\(\int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx\) [203]

Optimal result

Integrand size = 37, antiderivative size = 181 \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx=-\frac {\sqrt {2} (A+C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{\sqrt {a} d}+\frac {2 A \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}-\frac {2 A \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {2 (13 A+15 C) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \] Output:

-2^(1/2)*(A+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))/a^(1/2)/d+2/5*A*sin(d*x+c)/d/cos(d*x+c)^(5/2)/(a+a*cos 
(d*x+c))^(1/2)-2/15*A*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(1/2) 
+2/15*(13*A+15*C)*sin(d*x+c)/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 = 7.43 (sec) , antiderivative size = 1757, normalized size of antiderivative = 9.71 \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx =\text {Too large to display} \] Input:

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

Output:

(-4*C*Cos[c/2 + (d*x)/2]*Sin[c/2 + (d*x)/2])/(5*d*Sqrt[a*(1 + Cos[c + d*x] 
)]*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(5/2)) + (4*C*Cos[c/2 + (d*x)/2]*Sin[c/2 + 
 (d*x)/2])/(15*d*Sqrt[a*(1 + Cos[c + d*x])]*(1 - 2*Sin[c/2 + (d*x)/2]^2)^( 
3/2)) + (8*C*Cos[c/2 + (d*x)/2]*Sin[c/2 + (d*x)/2])/(15*d*Sqrt[a*(1 + Cos[ 
c + d*x])]*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]) - (2*(A + C)*Cot[c/2 + (d*x)/ 
2]*Csc[c/2 + (d*x)/2]^6*(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 + 655812 
*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*Cos[(c + 
 d*x)/2]^6*HypergeometricPFQ[{2, 2, 2, 9/2}, {1, 1, 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 - 518760*Sin[c/ 
2 + (d*x)/2]^12 + 1770*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]^12 + 226656*Sin[c/2 + 
(d*x)/2]^14 - 1500*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]^14 - 42048*Sin[c/2 + (d*x) 
/2]^16 + 440*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]^16 + 4725*ArcTanh[Sqrt[Sin[c/2 + 
 (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 
+ 2*Sin[c/2 + (d*x)/2]^2)] - 56700*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 
 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^2*Sqrt[Sin[c/2 + (d*x)/2]...
 

Rubi [A] (verified)

Time = 1.03 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.10, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.324, Rules used = {3042, 3523, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3523

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3463

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3463

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3261

\(\displaystyle \frac {2 A \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a A \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {30 a^3 (A+C) \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x) a^3}{\cos (c+d x) a+a}+2 a^2}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}\right )}{d}+\frac {2 a^2 (13 A+15 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}}{3 a}}{5 a}\)

\(\Big \downarrow \) 218

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

Input:

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

Output:

(2*A*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)*Sqrt[a + a*Cos[c + d*x]]) - ((2 
*a*A*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) - ((- 
15*Sqrt[2]*a^(3/2)*(A + 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^2*(13*A + 15*C)*Sin[c + d* 
x])/(d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]]))/(3*a))/(5*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 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 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 [A] (verified)

Time = 2.83 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.16

method	result	size
default	\(\frac {\sqrt {2}\, \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (A \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (15 \cos \left (d x +c \right )^{3}+15 \cos \left (d x +c \right )^{2}\right )+C \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (15 \cos \left (d x +c \right )^{3}+15 \cos \left (d x +c \right )^{2}\right )+\sin \left (d x +c \right ) \left (13 \cos \left (d x +c \right )^{2}-\cos \left (d x +c \right )+3\right ) \sqrt {2}\, A +15 C \cos \left (d x +c \right )^{2} \sin \left (d x +c \right ) \sqrt {2}\right )}{15 d a \cos \left (d x +c \right )^{\frac {5}{2}} \left (1+\cos \left (d x +c \right )\right )}\)	\(210\)
parts	\(\frac {A \sqrt {2}\, \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (\sin \left (d x +c \right ) \left (13 \cos \left (d x +c \right )^{2}-\cos \left (d x +c \right )+3\right ) \sqrt {2}+\arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (15 \cos \left (d x +c \right )^{3}+15 \cos \left (d x +c \right )^{2}\right )\right )}{15 d a \cos \left (d x +c \right )^{\frac {5}{2}} \left (1+\cos \left (d x +c \right )\right )}+\frac {C \sqrt {2}\, \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (\left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )+\sqrt {2}\, \sin \left (d x +c \right )\right )}{d a \sqrt {\cos \left (d x +c \right )}\, \left (1+\cos \left (d x +c \right )\right )}\)	\(228\)

Input:

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

Output:

