\(\int \frac {(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx\) [415]

Optimal result

Integrand size = 43, antiderivative size = 173 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=-\frac {(3 A-2 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{a^{3/2} d}+\frac {(9 A-5 B+C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(A-B+C) \tan (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(3 A-B+C) \tan (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)}} \] Output:

-(3*A-2*B)*arctanh(a^(1/2)*sin(d*x+c)/(a+a*cos(d*x+c))^(1/2))/a^(3/2)/d+1/ 
4*(9*A-5*B+C)*arctanh(1/2*a^(1/2)*sin(d*x+c)*2^(1/2)/(a+a*cos(d*x+c))^(1/2 
))*2^(1/2)/a^(3/2)/d-1/2*(A-B+C)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(3/2)+1/2*( 
3*A-B+C)*tan(d*x+c)/a/d/(a+a*cos(d*x+c))^(1/2)
 

Mathematica [A] (warning: unable to verify)

Time = 1.84 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.13 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {\cos ^3\left (\frac {1}{2} (c+d x)\right ) \cos ^2(c+d x) \left (C+B \sec (c+d x)+A \sec ^2(c+d x)\right ) \left (2 (9 A-5 B+C) \text {arctanh}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {4 \sqrt {2} (3 A-2 B) \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right )-2 (3 A-B+C+2 A \sec (c+d x)) \sin \left (\frac {1}{2} (c+d x)\right )}{-1+\sin ^2\left (\frac {1}{2} (c+d x)\right )}\right )}{d (a (1+\cos (c+d x)))^{3/2} (2 A+C+2 B \cos (c+d x)+C \cos (2 (c+d x)))} \] Input:

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

Output:

(Cos[(c + d*x)/2]^3*Cos[c + d*x]^2*(C + B*Sec[c + d*x] + A*Sec[c + d*x]^2) 
*(2*(9*A - 5*B + C)*ArcTanh[Sin[(c + d*x)/2]] + (4*Sqrt[2]*(3*A - 2*B)*Arc 
Tanh[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^2 - 2*(3*A - B + C + 2*A*S 
ec[c + d*x])*Sin[(c + d*x)/2])/(-1 + Sin[(c + d*x)/2]^2)))/(d*(a*(1 + Cos[ 
c + d*x]))^(3/2)*(2*A + C + 2*B*Cos[c + d*x] + C*Cos[2*(c + d*x)]))
 

Rubi [A] (verified)

Time = 1.18 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.05, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.302, Rules used = {3042, 3520, 27, 3042, 3463, 25, 3042, 3464, 3042, 3128, 219, 3252, 219}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3520

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3463

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3464

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3128

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 3252

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

-1/2*((A - B + C)*Tan[c + d*x])/(d*(a + a*Cos[c + d*x])^(3/2)) + (-(((4*a^ 
(3/2)*(3*A - 2*B)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]] 
)/d - (Sqrt[2]*a^(3/2)*(9*A - 5*B + C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqr 
t[2]*Sqrt[a + a*Cos[c + d*x]])])/d)/a) + (2*a*(3*A - B + C)*Tan[c + d*x])/ 
(d*Sqrt[a + a*Cos[c + d*x]]))/(4*a^2)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 3042

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

rule 3128

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2/d 
Subst[Int[1/(2*a - x^2), x], x, b*(Cos[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], 
 x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
 

rule 3252

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)]), x_Symbol] :> Simp[-2*(b/f)   Subst[Int[1/(b*c + a*d - d*x^2), 
x], x, b*(Cos[e + f*x]/Sqrt[a + b*Sin[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 3464

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + 
(f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(A 
*b - a*B)/(b*c - a*d)   Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Simp[(B*c 
- A*d)/(b*c - a*d)   Int[Sqrt[a + b*Sin[e + f*x]]/(c + d*Sin[e + f*x]), x], 
 x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - 
 b^2, 0] && NeQ[c^2 - d^2, 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. \(991\) vs. \(2(148)=296\).

