3.1023 \(\int \cos (e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx\)

Optimal. Leaf size=59 \[ \frac{B (a \sin (e+f x)+a)^{m+2}}{a^2 f (m+2)}+\frac{(A-B) (a \sin (e+f x)+a)^{m+1}}{a f (m+1)} \]

[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))

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Rubi [A]  time = 0.0705771, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.069, Rules used = {2833, 43} \[ \frac{B (a \sin (e+f x)+a)^{m+2}}{a^2 f (m+2)}+\frac{(A-B) (a \sin (e+f x)+a)^{m+1}}{a f (m+1)} \]

Antiderivative was successfully verified.

[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 2833

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*x)/b)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[
{a, b, c, d, e, f, m, n}, x]

Rule 43

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])

Rubi steps

\begin{align*} \int \cos (e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx &=\frac{\operatorname{Subst}\left (\int (a+x)^m \left (A+\frac{B x}{a}\right ) \, dx,x,a \sin (e+f x)\right )}{a f}\\ &=\frac{\operatorname{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]  time = 0.122547, size = 51, normalized size = 0.86 \[ \frac{(a (\sin (e+f x)+1))^{m+1} (A (m+2)+B (m+1) \sin (e+f x)-B)}{a f (m+1) (m+2)} \]

Antiderivative was successfully verified.

[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))

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Maple [F]  time = 1.661, size = 0, normalized size = 0. \begin{align*} \int \cos \left ( fx+e \right ) \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m} \left ( A+B\sin \left ( fx+e \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.49354, size = 169, normalized size = 2.86 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]  time = 8.30572, size = 428, normalized size = 7.25 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Giac [B]  time = 1.32114, size = 211, normalized size = 3.58 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)