\(\int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx\) [96]

Optimal result

Integrand size = 33, antiderivative size = 156 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {a^{5/2} (19 A+20 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{4 d}-\frac {a^3 (9 A-4 B) \sin (c+d x)}{4 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (7 A+4 B) \sqrt {a+a \cos (c+d x)} \tan (c+d x)}{4 d}+\frac {a A (a+a \cos (c+d x))^{3/2} \sec (c+d x) \tan (c+d x)}{2 d} \] Output:

1/4*a^(5/2)*(19*A+20*B)*arctanh(a^(1/2)*sin(d*x+c)/(a+a*cos(d*x+c))^(1/2)) 
/d-1/4*a^3*(9*A-4*B)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/4*a^2*(7*A+4*B) 
*(a+a*cos(d*x+c))^(1/2)*tan(d*x+c)/d+1/2*a*A*(a+a*cos(d*x+c))^(3/2)*sec(d* 
x+c)*tan(d*x+c)/d
 

Mathematica [A] (verified)

Time = 0.67 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.81 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {a^2 \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \left (\sqrt {2} (19 A+20 B) \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^2(c+d x)+2 ((11 A+4 B) \cos (c+d x)+2 (A+2 B+2 B \cos (2 (c+d x)))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{8 d} \] Input:

Integrate[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x])*Sec[c + d*x]^3,x 
]
 

Output:

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sec[c + d*x]^2*(Sqrt[2]*( 
19*A + 20*B)*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x]^2 + 2*((11*A + 
 4*B)*Cos[c + d*x] + 2*(A + 2*B + 2*B*Cos[2*(c + d*x)]))*Sin[(c + d*x)/2]) 
)/(8*d)
 

Rubi [A] (verified)

Time = 0.94 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.02, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3454, 27, 3042, 3454, 27, 3042, 3460, 3042, 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.

\(\displaystyle \int \sec ^3(c+d x) (a \cos (c+d x)+a)^{5/2} (A+B \cos (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx\)

\(\Big \downarrow \) 3454

\(\displaystyle \frac {1}{2} \int \frac {1}{2} (\cos (c+d x) a+a)^{3/2} (a (7 A+4 B)-a (A-4 B) \cos (c+d x)) \sec ^2(c+d x)dx+\frac {a A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^{3/2}}{2 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} \int (\cos (c+d x) a+a)^{3/2} (a (7 A+4 B)-a (A-4 B) \cos (c+d x)) \sec ^2(c+d x)dx+\frac {a A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^{3/2}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (a (7 A+4 B)-a (A-4 B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^{3/2}}{2 d}\)

\(\Big \downarrow \) 3454

\(\displaystyle \frac {1}{4} \left (\int \frac {1}{2} \sqrt {\cos (c+d x) a+a} \left (a^2 (19 A+20 B)-a^2 (9 A-4 B) \cos (c+d x)\right ) \sec (c+d x)dx+\frac {a^2 (7 A+4 B) \tan (c+d x) \sqrt {a \cos (c+d x)+a}}{d}\right )+\frac {a A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^{3/2}}{2 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \int \sqrt {\cos (c+d x) a+a} \left (a^2 (19 A+20 B)-a^2 (9 A-4 B) \cos (c+d x)\right ) \sec (c+d x)dx+\frac {a^2 (7 A+4 B) \tan (c+d x) \sqrt {a \cos (c+d x)+a}}{d}\right )+\frac {a A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^{3/2}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (a^2 (19 A+20 B)-a^2 (9 A-4 B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a^2 (7 A+4 B) \tan (c+d x) \sqrt {a \cos (c+d x)+a}}{d}\right )+\frac {a A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^{3/2}}{2 d}\)

\(\Big \downarrow \) 3460

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (a^2 (19 A+20 B) \int \sqrt {\cos (c+d x) a+a} \sec (c+d x)dx-\frac {2 a^3 (9 A-4 B) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 (7 A+4 B) \tan (c+d x) \sqrt {a \cos (c+d x)+a}}{d}\right )+\frac {a A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^{3/2}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (a^2 (19 A+20 B) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {2 a^3 (9 A-4 B) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 (7 A+4 B) \tan (c+d x) \sqrt {a \cos (c+d x)+a}}{d}\right )+\frac {a A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^{3/2}}{2 d}\)

