\(\int \cos ^2(c+d x) \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \, dx\) [75]

Optimal result

Integrand size = 33, antiderivative size = 144 \[ \int \cos ^2(c+d x) \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {2 a (7 A+6 B) \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a B \cos ^3(c+d x) \sin (c+d x)}{7 d \sqrt {a+a \cos (c+d x)}}-\frac {4 (7 A+6 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{105 d}+\frac {2 (7 A+6 B) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{35 a d} \]

[Out]

2/35*(7*A+6*B)*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/a/d+2/15*a*(7*A+6*B)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/7*
a*B*cos(d*x+c)^3*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)-4/105*(7*A+6*B)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {3060, 2838, 2830, 2725} \[ \int \cos ^2(c+d x) \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {2 (7 A+6 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{35 a d}-\frac {4 (7 A+6 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{105 d}+\frac {2 a (7 A+6 B) \sin (c+d x)}{15 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a B \sin (c+d x) \cos ^3(c+d x)}{7 d \sqrt {a \cos (c+d x)+a}} \]

[In]

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

[Out]

(2*a*(7*A + 6*B)*Sin[c + d*x])/(15*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*B*Cos[c + d*x]^3*Sin[c + d*x])/(7*d*Sqrt
[a + a*Cos[c + d*x]]) - (4*(7*A + 6*B)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(105*d) + (2*(7*A + 6*B)*(a + a*
Cos[c + d*x])^(3/2)*Sin[c + d*x])/(35*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 2838

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

Rule 3060

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] + Dist[(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]

Rubi steps \begin{align*} \text {integral}& = \frac {2 a B \cos ^3(c+d x) \sin (c+d x)}{7 d \sqrt {a+a \cos (c+d x)}}+\frac {1}{7} (7 A+6 B) \int \cos ^2(c+d x) \sqrt {a+a \cos (c+d x)} \, dx \\ & = \frac {2 a B \cos ^3(c+d x) \sin (c+d x)}{7 d \sqrt {a+a \cos (c+d x)}}+\frac {2 (7 A+6 B) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{35 a d}+\frac {(2 (7 A+6 B)) \int \left (\frac {3 a}{2}-a \cos (c+d x)\right ) \sqrt {a+a \cos (c+d x)} \, dx}{35 a} \\ & = \frac {2 a B \cos ^3(c+d x) \sin (c+d x)}{7 d \sqrt {a+a \cos (c+d x)}}-\frac {4 (7 A+6 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{105 d}+\frac {2 (7 A+6 B) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{35 a d}+\frac {1}{15} (7 A+6 B) \int \sqrt {a+a \cos (c+d x)} \, dx \\ & = \frac {2 a (7 A+6 B) \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a B \cos ^3(c+d x) \sin (c+d x)}{7 d \sqrt {a+a \cos (c+d x)}}-\frac {4 (7 A+6 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{105 d}+\frac {2 (7 A+6 B) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{35 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.56 \[ \int \cos ^2(c+d x) \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {\sqrt {a (1+\cos (c+d x))} (266 A+228 B+(112 A+141 B) \cos (c+d x)+6 (7 A+6 B) \cos (2 (c+d x))+15 B \cos (3 (c+d x))) \tan \left (\frac {1}{2} (c+d x)\right )}{210 d} \]

[In]

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

[Out]

(Sqrt[a*(1 + Cos[c + d*x])]*(266*A + 228*B + (112*A + 141*B)*Cos[c + d*x] + 6*(7*A + 6*B)*Cos[2*(c + d*x)] + 1
5*B*Cos[3*(c + d*x)])*Tan[(c + d*x)/2])/(210*d)

Maple [A] (verified)

Time = 2.76 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.71

method	result	size
default	\(\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (-120 B \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (84 A +252 B \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-140 A -210 B \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+105 A +105 B \right ) \sqrt {2}}{105 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\)	\(102\)
parts	\(\frac {2 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (12 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+7\right ) \sqrt {2}}{15 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}+\frac {2 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (40 \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-36 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+22 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+9\right ) \sqrt {2}}{35 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\)	\(157\)

[In]

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

[Out]

2/105*cos(1/2*d*x+1/2*c)*a*sin(1/2*d*x+1/2*c)*(-120*B*sin(1/2*d*x+1/2*c)^6+(84*A+252*B)*sin(1/2*d*x+1/2*c)^4+(
-140*A-210*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 = 82, normalized size of antiderivative = 0.57 \[ \int \cos ^2(c+d x) \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {2 \, {\left (15 \, B \cos \left (d x + c\right )^{3} + 3 \, {\left (7 \, A + 6 \, B\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (7 \, A + 6 \, B\right )} \cos \left (d x + c\right ) + 56 \, A + 48 \, B\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

[In]

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

[Out]

2/105*(15*B*cos(d*x + c)^3 + 3*(7*A + 6*B)*cos(d*x + c)^2 + 4*(7*A + 6*B)*cos(d*x + c) + 56*A + 48*B)*sqrt(a*c
os(d*x + c) + a)*sin(d*x + c)/(d*cos(d*x + c) + d)

Sympy [F]

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

[In]

integrate(cos(d*x+c)**2*(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)**2, x)

Maxima [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.82 \[ \int \cos ^2(c+d x) \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {14 \, {\left (3 \, \sqrt {2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, \sqrt {2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 30 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} A \sqrt {a} + 3 \, {\left (5 \, \sqrt {2} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 7 \, \sqrt {2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 35 \, \sqrt {2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 105 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} B \sqrt {a}}{420 \, d} \]

[In]

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

[Out]

1/420*(14*(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))*
A*sqrt(a) + 3*(5*sqrt(2)*sin(7/2*d*x + 7/2*c) + 7*sqrt(2)*sin(5/2*d*x + 5/2*c) + 35*sqrt(2)*sin(3/2*d*x + 3/2*
c) + 105*sqrt(2)*sin(1/2*d*x + 1/2*c))*B*sqrt(a))/d

Giac [A] (verification not implemented)

none

Time = 0.44 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.02 \[ \int \cos ^2(c+d x) \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {\sqrt {2} {\left (15 \, B \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 21 \, {\left (2 \, A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + B \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 35 \, {\left (2 \, A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 3 \, B \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 105 \, {\left (4 \, A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 3 \, B \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{420 \, d} \]

[In]

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

[Out]

1/420*sqrt(2)*(15*B*sgn(cos(1/2*d*x + 1/2*c))*sin(7/2*d*x + 7/2*c) + 21*(2*A*sgn(cos(1/2*d*x + 1/2*c)) + B*sgn
(cos(1/2*d*x + 1/2*c)))*sin(5/2*d*x + 5/2*c) + 35*(2*A*sgn(cos(1/2*d*x + 1/2*c)) + 3*B*sgn(cos(1/2*d*x + 1/2*c
)))*sin(3/2*d*x + 3/2*c) + 105*(4*A*sgn(cos(1/2*d*x + 1/2*c)) + 3*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. \[ \int \cos ^2(c+d x) \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\int {\cos \left (c+d\,x\right )}^2\,\left (A+B\,\cos \left (c+d\,x\right )\right )\,\sqrt {a+a\,\cos \left (c+d\,x\right )} \,d x \]

[In]

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

[Out]

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