\(\int \frac {(A+C \cos ^2(c+d x)) \sec ^4(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx\) [109]

Optimal result

Integrand size = 35, antiderivative size = 200 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=-\frac {(9 A+8 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{8 \sqrt {a} d}+\frac {\sqrt {2} (A+C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{\sqrt {a} d}+\frac {(7 A+8 C) \tan (c+d x)}{8 d \sqrt {a+a \cos (c+d x)}}-\frac {A \sec (c+d x) \tan (c+d x)}{12 d \sqrt {a+a \cos (c+d x)}}+\frac {A \sec ^2(c+d x) \tan (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}} \] Output:

-1/8*(9*A+8*C)*arctanh(a^(1/2)*sin(d*x+c)/(a+a*cos(d*x+c))^(1/2))/a^(1/2)/ 
d+2^(1/2)*(A+C)*arctanh(1/2*a^(1/2)*sin(d*x+c)*2^(1/2)/(a+a*cos(d*x+c))^(1 
/2))/a^(1/2)/d+1/8*(7*A+8*C)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)-1/12*A*se 
c(d*x+c)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/3*A*sec(d*x+c)^2*tan(d*x+c) 
/d/(a+a*cos(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 1.59 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.66 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \left (48 (A+C) \text {arctanh}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-3 \sqrt {2} (9 A+8 C) \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )+(37 A+24 C-4 A \cos (c+d x)+3 (7 A+8 C) \cos (2 (c+d x))) \sec ^3(c+d x) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{24 d \sqrt {a (1+\cos (c+d x))}} \] Input:

Integrate[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^4)/Sqrt[a + a*Cos[c + d*x]] 
,x]
 

Output:

(Cos[(c + d*x)/2]*(48*(A + C)*ArcTanh[Sin[(c + d*x)/2]] - 3*Sqrt[2]*(9*A + 
 8*C)*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]] + (37*A + 24*C - 4*A*Cos[c + d*x] 
+ 3*(7*A + 8*C)*Cos[2*(c + d*x)])*Sec[c + d*x]^3*Sin[(c + d*x)/2]))/(24*d* 
Sqrt[a*(1 + Cos[c + d*x])])
 

Rubi [A] (verified)

Time = 1.42 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.12, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.457, Rules used = {3042, 3523, 27, 3042, 3463, 27, 3042, 3463, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3523

\(\displaystyle \frac {\int -\frac {(a A-a (5 A+6 C) \cos (c+d x)) \sec ^3(c+d x)}{2 \sqrt {\cos (c+d x) a+a}}dx}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {(a A-a (5 A+6 C) \cos (c+d x)) \sec ^3(c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{6 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {a A-a (5 A+6 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{6 a}\)

\(\Big \downarrow \) 3463

\(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {\int -\frac {3 \left (a^2 (7 A+8 C)-a^2 A \cos (c+d x)\right ) \sec ^2(c+d x)}{2 \sqrt {\cos (c+d x) a+a}}dx}{2 a}+\frac {a A \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}}{6 a}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a A \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {3 \int \frac {\left (a^2 (7 A+8 C)-a^2 A \cos (c+d x)\right ) \sec ^2(c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{4 a}}{6 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a A \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {3 \int \frac {a^2 (7 A+8 C)-a^2 A \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a}}{6 a}\)

\(\Big \downarrow \) 3463

\(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a A \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {\int -\frac {\left (a^3 (9 A+8 C)-a^3 (7 A+8 C) \cos (c+d x)\right ) \sec (c+d x)}{2 \sqrt {\cos (c+d x) a+a}}dx}{a}+\frac {a^2 (7 A+8 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )}{4 a}}{6 a}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a A \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {a^2 (7 A+8 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {\left (a^3 (9 A+8 C)-a^3 (7 A+8 C) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{2 a}\right )}{4 a}}{6 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a A \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {a^2 (7 A+8 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {a^3 (9 A+8 C)-a^3 (7 A+8 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}\right )}{4 a}}{6 a}\)

\(\Big \downarrow \) 3464

\(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a A \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {a^2 (7 A+8 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {a^2 (9 A+8 C) \int \sqrt {\cos (c+d x) a+a} \sec (c+d x)dx-16 a^3 (A+C) \int \frac {1}{\sqrt {\cos (c+d x) a+a}}dx}{2 a}\right )}{4 a}}{6 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a A \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {a^2 (7 A+8 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {a^2 (9 A+8 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-16 a^3 (A+C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}\right )}{4 a}}{6 a}\)

\(\Big \downarrow \) 3128

\(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a A \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {a^2 (7 A+8 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {a^2 (9 A+8 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {32 a^3 (A+C) \int \frac {1}{2 a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}}{2 a}\right )}{4 a}}{6 a}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a A \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {a^2 (7 A+8 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {a^2 (9 A+8 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {16 \sqrt {2} a^{5/2} (A+C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{2 a}\right )}{4 a}}{6 a}\)

\(\Big \downarrow \) 3252

\(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a A \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {a^2 (7 A+8 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {-\frac {2 a^3 (9 A+8 C) \int \frac {1}{a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {16 \sqrt {2} a^{5/2} (A+C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{2 a}\right )}{4 a}}{6 a}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a A \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {a^2 (7 A+8 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^{5/2} (9 A+8 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {16 \sqrt {2} a^{5/2} (A+C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{2 a}\right )}{4 a}}{6 a}\)

Input:

Int[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^4)/Sqrt[a + a*Cos[c + d*x]],x]
 

Output:

(A*Sec[c + d*x]^2*Tan[c + d*x])/(3*d*Sqrt[a + a*Cos[c + d*x]]) - ((a*A*Sec 
[c + d*x]*Tan[c + d*x])/(2*d*Sqrt[a + a*Cos[c + d*x]]) - (3*(-1/2*((2*a^(5 
/2)*(9*A + 8*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/ 
d - (16*Sqrt[2]*a^(5/2)*(A + C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sq 
rt[a + a*Cos[c + d*x]])])/d)/a + (a^2*(7*A + 8*C)*Tan[c + d*x])/(d*Sqrt[a 
+ a*Cos[c + d*x]])))/(4*a))/(6*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 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 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 [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(1357\) vs. \(2(171)=342\).

