3.321 \(\int \frac{\cos ^2(e+f x) \sin (e+f x)}{(a+a \sin (e+f x))^6} \, dx\)

Optimal. Leaf size=144 \[ \frac{4 \cos (e+f x)}{315 f \left (a^6 \sin (e+f x)+a^6\right )}+\frac{4 \cos (e+f x)}{315 f \left (a^3 \sin (e+f x)+a^3\right )^2}+\frac{2 \cos (e+f x)}{105 f \left (a^2 \sin (e+f x)+a^2\right )^3}-\frac{19 \cos (e+f x)}{63 a^2 f (a \sin (e+f x)+a)^4}+\frac{2 \cos (e+f x)}{9 a f (a \sin (e+f x)+a)^5} \]

[Out]

(2*Cos[e + f*x])/(9*a*f*(a + a*Sin[e + f*x])^5) - (19*Cos[e + f*x])/(63*a^2*f*(a + a*Sin[e + f*x])^4) + (2*Cos
[e + f*x])/(105*f*(a^2 + a^2*Sin[e + f*x])^3) + (4*Cos[e + f*x])/(315*f*(a^3 + a^3*Sin[e + f*x])^2) + (4*Cos[e
+ f*x])/(315*f*(a^6 + a^6*Sin[e + f*x]))

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Rubi [A]  time = 0.162312, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {2857, 2750, 2650, 2648} \[ \frac{4 \cos (e+f x)}{315 f \left (a^6 \sin (e+f x)+a^6\right )}+\frac{4 \cos (e+f x)}{315 f \left (a^3 \sin (e+f x)+a^3\right )^2}+\frac{2 \cos (e+f x)}{105 f \left (a^2 \sin (e+f x)+a^2\right )^3}-\frac{19 \cos (e+f x)}{63 a^2 f (a \sin (e+f x)+a)^4}+\frac{2 \cos (e+f x)}{9 a f (a \sin (e+f x)+a)^5} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[e + f*x]^2*Sin[e + f*x])/(a + a*Sin[e + f*x])^6,x]

[Out]

(2*Cos[e + f*x])/(9*a*f*(a + a*Sin[e + f*x])^5) - (19*Cos[e + f*x])/(63*a^2*f*(a + a*Sin[e + f*x])^4) + (2*Cos
[e + f*x])/(105*f*(a^2 + a^2*Sin[e + f*x])^3) + (4*Cos[e + f*x])/(315*f*(a^3 + a^3*Sin[e + f*x])^2) + (4*Cos[e
+ f*x])/(315*f*(a^6 + a^6*Sin[e + f*x]))

Rule 2857

Int[cos[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_
)]), x_Symbol] :> Simp[(2*(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b^2*f*(2*m + 3)), x] + Dist[
1/(b^3*(2*m + 3)), Int[(a + b*Sin[e + f*x])^(m + 2)*(b*c + 2*a*d*(m + 1) - b*d*(2*m + 3)*Sin[e + f*x]), x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -3/2]

Rule 2750

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((b
*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(a*f*(2*m + 1)), x] + Dist[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1)
), Int[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 -
b^2, 0] && LtQ[m, -2^(-1)]