1/15/d/a*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)*(15*cos(d*x+c)^3+15*cos(d*x+c)^2)+C*ar 
csin(-csc(d*x+c)+cot(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(15*cos(d*x 
+c)^3+15*cos(d*x+c)^2)+sin(d*x+c)*(13*cos(d*x+c)^2-cos(d*x+c)+3)*2^(1/2)*A 
+15*C*cos(d*x+c)^2*sin(d*x+c)*2^(1/2))/cos(d*x+c)^(5/2)/(1+cos(d*x+c))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.95 \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx=\frac {2 \, {\left ({\left (13 \, A + 15 \, C\right )} \cos \left (d x + c\right )^{2} - A \cos \left (d x + c\right ) + 3 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - \frac {15 \, \sqrt {2} {\left ({\left (A + C\right )} a \cos \left (d x + c\right )^{4} + {\left (A + C\right )} a \cos \left (d x + c\right )^{3}\right )} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a}}\right )}{\sqrt {a}}}{15 \, {\left (a d \cos \left (d x + c\right )^{4} + a d \cos \left (d x + c\right )^{3}\right )}} \] Input:

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

Output:

1/15*(2*((13*A + 15*C)*cos(d*x + c)^2 - A*cos(d*x + c) + 3*A)*sqrt(a*cos(d 
*x + c) + a)*sqrt(cos(d*x + c))*sin(d*x + c) - 15*sqrt(2)*((A + C)*a*cos(d 
*x + c)^4 + (A + C)*a*cos(d*x + c)^3)*arctan(1/2*sqrt(2)*sqrt(a*cos(d*x + 
c) + a)*sqrt(cos(d*x + c))*sin(d*x + c)/((cos(d*x + c)^2 + cos(d*x + c))*s 
qrt(a)))/sqrt(a))/(a*d*cos(d*x + c)^4 + a*d*cos(d*x + c)^3)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.77 (sec) , antiderivative size = 1674, normalized size of antiderivative = 9.25 \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

1/15*(15*(2*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(d 
*x + c) - 2*(cos(d*x + c) - 1)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x 
 + 2*c) + 1)) - sqrt(2)*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2 
*d*x + 2*c) + 1)^(1/4)*arctan2(((abs(e^(I*d*x + I*c) + 1)^4 + cos(d*x + c) 
^4 + sin(d*x + c)^4 + 2*(cos(d*x + c)^2 - sin(d*x + c)^2 - 2*cos(d*x + c) 
+ 1)*abs(e^(I*d*x + I*c) + 1)^2 - 4*cos(d*x + c)^3 + 2*(cos(d*x + c)^2 - 2 
*cos(d*x + c) + 1)*sin(d*x + c)^2 + 6*cos(d*x + c)^2 - 4*cos(d*x + c) + 1) 
^(1/4)*sin(1/2*arctan2(2*(cos(d*x + c) - 1)*sin(d*x + c)/abs(e^(I*d*x + I* 
c) + 1)^2, (abs(e^(I*d*x + I*c) + 1)^2 + cos(d*x + c)^2 - sin(d*x + c)^2 - 
 2*cos(d*x + c) + 1)/abs(e^(I*d*x + I*c) + 1)^2)) + sin(d*x + c))/abs(e^(I 
*d*x + I*c) + 1), ((abs(e^(I*d*x + I*c) + 1)^4 + cos(d*x + c)^4 + sin(d*x 
+ c)^4 + 2*(cos(d*x + c)^2 - sin(d*x + c)^2 - 2*cos(d*x + c) + 1)*abs(e^(I 
*d*x + I*c) + 1)^2 - 4*cos(d*x + c)^3 + 2*(cos(d*x + c)^2 - 2*cos(d*x + c) 
 + 1)*sin(d*x + c)^2 + 6*cos(d*x + c)^2 - 4*cos(d*x + c) + 1)^(1/4)*sqrt(a 
)*cos(1/2*arctan2(2*(cos(d*x + c) - 1)*sin(d*x + c)/abs(e^(I*d*x + I*c) + 
1)^2, (abs(e^(I*d*x + I*c) + 1)^2 + cos(d*x + c)^2 - sin(d*x + c)^2 - 2*co 
s(d*x + c) + 1)/abs(e^(I*d*x + I*c) + 1)^2)) + sqrt(a)*cos(d*x + c) - sqrt 
(a))/(sqrt(a)*abs(e^(I*d*x + I*c) + 1))))*C/((cos(2*d*x + 2*c)^2 + sin(2*d 
*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sqrt(a)) - (15*(sqrt(2)*cos(2* 
d*x + 2*c)^2 + sqrt(2)*sin(2*d*x + 2*c)^2 + 2*sqrt(2)*cos(2*d*x + 2*c) ...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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