\(\int \cos (e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx\) [1023]

Optimal result

Integrand size = 29, antiderivative size = 59 \[ \int \cos (e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\frac {(A-B) (a+a \sin (e+f x))^{1+m}}{a f (1+m)}+\frac {B (a+a \sin (e+f x))^{2+m}}{a^2 f (2+m)} \]

[Out]

(A-B)*(a+a*sin(f*x+e))^(1+m)/a/f/(1+m)+B*(a+a*sin(f*x+e))^(2+m)/a^2/f/(2+m)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {2912, 45} \[ \int \cos (e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\frac {B (a \sin (e+f x)+a)^{m+2}}{a^2 f (m+2)}+\frac {(A-B) (a \sin (e+f x)+a)^{m+1}}{a f (m+1)} \]

[In]

Int[Cos[e + f*x]*(a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x]),x]

[Out]

((A - B)*(a + a*Sin[e + f*x])^(1 + m))/(a*f*(1 + m)) + (B*(a + a*Sin[e + f*x])^(2 + m))/(a^2*f*(2 + m))

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2912

Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)
])^(n_.), x_Symbol] :> Dist[1/(b*f), Subst[Int[(a + x)^m*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[
{a, b, c, d, e, f, m, n}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int (a+x)^m \left (A+\frac {B x}{a}\right ) \, dx,x,a \sin (e+f x)\right )}{a f} \\ & = \frac {\text {Subst}\left (\int \left ((A-B) (a+x)^m+\frac {B (a+x)^{1+m}}{a}\right ) \, dx,x,a \sin (e+f x)\right )}{a f} \\ & = \frac {(A-B) (a+a \sin (e+f x))^{1+m}}{a f (1+m)}+\frac {B (a+a \sin (e+f x))^{2+m}}{a^2 f (2+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.86 \[ \int \cos (e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\frac {(a (1+\sin (e+f x)))^{1+m} (-B+A (2+m)+B (1+m) \sin (e+f x))}{a f (1+m) (2+m)} \]

[In]

Integrate[Cos[e + f*x]*(a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x]),x]

[Out]

((a*(1 + Sin[e + f*x]))^(1 + m)*(-B + A*(2 + m) + B*(1 + m)*Sin[e + f*x]))/(a*f*(1 + m)*(2 + m))

Maple [A] (verified)

Time = 1.16 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.22

method	result	size
parallelrisch	\(\frac {\left (a \left (1+\sin \left (f x +e \right )\right )\right )^{m} \left (-\frac {B \left (1+m \right ) \cos \left (2 f x +2 e \right )}{2}+\left (\left (A +B \right ) m +2 A \right ) \sin \left (f x +e \right )+\left (A +\frac {B}{2}\right ) m +2 A -\frac {B}{2}\right )}{f \left (m^{2}+3 m +2\right )}\)	\(72\)
derivativedivides	\(\frac {\left (A m +2 A -B \right ) {\mathrm e}^{m \ln \left (a +a \sin \left (f x +e \right )\right )}}{f \left (m^{2}+3 m +2\right )}+\frac {B \left (\sin ^{2}\left (f x +e \right )\right ) {\mathrm e}^{m \ln \left (a +a \sin \left (f x +e \right )\right )}}{f \left (2+m \right )}+\frac {\left (A m +B m +2 A \right ) \sin \left (f x +e \right ) {\mathrm e}^{m \ln \left (a +a \sin \left (f x +e \right )\right )}}{f \left (m^{2}+3 m +2\right )}\)	\(116\)
default	\(\frac {\left (A m +2 A -B \right ) {\mathrm e}^{m \ln \left (a +a \sin \left (f x +e \right )\right )}}{f \left (m^{2}+3 m +2\right )}+\frac {B \left (\sin ^{2}\left (f x +e \right )\right ) {\mathrm e}^{m \ln \left (a +a \sin \left (f x +e \right )\right )}}{f \left (2+m \right )}+\frac {\left (A m +B m +2 A \right ) \sin \left (f x +e \right ) {\mathrm e}^{m \ln \left (a +a \sin \left (f x +e \right )\right )}}{f \left (m^{2}+3 m +2\right )}\)	\(116\)

[In]

int(cos(f*x+e)*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x,method=_RETURNVERBOSE)

[Out]

