\(\int \cos ^3(c+d x) (a+b \cos (c+d x))^2 \, dx\) [418]

Optimal result

Integrand size = 21, antiderivative size = 111 \[ \int \cos ^3(c+d x) (a+b \cos (c+d x))^2 \, dx=\frac {3 a b x}{4}+\frac {\left (a^2+b^2\right ) \sin (c+d x)}{d}+\frac {3 a b \cos (c+d x) \sin (c+d x)}{4 d}+\frac {a b \cos ^3(c+d x) \sin (c+d x)}{2 d}-\frac {\left (a^2+2 b^2\right ) \sin ^3(c+d x)}{3 d}+\frac {b^2 \sin ^5(c+d x)}{5 d} \] Output:

3/4*a*b*x+(a^2+b^2)*sin(d*x+c)/d+3/4*a*b*cos(d*x+c)*sin(d*x+c)/d+1/2*a*b*c 
os(d*x+c)^3*sin(d*x+c)/d-1/3*(a^2+2*b^2)*sin(d*x+c)^3/d+1/5*b^2*sin(d*x+c) 
^5/d
 

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.77 \[ \int \cos ^3(c+d x) (a+b \cos (c+d x))^2 \, dx=\frac {240 \left (a^2+b^2\right ) \sin (c+d x)-80 \left (a^2+2 b^2\right ) \sin ^3(c+d x)+48 b^2 \sin ^5(c+d x)+15 a b (12 (c+d x)+8 \sin (2 (c+d x))+\sin (4 (c+d x)))}{240 d} \] Input:

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

Output:

(240*(a^2 + b^2)*Sin[c + d*x] - 80*(a^2 + 2*b^2)*Sin[c + d*x]^3 + 48*b^2*S 
in[c + d*x]^5 + 15*a*b*(12*(c + d*x) + 8*Sin[2*(c + d*x)] + Sin[4*(c + d*x 
)]))/(240*d)
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.02, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 3268, 3042, 3115, 3042, 3115, 24, 3492, 290, 2009}

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.

\(\displaystyle \int \cos ^3(c+d x) (a+b \cos (c+d x))^2 \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3268

\(\displaystyle \int \cos ^3(c+d x) \left (a^2+b^2 \cos ^2(c+d x)\right )dx+2 a b \int \cos ^4(c+d x)dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a^2+b^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx+2 a b \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx\)

\(\Big \downarrow \) 3115

\(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a^2+b^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx+2 a b \left (\frac {3}{4} \int \cos ^2(c+d x)dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a^2+b^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx+2 a b \left (\frac {3}{4} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )\)

\(\Big \downarrow \) 3115

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

\(\Big \downarrow \) 24

\(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a^2+b^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx+2 a b \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )\)

\(\Big \downarrow \) 3492

\(\displaystyle 2 a b \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )-\frac {\int \left (1-\sin ^2(c+d x)\right ) \left (a^2+b^2-b^2 \sin ^2(c+d x)\right )d(-\sin (c+d x))}{d}\)

\(\Big \downarrow \) 290

\(\displaystyle 2 a b \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )-\frac {\int \left (b^2 \sin ^4(c+d x)-\left (a^2+2 b^2\right ) \sin ^2(c+d x)+a^2 \left (\frac {b^2}{a^2}+1\right )\right )d(-\sin (c+d x))}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle 2 a b \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )-\frac {\frac {1}{3} \left (a^2+2 b^2\right ) \sin ^3(c+d x)-\left (a^2+b^2\right ) \sin (c+d x)-\frac {1}{5} b^2 \sin ^5(c+d x)}{d}\)

Input:

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

Output:

-((-((a^2 + b^2)*Sin[c + d*x]) + ((a^2 + 2*b^2)*Sin[c + d*x]^3)/3 - (b^2*S 
in[c + d*x]^5)/5)/d) + 2*a*b*((Cos[c + d*x]^3*Sin[c + d*x])/(4*d) + (3*(x/ 
2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)))/4)
 

Defintions of rubi rules used

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 290

Int[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> I 
nt[ExpandIntegrand[(a + b*x^2)^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d 
}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && IGtQ[q, 0]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 3115

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n)   Int[(b*Sin 
[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 
2*n]
 

rule 3268

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)])^2, x_Symbol] :> Simp[2*c*(d/b)   Int[(b*Sin[e + f*x])^(m + 1), x], x] 
+ Int[(b*Sin[e + f*x])^m*(c^2 + d^2*Sin[e + f*x]^2), x] /; FreeQ[{b, c, d, 
e, f, m}, x]
 

rule 3492

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), 
 x_Symbol] :> Simp[-f^(-1)   Subst[Int[(1 - x^2)^((m - 1)/2)*(A + C - C*x^2 
), x], x, Cos[e + f*x]], x] /; FreeQ[{e, f, A, C}, x] && IGtQ[(m + 1)/2, 0]
 

