\(\int \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx\) [175]

Optimal result

Integrand size = 35, antiderivative size = 180 \[ \int \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx=\frac {a^{3/2} (14 A+11 B) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{8 d}+\frac {a^2 (14 A+11 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{8 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (6 A+7 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{12 d \sqrt {a+a \cos (c+d x)}}+\frac {a B \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{3 d} \] Output:

1/8*a^(3/2)*(14*A+11*B)*arcsin(a^(1/2)*sin(d*x+c)/(a+a*cos(d*x+c))^(1/2))/ 
d+1/8*a^2*(14*A+11*B)*cos(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2) 
+1/12*a^2*(6*A+7*B)*cos(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1 
/3*a*B*cos(d*x+c)^(3/2)*(a+a*cos(d*x+c))^(1/2)*sin(d*x+c)/d
 

Mathematica [A] (verified)

Time = 0.75 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.66 \[ \int \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx=\frac {a \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \left (3 \sqrt {2} (14 A+11 B) \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \sqrt {\cos (c+d x)} (42 A+37 B+2 (6 A+11 B) \cos (c+d x)+4 B \cos (2 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{48 d} \] Input:

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

Output:

(a*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*(3*Sqrt[2]*(14*A + 11*B)*Ar 
cSin[Sqrt[2]*Sin[(c + d*x)/2]] + 2*Sqrt[Cos[c + d*x]]*(42*A + 37*B + 2*(6* 
A + 11*B)*Cos[c + d*x] + 4*B*Cos[2*(c + d*x)])*Sin[(c + d*x)/2]))/(48*d)
 

Rubi [A] (verified)

Time = 0.86 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.98, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3455, 27, 3042, 3460, 3042, 3249, 3042, 3253, 223}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3455

\(\displaystyle \frac {1}{3} \int \frac {1}{2} \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a} (3 a (2 A+B)+a (6 A+7 B) \cos (c+d x))dx+\frac {a B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{6} \int \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a} (3 a (2 A+B)+a (6 A+7 B) \cos (c+d x))dx+\frac {a B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{6} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (3 a (2 A+B)+a (6 A+7 B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {a B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 3460

\(\displaystyle \frac {1}{6} \left (\frac {3}{4} a (14 A+11 B) \int \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}dx+\frac {a^2 (6 A+7 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{6} \left (\frac {3}{4} a (14 A+11 B) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a^2 (6 A+7 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 3249

\(\displaystyle \frac {1}{6} \left (\frac {3}{4} a (14 A+11 B) \left (\frac {1}{2} \int \frac {\sqrt {\cos (c+d x) a+a}}{\sqrt {\cos (c+d x)}}dx+\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 (6 A+7 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{6} \left (\frac {3}{4} a (14 A+11 B) \left (\frac {1}{2} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 (6 A+7 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 3253

\(\displaystyle \frac {1}{6} \left (\frac {3}{4} a (14 A+11 B) \left (\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {1}{\sqrt {1-\frac {a \sin ^2(c+d x)}{\cos (c+d x) a+a}}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}\right )+\frac {a^2 (6 A+7 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 223

\(\displaystyle \frac {1}{6} \left (\frac {a^2 (6 A+7 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}+\frac {3}{4} a (14 A+11 B) \left (\frac {\sqrt {a} \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )\right )+\frac {a B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\)

Input:

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

Output:

(a*B*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(3*d) + ((a 
^2*(6*A + 7*B)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(2*d*Sqrt[a + a*Cos[c + d* 
x]]) + (3*a*(14*A + 11*B)*((Sqrt[a]*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + 
 a*Cos[c + d*x]]])/d + (a*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*C 
os[c + d*x]])))/4)/6
 

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 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt 
[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3249

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])^(n_), x_Symbol] :> Simp[-2*b*Cos[e + f*x]*((c + d*Sin[e + f*x]) 
^n/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[2*n*((b*c + a*d)/(b*( 
2*n + 1)))   Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n - 1), x], 
 x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 
0] && NeQ[c^2 - d^2, 0] && GtQ[n, 0] && IntegerQ[2*n]
 

rule 3253

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.) 
*(x_)]], x_Symbol] :> Simp[-2/f   Subst[Int[1/Sqrt[1 - x^2/a], x], x, b*(Co 
s[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, d, e, f}, x] && E 
qQ[a^2 - b^2, 0] && EqQ[d, a/b]
 

rule 3455

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(-b)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^(n 
 + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1))   Int[(a + b*Sin[e + 
f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1 
) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + 
f*x], x], 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] && GtQ[m, 1/2] &&  !LtQ[n, -1 
] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
 

rule 3460

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] + Simp[(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]
 

Maple [A] (verified)

