\(\int (a+a \cos (c+d x))^{3/2} (B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\) [278]

Optimal result

Integrand size = 34, antiderivative size = 138 \[ \int (a+a \cos (c+d x))^{3/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {8 a^2 (21 B+19 C) \sin (c+d x)}{105 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a (21 B+19 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{105 d}+\frac {2 (7 B-2 C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac {2 C (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{7 a d} \] Output:

8/105*a^2*(21*B+19*C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/105*a*(21*B+19 
*C)*(a+a*cos(d*x+c))^(1/2)*sin(d*x+c)/d+2/35*(7*B-2*C)*(a+a*cos(d*x+c))^(3 
/2)*sin(d*x+c)/d+2/7*C*(a+a*cos(d*x+c))^(5/2)*sin(d*x+c)/a/d
 

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.59 \[ \int (a+a \cos (c+d x))^{3/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {a \sqrt {a (1+\cos (c+d x))} (546 B+494 C+(252 B+253 C) \cos (c+d x)+6 (7 B+13 C) \cos (2 (c+d x))+15 C \cos (3 (c+d x))) \tan \left (\frac {1}{2} (c+d x)\right )}{210 d} \] Input:

Integrate[(a + a*Cos[c + d*x])^(3/2)*(B*Cos[c + d*x] + C*Cos[c + d*x]^2),x 
]
 

Output:

(a*Sqrt[a*(1 + Cos[c + d*x])]*(546*B + 494*C + (252*B + 253*C)*Cos[c + d*x 
] + 6*(7*B + 13*C)*Cos[2*(c + d*x)] + 15*C*Cos[3*(c + d*x)])*Tan[(c + d*x) 
/2])/(210*d)
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.265, Rules used = {3042, 3502, 27, 3042, 3230, 3042, 3126, 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.

\(\displaystyle \int (a \cos (c+d x)+a)^{3/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3230

\(\displaystyle \frac {\frac {1}{5} a (21 B+19 C) \int (\cos (c+d x) a+a)^{3/2}dx+\frac {2 a (7 B-2 C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} a (21 B+19 C) \int \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}dx+\frac {2 a (7 B-2 C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 a d}\)

\(\Big \downarrow \) 3126

\(\displaystyle \frac {\frac {1}{5} a (21 B+19 C) \left (\frac {4}{3} a \int \sqrt {\cos (c+d x) a+a}dx+\frac {2 a \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {2 a (7 B-2 C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} a (21 B+19 C) \left (\frac {4}{3} a \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {2 a (7 B-2 C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 a d}\)

\(\Big \downarrow \) 3125

\(\displaystyle \frac {\frac {1}{5} a (21 B+19 C) \left (\frac {8 a^2 \sin (c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {2 a (7 B-2 C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 a d}\)

Input:

Int[(a + a*Cos[c + d*x])^(3/2)*(B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]
 

Output:

(2*C*(a + a*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(7*a*d) + ((2*a*(7*B - 2*C)* 
(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*d) + (a*(21*B + 19*C)*((8*a^2* 
Sin[c + d*x])/(3*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*Sqrt[a + a*Cos[c + d*x 
]]*Sin[c + d*x])/(3*d)))/5)/(7*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 3126

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] + Simp[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 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 = 0.48 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.75

method	result	size
default	\(\frac {4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (-60 C \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (42 B +168 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-105 B -175 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+105 B +105 C \right ) \sqrt {2}}{105 \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\)	\(104\)
parts	\(\frac {4 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2\right ) \sqrt {2}}{5 \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}+\frac {4 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (60 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-12 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+19 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+38\right ) \sqrt {2}}{105 \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\)	\(159\)

Input:

int((a+a*cos(d*x+c))^(3/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2),x,method=_RETURNV 
ERBOSE)
 

Output:

