3.1021 \(\int \cos ^5(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx\)

Optimal. Leaf size=123 \[ \frac{4 (A-B) (a \sin (e+f x)+a)^{m+3}}{a^3 f (m+3)}-\frac{4 (A-2 B) (a \sin (e+f x)+a)^{m+4}}{a^4 f (m+4)}+\frac{(A-5 B) (a \sin (e+f x)+a)^{m+5}}{a^5 f (m+5)}+\frac{B (a \sin (e+f x)+a)^{m+6}}{a^6 f (m+6)} \]

[Out]

(4*(A - B)*(a + a*Sin[e + f*x])^(3 + m))/(a^3*f*(3 + m)) - (4*(A - 2*B)*(a + a*Sin[e + f*x])^(4 + m))/(a^4*f*(
4 + m)) + ((A - 5*B)*(a + a*Sin[e + f*x])^(5 + m))/(a^5*f*(5 + m)) + (B*(a + a*Sin[e + f*x])^(6 + m))/(a^6*f*(
6 + m))

________________________________________________________________________________________

Rubi [A]  time = 0.135305, antiderivative size = 123, 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} \[ \frac{4 (A-B) (a \sin (e+f x)+a)^{m+3}}{a^3 f (m+3)}-\frac{4 (A-2 B) (a \sin (e+f x)+a)^{m+4}}{a^4 f (m+4)}+\frac{(A-5 B) (a \sin (e+f x)+a)^{m+5}}{a^5 f (m+5)}+\frac{B (a \sin (e+f x)+a)^{m+6}}{a^6 f (m+6)} \]

Antiderivative was successfully verified.

[In]

Int[Cos[e + f*x]^5*(a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x]),x]

[Out]

(4*(A - B)*(a + a*Sin[e + f*x])^(3 + m))/(a^3*f*(3 + m)) - (4*(A - 2*B)*(a + a*Sin[e + f*x])^(4 + m))/(a^4*f*(
4 + m)) + ((A - 5*B)*(a + a*Sin[e + f*x])^(5 + m))/(a^5*f*(5 + m)) + (B*(a + a*Sin[e + f*x])^(6 + m))/(a^6*f*(
6 + 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 ^5(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx &=\frac{\operatorname{Subst}\left (\int (a-x)^2 (a+x)^{2+m} \left (A+\frac{B x}{a}\right ) \, dx,x,a \sin (e+f x)\right )}{a^5 f}\\ &=\frac{\operatorname{Subst}\left (\int \left (4 a^2 (A-B) (a+x)^{2+m}-4 a (A-2 B) (a+x)^{3+m}+(A-5 B) (a+x)^{4+m}+\frac{B (a+x)^{5+m}}{a}\right ) \, dx,x,a \sin (e+f x)\right )}{a^5 f}\\ &=\frac{4 (A-B) (a+a \sin (e+f x))^{3+m}}{a^3 f (3+m)}-\frac{4 (A-2 B) (a+a \sin (e+f x))^{4+m}}{a^4 f (4+m)}+\frac{(A-5 B) (a+a \sin (e+f x))^{5+m}}{a^5 f (5+m)}+\frac{B (a+a \sin (e+f x))^{6+m}}{a^6 f (6+m)}\\ \end{align*}

Mathematica [A]  time = 0.382035, size = 103, normalized size = 0.84 \[ \frac{(a (\sin (e+f x)+1))^{m+3} \left (\frac{a^3 (A-5 B) (\sin (e+f x)+1)^2}{m+5}-\frac{4 a^3 (A-2 B) (\sin (e+f x)+1)}{m+4}+\frac{4 a^3 (A-B)}{m+3}+\frac{B (a \sin (e+f x)+a)^3}{m+6}\right )}{a^6 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[e + f*x]^5*(a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x]),x]

[Out]

((a*(1 + Sin[e + f*x]))^(3 + m)*((4*a^3*(A - B))/(3 + m) - (4*a^3*(A - 2*B)*(1 + Sin[e + f*x]))/(4 + m) + (a^3
*(A - 5*B)*(1 + Sin[e + f*x])^2)/(5 + m) + (B*(a + a*Sin[e + f*x])^3)/(6 + m)))/(a^6*f)

________________________________________________________________________________________

Maple [F]  time = 5.923, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \left ( fx+e \right ) \right ) ^{5} \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)^5*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x)

[Out]

int(cos(f*x+e)^5*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x)

