\(\int \cos (c+d x) (a+b \sin (c+d x))^m \, dx\) [633]

Optimal result

Integrand size = 19, antiderivative size = 26 \[ \int \cos (c+d x) (a+b \sin (c+d x))^m \, dx=\frac {(a+b \sin (c+d x))^{1+m}}{b d (1+m)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2747, 32} \[ \int \cos (c+d x) (a+b \sin (c+d x))^m \, dx=\frac {(a+b \sin (c+d x))^{m+1}}{b d (m+1)} \]

[In]

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

[Out]

(a + b*Sin[c + d*x])^(1 + m)/(b*d*(1 + m))

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 2747

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*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int (a+x)^m \, dx,x,b \sin (c+d x)\right )}{b d} \\ & = \frac {(a+b \sin (c+d x))^{1+m}}{b d (1+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \cos (c+d x) (a+b \sin (c+d x))^m \, dx=\frac {(a+b \sin (c+d x))^{1+m}}{b d (1+m)} \]

[In]

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

[Out]

(a + b*Sin[c + d*x])^(1 + m)/(b*d*(1 + m))

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04

method	result	size
derivativedivides	\(\frac {\left (a +b \sin \left (d x +c \right )\right )^{1+m}}{b d \left (1+m \right )}\)	\(27\)
default	\(\frac {\left (a +b \sin \left (d x +c \right )\right )^{1+m}}{b d \left (1+m \right )}\)	\(27\)
parallelrisch	\(\frac {\left (a +b \sin \left (d x +c \right )\right )^{1+m}}{b d \left (1+m \right )}\)	\(27\)
norman	\(\frac {\frac {a \,{\mathrm e}^{m \ln \left (a +\frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\right )}}{b d \left (1+m \right )}+\frac {a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) {\mathrm e}^{m \ln \left (a +\frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\right )}}{b d \left (1+m \right )}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) {\mathrm e}^{m \ln \left (a +\frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\right )}}{\left (1+m \right ) d}}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\)	\(173\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.27 \[ \int \cos (c+d x) (a+b \sin (c+d x))^m \, dx=\frac {{\left (b \sin \left (d x + c\right ) + a\right )} {\left (b \sin \left (d x + c\right ) + a\right )}^{m}}{b d m + b d} \]

[In]

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

[Out]

(b*sin(d*x + c) + a)*(b*sin(d*x + c) + a)^m/(b*d*m + b*d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 99 vs. \(2 (19) = 38\).

Time = 0.55 (sec) , antiderivative size = 99, normalized size of antiderivative = 3.81 \[ \int \cos (c+d x) (a+b \sin (c+d x))^m \, dx=\begin {cases} \frac {x \cos {\left (c \right )}}{a} & \text {for}\: b = 0 \wedge d = 0 \wedge m = -1 \\\frac {a^{m} \sin {\left (c + d x \right )}}{d} & \text {for}\: b = 0 \\x \left (a + b \sin {\left (c \right )}\right )^{m} \cos {\left (c \right )} & \text {for}\: d = 0 \\\frac {\log {\left (\frac {a}{b} + \sin {\left (c + d x \right )} \right )}}{b d} & \text {for}\: m = -1 \\\frac {a \left (a + b \sin {\left (c + d x \right )}\right )^{m}}{b d m + b d} + \frac {b \left (a + b \sin {\left (c + d x \right )}\right )^{m} \sin {\left (c + d x \right )}}{b d m + b d} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((x*cos(c)/a, Eq(b, 0) & Eq(d, 0) & Eq(m, -1)), (a**m*sin(c + d*x)/d, Eq(b, 0)), (x*(a + b*sin(c))**m
*cos(c), Eq(d, 0)), (log(a/b + sin(c + d*x))/(b*d), Eq(m, -1)), (a*(a + b*sin(c + d*x))**m/(b*d*m + b*d) + b*(
a + b*sin(c + d*x))**m*sin(c + d*x)/(b*d*m + b*d), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \cos (c+d x) (a+b \sin (c+d x))^m \, dx=\frac {{\left (b \sin \left (d x + c\right ) + a\right )}^{m + 1}}{b d {\left (m + 1\right )}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \cos (c+d x) (a+b \sin (c+d x))^m \, dx=\frac {{\left (b \sin \left (d x + c\right ) + a\right )}^{m + 1}}{b d {\left (m + 1\right )}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 6.16 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \cos (c+d x) (a+b \sin (c+d x))^m \, dx=\frac {{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{m+1}}{b\,d\,\left (m+1\right )} \]

[In]

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

[Out]

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