Time = 17.00 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.15

method	result	size
default	\(\frac {\left (42 A \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )+33 B \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )+\left (12 \cos \left (d x +c \right )+42\right ) \sin \left (d x +c \right ) A \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+\left (8 \cos \left (d x +c \right )^{2}+22 \cos \left (d x +c \right )+33\right ) \sin \left (d x +c \right ) B \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, a \sqrt {2}}{24 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\)	\(207\)
parts	\(\frac {A \left (7 \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )+\sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\sin \left (2 d x +2 c \right )+7 \sin \left (d x +c \right )\right )\right ) \sqrt {a \left (\cos \left (d x +c \right )+1\right )}\, \sqrt {\cos \left (d x +c \right )}\, a}{4 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}+\frac {B \left (33 \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )+\left (8 \cos \left (d x +c \right )^{2}+22 \cos \left (d x +c \right )+33\right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, a \sqrt {2}}{24 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\)	\(264\)

Input:

int(cos(d*x+c)^(1/2)*(a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)),x,method=_RET 
URNVERBOSE)
 

Output:

1/24/d*(42*A*arctan(tan(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+33*B*arc 
tan(tan(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+(12*cos(d*x+c)+42)*sin(d 
*x+c)*A*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+(8*cos(d*x+c)^2+22*cos(d*x+c)+33 
)*sin(d*x+c)*B*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^(1/2)*(a*cos( 
1/2*d*x+1/2*c)^2)^(1/2)/(cos(d*x+c)+1)/(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a 
*2^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.91 \[ \int \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx=\frac {{\left (8 \, B a \cos \left (d x + c\right )^{2} + 2 \, {\left (6 \, A + 11 \, B\right )} a \cos \left (d x + c\right ) + 3 \, {\left (14 \, A + 11 \, B\right )} a\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 3 \, {\left ({\left (14 \, A + 11 \, B\right )} a \cos \left (d x + c\right ) + {\left (14 \, A + 11 \, B\right )} a\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )}\right )}{24 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \] Input:

integrate(cos(d*x+c)^(1/2)*(a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)),x, algo 
rithm="fricas")
 

Output:

1/24*((8*B*a*cos(d*x + c)^2 + 2*(6*A + 11*B)*a*cos(d*x + c) + 3*(14*A + 11 
*B)*a)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*sin(d*x + c) + 3*((14*A 
 + 11*B)*a*cos(d*x + c) + (14*A + 11*B)*a)*sqrt(a)*arctan(sqrt(a*cos(d*x + 
 c) + a)*sqrt(a)*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 + a*cos 
(d*x + c))))/(d*cos(d*x + c) + d)
 

Sympy [F(-1)]

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

integrate(cos(d*x+c)**(1/2)*(a+a*cos(d*x+c))**(3/2)*(A+B*cos(d*x+c)),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3023 vs. \(2 (154) = 308\).

Time = 0.44 (sec) , antiderivative size = 3023, normalized size of antiderivative = 16.79 \[ \int \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)^(1/2)*(a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)),x, algo 
rithm="maxima")
 

Output:

1/96*(6*(2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 
 1)^(1/4)*((a*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(2*d 
*x + 2*c) + a*sin(2*d*x + 2*c) - (a*cos(2*d*x + 2*c) - 6*a)*sin(1/2*arctan 
2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*cos(1/2*arctan2(sin(2*d*x + 2*c), 
cos(2*d*x + 2*c) + 1)) + (a*sin(2*d*x + 2*c)*sin(1/2*arctan2(sin(2*d*x + 2 
*c), cos(2*d*x + 2*c))) - a*cos(2*d*x + 2*c) + (a*cos(2*d*x + 2*c) - 6*a)* 
cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 6*a)*sin(1/2*arctan 
2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*sqrt(a) + 7*(a*arctan2((cos(2* 
d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2 
*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(1/2*arctan2(sin(2*d*x + 
2*c), cos(2*d*x + 2*c) + 1)) - cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x 
 + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))), (cos( 
2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1 
/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*cos(1/2*arctan2(sin(2* 
d*x + 2*c), cos(2*d*x + 2*c))) + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d 
*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))) + 1) 
 - a*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) 
 + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(1/2* 
arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - cos(1/2*arctan2(sin(2*d 
*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos...
 

Giac [F]

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

integrate(cos(d*x+c)^(1/2)*(a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)),x, algo 
rithm="giac")
 

Output:

integrate((B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)^(3/2)*sqrt(cos(d*x + c 
)), x)
 

Mupad [F(-1)]

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

int(cos(c + d*x)^(1/2)*(A + B*cos(c + d*x))*(a + a*cos(c + d*x))^(3/2),x)
 

Output:

int(cos(c + d*x)^(1/2)*(A + B*cos(c + d*x))*(a + a*cos(c + d*x))^(3/2), x)
 

Reduce [F]

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

int(cos(d*x+c)^(1/2)*(a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)),x)
 

Output:

sqrt(a)*a*(int(sqrt(cos(c + d*x) + 1)*sqrt(cos(c + d*x))*cos(c + d*x),x)*a 
 + int(sqrt(cos(c + d*x) + 1)*sqrt(cos(c + d*x))*cos(c + d*x),x)*b + int(s 
qrt(cos(c + d*x) + 1)*sqrt(cos(c + d*x))*cos(c + d*x)**2,x)*b + int(sqrt(c 
os(c + d*x) + 1)*sqrt(cos(c + d*x)),x)*a)