Rule 2650

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^n)/(a*
d*(2*n + 1)), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int \frac{\cos ^2(e+f x) \sin (e+f x)}{(a+a \sin (e+f x))^6} \, dx &=\frac{2 \cos (e+f x)}{9 a f (a+a \sin (e+f x))^5}-\frac{\int \frac{-10 a+9 a \sin (e+f x)}{(a+a \sin (e+f x))^4} \, dx}{9 a^3}\\ &=\frac{2 \cos (e+f x)}{9 a f (a+a \sin (e+f x))^5}-\frac{19 \cos (e+f x)}{63 a^2 f (a+a \sin (e+f x))^4}-\frac{2 \int \frac{1}{(a+a \sin (e+f x))^3} \, dx}{21 a^3}\\ &=\frac{2 \cos (e+f x)}{9 a f (a+a \sin (e+f x))^5}-\frac{19 \cos (e+f x)}{63 a^2 f (a+a \sin (e+f x))^4}+\frac{2 \cos (e+f x)}{105 f \left (a^2+a^2 \sin (e+f x)\right )^3}-\frac{4 \int \frac{1}{(a+a \sin (e+f x))^2} \, dx}{105 a^4}\\ &=\frac{2 \cos (e+f x)}{9 a f (a+a \sin (e+f x))^5}-\frac{19 \cos (e+f x)}{63 a^2 f (a+a \sin (e+f x))^4}+\frac{2 \cos (e+f x)}{105 f \left (a^2+a^2 \sin (e+f x)\right )^3}+\frac{4 \cos (e+f x)}{315 f \left (a^3+a^3 \sin (e+f x)\right )^2}-\frac{4 \int \frac{1}{a+a \sin (e+f x)} \, dx}{315 a^5}\\ &=\frac{2 \cos (e+f x)}{9 a f (a+a \sin (e+f x))^5}-\frac{19 \cos (e+f x)}{63 a^2 f (a+a \sin (e+f x))^4}+\frac{2 \cos (e+f x)}{105 f \left (a^2+a^2 \sin (e+f x)\right )^3}+\frac{4 \cos (e+f x)}{315 f \left (a^3+a^3 \sin (e+f x)\right )^2}+\frac{4 \cos (e+f x)}{315 f \left (a^6+a^6 \sin (e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.920331, size = 171, normalized size = 1.19 \[ -\frac{2562 \sin \left (2 e+\frac{3 f x}{2}\right )-900 \sin \left (2 e+\frac{5 f x}{2}\right )-27 \sin \left (4 e+\frac{7 f x}{2}\right )+25 \sin \left (4 e+\frac{9 f x}{2}\right )+378 \cos \left (e+\frac{f x}{2}\right )+210 \cos \left (e+\frac{3 f x}{2}\right )-108 \cos \left (3 e+\frac{5 f x}{2}\right )+225 \cos \left (3 e+\frac{7 f x}{2}\right )+3 \cos \left (5 e+\frac{9 f x}{2}\right )+3150 \sin \left (\frac{f x}{2}\right )}{13860 a^6 f \left (\sin \left (\frac{e}{2}\right )+\cos \left (\frac{e}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^9} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[e + f*x]^2*Sin[e + f*x])/(a + a*Sin[e + f*x])^6,x]

[Out]

-(378*Cos[e + (f*x)/2] + 210*Cos[e + (3*f*x)/2] - 108*Cos[3*e + (5*f*x)/2] + 225*Cos[3*e + (7*f*x)/2] + 3*Cos[
5*e + (9*f*x)/2] + 3150*Sin[(f*x)/2] + 2562*Sin[2*e + (3*f*x)/2] - 900*Sin[2*e + (5*f*x)/2] - 27*Sin[4*e + (7*
f*x)/2] + 25*Sin[4*e + (9*f*x)/2])/(13860*a^6*f*(Cos[e/2] + Sin[e/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^9)

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Maple [A]  time = 0.132, size = 130, normalized size = 0.9 \begin{align*} 4\,{\frac{1}{f{a}^{6}} \left ({\frac{16}{9\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{9}}}-9\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{-4}-8\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{-8}+{\frac{116}{7\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{7}}}-{\frac{62}{3\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{6}}}-1/2\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{-2}+{\frac{84}{5\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{5}}}+3\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{-3} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(f*x+e)^2*sin(f*x+e)/(a+a*sin(f*x+e))^6,x)

[Out]

4/f/a^6*(16/9/(tan(1/2*f*x+1/2*e)+1)^9-9/(tan(1/2*f*x+1/2*e)+1)^4-8/(tan(1/2*f*x+1/2*e)+1)^8+116/7/(tan(1/2*f*
x+1/2*e)+1)^7-62/3/(tan(1/2*f*x+1/2*e)+1)^6-1/2/(tan(1/2*f*x+1/2*e)+1)^2+84/5/(tan(1/2*f*x+1/2*e)+1)^5+3/(tan(
1/2*f*x+1/2*e)+1)^3)

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Maxima [B]  time = 1.13211, size = 479, normalized size = 3.33 \begin{align*} -\frac{2 \,{\left (\frac{99 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{81 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{609 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{441 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac{945 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac{315 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac{315 \, \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} + 11\right )}}{315 \,{\left (a^{6} + \frac{9 \, a^{6} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{36 \, a^{6} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{84 \, a^{6} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{126 \, a^{6} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac{126 \, a^{6} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac{84 \, a^{6} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac{36 \, a^{6} \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} + \frac{9 \, a^{6} \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}} + \frac{a^{6} \sin \left (f x + e\right )^{9}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{9}}\right )} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*sin(f*x+e)/(a+a*sin(f*x+e))^6,x, algorithm="maxima")

