3.1022 \(\int \cos ^3(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx\)

Optimal. Leaf size=93 \[ \frac{2 (A-B) (a \sin (e+f x)+a)^{m+2}}{a^2 f (m+2)}-\frac{(A-3 B) (a \sin (e+f x)+a)^{m+3}}{a^3 f (m+3)}-\frac{B (a \sin (e+f x)+a)^{m+4}}{a^4 f (m+4)} \]

[Out]

(2*(A - B)*(a + a*Sin[e + f*x])^(2 + m))/(a^2*f*(2 + m)) - ((A - 3*B)*(a + a*Sin[e + f*x])^(3 + m))/(a^3*f*(3
+ m)) - (B*(a + a*Sin[e + f*x])^(4 + m))/(a^4*f*(4 + m))

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Rubi [A]  time = 0.117641, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.065, Rules used = {2836, 77} \[ \frac{2 (A-B) (a \sin (e+f x)+a)^{m+2}}{a^2 f (m+2)}-\frac{(A-3 B) (a \sin (e+f x)+a)^{m+3}}{a^3 f (m+3)}-\frac{B (a \sin (e+f x)+a)^{m+4}}{a^4 f (m+4)} \]

Antiderivative was successfully verified.

[In]

Int[Cos[e + f*x]^3*(a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x]),x]

[Out]

(2*(A - B)*(a + a*Sin[e + f*x])^(2 + m))/(a^2*f*(2 + m)) - ((A - 3*B)*(a + a*Sin[e + f*x])^(3 + m))/(a^3*f*(3
+ m)) - (B*(a + a*Sin[e + f*x])^(4 + m))/(a^4*f*(4 + m))

Rule 2836

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d*x)/b
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin{align*} \int \cos ^3(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx &=\frac{\operatorname{Subst}\left (\int (a-x) (a+x)^{1+m} \left (A+\frac{B x}{a}\right ) \, dx,x,a \sin (e+f x)\right )}{a^3 f}\\ &=\frac{\operatorname{Subst}\left (\int \left (2 a (A-B) (a+x)^{1+m}+(-A+3 B) (a+x)^{2+m}-\frac{B (a+x)^{3+m}}{a}\right ) \, dx,x,a \sin (e+f x)\right )}{a^3 f}\\ &=\frac{2 (A-B) (a+a \sin (e+f x))^{2+m}}{a^2 f (2+m)}-\frac{(A-3 B) (a+a \sin (e+f x))^{3+m}}{a^3 f (3+m)}-\frac{B (a+a \sin (e+f x))^{4+m}}{a^4 f (4+m)}\\ \end{align*}

Mathematica [A]  time = 0.297884, size = 93, normalized size = 1. \[ \frac{2 (A-B) (a \sin (e+f x)+a)^{m+2}}{a^2 f (m+2)}-\frac{(A-3 B) (a \sin (e+f x)+a)^{m+3}}{a^3 f (m+3)}-\frac{B (a \sin (e+f x)+a)^{m+4}}{a^4 f (m+4)} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[e + f*x]^3*(a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x]),x]

[Out]

(2*(A - B)*(a + a*Sin[e + f*x])^(2 + m))/(a^2*f*(2 + m)) - ((A - 3*B)*(a + a*Sin[e + f*x])^(3 + m))/(a^3*f*(3
+ m)) - (B*(a + a*Sin[e + f*x])^(4 + m))/(a^4*f*(4 + m))

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Maple [F]  time = 2.612, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \left ( fx+e \right ) \right ) ^{3} \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)^3*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x)

[Out]

int(cos(f*x+e)^3*(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)^3*(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 = 1.5578, size = 333, normalized size = 3.58 \begin{align*} -\frac{{\left ({\left (B m^{2} + 5 \, B m + 6 \, B\right )} \cos \left (f x + e\right )^{4} -{\left ({\left (A + B\right )} m^{2} + 4 \, A m\right )} \cos \left (f x + e\right )^{2} - 4 \,{\left (A + B\right )} m -{\left ({\left ({\left (A + B\right )} m^{2} + 2 \,{\left (3 \, A + B\right )} m + 8 \, A\right )} \cos \left (f x + e\right )^{2} + 4 \,{\left (A + B\right )} m + 16 \, A\right )} \sin \left (f x + e\right ) - 16 \, A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{f m^{3} + 9 \, f m^{2} + 26 \, f m + 24 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^3*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x, algorithm="fricas")

[Out]