\(\Big \downarrow \) 3252

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (-\frac {2 a^3 (19 A+20 B) \int \frac {1}{a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {2 a^3 (9 A-4 B) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 (7 A+4 B) \tan (c+d x) \sqrt {a \cos (c+d x)+a}}{d}\right )+\frac {a A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^{3/2}}{2 d}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{4} \left (\frac {a^2 (7 A+4 B) \tan (c+d x) \sqrt {a \cos (c+d x)+a}}{d}+\frac {1}{2} \left (\frac {2 a^{5/2} (19 A+20 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {2 a^3 (9 A-4 B) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )\right )+\frac {a A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^{3/2}}{2 d}\)

Input:

Int[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x])*Sec[c + d*x]^3,x]
 

Output:

(a*A*(a + a*Cos[c + d*x])^(3/2)*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (((2*a^ 
(5/2)*(19*A + 20*B)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x] 
]])/d - (2*a^3*(9*A - 4*B)*Sin[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]]))/2 + 
 (a^2*(7*A + 4*B)*Sqrt[a + a*Cos[c + d*x]]*Tan[c + d*x])/d)/4
 

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 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 3454

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^2)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[ 
e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a*d))), x] - Simp[b/(d*(n + 1)*(b*c + 
 a*d))   Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp 
[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n + 1) - B 
*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], 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] 
&& GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0 
])
 

rule 3460

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + ( 
f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp 
[-2*b*B*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(2*n + 3)*Sqrt[a + 
b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b 
*d*(2*n + 3))   Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, 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[n, -1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(994\) vs. \(2(136)=272\).

Time = 59.41 (sec) , antiderivative size = 995, normalized size of antiderivative = 6.38

method	result	size
parts	\(\text {Expression too large to display}\)	\(995\)
default	\(\text {Expression too large to display}\)	\(1028\)

Input:

int((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^3,x,method=_RETURNV 
ERBOSE)
 

Output:

1/2*A*a^(3/2)*cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(76*a*(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)-2^(1/2)*(a 
*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))+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)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)* 
a^(1/2)+2*a)))*sin(1/2*d*x+1/2*c)^4+(-44*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^ 
(1/2)*a^(1/2)-76*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)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))*a-76*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)+2^(1/2)*(a*sin 
(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*a))*a)*sin(1/2*d*x+1/2*c)^2+26*2^(1/2)* 
(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+19*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)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)* 
a^(1/2)-2*a))*a+19*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)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*a))*a)/(2*cos( 
1/2*d*x+1/2*c)-2^(1/2))^2/(2*cos(1/2*d*x+1/2*c)+2^(1/2))^2/sin(1/2*d*x+1/2 
*c)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d+1/2*B*a^(3/2)*cos(1/2*d*x+1/2*c)*(a*s 
in(1/2*d*x+1/2*c)^2)^(1/2)*(-10*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)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^( 
1/2)+2*a))*sin(1/2*d*x+1/2*c)^2*a-10*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)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/ 
2)*a^(1/2)-2*a))*sin(1/2*d*x+1/2*c)^2*a-16*a^(1/2)*(a*sin(1/2*d*x+1/2*c...
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.31 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {{\left ({\left (19 \, A + 20 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + {\left (19 \, A + 20 \, B\right )} a^{2} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left (8 \, B a^{2} \cos \left (d x + c\right )^{2} + {\left (11 \, A + 4 \, B\right )} a^{2} \cos \left (d x + c\right ) + 2 \, A a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{16 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}} \] Input:

integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^3,x, algorith 
m="fricas")
 

Output:

1/16*(((19*A + 20*B)*a^2*cos(d*x + c)^3 + (19*A + 20*B)*a^2*cos(d*x + c)^2 
)*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 + c 
os(d*x + c)^2)) + 4*(8*B*a^2*cos(d*x + c)^2 + (11*A + 4*B)*a^2*cos(d*x + c 
) + 2*A*a^2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(d*cos(d*x + c)^3 + d* 
cos(d*x + c)^2)
 

Sympy [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\text {Timed out} \] Input:

integrate((a+a*cos(d*x+c))**(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)**3,x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 11782 vs. \(2 (136) = 272\).

