\(\int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx\) [93]

Optimal result

Integrand size = 25, antiderivative size = 138 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx=\frac {64 a^3 (7 A+5 B) \sin (c+d x)}{105 d \sqrt {a+a \cos (c+d x)}}+\frac {16 a^2 (7 A+5 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{105 d}+\frac {2 a (7 A+5 B) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac {2 B (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{7 d} \] Output:

64/105*a^3*(7*A+5*B)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+16/105*a^2*(7*A+5 
*B)*(a+a*cos(d*x+c))^(1/2)*sin(d*x+c)/d+2/35*a*(7*A+5*B)*(a+a*cos(d*x+c))^ 
(3/2)*sin(d*x+c)/d+2/7*B*(a+a*cos(d*x+c))^(5/2)*sin(d*x+c)/d
 

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.60 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx=\frac {a^2 \sqrt {a (1+\cos (c+d x))} (1246 A+1040 B+(392 A+505 B) \cos (c+d x)+6 (7 A+20 B) \cos (2 (c+d x))+15 B \cos (3 (c+d x))) \tan \left (\frac {1}{2} (c+d x)\right )}{210 d} \] Input:

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

Output:

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*(1246*A + 1040*B + (392*A + 505*B)*Cos[c + 
 d*x] + 6*(7*A + 20*B)*Cos[2*(c + d*x)] + 15*B*Cos[3*(c + d*x)])*Tan[(c + 
d*x)/2])/(210*d)
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.96, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3230, 3042, 3126, 3042, 3126, 3042, 3125}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3230

\(\displaystyle \frac {1}{7} (7 A+5 B) \int (\cos (c+d x) a+a)^{5/2}dx+\frac {2 B \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} (7 A+5 B) \int \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2}dx+\frac {2 B \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 d}\)

\(\Big \downarrow \) 3126

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3126

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3125

\(\displaystyle \frac {1}{7} (7 A+5 B) \left (\frac {8}{5} a \left (\frac {8 a^2 \sin (c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {2 a \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\right )+\frac {2 B \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 d}\)

Input:

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

Output:

(2*B*(a + a*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(7*d) + ((7*A + 5*B)*((2*a*( 
a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*d) + (8*a*((8*a^2*Sin[c + d*x]) 
/(3*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d* 
x])/(3*d)))/5))/7
 

Defintions of rubi rules used

rule 3042

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

rule 3125

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

rule 3126

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

rule 3230

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[(-d)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/( 
f*(m + 1))), x] + Simp[(a*d*m + b*c*(m + 1))/(b*(m + 1))   Int[(a + b*Sin[e 
 + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] 
&& EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)]
 

Maple [A] (verified)

Time = 3.32 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.75

method	result	size
default	\(\frac {8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{3} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (-30 B \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (21 A +105 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-70 A -140 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+105 A +105 B \right ) \sqrt {2}}{105 \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\)	\(104\)
parts	\(\frac {8 A \,a^{3} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (3 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+8\right ) \sqrt {2}}{15 \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}+\frac {8 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{3} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (6 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+3 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+8\right ) \sqrt {2}}{21 \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\)	\(161\)

Input:

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

Output:

8/105*cos(1/2*d*x+1/2*c)*a^3*sin(1/2*d*x+1/2*c)*(-30*B*sin(1/2*d*x+1/2*c)^ 
6+(21*A+105*B)*sin(1/2*d*x+1/2*c)^4+(-70*A-140*B)*sin(1/2*d*x+1/2*c)^2+105 
*A+105*B)*2^(1/2)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.69 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx=\frac {2 \, {\left (15 \, B a^{2} \cos \left (d x + c\right )^{3} + 3 \, {\left (7 \, A + 20 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (98 \, A + 115 \, B\right )} a^{2} \cos \left (d x + c\right ) + {\left (301 \, A + 230 \, B\right )} a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \] Input:

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

Output:

2/105*(15*B*a^2*cos(d*x + c)^3 + 3*(7*A + 20*B)*a^2*cos(d*x + c)^2 + (98*A 
 + 115*B)*a^2*cos(d*x + c) + (301*A + 230*B)*a^2)*sqrt(a*cos(d*x + c) + a) 
*sin(d*x + c)/(d*cos(d*x + c) + d)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.01 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx=\frac {14 \, {\left (3 \, \sqrt {2} a^{2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 25 \, \sqrt {2} a^{2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 150 \, \sqrt {2} a^{2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} A \sqrt {a} + 5 \, {\left (3 \, \sqrt {2} a^{2} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 21 \, \sqrt {2} a^{2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 77 \, \sqrt {2} a^{2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 315 \, \sqrt {2} a^{2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} B \sqrt {a}}{420 \, d} \] Input:

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

Output:

1/420*(14*(3*sqrt(2)*a^2*sin(5/2*d*x + 5/2*c) + 25*sqrt(2)*a^2*sin(3/2*d*x 
 + 3/2*c) + 150*sqrt(2)*a^2*sin(1/2*d*x + 1/2*c))*A*sqrt(a) + 5*(3*sqrt(2) 
*a^2*sin(7/2*d*x + 7/2*c) + 21*sqrt(2)*a^2*sin(5/2*d*x + 5/2*c) + 77*sqrt( 
2)*a^2*sin(3/2*d*x + 3/2*c) + 315*sqrt(2)*a^2*sin(1/2*d*x + 1/2*c))*B*sqrt 
(a))/d
 

Giac [A] (verification not implemented)

Time = 0.58 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.22 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx=\frac {\sqrt {2} {\left (15 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 21 \, {\left (2 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 5 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 35 \, {\left (10 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 11 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 525 \, {\left (4 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 3 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{420 \, d} \] Input:

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

Output:

1/420*sqrt(2)*(15*B*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(7/2*d*x + 7/2*c) + 2 
1*(2*A*a^2*sgn(cos(1/2*d*x + 1/2*c)) + 5*B*a^2*sgn(cos(1/2*d*x + 1/2*c)))* 
sin(5/2*d*x + 5/2*c) + 35*(10*A*a^2*sgn(cos(1/2*d*x + 1/2*c)) + 11*B*a^2*s 
gn(cos(1/2*d*x + 1/2*c)))*sin(3/2*d*x + 3/2*c) + 525*(4*A*a^2*sgn(cos(1/2* 
d*x + 1/2*c)) + 3*B*a^2*sgn(cos(1/2*d*x + 1/2*c)))*sin(1/2*d*x + 1/2*c))*s 
qrt(a)/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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