Time = 0.49 (sec) , antiderivative size = 1358, normalized size of antiderivative = 6.79

method	result	size
parts	\(\text {Expression too large to display}\)	\(1358\)
default	\(\text {Expression too large to display}\)	\(1661\)

Input:

int((A+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+a*cos(d*x+c))^(1/2),x,method=_RETUR 
NVERBOSE)
 

Output:

1/6*A*cos(1/2*d*x+1/2*c)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*(-24*a*(16*2^(1/2) 
*ln(4/cos(1/2*d*x+1/2*c)*(a^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+a))-9*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))-9*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)))*sin(1/2*d*x+1/2*c)^6+(576*2^(1/2)*ln(4/cos(1/2*d*x+1/2*c) 
*(a^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+a))*a+168*a^(1/2)*2^(1/2)*(sin(1/ 
2*d*x+1/2*c)^2*a)^(1/2)-324*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))* 
a-324*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))*a)*sin(1/2*d*x+1/2*c)^4 
+(-288*2^(1/2)*ln(4/cos(1/2*d*x+1/2*c)*(a^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^( 
1/2)+a))*a-160*a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+162*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))*a+162*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))*a)*sin(1/2*d*x+1/2*c)^2+48*2^(1/2)*ln(4/cos(1/2*d*x+1/2*c)* 
(a^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+a))*a+54*a^(1/2)*2^(1/2)*(sin(1/2* 
d*x+1/2*c)^2*a)^(1/2)-27*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(a*2^(1/2)*c 
os(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))*...
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.48 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {3 \, {\left ({\left (9 \, A + 8 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (9 \, A + 8 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} + 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left (3 \, {\left (7 \, A + 8 \, C\right )} \cos \left (d x + c\right )^{2} - 2 \, A \cos \left (d x + c\right ) + 8 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right ) + \frac {48 \, \sqrt {2} {\left ({\left (A + C\right )} a \cos \left (d x + c\right )^{4} + {\left (A + C\right )} a \cos \left (d x + c\right )^{3}\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} - \frac {2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{\sqrt {a}}}{96 \, {\left (a d \cos \left (d x + c\right )^{4} + a d \cos \left (d x + c\right )^{3}\right )}} \] Input:

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+a*cos(d*x+c))^(1/2),x, algori 
thm="fricas")
 

Output:

1/96*(3*((9*A + 8*C)*cos(d*x + c)^4 + (9*A + 8*C)*cos(d*x + c)^3)*sqrt(a)* 
log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 + 4*sqrt(a*cos(d*x + c) + a)*sq 
rt(a)*(cos(d*x + c) - 2)*sin(d*x + c) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c 
)^2)) + 4*(3*(7*A + 8*C)*cos(d*x + c)^2 - 2*A*cos(d*x + c) + 8*A)*sqrt(a*c 
os(d*x + c) + a)*sin(d*x + c) + 48*sqrt(2)*((A + C)*a*cos(d*x + c)^4 + (A 
+ C)*a*cos(d*x + c)^3)*log(-(cos(d*x + c)^2 - 2*sqrt(2)*sqrt(a*cos(d*x + c 
) + a)*sin(d*x + c)/sqrt(a) - 2*cos(d*x + c) - 3)/(cos(d*x + c)^2 + 2*cos( 
d*x + c) + 1))/sqrt(a))/(a*d*cos(d*x + c)^4 + a*d*cos(d*x + c)^3)
 

Sympy [F]

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

integrate((A+C*cos(d*x+c)**2)*sec(d*x+c)**4/(a+a*cos(d*x+c))**(1/2),x)
 

Output:

Integral((A + C*cos(c + d*x)**2)*sec(c + d*x)**4/sqrt(a*(cos(c + d*x) + 1) 
), x)
 

Maxima [F(-1)]

Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\text {Timed out} \] Input:

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+a*cos(d*x+c))^(1/2),x, algori 
thm="maxima")
 

Output:

Timed out
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+a*cos(d*x+c))^(1/2),x, algori 
thm="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. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+A}{{\cos \left (c+d\,x\right )}^4\,\sqrt {a+a\,\cos \left (c+d\,x\right )}} \,d x \] Input:

int((A + C*cos(c + d*x)^2)/(cos(c + d*x)^4*(a + a*cos(c + d*x))^(1/2)),x)
 

Output:

int((A + C*cos(c + d*x)^2)/(cos(c + d*x)^4*(a + a*cos(c + d*x))^(1/2)), x)
 

Reduce [F]

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

int((A+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+a*cos(d*x+c))^(1/2),x)
 

Output:

(sqrt(a)*(int((sqrt(cos(c + d*x) + 1)*cos(c + d*x)**2*sec(c + d*x)**4)/(co 
s(c + d*x) + 1),x)*c + int((sqrt(cos(c + d*x) + 1)*sec(c + d*x)**4)/(cos(c 
 + d*x) + 1),x)*a))/a