\(\int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}} \, dx\) [519]

Optimal result
Mathematica [B] (warning: unable to verify)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [A] (verification not implemented)
Sympy [F(-1)]
Maxima [F]
Giac [F(-1)]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 45, antiderivative size = 163 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}} \, dx=\frac {(19 A+5 B+3 C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B+C) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(9 A-B-7 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}} \] Output:

1/32*(19*A+5*B+3*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^(5/2)/d-1/4*(A-B+C)*cos(d*x+c)^(1/2)*si 
n(d*x+c)/d/(a+a*cos(d*x+c))^(5/2)-1/16*(9*A-B-7*C)*cos(d*x+c)^(1/2)*sin(d* 
x+c)/a/d/(a+a*cos(d*x+c))^(3/2)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(711\) vs. \(2(163)=326\).

Time = 6.47 (sec) , antiderivative size = 711, normalized size of antiderivative = 4.36 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

(3*C*ArcSin[Sin[c/2 + (d*x)/2]/Sqrt[Cos[(c + d*x)/2]^2]]*Cos[c/2 + (d*x)/2 
]^5)/(4*d*(a*(1 + Cos[c + d*x]))^(5/2)) + (5*C*Cos[c/2 + (d*x)/2]^5*Sin[c/ 
2 + (d*x)/2]*Sqrt[1 - Sec[(c + d*x)/2]^2*Sin[c/2 + (d*x)/2]^2])/(4*d*Sqrt[ 
Cos[(c + d*x)/2]^2]*(a*(1 + Cos[c + d*x]))^(5/2)) - (C*Cos[c/2 + (d*x)/2]^ 
5*Sin[c/2 + (d*x)/2]^3*Sqrt[1 - Sec[(c + d*x)/2]^2*Sin[c/2 + (d*x)/2]^2])/ 
(2*d*(Cos[(c + d*x)/2]^2)^(3/2)*(a*(1 + Cos[c + d*x]))^(5/2)) - (A*Cos[c/2 
 + (d*x)/2]^5*Sec[(c + d*x)/2]^4*Sin[c/2 + (d*x)/2]*(11 - 31*Sin[c/2 + (d* 
x)/2]^2 + 18*Sin[c/2 + (d*x)/2]^4 - (19*ArcTanh[Sqrt[-(Sin[c/2 + (d*x)/2]^ 
2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]]*Cos[(c + d*x)/2]^4)/Sqrt[-(Sin[c/2 + (d* 
x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]))/(4*d*(a*(1 + Cos[c + d*x]))^(5/2) 
*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]) + (B*Cos[c/2 + (d*x)/2]^5*Sec[(c + d*x) 
/2]^4*Sin[c/2 + (d*x)/2]*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]*(3 - Sin[c/2 + ( 
d*x)/2]^2 - 5*ArcTanh[Sqrt[-(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2 
]^2))]]*Cos[(c + d*x)/2]^4*Csc[c/2 + (d*x)/2]^2*Sqrt[-(Sin[c/2 + (d*x)/2]^ 
2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]))/(4*d*(a*(1 + Cos[c + d*x]))^(5/2))
 

Rubi [A] (verified)

Time = 0.81 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 3520, 27, 3042, 3457, 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.

\(\displaystyle \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3520

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

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {a (7 A+B-C)-2 a (A-B-3 C) \cos (c+d x)}{\sqrt {\cos (c+d x)} (\cos (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {a (7 A+B-C)-2 a (A-B-3 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3457

\(\displaystyle \frac {\frac {\int \frac {a^2 (19 A+5 B+3 C)}{2 \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}-\frac {a (9 A-B-7 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{4} (19 A+5 B+3 C) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx-\frac {a (9 A-B-7 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{4} (19 A+5 B+3 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {a (9 A-B-7 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3261

\(\displaystyle \frac {-\frac {a (19 A+5 B+3 C) \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x) a^3}{\cos (c+d x) a+a}+2 a^2}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}\right )}{2 d}-\frac {a (9 A-B-7 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {(19 A+5 B+3 C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} \sqrt {a} d}-\frac {a (9 A-B-7 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\)

Input:

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

Output:

-1/4*((A - B + C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*(a + a*Cos[c + d*x]) 
^(5/2)) + (((19*A + 5*B + 3*C)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt 
[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(2*Sqrt[2]*Sqrt[a]*d) - (a*(9*A 
 - B - 7*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/ 
2)))/(8*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 3457
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*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^( 
n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/(a*(2*m + 1)*(b*c - a*d)) 
  Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b 
*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*(m + n + 
2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ 
[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] 
 &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
 

rule 3520
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*A - b*B + 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))   Int[(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)) + B*(b*c*m + a 
*d*(n + 1)) - C*(a*c*m + b*d*(n + 1)) + (d*(a*A - b*B)*(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, B, 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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(315\) vs. \(2(138)=276\).

