3.1.76 \(\int \cos (c+d x) \sqrt {a+a \cos (c+d x)} (A+C \cos ^2(c+d x)) \, dx\) [76]

3.1.76.1 Optimal result
3.1.76.2 Mathematica [A] (verified)
3.1.76.3 Rubi [A] (verified)
3.1.76.4 Maple [A] (verified)
3.1.76.5 Fricas [A] (verification not implemented)
3.1.76.6 Sympy [F]
3.1.76.7 Maxima [A] (verification not implemented)
3.1.76.8 Giac [A] (verification not implemented)
3.1.76.9 Mupad [F(-1)]

3.1.76.1 Optimal result

Integrand size = 33, antiderivative size = 137 \[ \int \cos (c+d x) \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {2 a (35 A+27 C) \sin (c+d x)}{105 d \sqrt {a+a \cos (c+d x)}}+\frac {2 (35 A+18 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{105 d}+\frac {2 C \cos ^2(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{7 d}+\frac {2 C (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{35 a d} \]

output
2/35*C*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/a/d+2/105*a*(35*A+27*C)*sin(d*x+c 
)/d/(a+a*cos(d*x+c))^(1/2)+2/105*(35*A+18*C)*sin(d*x+c)*(a+a*cos(d*x+c))^( 
1/2)/d+2/7*C*cos(d*x+c)^2*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d
 
3.1.76.2 Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.54 \[ \int \cos (c+d x) \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {\sqrt {a (1+\cos (c+d x))} (280 A+228 C+(140 A+141 C) \cos (c+d x)+36 C \cos (2 (c+d x))+15 C \cos (3 (c+d x))) \tan \left (\frac {1}{2} (c+d x)\right )}{210 d} \]

input
Integrate[Cos[c + d*x]*Sqrt[a + a*Cos[c + d*x]]*(A + C*Cos[c + d*x]^2),x]
 
output
(Sqrt[a*(1 + Cos[c + d*x])]*(280*A + 228*C + (140*A + 141*C)*Cos[c + d*x] 
+ 36*C*Cos[2*(c + d*x)] + 15*C*Cos[3*(c + d*x)])*Tan[(c + d*x)/2])/(210*d)
 
3.1.76.3 Rubi [A] (verified)

Time = 0.84 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.13, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3042, 3525, 27, 3042, 3447, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3525

\(\displaystyle \frac {2 \int \frac {1}{2} \cos (c+d x) \sqrt {\cos (c+d x) a+a} (a (7 A+4 C)+a C \cos (c+d x))dx}{7 a}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{7 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \cos (c+d x) \sqrt {\cos (c+d x) a+a} (a (7 A+4 C)+a C \cos (c+d x))dx}{7 a}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (a (7 A+4 C)+a C \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{7 a}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{7 d}\)

\(\Big \downarrow \) 3447

\(\displaystyle \frac {\int \sqrt {\cos (c+d x) a+a} \left (a C \cos ^2(c+d x)+a (7 A+4 C) \cos (c+d x)\right )dx}{7 a}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (a C \sin \left (c+d x+\frac {\pi }{2}\right )^2+a (7 A+4 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{7 a}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{7 d}\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {\frac {2 \int \frac {1}{2} \sqrt {\cos (c+d x) a+a} \left (3 C a^2+(35 A+18 C) \cos (c+d x) a^2\right )dx}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{7 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \sqrt {\cos (c+d x) a+a} \left (3 C a^2+(35 A+18 C) \cos (c+d x) a^2\right )dx}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (3 C a^2+(35 A+18 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )dx}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{7 d}\)

\(\Big \downarrow \) 3230

\(\displaystyle \frac {\frac {\frac {1}{3} a^2 (35 A+27 C) \int \sqrt {\cos (c+d x) a+a}dx+\frac {2 a^2 (35 A+18 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {1}{3} a^2 (35 A+27 C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a^2 (35 A+18 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{7 d}\)

\(\Big \downarrow \) 3125

\(\displaystyle \frac {\frac {\frac {2 a^3 (35 A+27 C) \sin (c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a^2 (35 A+18 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{7 d}\)

input
Int[Cos[c + d*x]*Sqrt[a + a*Cos[c + d*x]]*(A + C*Cos[c + d*x]^2),x]
 
output
(2*C*Cos[c + d*x]^2*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(7*d) + ((2*C*( 
a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*d) + ((2*a^3*(35*A + 27*C)*Sin[ 
c + d*x])/(3*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(35*A + 18*C)*Sqrt[a + a 
*Cos[c + d*x]]*Sin[c + d*x])/(3*d))/(5*a))/(7*a)
 