Maple [A] (verified)

Time = 13.48 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.86

method	result	size
derivativedivides	\(\frac {\frac {a^{2} \left (\cos \left (d x +c \right )^{2}+2\right ) \sin \left (d x +c \right )}{3}+2 a b \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {b^{2} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}}{d}\)	\(95\)
default	\(\frac {\frac {a^{2} \left (\cos \left (d x +c \right )^{2}+2\right ) \sin \left (d x +c \right )}{3}+2 a b \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {b^{2} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}}{d}\)	\(95\)
parts	\(\frac {a^{2} \left (\cos \left (d x +c \right )^{2}+2\right ) \sin \left (d x +c \right )}{3 d}+\frac {b^{2} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5 d}+\frac {2 a b \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\)	\(100\)
parallelrisch	\(\frac {180 a b x d +180 a^{2} \sin \left (d x +c \right )+150 \sin \left (d x +c \right ) b^{2}+3 b^{2} \sin \left (5 d x +5 c \right )+15 a b \sin \left (4 d x +4 c \right )+20 \sin \left (3 d x +3 c \right ) a^{2}+25 \sin \left (3 d x +3 c \right ) b^{2}+120 a b \sin \left (2 d x +2 c \right )}{240 d}\)	\(103\)
risch	\(\frac {3 a b x}{4}+\frac {3 a^{2} \sin \left (d x +c \right )}{4 d}+\frac {5 b^{2} \sin \left (d x +c \right )}{8 d}+\frac {b^{2} \sin \left (5 d x +5 c \right )}{80 d}+\frac {a b \sin \left (4 d x +4 c \right )}{16 d}+\frac {\sin \left (3 d x +3 c \right ) a^{2}}{12 d}+\frac {5 \sin \left (3 d x +3 c \right ) b^{2}}{48 d}+\frac {a b \sin \left (2 d x +2 c \right )}{2 d}\)	\(118\)
norman	\(\frac {\frac {3 a b x}{4}+\frac {4 \left (25 a^{2}+29 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}+\frac {\left (4 a^{2}-5 a b +4 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{2 d}+\frac {\left (4 a^{2}+5 a b +4 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d}+\frac {\left (16 a^{2}-3 a b +8 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}+\frac {\left (16 a^{2}+3 a b +8 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {15 a b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{4}+\frac {15 a b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{2}+\frac {15 a b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{2}+\frac {15 a b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{4}+\frac {3 a b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{4}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5}}\)	\(252\)
orering	\(\text {Expression too large to display}\)	\(1770\)

Input:

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

Output:

1/d*(1/3*a^2*(cos(d*x+c)^2+2)*sin(d*x+c)+2*a*b*(1/4*(cos(d*x+c)^3+3/2*cos( 
d*x+c))*sin(d*x+c)+3/8*d*x+3/8*c)+1/5*b^2*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c) 
^2)*sin(d*x+c))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.77 \[ \int \cos ^3(c+d x) (a+b \cos (c+d x))^2 \, dx=\frac {45 \, a b d x + {\left (12 \, b^{2} \cos \left (d x + c\right )^{4} + 30 \, a b \cos \left (d x + c\right )^{3} + 45 \, a b \cos \left (d x + c\right ) + 4 \, {\left (5 \, a^{2} + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 40 \, a^{2} + 32 \, b^{2}\right )} \sin \left (d x + c\right )}{60 \, d} \] Input:

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

Output:

1/60*(45*a*b*d*x + (12*b^2*cos(d*x + c)^4 + 30*a*b*cos(d*x + c)^3 + 45*a*b 
*cos(d*x + c) + 4*(5*a^2 + 4*b^2)*cos(d*x + c)^2 + 40*a^2 + 32*b^2)*sin(d* 
x + c))/d
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 221 vs. \(2 (104) = 208\).

