\(\int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx\) [210]

Optimal result

Integrand size = 37, antiderivative size = 248 \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx=-\frac {(15 A+7 C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(A+C) \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac {(9 A+5 C) \sin (c+d x)}{10 a d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}-\frac {(13 A+5 C) \sin (c+d x)}{10 a d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {(49 A+25 C) \sin (c+d x)}{10 a d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \] 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))*2^(1/2)/a^(3/2)/d-1/2*(A+C)*sin(d*x+c)/d/cos(d*x+c)^( 
5/2)/(a+a*cos(d*x+c))^(3/2)+1/10*(9*A+5*C)*sin(d*x+c)/a/d/cos(d*x+c)^(5/2) 
/(a+a*cos(d*x+c))^(1/2)-1/10*(13*A+5*C)*sin(d*x+c)/a/d/cos(d*x+c)^(3/2)/(a 
+a*cos(d*x+c))^(1/2)+1/10*(49*A+25*C)*sin(d*x+c)/a/d/cos(d*x+c)^(1/2)/(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 = 7.45 (sec) , antiderivative size = 2455, normalized size of antiderivative = 9.90 \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx=\text {Result too large to show} \] Input:

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

Output:

(8*C*Cos[c/2 + (d*x)/2]^3*Sin[c/2 + (d*x)/2])/(5*d*(a*(1 + Cos[c + d*x]))^ 
(3/2)*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(5/2)) - ((A + C)*Cos[c/2 + (d*x)/2]^3* 
(1 - 2*Sin[c/2 + (d*x)/2]))/(10*d*(a*(1 + Cos[c + d*x]))^(3/2)*(1 + Sin[c/ 
2 + (d*x)/2])*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(5/2)) + ((A + C)*Cos[c/2 + (d* 
x)/2]^3*(1 + 2*Sin[c/2 + (d*x)/2]))/(10*d*(a*(1 + Cos[c + d*x]))^(3/2)*(1 
- Sin[c/2 + (d*x)/2])*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(5/2)) + (32*C*Cos[c/2 
+ (d*x)/2]^3*Sin[c/2 + (d*x)/2])/(15*d*(a*(1 + Cos[c + d*x]))^(3/2)*(1 - 2 
*Sin[c/2 + (d*x)/2]^2)^(3/2)) + (64*C*Cos[c/2 + (d*x)/2]^3*Sin[c/2 + (d*x) 
/2])/(15*d*(a*(1 + Cos[c + d*x]))^(3/2)*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]) 
+ ((A + C)*Cos[c/2 + (d*x)/2]^3*(105*ArcTan[(1 - 2*Sin[c/2 + (d*x)/2])/Sqr 
t[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])*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]) + (67*Sqr 
t[1 - 2*Sin[c/2 + (d*x)/2]^2])/(1 - Sin[c/2 + (d*x)/2])))/(15*d*(a*(1 + Co 
s[c + d*x]))^(3/2)) - ((A + C)*Cos[c/2 + (d*x)/2]^3*(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 + 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])*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] 
)))/(15*d*(a*(1 + Cos[c + d*x]))^(3/2)) + ((-A + 7*C)*Cot[c/2 + (d*x)/2...
 

Rubi [A] (verified)

Time = 1.43 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.06, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.405, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3521

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3463

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3463

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3463

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3261

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

-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*C 
os[c + d*x]]) - ((2*a^2*(13*A + 5*C)*Sin[c + d*x])/(d*Cos[c + d*x]^(3/2)*S 
qrt[a + a*Cos[c + d*x]]) - ((-5*Sqrt[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)]
 

Maple [A] (verified)