4/105*cos(1/2*d*x+1/2*c)*a^2*sin(1/2*d*x+1/2*c)*(-60*C*sin(1/2*d*x+1/2*c)^ 
6+(42*B+168*C)*sin(1/2*d*x+1/2*c)^4+(-105*B-175*C)*sin(1/2*d*x+1/2*c)^2+10 
5*B+105*C)*2^(1/2)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.64 \[ \int (a+a \cos (c+d x))^{3/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (15 \, C a \cos \left (d x + c\right )^{3} + 3 \, {\left (7 \, B + 13 \, C\right )} a \cos \left (d x + c\right )^{2} + {\left (63 \, B + 52 \, C\right )} a \cos \left (d x + c\right ) + 2 \, {\left (63 \, B + 52 \, C\right )} a\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \] Input:

integrate((a+a*cos(d*x+c))^(3/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorith 
m="fricas")
 

Output:

2/105*(15*C*a*cos(d*x + c)^3 + 3*(7*B + 13*C)*a*cos(d*x + c)^2 + (63*B + 5 
2*C)*a*cos(d*x + c) + 2*(63*B + 52*C)*a)*sqrt(a*cos(d*x + c) + a)*sin(d*x 
+ c)/(d*cos(d*x + c) + d)
 

Sympy [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^{3/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:

integrate((a+a*cos(d*x+c))**(3/2)*(B*cos(d*x+c)+C*cos(d*x+c)**2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.89 \[ \int (a+a \cos (c+d x))^{3/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {42 \, {\left (\sqrt {2} a \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, \sqrt {2} a \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 20 \, \sqrt {2} a \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} B \sqrt {a} + {\left (15 \, \sqrt {2} a \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 63 \, \sqrt {2} a \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 175 \, \sqrt {2} a \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 735 \, \sqrt {2} a \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} C \sqrt {a}}{420 \, d} \] Input:

integrate((a+a*cos(d*x+c))^(3/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorith 
m="maxima")
 

Output:

1/420*(42*(sqrt(2)*a*sin(5/2*d*x + 5/2*c) + 5*sqrt(2)*a*sin(3/2*d*x + 3/2* 
c) + 20*sqrt(2)*a*sin(1/2*d*x + 1/2*c))*B*sqrt(a) + (15*sqrt(2)*a*sin(7/2* 
d*x + 7/2*c) + 63*sqrt(2)*a*sin(5/2*d*x + 5/2*c) + 175*sqrt(2)*a*sin(3/2*d 
*x + 3/2*c) + 735*sqrt(2)*a*sin(1/2*d*x + 1/2*c))*C*sqrt(a))/d
 

Giac [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.12 \[ \int (a+a \cos (c+d x))^{3/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {\sqrt {2} {\left (15 \, C a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 21 \, {\left (2 \, B a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 3 \, C a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 35 \, {\left (6 \, B a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 5 \, C a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 105 \, {\left (8 \, B a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 7 \, C a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{420 \, d} \] Input:

integrate((a+a*cos(d*x+c))^(3/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorith 
m="giac")
 

Output:

1/420*sqrt(2)*(15*C*a*sgn(cos(1/2*d*x + 1/2*c))*sin(7/2*d*x + 7/2*c) + 21* 
(2*B*a*sgn(cos(1/2*d*x + 1/2*c)) + 3*C*a*sgn(cos(1/2*d*x + 1/2*c)))*sin(5/ 
2*d*x + 5/2*c) + 35*(6*B*a*sgn(cos(1/2*d*x + 1/2*c)) + 5*C*a*sgn(cos(1/2*d 
*x + 1/2*c)))*sin(3/2*d*x + 3/2*c) + 105*(8*B*a*sgn(cos(1/2*d*x + 1/2*c)) 
+ 7*C*a*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. \[ \int (a+a \cos (c+d x))^{3/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int \left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )\right )\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2} \,d x \] Input:

int((B*cos(c + d*x) + C*cos(c + d*x)^2)*(a + a*cos(c + d*x))^(3/2),x)
 

Output:

int((B*cos(c + d*x) + C*cos(c + d*x)^2)*(a + a*cos(c + d*x))^(3/2), x)
 

Reduce [F]

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

int((a+a*cos(d*x+c))^(3/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2),x)
 

Output:

sqrt(a)*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)**3,x)*c + int(sqrt(cos(c + d*x) + 1)*cos(c + d*x)* 
*2,x)*b + int(sqrt(cos(c + d*x) + 1)*cos(c + d*x)**2,x)*c)