(a*(1+sin(f*x+e)))^m*(-1/2*B*(1+m)*cos(2*f*x+2*e)+((A+B)*m+2*A)*sin(f*x+e)+(A+1/2*B)*m+2*A-1/2*B)/f/(m^2+3*m+2
)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.19 \[ \int \cos (e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=-\frac {{\left ({\left (B m + B\right )} \cos \left (f x + e\right )^{2} - {\left (A + B\right )} m - {\left ({\left (A + B\right )} m + 2 \, A\right )} \sin \left (f x + e\right ) - 2 \, A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{f m^{2} + 3 \, f m + 2 \, f} \]

[In]

integrate(cos(f*x+e)*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x, algorithm="fricas")

[Out]

-((B*m + B)*cos(f*x + e)^2 - (A + B)*m - ((A + B)*m + 2*A)*sin(f*x + e) - 2*A)*(a*sin(f*x + e) + a)^m/(f*m^2 +
3*f*m + 2*f)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 428 vs. \(2 (46) = 92\).

Time = 1.08 (sec) , antiderivative size = 428, normalized size of antiderivative = 7.25 \[ \int \cos (e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\begin {cases} x \left (A + B \sin {\left (e \right )}\right ) \left (a \sin {\left (e \right )} + a\right )^{m} \cos {\left (e \right )} & \text {for}\: f = 0 \\- \frac {A}{a^{2} f \sin {\left (e + f x \right )} + a^{2} f} + \frac {B \log {\left (\sin {\left (e + f x \right )} + 1 \right )} \sin {\left (e + f x \right )}}{a^{2} f \sin {\left (e + f x \right )} + a^{2} f} + \frac {B \log {\left (\sin {\left (e + f x \right )} + 1 \right )}}{a^{2} f \sin {\left (e + f x \right )} + a^{2} f} + \frac {B}{a^{2} f \sin {\left (e + f x \right )} + a^{2} f} & \text {for}\: m = -2 \\\frac {A \log {\left (\sin {\left (e + f x \right )} + 1 \right )}}{a f} - \frac {B \log {\left (\sin {\left (e + f x \right )} + 1 \right )}}{a f} + \frac {B \sin {\left (e + f x \right )}}{a f} & \text {for}\: m = -1 \\\frac {A m \left (a \sin {\left (e + f x \right )} + a\right )^{m} \sin {\left (e + f x \right )}}{f m^{2} + 3 f m + 2 f} + \frac {A m \left (a \sin {\left (e + f x \right )} + a\right )^{m}}{f m^{2} + 3 f m + 2 f} + \frac {2 A \left (a \sin {\left (e + f x \right )} + a\right )^{m} \sin {\left (e + f x \right )}}{f m^{2} + 3 f m + 2 f} + \frac {2 A \left (a \sin {\left (e + f x \right )} + a\right )^{m}}{f m^{2} + 3 f m + 2 f} + \frac {B m \left (a \sin {\left (e + f x \right )} + a\right )^{m} \sin ^{2}{\left (e + f x \right )}}{f m^{2} + 3 f m + 2 f} + \frac {B m \left (a \sin {\left (e + f x \right )} + a\right )^{m} \sin {\left (e + f x \right )}}{f m^{2} + 3 f m + 2 f} + \frac {B \left (a \sin {\left (e + f x \right )} + a\right )^{m} \sin ^{2}{\left (e + f x \right )}}{f m^{2} + 3 f m + 2 f} - \frac {B \left (a \sin {\left (e + f x \right )} + a\right )^{m}}{f m^{2} + 3 f m + 2 f} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(f*x+e)*(a+a*sin(f*x+e))**m*(A+B*sin(f*x+e)),x)

[Out]