Time = 3.82 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.00

method	result	size
default	\(\frac {\sqrt {2}\, \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (A \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (75 \cos \left (d x +c \right )^{4}+150 \cos \left (d x +c \right )^{3}+75 \cos \left (d x +c \right )^{2}\right )+C \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (35 \cos \left (d x +c \right )^{4}+70 \cos \left (d x +c \right )^{3}+35 \cos \left (d x +c \right )^{2}\right )+\sin \left (d x +c \right ) \left (49 \cos \left (d x +c \right )^{3}+36 \cos \left (d x +c \right )^{2}-4 \cos \left (d x +c \right )+4\right ) \sqrt {2}\, A +\sin \left (d x +c \right ) \cos \left (d x +c \right )^{2} \left (25 \cos \left (d x +c \right )+20\right ) \sqrt {2}\, C \right )}{20 d \,a^{2} \cos \left (d x +c \right )^{\frac {5}{2}} \left (1+\cos \left (d x +c \right )\right )^{2}}\)	\(249\)
parts	\(\frac {A \sqrt {2}\, \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (\sin \left (d x +c \right ) \left (49 \cos \left (d x +c \right )^{3}+36 \cos \left (d x +c \right )^{2}-4 \cos \left (d x +c \right )+4\right ) \sqrt {2}+\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) \left (75 \cos \left (d x +c \right )^{4}+150 \cos \left (d x +c \right )^{3}+75 \cos \left (d x +c \right )^{2}\right )\right )}{20 d \,a^{2} \cos \left (d x +c \right )^{\frac {5}{2}} \left (1+\cos \left (d x +c \right )\right )^{2}}+\frac {C \sqrt {2}\, \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (\sin \left (d x +c \right ) \left (5 \cos \left (d x +c \right )+4\right ) \sqrt {2}+\frac {7 \left (\cos \left (2 d x +2 c \right )+4 \cos \left (d x +c \right )+3\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2}\right )}{4 d \,a^{2} \sqrt {\cos \left (d x +c \right )}\, \left (1+\cos \left (d x +c \right )\right )^{2}}\)	\(271\)

Input:

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

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 234, normalized size of antiderivative = 0.94 \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx=-\frac {5 \, \sqrt {2} {\left ({\left (15 \, A + 7 \, C\right )} \cos \left (d x + c\right )^{5} + 2 \, {\left (15 \, A + 7 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (15 \, A + 7 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )\right )}}\right ) - 2 \, {\left ({\left (49 \, A + 25 \, C\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (9 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{2} - 4 \, A \cos \left (d x + c\right ) + 4 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{20 \, {\left (a^{2} d \cos \left (d x + c\right )^{5} + 2 \, a^{2} d \cos \left (d x + c\right )^{4} + a^{2} d \cos \left (d x + c\right )^{3}\right )}} \] Input:

integrate((A+C*cos(d*x+c)^2)/cos(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)^5 + 2*(15*A + 7*C)*cos(d*x + c 
)^4 + (15*A + 7*C)*cos(d*x + c)^3)*sqrt(a)*arctan(1/2*sqrt(2)*sqrt(a*cos(d 
*x + c) + a)*sqrt(a)*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 + a 
*cos(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)*sqrt(cos(d*x + c 
))*sin(d*x + c))/(a^2*d*cos(d*x + c)^5 + 2*a^2*d*cos(d*x + c)^4 + a^2*d*co 
s(d*x + c)^3)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F]

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

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{6}+2 \cos \left (d x +c \right )^{5}+\cos \left (d x +c \right )^{4}}d x \right ) a +\left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{4}+2 \cos \left (d x +c \right )^{3}+\cos \left (d x +c \right )^{2}}d x \right ) c \right )}{a^{2}} \] Input:

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

Output:

(sqrt(a)*(int((sqrt(cos(c + d*x) + 1)*sqrt(cos(c + d*x)))/(cos(c + d*x)**6 
 + 2*cos(c + d*x)**5 + cos(c + d*x)**4),x)*a + int((sqrt(cos(c + d*x) + 1) 
*sqrt(cos(c + d*x)))/(cos(c + d*x)**4 + 2*cos(c + d*x)**3 + cos(c + d*x)** 
2),x)*c))/a**2