\(\int \cos (c+d x) (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx\) [92]

Optimal result

Integrand size = 31, antiderivative size = 175 \[ \int \cos (c+d x) (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx=\frac {64 a^3 (15 A+13 B) \sin (c+d x)}{315 d \sqrt {a+a \cos (c+d x)}}+\frac {16 a^2 (15 A+13 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{315 d}+\frac {2 a (15 A+13 B) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{105 d}+\frac {2 (9 A-2 B) (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{63 d}+\frac {2 B (a+a \cos (c+d x))^{7/2} \sin (c+d x)}{9 a d} \]

[Out]

2/105*a*(15*A+13*B)*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/d+2/63*(9*A-2*B)*(a+a*cos(d*x+c))^(5/2)*sin(d*x+c)/d+2/9
*B*(a+a*cos(d*x+c))^(7/2)*sin(d*x+c)/a/d+64/315*a^3*(15*A+13*B)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+16/315*a^2
*(15*A+13*B)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3047, 3102, 2830, 2726, 2725} \[ \int \cos (c+d x) (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx=\frac {64 a^3 (15 A+13 B) \sin (c+d x)}{315 d \sqrt {a \cos (c+d x)+a}}+\frac {16 a^2 (15 A+13 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{315 d}+\frac {2 (9 A-2 B) \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{63 d}+\frac {2 a (15 A+13 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{105 d}+\frac {2 B \sin (c+d x) (a \cos (c+d x)+a)^{7/2}}{9 a d} \]

[In]

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

[Out]

(64*a^3*(15*A + 13*B)*Sin[c + d*x])/(315*d*Sqrt[a + a*Cos[c + d*x]]) + (16*a^2*(15*A + 13*B)*Sqrt[a + a*Cos[c
+ d*x]]*Sin[c + d*x])/(315*d) + (2*a*(15*A + 13*B)*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(105*d) + (2*(9*A
- 2*B)*(a + a*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(63*d) + (2*B*(a + a*Cos[c + d*x])^(7/2)*Sin[c + d*x])/(9*a*d)

Rule 2725

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] && EqQ[a^2 - b^2, 0]

Rule 2726

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] + Dist[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 2830

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] + Dist[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*S
in[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)]

Rule 3047

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

Rule 3102

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[1/(
b*(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x],
x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \int (a+a \cos (c+d x))^{5/2} \left (A \cos (c+d x)+B \cos ^2(c+d x)\right ) \, dx \\ & = \frac {2 B (a+a \cos (c+d x))^{7/2} \sin (c+d x)}{9 a d}+\frac {2 \int (a+a \cos (c+d x))^{5/2} \left (\frac {7 a B}{2}+\frac {1}{2} a (9 A-2 B) \cos (c+d x)\right ) \, dx}{9 a} \\ & = \frac {2 (9 A-2 B) (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{63 d}+\frac {2 B (a+a \cos (c+d x))^{7/2} \sin (c+d x)}{9 a d}+\frac {1}{21} (15 A+13 B) \int (a+a \cos (c+d x))^{5/2} \, dx \\ & = \frac {2 a (15 A+13 B) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{105 d}+\frac {2 (9 A-2 B) (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{63 d}+\frac {2 B (a+a \cos (c+d x))^{7/2} \sin (c+d x)}{9 a d}+\frac {1}{105} (8 a (15 A+13 B)) \int (a+a \cos (c+d x))^{3/2} \, dx \\ & = \frac {16 a^2 (15 A+13 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{315 d}+\frac {2 a (15 A+13 B) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{105 d}+\frac {2 (9 A-2 B) (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{63 d}+\frac {2 B (a+a \cos (c+d x))^{7/2} \sin (c+d x)}{9 a d}+\frac {1}{315} \left (32 a^2 (15 A+13 B)\right ) \int \sqrt {a+a \cos (c+d x)} \, dx \\ & = \frac {64 a^3 (15 A+13 B) \sin (c+d x)}{315 d \sqrt {a+a \cos (c+d x)}}+\frac {16 a^2 (15 A+13 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{315 d}+\frac {2 a (15 A+13 B) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{105 d}+\frac {2 (9 A-2 B) (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{63 d}+\frac {2 B (a+a \cos (c+d x))^{7/2} \sin (c+d x)}{9 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.60 \[ \int \cos (c+d x) (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx=\frac {a^2 \sqrt {a (1+\cos (c+d x))} (6240 A+5653 B+(3030 A+3116 B) \cos (c+d x)+8 (90 A+127 B) \cos (2 (c+d x))+90 A \cos (3 (c+d x))+260 B \cos (3 (c+d x))+35 B \cos (4 (c+d x))) \tan \left (\frac {1}{2} (c+d x)\right )}{1260 d} \]

[In]

