3.235 \(\int \frac{a+a \sin (e+f x)}{(c-c \sin (e+f x))^5} \, dx\)

Optimal. Leaf size=126 \[ \frac{2 a c \cos ^3(e+f x)}{315 f \left (c^2-c^2 \sin (e+f x)\right )^3}+\frac{2 a \cos ^3(e+f x)}{105 c f (c-c \sin (e+f x))^4}+\frac{a \cos ^3(e+f x)}{21 f (c-c \sin (e+f x))^5}+\frac{a c \cos ^3(e+f x)}{9 f (c-c \sin (e+f x))^6} \]

[Out]

(a*c*Cos[e + f*x]^3)/(9*f*(c - c*Sin[e + f*x])^6) + (a*Cos[e + f*x]^3)/(21*f*(c - c*Sin[e + f*x])^5) + (2*a*Co
s[e + f*x]^3)/(105*c*f*(c - c*Sin[e + f*x])^4) + (2*a*c*Cos[e + f*x]^3)/(315*f*(c^2 - c^2*Sin[e + f*x])^3)

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Rubi [A]  time = 0.219617, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {2736, 2672, 2671} \[ \frac{2 a c \cos ^3(e+f x)}{315 f \left (c^2-c^2 \sin (e+f x)\right )^3}+\frac{2 a \cos ^3(e+f x)}{105 c f (c-c \sin (e+f x))^4}+\frac{a \cos ^3(e+f x)}{21 f (c-c \sin (e+f x))^5}+\frac{a c \cos ^3(e+f x)}{9 f (c-c \sin (e+f x))^6} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])/(c - c*Sin[e + f*x])^5,x]

[Out]

(a*c*Cos[e + f*x]^3)/(9*f*(c - c*Sin[e + f*x])^6) + (a*Cos[e + f*x]^3)/(21*f*(c - c*Sin[e + f*x])^5) + (2*a*Co
s[e + f*x]^3)/(105*c*f*(c - c*Sin[e + f*x])^4) + (2*a*c*Cos[e + f*x]^3)/(315*f*(c^2 - c^2*Sin[e + f*x])^3)

Rule 2736

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0,
n, m] || LtQ[m, n, 0]))

Rule 2672

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*(g*
Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*Simplify[2*m + p + 1]), x] + Dist[Simplify[m + p + 1]/(a*
Simplify[2*m + p + 1]), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g, m
, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[Simplify[m + p + 1], 0] && NeQ[2*m + p + 1, 0] &&  !IGtQ[m, 0]

Rule 2671

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*(g*
Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*m), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^
2, 0] && EqQ[Simplify[m + p + 1], 0] &&  !ILtQ[p, 0]

Rubi steps

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

Mathematica [A]  time = 0.584732, size = 124, normalized size = 0.98 \[ \frac{a \left (36 \sin \left (2 e+\frac{5 f x}{2}\right )-\sin \left (4 e+\frac{9 f x}{2}\right )+315 \cos \left (e+\frac{f x}{2}\right )-84 \cos \left (e+\frac{3 f x}{2}\right )+9 \cos \left (3 e+\frac{7 f x}{2}\right )+189 \sin \left (\frac{f x}{2}\right )\right )}{1260 c^5 f \left (\cos \left (\frac{e}{2}\right )-\sin \left (\frac{e}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^9} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])/(c - c*Sin[e + f*x])^5,x]

[Out]

(a*(315*Cos[e + (f*x)/2] - 84*Cos[e + (3*f*x)/2] + 9*Cos[3*e + (7*f*x)/2] + 189*Sin[(f*x)/2] + 36*Sin[2*e + (5
*f*x)/2] - Sin[4*e + (9*f*x)/2]))/(1260*c^5*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 = 0.096, size = 146, normalized size = 1.2 \begin{align*} 2\,{\frac{a}{f{c}^{5}} \left ( -{\frac{148}{3\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{6}}}-16\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-8}-{\frac{46}{3\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{3}}}-5\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-2}-{\frac{236}{5\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{5}}}- \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-1}-{\frac{32}{9\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{9}}}-32\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-4}-{\frac{248}{7\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{7}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))/(c-c*sin(f*x+e))^5,x)

[Out]