________________________________________________________________________________________

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)^5*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.71917, size = 563, normalized size = 4.58 \begin{align*} -\frac{{\left ({\left (B m^{3} + 12 \, B m^{2} + 47 \, B m + 60 \, B\right )} \cos \left (f x + e\right )^{6} -{\left ({\left (A + B\right )} m^{3} + 3 \,{\left (3 \, A + B\right )} m^{2} + 18 \, A m\right )} \cos \left (f x + e\right )^{4} - 8 \,{\left ({\left (A + B\right )} m^{2} + 6 \, A m\right )} \cos \left (f x + e\right )^{2} - 32 \,{\left (A + B\right )} m -{\left ({\left ({\left (A + B\right )} m^{3} +{\left (13 \, A + 7 \, B\right )} m^{2} + 6 \,{\left (9 \, A + 2 \, B\right )} m + 72 \, A\right )} \cos \left (f x + e\right )^{4} + 8 \,{\left ({\left (A + B\right )} m^{2} + 2 \,{\left (4 \, A + B\right )} m + 12 \, A\right )} \cos \left (f x + e\right )^{2} + 32 \,{\left (A + B\right )} m + 192 \, A\right )} \sin \left (f x + e\right ) - 192 \, A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{f m^{4} + 18 \, f m^{3} + 119 \, f m^{2} + 342 \, f m + 360 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^5*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x, algorithm="fricas")

[Out]

-((B*m^3 + 12*B*m^2 + 47*B*m + 60*B)*cos(f*x + e)^6 - ((A + B)*m^3 + 3*(3*A + B)*m^2 + 18*A*m)*cos(f*x + e)^4
- 8*((A + B)*m^2 + 6*A*m)*cos(f*x + e)^2 - 32*(A + B)*m - (((A + B)*m^3 + (13*A + 7*B)*m^2 + 6*(9*A + 2*B)*m +
72*A)*cos(f*x + e)^4 + 8*((A + B)*m^2 + 2*(4*A + B)*m + 12*A)*cos(f*x + e)^2 + 32*(A + B)*m + 192*A)*sin(f*x
+ e) - 192*A)*(a*sin(f*x + e) + a)^m/(f*m^4 + 18*f*m^3 + 119*f*m^2 + 342*f*m + 360*f)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.2915, size = 1162, normalized size = 9.45 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^5*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x, algorithm="giac")

[Out]

(((a*sin(f*x + e) + a)^5*(a*sin(f*x + e) + a)^m*m^2 - 4*(a*sin(f*x + e) + a)^4*(a*sin(f*x + e) + a)^m*a*m^2 +
4*(a*sin(f*x + e) + a)^3*(a*sin(f*x + e) + a)^m*a^2*m^2 + 7*(a*sin(f*x + e) + a)^5*(a*sin(f*x + e) + a)^m*m -
32*(a*sin(f*x + e) + a)^4*(a*sin(f*x + e) + a)^m*a*m + 36*(a*sin(f*x + e) + a)^3*(a*sin(f*x + e) + a)^m*a^2*m
+ 12*(a*sin(f*x + e) + a)^5*(a*sin(f*x + e) + a)^m - 60*(a*sin(f*x + e) + a)^4*(a*sin(f*x + e) + a)^m*a + 80*(
a*sin(f*x + e) + a)^3*(a*sin(f*x + e) + a)^m*a^2)*A/(a^4*m^3 + 12*a^4*m^2 + 47*a^4*m + 60*a^4) + ((a*sin(f*x +
e) + a)^6*(a*sin(f*x + e) + a)^m*m^3 - 5*(a*sin(f*x + e) + a)^5*(a*sin(f*x + e) + a)^m*a*m^3 + 8*(a*sin(f*x +
e) + a)^4*(a*sin(f*x + e) + a)^m*a^2*m^3 - 4*(a*sin(f*x + e) + a)^3*(a*sin(f*x + e) + a)^m*a^3*m^3 + 12*(a*si
n(f*x + e) + a)^6*(a*sin(f*x + e) + a)^m*m^2 - 65*(a*sin(f*x + e) + a)^5*(a*sin(f*x + e) + a)^m*a*m^2 + 112*(a
*sin(f*x + e) + a)^4*(a*sin(f*x + e) + a)^m*a^2*m^2 - 60*(a*sin(f*x + e) + a)^3*(a*sin(f*x + e) + a)^m*a^3*m^2
+ 47*(a*sin(f*x + e) + a)^6*(a*sin(f*x + e) + a)^m*m - 270*(a*sin(f*x + e) + a)^5*(a*sin(f*x + e) + a)^m*a*m
+ 504*(a*sin(f*x + e) + a)^4*(a*sin(f*x + e) + a)^m*a^2*m - 296*(a*sin(f*x + e) + a)^3*(a*sin(f*x + e) + a)^m*
a^3*m + 60*(a*sin(f*x + e) + a)^6*(a*sin(f*x + e) + a)^m - 360*(a*sin(f*x + e) + a)^5*(a*sin(f*x + e) + a)^m*a
+ 720*(a*sin(f*x + e) + a)^4*(a*sin(f*x + e) + a)^m*a^2 - 480*(a*sin(f*x + e) + a)^3*(a*sin(f*x + e) + a)^m*a
^3)*B/((a^4*m^4 + 18*a^4*m^3 + 119*a^4*m^2 + 342*a^4*m + 360*a^4)*a))/(a*f)