Time = 1.46 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.94

method result size
default \(-\frac {\left (\sin \left (d x +c \right ) \left (9 \cos \left (d x +c \right )+13\right ) \sqrt {2}\, A \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+\sin \left (d x +c \right ) \left (-\cos \left (d x +c \right )-5\right ) \sqrt {2}\, B \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+\sin \left (d x +c \right ) \left (-7 \cos \left (d x +c \right )-3\right ) \sqrt {2}\, C \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+\left (19 \cos \left (d x +c \right )^{2}+38 \cos \left (d x +c \right )+19\right ) A \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+\left (5 \cos \left (d x +c \right )^{2}+10 \cos \left (d x +c \right )+5\right ) B \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+\left (3 \cos \left (d x +c \right )^{2}+6 \cos \left (d x +c \right )+3\right ) C \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {2}\, \sqrt {a \left (1+\cos \left (d x +c \right )\right )}}{32 d \,a^{3} \left (\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\) \(316\)
parts \(\frac {A \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}\, \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (-2 \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}-11 \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )-19 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right )}{32 d \,a^{3} \sqrt {\cos \left (d x +c \right )}}+\frac {B \left (\sin \left (d x +c \right ) \left (\cos \left (d x +c \right )+5\right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+\left (-5 \cos \left (d x +c \right )^{2}-10 \cos \left (d x +c \right )-5\right ) \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {2}\, \sqrt {a \left (1+\cos \left (d x +c \right )\right )}}{32 d \,a^{3} \left (\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}+\frac {C \left (\sin \left (d x +c \right ) \left (7 \cos \left (d x +c \right )+3\right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+\left (-3 \cos \left (d x +c \right )^{2}-6 \cos \left (d x +c \right )-3\right ) \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {2}\, \sqrt {a \left (1+\cos \left (d x +c \right )\right )}}{32 d \,a^{3} \left (\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\) \(469\)

Input:

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(5/2 
),x,method=_RETURNVERBOSE)
 

Output:

-1/32/d/a^3*(sin(d*x+c)*(9*cos(d*x+c)+13)*2^(1/2)*A*(cos(d*x+c)/(1+cos(d*x 
+c)))^(1/2)+sin(d*x+c)*(-cos(d*x+c)-5)*2^(1/2)*B*(cos(d*x+c)/(1+cos(d*x+c) 
))^(1/2)+sin(d*x+c)*(-7*cos(d*x+c)-3)*2^(1/2)*C*(cos(d*x+c)/(1+cos(d*x+c)) 
)^(1/2)+(19*cos(d*x+c)^2+38*cos(d*x+c)+19)*A*arcsin(cot(d*x+c)-csc(d*x+c)) 
+(5*cos(d*x+c)^2+10*cos(d*x+c)+5)*B*arcsin(cot(d*x+c)-csc(d*x+c))+(3*cos(d 
*x+c)^2+6*cos(d*x+c)+3)*C*arcsin(cot(d*x+c)-csc(d*x+c)))*cos(d*x+c)^(1/2)* 
2^(1/2)*(a*(1+cos(d*x+c)))^(1/2)/(cos(d*x+c)^3+3*cos(d*x+c)^2+3*cos(d*x+c) 
+1)/(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.44 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {2} {\left ({\left (19 \, A + 5 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (19 \, A + 5 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (19 \, A + 5 \, B + 3 \, C\right )} \cos \left (d x + c\right ) + 19 \, A + 5 \, B + 3 \, C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )\right )}}\right ) - 2 \, {\left ({\left (9 \, A - B - 7 \, C\right )} \cos \left (d x + c\right ) + 13 \, A - 5 \, B - 3 \, C\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \] Input:

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c) 
)^(5/2),x, algorithm="fricas")
 

Output:

1/32*(sqrt(2)*((19*A + 5*B + 3*C)*cos(d*x + c)^3 + 3*(19*A + 5*B + 3*C)*co 
s(d*x + c)^2 + 3*(19*A + 5*B + 3*C)*cos(d*x + c) + 19*A + 5*B + 3*C)*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*((9*A - B - 7*C)*cos 
(d*x + c) + 13*A - 5*B - 3*C)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))* 
sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos 
(d*x + c) + a^3*d)
 

Sympy [F(-1)]

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

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)**(1/2)/(a+a*cos(d*x+ 
c))**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c) 
)^(5/2),x, algorithm="maxima")
 

Output:

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)/((a*cos(d*x + c) + a)^(5 
/2)*sqrt(cos(d*x + c))), x)
 

Giac [F(-1)]

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

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c) 
)^(5/2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

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

int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^(1/2)*(a + a*cos 
(c + d*x))^(5/2)),x)
 

Output:

int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^(1/2)*(a + a*cos 
(c + d*x))^(5/2)), x)
                                                                                    
                                                                                    
 

Reduce [F]

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

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(5/2 
),x)
 

Output:

(sqrt(a)*(int((sqrt(cos(c + d*x) + 1)*sqrt(cos(c + d*x))*cos(c + d*x))/(co 
s(c + d*x)**3 + 3*cos(c + d*x)**2 + 3*cos(c + d*x) + 1),x)*c + int((sqrt(c 
os(c + d*x) + 1)*sqrt(cos(c + d*x)))/(cos(c + d*x)**4 + 3*cos(c + d*x)**3 
+ 3*cos(c + d*x)**2 + cos(c + d*x)),x)*a + int((sqrt(cos(c + d*x) + 1)*sqr 
t(cos(c + d*x)))/(cos(c + d*x)**3 + 3*cos(c + d*x)**2 + 3*cos(c + d*x) + 1 
),x)*b))/a**3