\(\int \cos (c+d x) \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \, dx\) [76]

Optimal result

Integrand size = 31, antiderivative size = 101 \[ \int \cos (c+d x) \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {2 a (5 A+7 B) \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}+\frac {2 (5 A-2 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 B (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 a d} \]

[Out]

2/5*B*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/a/d+2/15*a*(5*A+7*B)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/15*(5*A-2*B
)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {3047, 3102, 2830, 2725} \[ \int \cos (c+d x) \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {2 (5 A-2 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{15 d}+\frac {2 a (5 A+7 B) \sin (c+d x)}{15 d \sqrt {a \cos (c+d x)+a}}+\frac {2 B \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 a d} \]

[In]

Int[Cos[c + d*x]*Sqrt[a + a*Cos[c + d*x]]*(A + B*Cos[c + d*x]),x]

[Out]

(2*a*(5*A + 7*B)*Sin[c + d*x])/(15*d*Sqrt[a + a*Cos[c + d*x]]) + (2*(5*A - 2*B)*Sqrt[a + a*Cos[c + d*x]]*Sin[c
+ d*x])/(15*d) + (2*B*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*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 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 \sqrt {a+a \cos (c+d x)} \left (A \cos (c+d x)+B \cos ^2(c+d x)\right ) \, dx \\ & = \frac {2 B (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 a d}+\frac {2 \int \sqrt {a+a \cos (c+d x)} \left (\frac {3 a B}{2}+\frac {1}{2} a (5 A-2 B) \cos (c+d x)\right ) \, dx}{5 a} \\ & = \frac {2 (5 A-2 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 B (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 a d}+\frac {1}{15} (5 A+7 B) \int \sqrt {a+a \cos (c+d x)} \, dx \\ & = \frac {2 a (5 A+7 B) \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}+\frac {2 (5 A-2 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 B (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.63 \[ \int \cos (c+d x) \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {\sqrt {a (1+\cos (c+d x))} (20 A+19 B+2 (5 A+4 B) \cos (c+d x)+3 B \cos (2 (c+d x))) \tan \left (\frac {1}{2} (c+d x)\right )}{15 d} \]

[In]

Integrate[Cos[c + d*x]*Sqrt[a + a*Cos[c + d*x]]*(A + B*Cos[c + d*x]),x]

[Out]

(Sqrt[a*(1 + Cos[c + d*x])]*(20*A + 19*B + 2*(5*A + 4*B)*Cos[c + d*x] + 3*B*Cos[2*(c + d*x)])*Tan[(c + d*x)/2]
)/(15*d)

Maple [A] (verified)

Time = 2.71 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.82

method	result	size
default	\(\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (12 B \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-10 A -20 B \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+15 A +15 B \right ) \sqrt {2}}{15 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\)	\(83\)
parts	\(\frac {2 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (1+2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right ) \sqrt {2}}{3 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}+\frac {2 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (12 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+7\right ) \sqrt {2}}{15 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\)	\(131\)

[In]

int(cos(d*x+c)*(a+cos(d*x+c)*a)^(1/2)*(A+B*cos(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

2/15*cos(1/2*d*x+1/2*c)*a*sin(1/2*d*x+1/2*c)*(12*B*sin(1/2*d*x+1/2*c)^4+(-10*A-20*B)*sin(1/2*d*x+1/2*c)^2+15*A
+15*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 = 64, normalized size of antiderivative = 0.63 \[ \int \cos (c+d x) \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {2 \, {\left (3 \, B \cos \left (d x + c\right )^{2} + {\left (5 \, A + 4 \, B\right )} \cos \left (d x + c\right ) + 10 \, A + 8 \, B\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

[In]

integrate(cos(d*x+c)*(a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)),x, algorithm="fricas")

[Out]

2/15*(3*B*cos(d*x + c)^2 + (5*A + 4*B)*cos(d*x + c) + 10*A + 8*B)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/(d*cos
(d*x + c) + d)

Sympy [F]

\[ \int \cos (c+d x) \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\int \sqrt {a \left (\cos {\left (c + d x \right )} + 1\right )} \left (A + B \cos {\left (c + d x \right )}\right ) \cos {\left (c + d x \right )}\, dx \]

[In]

integrate(cos(d*x+c)*(a+a*cos(d*x+c))**(1/2)*(A+B*cos(d*x+c)),x)

[Out]

Integral(sqrt(a*(cos(c + d*x) + 1))*(A + B*cos(c + d*x))*cos(c + d*x), x)

Maxima [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.87 \[ \int \cos (c+d x) \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {10 \, {\left (\sqrt {2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 3 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} A \sqrt {a} + {\left (3 \, \sqrt {2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, \sqrt {2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 30 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} B \sqrt {a}}{30 \, d} \]

[In]

integrate(cos(d*x+c)*(a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)),x, algorithm="maxima")

[Out]

1/30*(10*(sqrt(2)*sin(3/2*d*x + 3/2*c) + 3*sqrt(2)*sin(1/2*d*x + 1/2*c))*A*sqrt(a) + (3*sqrt(2)*sin(5/2*d*x +
5/2*c) + 5*sqrt(2)*sin(3/2*d*x + 3/2*c) + 30*sqrt(2)*sin(1/2*d*x + 1/2*c))*B*sqrt(a))/d

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.06 \[ \int \cos (c+d x) \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {\sqrt {2} {\left (3 \, B \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, {\left (2 \, A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + B \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 30 \, {\left (A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + B \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{30 \, d} \]

[In]

integrate(cos(d*x+c)*(a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)),x, algorithm="giac")

[Out]

1/30*sqrt(2)*(3*B*sgn(cos(1/2*d*x + 1/2*c))*sin(5/2*d*x + 5/2*c) + 5*(2*A*sgn(cos(1/2*d*x + 1/2*c)) + B*sgn(co
s(1/2*d*x + 1/2*c)))*sin(3/2*d*x + 3/2*c) + 30*(A*sgn(cos(1/2*d*x + 1/2*c)) + B*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. \[ \int \cos (c+d x) \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\int \cos \left (c+d\,x\right )\,\left (A+B\,\cos \left (c+d\,x\right )\right )\,\sqrt {a+a\,\cos \left (c+d\,x\right )} \,d x \]

[In]

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

[Out]

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