-((B*m^2 + 5*B*m + 6*B)*cos(f*x + e)^4 - ((A + B)*m^2 + 4*A*m)*cos(f*x + e)^2 - 4*(A + B)*m - (((A + B)*m^2 +
2*(3*A + B)*m + 8*A)*cos(f*x + e)^2 + 4*(A + B)*m + 16*A)*sin(f*x + e) - 16*A)*(a*sin(f*x + e) + a)^m/(f*m^3 +
9*f*m^2 + 26*f*m + 24*f)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)**3*(a+a*sin(f*x+e))**m*(A+B*sin(f*x+e)),x)

[Out]

Timed out

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Giac [B]  time = 1.27483, size = 618, normalized size = 6.65 \begin{align*} -\frac{\frac{{\left ({\left (a \sin \left (f x + e\right ) + a\right )}^{3}{\left (a \sin \left (f x + e\right ) + a\right )}^{m} m - 2 \,{\left (a \sin \left (f x + e\right ) + a\right )}^{2}{\left (a \sin \left (f x + e\right ) + a\right )}^{m} a m + 2 \,{\left (a \sin \left (f x + e\right ) + a\right )}^{3}{\left (a \sin \left (f x + e\right ) + a\right )}^{m} - 6 \,{\left (a \sin \left (f x + e\right ) + a\right )}^{2}{\left (a \sin \left (f x + e\right ) + a\right )}^{m} a\right )} A}{a^{2} m^{2} + 5 \, a^{2} m + 6 \, a^{2}} + \frac{{\left ({\left (a \sin \left (f x + e\right ) + a\right )}^{4}{\left (a \sin \left (f x + e\right ) + a\right )}^{m} m^{2} - 3 \,{\left (a \sin \left (f x + e\right ) + a\right )}^{3}{\left (a \sin \left (f x + e\right ) + a\right )}^{m} a m^{2} + 2 \,{\left (a \sin \left (f x + e\right ) + a\right )}^{2}{\left (a \sin \left (f x + e\right ) + a\right )}^{m} a^{2} m^{2} + 5 \,{\left (a \sin \left (f x + e\right ) + a\right )}^{4}{\left (a \sin \left (f x + e\right ) + a\right )}^{m} m - 18 \,{\left (a \sin \left (f x + e\right ) + a\right )}^{3}{\left (a \sin \left (f x + e\right ) + a\right )}^{m} a m + 14 \,{\left (a \sin \left (f x + e\right ) + a\right )}^{2}{\left (a \sin \left (f x + e\right ) + a\right )}^{m} a^{2} m + 6 \,{\left (a \sin \left (f x + e\right ) + a\right )}^{4}{\left (a \sin \left (f x + e\right ) + a\right )}^{m} - 24 \,{\left (a \sin \left (f x + e\right ) + a\right )}^{3}{\left (a \sin \left (f x + e\right ) + a\right )}^{m} a + 24 \,{\left (a \sin \left (f x + e\right ) + a\right )}^{2}{\left (a \sin \left (f x + e\right ) + a\right )}^{m} a^{2}\right )} B}{{\left (a^{2} m^{3} + 9 \, a^{2} m^{2} + 26 \, a^{2} m + 24 \, a^{2}\right )} a}}{a f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^3*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x, algorithm="giac")

[Out]

-(((a*sin(f*x + e) + a)^3*(a*sin(f*x + e) + a)^m*m - 2*(a*sin(f*x + e) + a)^2*(a*sin(f*x + e) + a)^m*a*m + 2*(
a*sin(f*x + e) + a)^3*(a*sin(f*x + e) + a)^m - 6*(a*sin(f*x + e) + a)^2*(a*sin(f*x + e) + a)^m*a)*A/(a^2*m^2 +
5*a^2*m + 6*a^2) + ((a*sin(f*x + e) + a)^4*(a*sin(f*x + e) + a)^m*m^2 - 3*(a*sin(f*x + e) + a)^3*(a*sin(f*x +
e) + a)^m*a*m^2 + 2*(a*sin(f*x + e) + a)^2*(a*sin(f*x + e) + a)^m*a^2*m^2 + 5*(a*sin(f*x + e) + a)^4*(a*sin(f
*x + e) + a)^m*m - 18*(a*sin(f*x + e) + a)^3*(a*sin(f*x + e) + a)^m*a*m + 14*(a*sin(f*x + e) + a)^2*(a*sin(f*x
+ e) + a)^m*a^2*m + 6*(a*sin(f*x + e) + a)^4*(a*sin(f*x + e) + a)^m - 24*(a*sin(f*x + e) + a)^3*(a*sin(f*x +
e) + a)^m*a + 24*(a*sin(f*x + e) + a)^2*(a*sin(f*x + e) + a)^m*a^2)*B/((a^2*m^3 + 9*a^2*m^2 + 26*a^2*m + 24*a^
2)*a))/(a*f)