\(\int \sqrt {a+a \cos (c+d x)} (B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\) [277]

Optimal result

Integrand size = 34, antiderivative size = 101 \[ \int \sqrt {a+a \cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {2 a (5 B+7 C) \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}+\frac {2 (5 B-2 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 C (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 a d} \] Output:

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

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.63 \[ \int \sqrt {a+a \cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {\sqrt {a (1+\cos (c+d x))} (20 B+19 C+2 (5 B+4 C) \cos (c+d x)+3 C \cos (2 (c+d x))) \tan \left (\frac {1}{2} (c+d x)\right )}{15 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.206, Rules used = {3042, 3502, 27, 3042, 3230, 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.

\(\displaystyle \int \sqrt {a \cos (c+d x)+a} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

\(\displaystyle \frac {2 \int \frac {1}{2} \sqrt {\cos (c+d x) a+a} (3 a C+a (5 B-2 C) \cos (c+d x))dx}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 a d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \sqrt {\cos (c+d x) a+a} (3 a C+a (5 B-2 C) \cos (c+d x))dx}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (3 a C+a (5 B-2 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 a d}\)

\(\Big \downarrow \) 3230

\(\displaystyle \frac {\frac {1}{3} a (5 B+7 C) \int \sqrt {\cos (c+d x) a+a}dx+\frac {2 a (5 B-2 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 a d}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3125

\(\displaystyle \frac {\frac {2 a^2 (5 B+7 C) \sin (c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a (5 B-2 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 a d}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 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)]
 

rule 3502

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[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]
 

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.82

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

Input:

int((a+a*cos(d*x+c))^(1/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2),x,method=_RETURNV 
ERBOSE)
 

Output:

2/15*cos(1/2*d*x+1/2*c)*a*sin(1/2*d*x+1/2*c)*(12*C*sin(1/2*d*x+1/2*c)^4+(- 
10*B-20*C)*sin(1/2*d*x+1/2*c)^2+15*B+15*C)*2^(1/2)/(a*cos(1/2*d*x+1/2*c)^2 
)^(1/2)/d
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.63 \[ \int \sqrt {a+a \cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (3 \, C \cos \left (d x + c\right )^{2} + {\left (5 \, B + 4 \, C\right )} \cos \left (d x + c\right ) + 10 \, B + 8 \, C\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \] Input:

integrate((a+a*cos(d*x+c))^(1/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorith 
m="fricas")
 

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.87 \[ \int \sqrt {a+a \cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {10 \, {\left (\sqrt {2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 3 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} B \sqrt {a} + {\left (3 \, \sqrt {2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, \sqrt {2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 30 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} C \sqrt {a}}{30 \, d} \] Input:

integrate((a+a*cos(d*x+c))^(1/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorith 
m="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.06 \[ \int \sqrt {a+a \cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {\sqrt {2} {\left (3 \, C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, {\left (2 \, B \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 30 \, {\left (B \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{30 \, d} \] Input:

integrate((a+a*cos(d*x+c))^(1/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorith 
m="giac")
 

Output:

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

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

Output:

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

Reduce [F]

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

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

Output:

sqrt(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)**2,x)*c)