Time = 3.31 (sec) , antiderivative size = 11782, normalized size of antiderivative = 75.53 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\text {Too large to display} \] Input:

integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^3,x, algorith 
m="maxima")
                                                                                    
                                                                                    
 

Output:

-1/1008*(63*(150*sqrt(2)*a^2*cos(7/2*d*x + 7/2*c)*sin(2*d*x + 2*c) + 154*s 
qrt(2)*a^2*cos(5/2*d*x + 5/2*c)*sin(2*d*x + 2*c) - 28*sqrt(2)*a^2*sin(3/2* 
d*x + 3/2*c) + 44*sqrt(2)*a^2*sin(1/2*d*x + 1/2*c) - (3*sqrt(2)*a^2*sin(7/ 
2*d*x + 7/2*c) + 5*sqrt(2)*a^2*sin(5/2*d*x + 5/2*c) - 17*sqrt(2)*a^2*sin(3 
/2*d*x + 3/2*c) - 55*sqrt(2)*a^2*sin(1/2*d*x + 1/2*c) + 19*a^2*log(2*cos(1 
/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2 
*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - 19*a^2*log(2*cos(1/2*d*x + 1/2 
*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt 
(2)*sin(1/2*d*x + 1/2*c) + 2) + 19*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*si 
n(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2* 
d*x + 1/2*c) + 2) - 19*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 
1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) 
 + 2))*cos(4*d*x + 4*c)^2 + 4*(17*sqrt(2)*a^2*sin(3/2*d*x + 3/2*c) + 55*sq 
rt(2)*a^2*sin(1/2*d*x + 1/2*c) - 19*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*s 
in(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2 
*d*x + 1/2*c) + 2) + 19*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 
 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c 
) + 2) - 19*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 
2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + 19* 
a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)...
 

Giac [A] (verification not implemented)

Time = 0.69 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.53 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {\sqrt {2} {\left (32 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {2} {\left (19 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 20 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right ) - \frac {4 \, {\left (22 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 8 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 13 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}\right )} \sqrt {a}}{16 \, d} \] Input:

integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^3,x, algorith 
m="giac")
 

Output:

1/16*sqrt(2)*(32*B*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c) - sq 
rt(2)*(19*A*a^2*sgn(cos(1/2*d*x + 1/2*c)) + 20*B*a^2*sgn(cos(1/2*d*x + 1/2 
*c)))*log(abs(-2*sqrt(2) + 4*sin(1/2*d*x + 1/2*c))/abs(2*sqrt(2) + 4*sin(1 
/2*d*x + 1/2*c))) - 4*(22*A*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/ 
2*c)^3 + 8*B*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^3 - 13*A*a 
^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c) - 4*B*a^2*sgn(cos(1/2*d* 
x + 1/2*c))*sin(1/2*d*x + 1/2*c))/(2*sin(1/2*d*x + 1/2*c)^2 - 1)^2)*sqrt(a 
)/d
 

Mupad [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\int \frac {\left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^3} \,d x \] Input:

int(((A + B*cos(c + d*x))*(a + a*cos(c + d*x))^(5/2))/cos(c + d*x)^3,x)
 

Output:

int(((A + B*cos(c + d*x))*(a + a*cos(c + d*x))^(5/2))/cos(c + d*x)^3, x)
 

Reduce [F]

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

int((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^3,x)
 

Output:

sqrt(a)*a**2*(2*int(sqrt(cos(c + d*x) + 1)*cos(c + d*x)*sec(c + d*x)**3,x) 
*a + int(sqrt(cos(c + d*x) + 1)*cos(c + d*x)*sec(c + d*x)**3,x)*b + int(sq 
rt(cos(c + d*x) + 1)*cos(c + d*x)**3*sec(c + d*x)**3,x)*b + int(sqrt(cos(c 
 + d*x) + 1)*cos(c + d*x)**2*sec(c + d*x)**3,x)*a + 2*int(sqrt(cos(c + d*x 
) + 1)*cos(c + d*x)**2*sec(c + d*x)**3,x)*b + int(sqrt(cos(c + d*x) + 1)*s 
ec(c + d*x)**3,x)*a)