3.3.82 \(\int \frac {\cos ^2(c+d x) (a B+b B \cos (c+d x))}{a+b \cos (c+d x)} \, dx\) [282]

3.3.82.1 Optimal result

Integrand size = 34, antiderivative size = 27 \[ \int \frac {\cos ^2(c+d x) (a B+b B \cos (c+d x))}{a+b \cos (c+d x)} \, dx=\frac {B x}{2}+\frac {B \cos (c+d x) \sin (c+d x)}{2 d} \]

output

1/2*B*x+1/2*B*cos(d*x+c)*sin(d*x+c)/d
 

3.3.82.2 Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {\cos ^2(c+d x) (a B+b B \cos (c+d x))}{a+b \cos (c+d x)} \, dx=\frac {B (2 (c+d x)+\sin (2 (c+d x)))}{4 d} \]

input

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

output

(B*(2*(c + d*x) + Sin[2*(c + d*x)]))/(4*d)
 

3.3.82.3 Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2011, 3042, 3115, 24}

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.

\(\displaystyle \int \frac {\cos ^2(c+d x) (a B+b B \cos (c+d x))}{a+b \cos (c+d x)} \, dx\)

\(\Big \downarrow \) 2011

\(\displaystyle B \int \cos ^2(c+d x)dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3115

\(\displaystyle B \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )\)

\(\Big \downarrow \) 24

\(\displaystyle B \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\)

input

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

output

B*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d))
 

3.3.82.3.1 Defintions of rubi rules used

rule 24

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

rule 2011

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> 
 Simp[(b/d)^m   Int[u*(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c + d*x 
, a + b*x])
 

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]
 

3.3.82.4 Maple [A] (verified)

Time = 0.81 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78

method	result	size
risch	\(\frac {x B}{2}+\frac {B \sin \left (2 d x +2 c \right )}{4 d}\)	\(21\)
parallelrisch	\(\frac {B \left (2 d x +\sin \left (2 d x +2 c \right )\right )}{4 d}\)	\(21\)
derivativedivides	\(\frac {B \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(28\)
default	\(\frac {B \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(28\)
norman	\(\frac {\frac {B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {x B}{2}-\frac {B \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {3 x B \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {3 x B \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {x B \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}\)	\(98\)

input

int(cos(d*x+c)^2*(B*a+b*B*cos(d*x+c))/(a+cos(d*x+c)*b),x,method=_RETURNVER 
BOSE)
 

output

1/2*x*B+1/4*B/d*sin(2*d*x+2*c)
 

3.3.82.5 Fricas [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {\cos ^2(c+d x) (a B+b B \cos (c+d x))}{a+b \cos (c+d x)} \, dx=\frac {B d x + B \cos \left (d x + c\right ) \sin \left (d x + c\right )}{2 \, d} \]

input

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

output

1/2*(B*d*x + B*cos(d*x + c)*sin(d*x + c))/d
 

3.3.82.6 Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (22) = 44\).

Time = 0.39 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.52 \[ \int \frac {\cos ^2(c+d x) (a B+b B \cos (c+d x))}{a+b \cos (c+d x)} \, dx=\begin {cases} \frac {B x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {B x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {B \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\\frac {x \left (B a + B b \cos {\left (c \right )}\right ) \cos ^{2}{\left (c \right )}}{a + b \cos {\left (c \right )}} & \text {otherwise} \end {cases} \]

input

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

output

Piecewise((B*x*sin(c + d*x)**2/2 + B*x*cos(c + d*x)**2/2 + B*sin(c + d*x)* 
cos(c + d*x)/(2*d), Ne(d, 0)), (x*(B*a + B*b*cos(c))*cos(c)**2/(a + b*cos( 
c)), True))
 

3.3.82.7 Maxima [F(-2)]

Exception generated. \[ \int \frac {\cos ^2(c+d x) (a B+b B \cos (c+d x))}{a+b \cos (c+d x)} \, dx=\text {Exception raised: ValueError} \]

input

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

output

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f 
or more de
 

3.3.82.8 Giac [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {\cos ^2(c+d x) (a B+b B \cos (c+d x))}{a+b \cos (c+d x)} \, dx=\frac {{\left (d x + c\right )} B + \frac {B \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1}}{2 \, d} \]

input

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

output

1/2*((d*x + c)*B + B*tan(d*x + c)/(tan(d*x + c)^2 + 1))/d
 

3.3.82.9 Mupad [B] (verification not implemented)

Time = 0.91 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.85 \[ \int \frac {\cos ^2(c+d x) (a B+b B \cos (c+d x))}{a+b \cos (c+d x)} \, dx=\frac {B\,x}{2}+\frac {B\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-B\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^2} \]

input

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

output

(B*x)/2 + (B*tan(c/2 + (d*x)/2) - B*tan(c/2 + (d*x)/2)^3)/(d*(tan(c/2 + (d 
*x)/2)^2 + 1)^2)