\(\int \cos ^2(c+d x) (a+a \sin (c+d x)) \, dx\) [6]

Optimal result

Integrand size = 19, antiderivative size = 43 \[ \int \cos ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a x}{2}-\frac {a \cos ^3(c+d x)}{3 d}+\frac {a \cos (c+d x) \sin (c+d x)}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2748, 2715, 8} \[ \int \cos ^2(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \cos ^3(c+d x)}{3 d}+\frac {a \sin (c+d x) \cos (c+d x)}{2 d}+\frac {a x}{2} \]

[In]

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

[Out]

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

Rule 8

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

Rule 2715

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] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 2748

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-b)*((g*Co
s[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + Dist[a, Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x
] && (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])

Rubi steps \begin{align*} \text {integral}& = -\frac {a \cos ^3(c+d x)}{3 d}+a \int \cos ^2(c+d x) \, dx \\ & = -\frac {a \cos ^3(c+d x)}{3 d}+\frac {a \cos (c+d x) \sin (c+d x)}{2 d}+\frac {1}{2} a \int 1 \, dx \\ & = \frac {a x}{2}-\frac {a \cos ^3(c+d x)}{3 d}+\frac {a \cos (c+d x) \sin (c+d x)}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.07 \[ \int \cos ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a (c+d x)}{2 d}-\frac {a \cos ^3(c+d x)}{3 d}+\frac {a \sin (2 (c+d x))}{4 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.95

method	result	size
derivativedivides	\(\frac {-\frac {a \left (\cos ^{3}\left (d x +c \right )\right )}{3}+a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(41\)
default	\(\frac {-\frac {a \left (\cos ^{3}\left (d x +c \right )\right )}{3}+a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(41\)
parallelrisch	\(-\frac {a \left (-6 d x +3 \cos \left (d x +c \right )+\cos \left (3 d x +3 c \right )-3 \sin \left (2 d x +2 c \right )+4\right )}{12 d}\)	\(41\)
risch	\(\frac {a x}{2}-\frac {a \cos \left (d x +c \right )}{4 d}-\frac {a \cos \left (3 d x +3 c \right )}{12 d}+\frac {a \sin \left (2 d x +2 c \right )}{4 d}\)	\(48\)
norman	\(\frac {\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {a x}{2}-\frac {2 a}{3 d}-\frac {a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {3 a x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {3 a x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {a x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-\frac {2 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}\)	\(121\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.86 \[ \int \cos ^2(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {2 \, a \cos \left (d x + c\right )^{3} - 3 \, a d x - 3 \, a \cos \left (d x + c\right ) \sin \left (d x + c\right )}{6 \, d} \]

[In]

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

[Out]

-1/6*(2*a*cos(d*x + c)^3 - 3*a*d*x - 3*a*cos(d*x + c)*sin(d*x + c))/d

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.65 \[ \int \cos ^2(c+d x) (a+a \sin (c+d x)) \, dx=\begin {cases} \frac {a x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {a x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {a \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} - \frac {a \cos ^{3}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right ) \cos ^{2}{\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((a*x*sin(c + d*x)**2/2 + a*x*cos(c + d*x)**2/2 + a*sin(c + d*x)*cos(c + d*x)/(2*d) - a*cos(c + d*x)*
*3/(3*d), Ne(d, 0)), (x*(a*sin(c) + a)*cos(c)**2, True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.86 \[ \int \cos ^2(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {4 \, a \cos \left (d x + c\right )^{3} - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a}{12 \, d} \]

[In]

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

[Out]

-1/12*(4*a*cos(d*x + c)^3 - 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*a)/d

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.09 \[ \int \cos ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {1}{2} \, a x - \frac {a \cos \left (3 \, d x + 3 \, c\right )}{12 \, d} - \frac {a \cos \left (d x + c\right )}{4 \, d} + \frac {a \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} \]

[In]

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

[Out]

1/2*a*x - 1/12*a*cos(3*d*x + 3*c)/d - 1/4*a*cos(d*x + c)/d + 1/4*a*sin(2*d*x + 2*c)/d

Mupad [B] (verification not implemented)

Time = 8.90 (sec) , antiderivative size = 103, normalized size of antiderivative = 2.40 \[ \int \cos ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a\,x}{2}+\frac {-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {a\,\left (9\,c+9\,d\,x-12\right )}{6}-\frac {3\,a\,\left (c+d\,x\right )}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a\,\left (3\,c+3\,d\,x-4\right )}{6}-\frac {a\,\left (c+d\,x\right )}{2}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^3} \]

[In]

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

[Out]

(a*x)/2 + ((a*(3*c + 3*d*x - 4))/6 + a*tan(c/2 + (d*x)/2) - (a*(c + d*x))/2 + tan(c/2 + (d*x)/2)^4*((a*(9*c +
9*d*x - 12))/6 - (3*a*(c + d*x))/2) - a*tan(c/2 + (d*x)/2)^5)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^3)