3.1.76.3.1 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 3447
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 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]
 

rule 3525
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_.)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> 
 Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1 
)/(d*f*(m + n + 2))), x] + Simp[1/(b*d*(m + n + 2))   Int[(a + b*Sin[e + f* 
x])^m*(c + d*Sin[e + f*x])^n*Simp[A*b*d*(m + n + 2) + C*(a*c*m + b*d*(n + 1 
)) + C*(a*d*m - b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, 
 e, f, A, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 
 - d^2, 0] &&  !LtQ[m, -2^(-1)] && NeQ[m + n + 2, 0]
 
3.1.76.4 Maple [A] (verified)

Time = 4.96 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.71

method result size
default \(\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (-120 C \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+252 C \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-70 A -210 C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+105 A +105 C \right ) \sqrt {2}}{105 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\) \(97\)
parts \(\frac {2 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (1+2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right ) \sqrt {2}}{3 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}+\frac {2 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (40 \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-36 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+22 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+9\right ) \sqrt {2}}{35 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\) \(144\)

input
int(cos(d*x+c)*(A+C*cos(d*x+c)^2)*(a+cos(d*x+c)*a)^(1/2),x,method=_RETURNV 
ERBOSE)
 
output
2/105*cos(1/2*d*x+1/2*c)*a*sin(1/2*d*x+1/2*c)*(-120*C*sin(1/2*d*x+1/2*c)^6 
+252*C*sin(1/2*d*x+1/2*c)^4+(-70*A-210*C)*sin(1/2*d*x+1/2*c)^2+105*A+105*C 
)*2^(1/2)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d
 
3.1.76.5 Fricas [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.55 \[ \int \cos (c+d x) \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (15 \, C \cos \left (d x + c\right )^{3} + 18 \, C \cos \left (d x + c\right )^{2} + {\left (35 \, A + 24 \, C\right )} \cos \left (d x + c\right ) + 70 \, A + 48 \, C\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

input
integrate(cos(d*x+c)*(A+C*cos(d*x+c)^2)*(a+a*cos(d*x+c))^(1/2),x, algorith 
m="fricas")
 
output
2/105*(15*C*cos(d*x + c)^3 + 18*C*cos(d*x + c)^2 + (35*A + 24*C)*cos(d*x + 
 c) + 70*A + 48*C)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/(d*cos(d*x + c) + 
 d)
 
3.1.76.6 Sympy [F]

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

input
integrate(cos(d*x+c)*(A+C*cos(d*x+c)**2)*(a+a*cos(d*x+c))**(1/2),x)
 
output
Integral(sqrt(a*(cos(c + d*x) + 1))*(A + C*cos(c + d*x)**2)*cos(c + d*x), 
x)
 
3.1.76.7 Maxima [A] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.75 \[ \int \cos (c+d x) \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {140 \, {\left (\sqrt {2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 3 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} A \sqrt {a} + 3 \, {\left (5 \, \sqrt {2} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 7 \, \sqrt {2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 35 \, \sqrt {2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 105 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} C \sqrt {a}}{420 \, d} \]

input
integrate(cos(d*x+c)*(A+C*cos(d*x+c)^2)*(a+a*cos(d*x+c))^(1/2),x, algorith 
m="maxima")
 
output
1/420*(140*(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*(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) 
)*C*sqrt(a))/d
 
3.1.76.8 Giac [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.96 \[ \int \cos (c+d x) \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {\sqrt {2} {\left (15 \, C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 21 \, C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 35 \, {\left (4 \, A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 3 \, C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 105 \, {\left (4 \, A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 3 \, C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{420 \, d} \]

input
integrate(cos(d*x+c)*(A+C*cos(d*x+c)^2)*(a+a*cos(d*x+c))^(1/2),x, algorith 
m="giac")
 
output
1/420*sqrt(2)*(15*C*sgn(cos(1/2*d*x + 1/2*c))*sin(7/2*d*x + 7/2*c) + 21*C* 
sgn(cos(1/2*d*x + 1/2*c))*sin(5/2*d*x + 5/2*c) + 35*(4*A*sgn(cos(1/2*d*x + 
 1/2*c)) + 3*C*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*C*sgn(cos(1/2*d*x + 1/2*c)))*sin(1/2*d*x + 1 
/2*c))*sqrt(a)/d
 
3.1.76.9 Mupad [F(-1)]

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

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