Time = 1.02 (sec) , antiderivative size = 992, normalized size of antiderivative = 5.73

method	result	size
parts	\(\text {Expression too large to display}\)	\(992\)
default	\(\text {Expression too large to display}\)	\(1193\)

Input:

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

Output:

1/2*A*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*(18*2^(1/2)*ln(2*(2*a^(1/2)*(sin(1/2* 
d*x+1/2*c)^2*a)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*cos(1/2*d*x+1/2*c)^4*a-12*l 
n(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/2)* 
2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-2*a))*cos(1/2*d*x+1/2*c)^4*a-12*ln( 
4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^( 
1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+2*a))*cos(1/2*d*x+1/2*c)^4*a-9*2^(1/2) 
*ln(2*(2*a^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*c 
os(1/2*d*x+1/2*c)^2*a+6*a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*cos 
(1/2*d*x+1/2*c)^2+6*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(a*2^(1/2)*cos(1/ 
2*d*x+1/2*c)-a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-2*a))*cos(1/2* 
d*x+1/2*c)^2*a+6*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d* 
x+1/2*c)+a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+2*a))*cos(1/2*d*x+ 
1/2*c)^2*a-a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2))/a^(5/2)/cos(1/2 
*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)-2^(1/2))/(2*cos(1/2*d*x+1/2*c)+2^(1/2))/ 
sin(1/2*d*x+1/2*c)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d+1/4*B/a^(5/2)/cos(1/2* 
d*x+1/2*c)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*(2*2^(1/2)*ln(-2/(2*cos(1/2*d*x+ 
1/2*c)-2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2)*(sin(1/2*d*x 
+1/2*c)^2*a)^(1/2)-2*a))*cos(1/2*d*x+1/2*c)^2*a+2*2^(1/2)*ln(2/(2*cos(1/2* 
d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(sin(1/2 
*d*x+1/2*c)^2*a)^(1/2)+2*a))*cos(1/2*d*x+1/2*c)^2*a-5*ln(2/cos(1/2*d*x+...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 343 vs. \(2 (148) = 296\).

Time = 0.14 (sec) , antiderivative size = 343, normalized size of antiderivative = 1.98 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {\sqrt {2} {\left ({\left (9 \, A - 5 \, B + C\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (9 \, A - 5 \, B + C\right )} \cos \left (d x + c\right )^{2} + {\left (9 \, A - 5 \, B + C\right )} \cos \left (d x + c\right )\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 2 \, {\left ({\left (3 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (3 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (3 \, A - 2 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left ({\left (3 \, A - B + C\right )} \cos \left (d x + c\right ) + 2 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{8 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\right )}} \] Input:

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

Output:

1/8*(sqrt(2)*((9*A - 5*B + C)*cos(d*x + c)^3 + 2*(9*A - 5*B + C)*cos(d*x + 
 c)^2 + (9*A - 5*B + C)*cos(d*x + c))*sqrt(a)*log(-(a*cos(d*x + c)^2 - 2*s 
qrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(a)*sin(d*x + c) - 2*a*cos(d*x + c) - 
3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) - 2*((3*A - 2*B)*cos(d*x + c)^ 
3 + 2*(3*A - 2*B)*cos(d*x + c)^2 + (3*A - 2*B)*cos(d*x + c))*sqrt(a)*log(( 
a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*sqrt(a*cos(d*x + c) + a)*sqrt(a) 
*(cos(d*x + c) - 2)*sin(d*x + c) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) 
 + 4*((3*A - B + C)*cos(d*x + c) + 2*A)*sqrt(a*cos(d*x + c) + a)*sin(d*x + 
 c))/(a^2*d*cos(d*x + c)^3 + 2*a^2*d*cos(d*x + c)^2 + a^2*d*cos(d*x + c))
 

Sympy [F]

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

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

Output:

Integral((A + B*cos(c + d*x) + C*cos(c + d*x)**2)*sec(c + d*x)**2/(a*(cos( 
c + d*x) + 1))**(3/2), x)
                                                                                    
                                                                                    
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 63653 vs. \(2 (148) = 296\).