Piecewise((x*(A + B*sin(e))*(a*sin(e) + a)**m*cos(e), Eq(f, 0)), (-A/(a**2*f*sin(e + f*x) + a**2*f) + B*log(si
n(e + f*x) + 1)*sin(e + f*x)/(a**2*f*sin(e + f*x) + a**2*f) + B*log(sin(e + f*x) + 1)/(a**2*f*sin(e + f*x) + a
**2*f) + B/(a**2*f*sin(e + f*x) + a**2*f), Eq(m, -2)), (A*log(sin(e + f*x) + 1)/(a*f) - B*log(sin(e + f*x) + 1
)/(a*f) + B*sin(e + f*x)/(a*f), Eq(m, -1)), (A*m*(a*sin(e + f*x) + a)**m*sin(e + f*x)/(f*m**2 + 3*f*m + 2*f) +
A*m*(a*sin(e + f*x) + a)**m/(f*m**2 + 3*f*m + 2*f) + 2*A*(a*sin(e + f*x) + a)**m*sin(e + f*x)/(f*m**2 + 3*f*m
+ 2*f) + 2*A*(a*sin(e + f*x) + a)**m/(f*m**2 + 3*f*m + 2*f) + B*m*(a*sin(e + f*x) + a)**m*sin(e + f*x)**2/(f*
m**2 + 3*f*m + 2*f) + B*m*(a*sin(e + f*x) + a)**m*sin(e + f*x)/(f*m**2 + 3*f*m + 2*f) + B*(a*sin(e + f*x) + a)
**m*sin(e + f*x)**2/(f*m**2 + 3*f*m + 2*f) - B*(a*sin(e + f*x) + a)**m/(f*m**2 + 3*f*m + 2*f), True))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.41 \[ \int \cos (e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\frac {\frac {{\left (a^{m} {\left (m + 1\right )} \sin \left (f x + e\right )^{2} + a^{m} m \sin \left (f x + e\right ) - a^{m}\right )} B {\left (\sin \left (f x + e\right ) + 1\right )}^{m}}{m^{2} + 3 \, m + 2} + \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{m + 1} A}{a {\left (m + 1\right )}}}{f} \]

[In]

integrate(cos(f*x+e)*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x, algorithm="maxima")

[Out]

((a^m*(m + 1)*sin(f*x + e)^2 + a^m*m*sin(f*x + e) - a^m)*B*(sin(f*x + e) + 1)^m/(m^2 + 3*m + 2) + (a*sin(f*x +
e) + a)^(m + 1)*A/(a*(m + 1)))/f

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 147 vs. \(2 (59) = 118\).

Time = 0.43 (sec) , antiderivative size = 147, normalized size of antiderivative = 2.49 \[ \int \cos (e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\frac {\frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{m + 1} A}{m + 1} + \frac {{\left ({\left (a \sin \left (f x + e\right ) + a\right )}^{2} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} m - {\left (a \sin \left (f x + e\right ) + a\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} a m + {\left (a \sin \left (f x + e\right ) + a\right )}^{2} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} - 2 \, {\left (a \sin \left (f x + e\right ) + a\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} a\right )} B}{{\left (m^{2} + 3 \, m + 2\right )} a}}{a f} \]

[In]

integrate(cos(f*x+e)*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x, algorithm="giac")

[Out]

((a*sin(f*x + e) + a)^(m + 1)*A/(m + 1) + ((a*sin(f*x + e) + a)^2*(a*sin(f*x + e) + a)^m*m - (a*sin(f*x + e) +
a)*(a*sin(f*x + e) + a)^m*a*m + (a*sin(f*x + e) + a)^2*(a*sin(f*x + e) + a)^m - 2*(a*sin(f*x + e) + a)*(a*sin
(f*x + e) + a)^m*a)*B/((m^2 + 3*m + 2)*a))/(a*f)

Mupad [B] (verification not implemented)

Time = 10.32 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.68 \[ \int \cos (e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\frac {{\left (a\,\left (\sin \left (e+f\,x\right )+1\right )\right )}^m\,\left (4\,A-B+2\,A\,m+B\,m+4\,A\,\sin \left (e+f\,x\right )+B\,\left (2\,{\sin \left (e+f\,x\right )}^2-1\right )+2\,A\,m\,\sin \left (e+f\,x\right )+2\,B\,m\,\sin \left (e+f\,x\right )+B\,m\,\left (2\,{\sin \left (e+f\,x\right )}^2-1\right )\right )}{2\,f\,\left (m^2+3\,m+2\right )} \]

[In]

int(cos(e + f*x)*(A + B*sin(e + f*x))*(a + a*sin(e + f*x))^m,x)

[Out]

((a*(sin(e + f*x) + 1))^m*(4*A - B + 2*A*m + B*m + 4*A*sin(e + f*x) + B*(2*sin(e + f*x)^2 - 1) + 2*A*m*sin(e +
f*x) + 2*B*m*sin(e + f*x) + B*m*(2*sin(e + f*x)^2 - 1)))/(2*f*(3*m + m^2 + 2))