Time = 0.22 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.99 \[ \int \cos ^3(c+d x) (a+b \cos (c+d x))^2 \, dx=\begin {cases} \frac {2 a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {a^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {3 a b x \sin ^{4}{\left (c + d x \right )}}{4} + \frac {3 a b x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2} + \frac {3 a b x \cos ^{4}{\left (c + d x \right )}}{4} + \frac {3 a b \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{4 d} + \frac {5 a b \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{4 d} + \frac {8 b^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {b^{2} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \cos {\left (c \right )}\right )^{2} \cos ^{3}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:

integrate(cos(d*x+c)**3*(a+b*cos(d*x+c))**2,x)
 

Output:

Piecewise((2*a**2*sin(c + d*x)**3/(3*d) + a**2*sin(c + d*x)*cos(c + d*x)** 
2/d + 3*a*b*x*sin(c + d*x)**4/4 + 3*a*b*x*sin(c + d*x)**2*cos(c + d*x)**2/ 
2 + 3*a*b*x*cos(c + d*x)**4/4 + 3*a*b*sin(c + d*x)**3*cos(c + d*x)/(4*d) + 
 5*a*b*sin(c + d*x)*cos(c + d*x)**3/(4*d) + 8*b**2*sin(c + d*x)**5/(15*d) 
+ 4*b**2*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + b**2*sin(c + d*x)*cos(c + 
 d*x)**4/d, Ne(d, 0)), (x*(a + b*cos(c))**2*cos(c)**3, True))
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.85 \[ \int \cos ^3(c+d x) (a+b \cos (c+d x))^2 \, dx=-\frac {80 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{2} - 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a b - 16 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} b^{2}}{240 \, d} \] Input:

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

Output:

-1/240*(80*(sin(d*x + c)^3 - 3*sin(d*x + c))*a^2 - 15*(12*d*x + 12*c + sin 
(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a*b - 16*(3*sin(d*x + c)^5 - 10*sin(d* 
x + c)^3 + 15*sin(d*x + c))*b^2)/d
 

Giac [A] (verification not implemented)

Time = 0.50 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.92 \[ \int \cos ^3(c+d x) (a+b \cos (c+d x))^2 \, dx=\frac {3}{4} \, a b x + \frac {b^{2} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {a b \sin \left (4 \, d x + 4 \, c\right )}{16 \, d} + \frac {a b \sin \left (2 \, d x + 2 \, c\right )}{2 \, d} + \frac {{\left (4 \, a^{2} + 5 \, b^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {{\left (6 \, a^{2} + 5 \, b^{2}\right )} \sin \left (d x + c\right )}{8 \, d} \] Input:

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

Output:

3/4*a*b*x + 1/80*b^2*sin(5*d*x + 5*c)/d + 1/16*a*b*sin(4*d*x + 4*c)/d + 1/ 
2*a*b*sin(2*d*x + 2*c)/d + 1/48*(4*a^2 + 5*b^2)*sin(3*d*x + 3*c)/d + 1/8*( 
6*a^2 + 5*b^2)*sin(d*x + c)/d
 

Mupad [B] (verification not implemented)

Time = 39.70 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.05 \[ \int \cos ^3(c+d x) (a+b \cos (c+d x))^2 \, dx=\frac {3\,a^2\,\sin \left (c+d\,x\right )}{4\,d}+\frac {5\,b^2\,\sin \left (c+d\,x\right )}{8\,d}+\frac {3\,a\,b\,x}{4}+\frac {a^2\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {5\,b^2\,\sin \left (3\,c+3\,d\,x\right )}{48\,d}+\frac {b^2\,\sin \left (5\,c+5\,d\,x\right )}{80\,d}+\frac {a\,b\,\sin \left (2\,c+2\,d\,x\right )}{2\,d}+\frac {a\,b\,\sin \left (4\,c+4\,d\,x\right )}{16\,d} \] Input:

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

Output:

(3*a^2*sin(c + d*x))/(4*d) + (5*b^2*sin(c + d*x))/(8*d) + (3*a*b*x)/4 + (a 
^2*sin(3*c + 3*d*x))/(12*d) + (5*b^2*sin(3*c + 3*d*x))/(48*d) + (b^2*sin(5 
*c + 5*d*x))/(80*d) + (a*b*sin(2*c + 2*d*x))/(2*d) + (a*b*sin(4*c + 4*d*x) 
)/(16*d)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.96 \[ \int \cos ^3(c+d x) (a+b \cos (c+d x))^2 \, dx=\frac {-30 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a b +75 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a b +12 \sin \left (d x +c \right )^{5} b^{2}-20 \sin \left (d x +c \right )^{3} a^{2}-40 \sin \left (d x +c \right )^{3} b^{2}+60 \sin \left (d x +c \right ) a^{2}+60 \sin \left (d x +c \right ) b^{2}+45 a b d x}{60 d} \] Input:

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

Output:

( - 30*cos(c + d*x)*sin(c + d*x)**3*a*b + 75*cos(c + d*x)*sin(c + d*x)*a*b 
 + 12*sin(c + d*x)**5*b**2 - 20*sin(c + d*x)**3*a**2 - 40*sin(c + d*x)**3* 
b**2 + 60*sin(c + d*x)*a**2 + 60*sin(c + d*x)*b**2 + 45*a*b*d*x)/(60*d)