[Out]

-2/315*(99*sin(f*x + e)/(cos(f*x + e) + 1) + 81*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 609*sin(f*x + e)^3/(cos(
f*x + e) + 1)^3 + 441*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 945*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 315*sin(
f*x + e)^6/(cos(f*x + e) + 1)^6 + 315*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 11)/((a^6 + 9*a^6*sin(f*x + e)/(co
s(f*x + e) + 1) + 36*a^6*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 84*a^6*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 12
6*a^6*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 126*a^6*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 84*a^6*sin(f*x + e)^
6/(cos(f*x + e) + 1)^6 + 36*a^6*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 9*a^6*sin(f*x + e)^8/(cos(f*x + e) + 1)^
8 + a^6*sin(f*x + e)^9/(cos(f*x + e) + 1)^9)*f)

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Fricas [A]  time = 1.65385, size = 624, normalized size = 4.33 \begin{align*} \frac{4 \, \cos \left (f x + e\right )^{5} - 16 \, \cos \left (f x + e\right )^{4} - 50 \, \cos \left (f x + e\right )^{3} - 65 \, \cos \left (f x + e\right )^{2} -{\left (4 \, \cos \left (f x + e\right )^{4} + 20 \, \cos \left (f x + e\right )^{3} - 30 \, \cos \left (f x + e\right )^{2} + 35 \, \cos \left (f x + e\right ) + 70\right )} \sin \left (f x + e\right ) + 35 \, \cos \left (f x + e\right ) + 70}{315 \,{\left (a^{6} f \cos \left (f x + e\right )^{5} + 5 \, a^{6} f \cos \left (f x + e\right )^{4} - 8 \, a^{6} f \cos \left (f x + e\right )^{3} - 20 \, a^{6} f \cos \left (f x + e\right )^{2} + 8 \, a^{6} f \cos \left (f x + e\right ) + 16 \, a^{6} f +{\left (a^{6} f \cos \left (f x + e\right )^{4} - 4 \, a^{6} f \cos \left (f x + e\right )^{3} - 12 \, a^{6} f \cos \left (f x + e\right )^{2} + 8 \, a^{6} f \cos \left (f x + e\right ) + 16 \, a^{6} f\right )} \sin \left (f x + e\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*sin(f*x+e)/(a+a*sin(f*x+e))^6,x, algorithm="fricas")

[Out]

1/315*(4*cos(f*x + e)^5 - 16*cos(f*x + e)^4 - 50*cos(f*x + e)^3 - 65*cos(f*x + e)^2 - (4*cos(f*x + e)^4 + 20*c
os(f*x + e)^3 - 30*cos(f*x + e)^2 + 35*cos(f*x + e) + 70)*sin(f*x + e) + 35*cos(f*x + e) + 70)/(a^6*f*cos(f*x
+ e)^5 + 5*a^6*f*cos(f*x + e)^4 - 8*a^6*f*cos(f*x + e)^3 - 20*a^6*f*cos(f*x + e)^2 + 8*a^6*f*cos(f*x + e) + 16
*a^6*f + (a^6*f*cos(f*x + e)^4 - 4*a^6*f*cos(f*x + e)^3 - 12*a^6*f*cos(f*x + e)^2 + 8*a^6*f*cos(f*x + e) + 16*
a^6*f)*sin(f*x + e))

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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)**2*sin(f*x+e)/(a+a*sin(f*x+e))**6,x)

[Out]

Timed out

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Giac [A]  time = 1.32431, size = 162, normalized size = 1.12 \begin{align*} -\frac{2 \,{\left (315 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{7} + 315 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} + 945 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 441 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 609 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 81 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 99 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 11\right )}}{315 \, a^{6} f{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*sin(f*x+e)/(a+a*sin(f*x+e))^6,x, algorithm="giac")

[Out]

-2/315*(315*tan(1/2*f*x + 1/2*e)^7 + 315*tan(1/2*f*x + 1/2*e)^6 + 945*tan(1/2*f*x + 1/2*e)^5 + 441*tan(1/2*f*x
+ 1/2*e)^4 + 609*tan(1/2*f*x + 1/2*e)^3 + 81*tan(1/2*f*x + 1/2*e)^2 + 99*tan(1/2*f*x + 1/2*e) + 11)/(a^6*f*(t
an(1/2*f*x + 1/2*e) + 1)^9)