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

[Out]

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*(6240*A + 5653*B + (3030*A + 3116*B)*Cos[c + d*x] + 8*(90*A + 127*B)*Cos[2*(c
+ d*x)] + 90*A*Cos[3*(c + d*x)] + 260*B*Cos[3*(c + d*x)] + 35*B*Cos[4*(c + d*x)])*Tan[(c + d*x)/2])/(1260*d)

Maple [A] (verified)

Time = 4.78 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.70

method	result	size
default	\(\frac {8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{3} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (140 B \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-90 A -540 B \right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (315 A +819 B \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-420 A -630 B \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+315 A +315 B \right ) \sqrt {2}}{315 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\)	\(123\)
parts	\(\frac {8 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{3} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (6 \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+4 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8\right ) \sqrt {2}}{21 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}+\frac {8 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{3} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (140 \left (\cos ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-20 \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+39 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+52 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+104\right ) \sqrt {2}}{315 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\)	\(187\)

[In]

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

[Out]

8/315*cos(1/2*d*x+1/2*c)*a^3*sin(1/2*d*x+1/2*c)*(140*B*sin(1/2*d*x+1/2*c)^8+(-90*A-540*B)*sin(1/2*d*x+1/2*c)^6
+(315*A+819*B)*sin(1/2*d*x+1/2*c)^4+(-420*A-630*B)*sin(1/2*d*x+1/2*c)^2+315*A+315*B)*2^(1/2)/(a*cos(1/2*d*x+1/
2*c)^2)^(1/2)/d

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.66 \[ \int \cos (c+d x) (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx=\frac {2 \, {\left (35 \, B a^{2} \cos \left (d x + c\right )^{4} + 5 \, {\left (9 \, A + 26 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 3 \, {\left (60 \, A + 73 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (345 \, A + 292 \, B\right )} a^{2} \cos \left (d x + c\right ) + 2 \, {\left (345 \, A + 292 \, B\right )} a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

[In]

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

[Out]

2/315*(35*B*a^2*cos(d*x + c)^4 + 5*(9*A + 26*B)*a^2*cos(d*x + c)^3 + 3*(60*A + 73*B)*a^2*cos(d*x + c)^2 + (345
*A + 292*B)*a^2*cos(d*x + c) + 2*(345*A + 292*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. \[ \int \cos (c+d x) (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.98 \[ \int \cos (c+d x) (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx=\frac {30 \, {\left (3 \, \sqrt {2} a^{2} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 21 \, \sqrt {2} a^{2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 77 \, \sqrt {2} a^{2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 315 \, \sqrt {2} a^{2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} A \sqrt {a} + {\left (35 \, \sqrt {2} a^{2} \sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) + 225 \, \sqrt {2} a^{2} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 756 \, \sqrt {2} a^{2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 2100 \, \sqrt {2} a^{2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 8190 \, \sqrt {2} a^{2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} B \sqrt {a}}{2520 \, d} \]

[In]

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

[Out]

1/2520*(30*(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))*A*sqrt(a) + (35*sqrt(2)*a^2*sin(9/2*d*x + 9/2*c) + 225*sq
rt(2)*a^2*sin(7/2*d*x + 7/2*c) + 756*sqrt(2)*a^2*sin(5/2*d*x + 5/2*c) + 2100*sqrt(2)*a^2*sin(3/2*d*x + 3/2*c)
+ 8190*sqrt(2)*a^2*sin(1/2*d*x + 1/2*c))*B*sqrt(a))/d

Giac [A] (verification not implemented)

none

Time = 1.13 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.22 \[ \int \cos (c+d x) (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx=\frac {\sqrt {2} {\left (35 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) + 45 \, {\left (2 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 5 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 126 \, {\left (5 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 6 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 210 \, {\left (11 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 10 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 630 \, {\left (15 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 13 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{2520 \, d} \]

[In]

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

[Out]

1/2520*sqrt(2)*(35*B*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(9/2*d*x + 9/2*c) + 45*(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(7/2*d*x + 7/2*c) + 126*(5*A*a^2*sgn(cos(1/2*d*x + 1/2*c)) + 6*B*a^2
*sgn(cos(1/2*d*x + 1/2*c)))*sin(5/2*d*x + 5/2*c) + 210*(11*A*a^2*sgn(cos(1/2*d*x + 1/2*c)) + 10*B*a^2*sgn(cos(
1/2*d*x + 1/2*c)))*sin(3/2*d*x + 3/2*c) + 630*(15*A*a^2*sgn(cos(1/2*d*x + 1/2*c)) + 13*B*a^2*sgn(cos(1/2*d*x +
1/2*c)))*sin(1/2*d*x + 1/2*c))*sqrt(a)/d

Mupad [F(-1)]

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

[In]

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

[Out]

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