\(\int (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx\) [85]

Optimal result

Integrand size = 25, antiderivative size = 101 \[ \int (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx=\frac {8 a^2 (5 A+3 B) \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a (5 A+3 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 B (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 d} \] Output:

8/15*a^2*(5*A+3*B)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/15*a*(5*A+3*B)*(a 
+a*cos(d*x+c))^(1/2)*sin(d*x+c)/d+2/5*B*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/ 
d
 

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.64 \[ \int (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx=\frac {a \sqrt {a (1+\cos (c+d x))} (50 A+39 B+2 (5 A+9 B) \cos (c+d x)+3 B \cos (2 (c+d x))) \tan \left (\frac {1}{2} (c+d x)\right )}{15 d} \] Input:

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

Output:

(a*Sqrt[a*(1 + Cos[c + d*x])]*(50*A + 39*B + 2*(5*A + 9*B)*Cos[c + d*x] + 
3*B*Cos[2*(c + d*x)])*Tan[(c + d*x)/2])/(15*d)
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3230

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3126

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3125

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

Input:

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

Output:

(2*B*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*d) + ((5*A + 3*B)*((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
 

Defintions of rubi rules used

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)]
 

Maple [A] (verified)

Time = 2.68 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.84

method	result	size
default	\(\frac {4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (6 B \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-5 A -15 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+15 A +15 B \right ) \sqrt {2}}{15 \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\)	\(85\)
parts	\(\frac {4 A \,a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2\right ) \sqrt {2}}{3 \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}+\frac {4 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2\right ) \sqrt {2}}{5 \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\)	\(131\)

Input:

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

Output:

4/15*cos(1/2*d*x+1/2*c)*a^2*sin(1/2*d*x+1/2*c)*(6*B*sin(1/2*d*x+1/2*c)^4+( 
-5*A-15*B)*sin(1/2*d*x+1/2*c)^2+15*A+15*B)*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 = 69, normalized size of antiderivative = 0.68 \[ \int (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx=\frac {2 \, {\left (3 \, B a \cos \left (d x + c\right )^{2} + {\left (5 \, A + 9 \, B\right )} a \cos \left (d x + c\right ) + {\left (25 \, A + 18 \, B\right )} a\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \] Input:

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

Output:

2/15*(3*B*a*cos(d*x + c)^2 + (5*A + 9*B)*a*cos(d*x + c) + (25*A + 18*B)*a) 
*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/(d*cos(d*x + c) + d)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.92 \[ \int (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx=\frac {10 \, {\left (\sqrt {2} a \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 9 \, \sqrt {2} a \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} A \sqrt {a} + 3 \, {\left (\sqrt {2} a \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, \sqrt {2} a \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 20 \, \sqrt {2} a \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} B \sqrt {a}}{30 \, d} \] Input:

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

Output:

1/30*(10*(sqrt(2)*a*sin(3/2*d*x + 3/2*c) + 9*sqrt(2)*a*sin(1/2*d*x + 1/2*c 
))*A*sqrt(a) + 3*(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))/d
 

Giac [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.14 \[ \int (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx=\frac {\sqrt {2} {\left (3 \, B a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, {\left (2 \, A a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 3 \, B a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 30 \, {\left (3 \, A a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 2 \, B a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{30 \, d} \] Input:

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

Output:

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

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

Output:

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

Reduce [F]

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

int((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),x)*a + int(sqrt(cos(c + d*x) + 1)*co 
s(c + d*x),x)*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)*b)