Time = 2.23 (sec) , antiderivative size = 63653, normalized size of antiderivative = 367.94 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

1/4*((128*cos(3/2*d*x + 3/2*c)*sin(3*d*x + 3*c)^3 + 1152*cos(3/2*d*x + 3/2 
*c)*sin(2*d*x + 2*c)^3 - 128*cos(3*d*x + 3*c)^3*sin(3/2*d*x + 3/2*c) - 115 
2*cos(2*d*x + 2*c)^3*sin(3/2*d*x + 3/2*c) + 32*(4*cos(3/2*d*x + 3/2*c)*sin 
(2*d*x + 2*c) - 9*(3*cos(d*x + c) + 1)*sin(3/2*d*x + 3/2*c) - 28*cos(2*d*x 
 + 2*c)*sin(3/2*d*x + 3/2*c) + 3*cos(3/2*d*x + 3/2*c)*sin(d*x + c))*cos(3* 
d*x + 3*c)^2 - 96*(11*(3*cos(d*x + c) + 1)*sin(3/2*d*x + 3/2*c) - 9*cos(3/ 
2*d*x + 3/2*c)*sin(d*x + c))*cos(2*d*x + 2*c)^2 + 32*(28*cos(3/2*d*x + 3/2 
*c)*sin(2*d*x + 2*c) - (3*cos(d*x + c) + 1)*sin(3/2*d*x + 3/2*c) - 4*cos(3 
*d*x + 3*c)*sin(3/2*d*x + 3/2*c) - 4*cos(2*d*x + 2*c)*sin(3/2*d*x + 3/2*c) 
 + 27*cos(3/2*d*x + 3/2*c)*sin(d*x + c))*sin(3*d*x + 3*c)^2 - 288*((3*cos( 
d*x + c) + 1)*sin(3/2*d*x + 3/2*c) + 4*cos(2*d*x + 2*c)*sin(3/2*d*x + 3/2* 
c) - 11*cos(3/2*d*x + 3/2*c)*sin(d*x + c))*sin(2*d*x + 2*c)^2 - 32*(cos(3* 
d*x + 3*c)^2*sin(3/2*d*x + 3/2*c) + 6*(3*cos(d*x + c) + 1)*cos(2*d*x + 2*c 
)*sin(3/2*d*x + 3/2*c) + 9*cos(2*d*x + 2*c)^2*sin(3/2*d*x + 3/2*c) + sin(3 
*d*x + 3*c)^2*sin(3/2*d*x + 3/2*c) + 9*sin(2*d*x + 2*c)^2*sin(3/2*d*x + 3/ 
2*c) + 18*sin(2*d*x + 2*c)*sin(3/2*d*x + 3/2*c)*sin(d*x + c) + 2*((3*cos(d 
*x + c) + 1)*sin(3/2*d*x + 3/2*c) + 3*cos(2*d*x + 2*c)*sin(3/2*d*x + 3/2*c 
))*cos(3*d*x + 3*c) + 6*(sin(2*d*x + 2*c)*sin(3/2*d*x + 3/2*c) + sin(3/2*d 
*x + 3/2*c)*sin(d*x + c))*sin(3*d*x + 3*c) + (9*cos(d*x + c)^2 + 9*sin(d*x 
 + c)^2 + 6*cos(d*x + c) + 1)*sin(3/2*d*x + 3/2*c))*cos(5*d*x + 5*c) - ...
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(a)*(int((sqrt(cos(c + d*x) + 1)*cos(c + d*x)*sec(c + d*x)**2)/(cos(c 
 + d*x)**2 + 2*cos(c + d*x) + 1),x)*b + int((sqrt(cos(c + d*x) + 1)*cos(c 
+ d*x)**2*sec(c + d*x)**2)/(cos(c + d*x)**2 + 2*cos(c + d*x) + 1),x)*c + i 
nt((sqrt(cos(c + d*x) + 1)*sec(c + d*x)**2)/(cos(c + d*x)**2 + 2*cos(c + d 
*x) + 1),x)*a))/a**2