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

Optimal result

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

[Out]

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+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)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2830, 2726, 2725} \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx=\frac {64 a^3 (7 A+5 B) \sin (c+d x)}{105 d \sqrt {a \cos (c+d x)+a}}+\frac {16 a^2 (7 A+5 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{105 d}+\frac {2 a (7 A+5 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{35 d}+\frac {2 B \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 d} \]

[In]

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

[Out]

(64*a^3*(7*A + 5*B)*Sin[c + d*x])/(105*d*Sqrt[a + a*Cos[c + d*x]]) + (16*a^2*(7*A + 5*B)*Sqrt[a + a*Cos[c + d*
x]]*Sin[c + d*x])/(105*d) + (2*a*(7*A + 5*B)*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(35*d) + (2*B*(a + a*Cos
[c + d*x])^(5/2)*Sin[c + d*x])/(7*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)]

Rubi steps \begin{align*} \text {integral}& = \frac {2 B (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac {1}{7} (7 A+5 B) \int (a+a \cos (c+d x))^{5/2} \, dx \\ & = \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}+\frac {1}{35} (8 a (7 A+5 B)) \int (a+a \cos (c+d x))^{3/2} \, dx \\ & = \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}+\frac {1}{105} \left (32 a^2 (7 A+5 B)\right ) \int \sqrt {a+a \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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.60 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx=\frac {a^2 \sqrt {a (1+\cos (c+d x))} (1246 A+1040 B+(392 A+505 B) \cos (c+d x)+6 (7 A+20 B) \cos (2 (c+d x))+15 B \cos (3 (c+d x))) \tan \left (\frac {1}{2} (c+d x)\right )}{210 d} \]

[In]

Integrate[(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])]*(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)

Maple [A] (verified)

Time = 3.23 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.75

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

[In]

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

[Out]

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)

none

Time = 0.30 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.69 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx=\frac {2 \, {\left (15 \, B a^{2} \cos \left (d x + c\right )^{3} + 3 \, {\left (7 \, A + 20 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (98 \, A + 115 \, B\right )} a^{2} \cos \left (d x + c\right ) + {\left (301 \, A + 230 \, B\right )} a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

[In]

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

[Out]

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

[In]

integrate((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 = 0.36 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.01 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx=\frac {14 \, {\left (3 \, \sqrt {2} a^{2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 25 \, \sqrt {2} a^{2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 150 \, \sqrt {2} a^{2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} A \sqrt {a} + 5 \, {\left (3 \, \sqrt {2} a^{2} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 21 \, \sqrt {2} a^{2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 77 \, \sqrt {2} a^{2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 315 \, \sqrt {2} a^{2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} B \sqrt {a}}{420 \, d} \]

[In]

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

[Out]

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*sqr
t(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)

none

Time = 0.57 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.22 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx=\frac {\sqrt {2} {\left (15 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 21 \, {\left (2 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 5 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 35 \, {\left (10 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 11 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 525 \, {\left (4 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 3 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{420 \, d} \]

[In]

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

[Out]

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) + 21*(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
*sgn(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))*sqrt(a)/d

Mupad [F(-1)]

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

[In]

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

[Out]

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