/2 
)*C)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 214, normalized size of antiderivative = 0.80 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {5 \, \sqrt {2} {\left ({\left (15 \, A + 7 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (15 \, A + 7 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (15 \, A + 7 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) + \frac {2 \, {\left ({\left (49 \, A + 25 \, C\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (9 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{2} - 4 \, A \cos \left (d x + c\right ) + 4 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{20 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}} \] Input:

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2)/(a+a*cos(d*x+c))^(3/2),x, al 
gorithm="fricas")
 

Output:

1/20*(5*sqrt(2)*((15*A + 7*C)*cos(d*x + c)^4 + 2*(15*A + 7*C)*cos(d*x + c) 
^3 + (15*A + 7*C)*cos(d*x + c)^2)*sqrt(a)*arctan(sqrt(2)*sqrt(a*cos(d*x + 
c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) + 2*((49*A + 25*C)*cos( 
d*x + c)^3 + 4*(9*A + 5*C)*cos(d*x + c)^2 - 4*A*cos(d*x + c) + 4*A)*sqrt(a 
*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^2*d*cos(d*x + c)^4 
+ 2*a^2*d*cos(d*x + c)^3 + a^2*d*cos(d*x + c)^2)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-1)]

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

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2)/(a+a*cos(d*x+c))^(3/2),x, al 
gorithm="maxima")
 

Output:

Timed out
 

Giac [F]

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

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2)/(a+a*cos(d*x+c))^(3/2),x, al 
gorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

int(((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(7/2))/(a + a*cos(c + d*x))^( 
3/2),x)
 

Output:

int(((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(7/2))/(a + a*cos(c + d*x))^( 
3/2), x)
 

Reduce [F]

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

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

Output:

(sqrt(a)*(int((sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*cos(c + d*x)**2*s 
ec(c + d*x)**3)/(cos(c + d*x)**2 + 2*cos(c + d*x) + 1),x)*c + int((sqrt(se 
c(c + d*x))*sqrt(cos(c + d*x) + 1)*sec(c + d*x)**3)/(cos(c + d*x)**2 + 2*c 
os(c + d*x) + 1),x)*a))/a**2