\(\int \cos (c+d x) (a+a \sin (c+d x)) \, dx\) [7]

Optimal result

Integrand size = 17, antiderivative size = 22 \[ \int \cos (c+d x) (a+a \sin (c+d x)) \, dx=\frac {(a+a \sin (c+d x))^2}{2 a d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {2746} \[ \int \cos (c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \sin ^2(c+d x)}{2 d}+\frac {a \sin (c+d x)}{d} \]

[In]

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

[Out]

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

Rule 2746

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.77 \[ \int \cos (c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \cos ^2(c+d x)}{2 d}+\frac {a \cos (d x) \sin (c)}{d}+\frac {a \cos (c) \sin (d x)}{d} \]

[In]

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

[Out]

-1/2*(a*Cos[c + d*x]^2)/d + (a*Cos[d*x]*Sin[c])/d + (a*Cos[c]*Sin[d*x])/d

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05

method	result	size
derivativedivides	\(\frac {a \left (\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}+\sin \left (d x +c \right )\right )}{d}\)	\(23\)
default	\(\frac {a \left (\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}+\sin \left (d x +c \right )\right )}{d}\)	\(23\)
parallelrisch	\(-\frac {a \left (-1+\cos \left (2 d x +2 c \right )-4 \sin \left (d x +c \right )\right )}{4 d}\)	\(26\)
risch	\(\frac {a \sin \left (d x +c \right )}{d}-\frac {a \cos \left (2 d x +2 c \right )}{4 d}\)	\(28\)
norman	\(\frac {\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}\)	\(67\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \cos (c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \cos \left (d x + c\right )^{2} - 2 \, a \sin \left (d x + c\right )}{2 \, d} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 34 vs. \(2 (15) = 30\).

Time = 0.08 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.55 \[ \int \cos (c+d x) (a+a \sin (c+d x)) \, dx=\begin {cases} \frac {a \sin ^{2}{\left (c + d x \right )}}{2 d} + \frac {a \sin {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right ) \cos {\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \cos (c+d x) (a+a \sin (c+d x)) \, dx=\frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{2}}{2 \, a d} \]

[In]

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

[Out]

1/2*(a*sin(d*x + c) + a)^2/(a*d)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \cos (c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \sin \left (d x + c\right )^{2} + 2 \, a \sin \left (d x + c\right )}{2 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \cos (c+d x) (a+a \sin (c+d x)) \, dx=\frac {a\,\sin \left (c+d\,x\right )\,\left (\sin \left (c+d\,x\right )+2\right )}{2\,d} \]

[In]

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

[Out]

(a*sin(c + d*x)*(sin(c + d*x) + 2))/(2*d)