2/f*a/c^5*(-148/3/(tan(1/2*f*x+1/2*e)-1)^6-16/(tan(1/2*f*x+1/2*e)-1)^8-46/3/(tan(1/2*f*x+1/2*e)-1)^3-5/(tan(1/
2*f*x+1/2*e)-1)^2-236/5/(tan(1/2*f*x+1/2*e)-1)^5-1/(tan(1/2*f*x+1/2*e)-1)-32/9/(tan(1/2*f*x+1/2*e)-1)^9-32/(ta
n(1/2*f*x+1/2*e)-1)^4-248/7/(tan(1/2*f*x+1/2*e)-1)^7)

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Maxima [B]  time = 1.26822, size = 990, normalized size = 7.86 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))/(c-c*sin(f*x+e))^5,x, algorithm="maxima")

[Out]

-2/315*(a*(432*sin(f*x + e)/(cos(f*x + e) + 1) - 1728*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 3612*sin(f*x + e)^
3/(cos(f*x + e) + 1)^3 - 5418*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 5040*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 -
3360*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 1260*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - 315*sin(f*x + e)^8/(cos
(f*x + e) + 1)^8 - 83)/(c^5 - 9*c^5*sin(f*x + e)/(cos(f*x + e) + 1) + 36*c^5*sin(f*x + e)^2/(cos(f*x + e) + 1)
^2 - 84*c^5*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 126*c^5*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 126*c^5*sin(f*
x + e)^5/(cos(f*x + e) + 1)^5 + 84*c^5*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 36*c^5*sin(f*x + e)^7/(cos(f*x +
e) + 1)^7 + 9*c^5*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 - c^5*sin(f*x + e)^9/(cos(f*x + e) + 1)^9) - 5*a*(45*sin
(f*x + e)/(cos(f*x + e) + 1) - 117*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 273*sin(f*x + e)^3/(cos(f*x + e) + 1)
^3 - 315*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 315*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 147*sin(f*x + e)^6/(c
os(f*x + e) + 1)^6 + 63*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - 5)/(c^5 - 9*c^5*sin(f*x + e)/(cos(f*x + e) + 1)
+ 36*c^5*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 84*c^5*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 126*c^5*sin(f*x +
e)^4/(cos(f*x + e) + 1)^4 - 126*c^5*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 84*c^5*sin(f*x + e)^6/(cos(f*x + e)
+ 1)^6 - 36*c^5*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 9*c^5*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 - c^5*sin(f*x
+ e)^9/(cos(f*x + e) + 1)^9))/f

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Fricas [B]  time = 1.38339, size = 653, normalized size = 5.18 \begin{align*} -\frac{2 \, a \cos \left (f x + e\right )^{5} - 8 \, a \cos \left (f x + e\right )^{4} - 25 \, a \cos \left (f x + e\right )^{3} + 20 \, a \cos \left (f x + e\right )^{2} - 35 \, a \cos \left (f x + e\right ) +{\left (2 \, a \cos \left (f x + e\right )^{4} + 10 \, a \cos \left (f x + e\right )^{3} - 15 \, a \cos \left (f x + e\right )^{2} - 35 \, a \cos \left (f x + e\right ) - 70 \, a\right )} \sin \left (f x + e\right ) - 70 \, a}{315 \,{\left (c^{5} f \cos \left (f x + e\right )^{5} + 5 \, c^{5} f \cos \left (f x + e\right )^{4} - 8 \, c^{5} f \cos \left (f x + e\right )^{3} - 20 \, c^{5} f \cos \left (f x + e\right )^{2} + 8 \, c^{5} f \cos \left (f x + e\right ) + 16 \, c^{5} f -{\left (c^{5} f \cos \left (f x + e\right )^{4} - 4 \, c^{5} f \cos \left (f x + e\right )^{3} - 12 \, c^{5} f \cos \left (f x + e\right )^{2} + 8 \, c^{5} f \cos \left (f x + e\right ) + 16 \, c^{5} f\right )} \sin \left (f x + e\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))/(c-c*sin(f*x+e))^5,x, algorithm="fricas")

[Out]

-1/315*(2*a*cos(f*x + e)^5 - 8*a*cos(f*x + e)^4 - 25*a*cos(f*x + e)^3 + 20*a*cos(f*x + e)^2 - 35*a*cos(f*x + e
) + (2*a*cos(f*x + e)^4 + 10*a*cos(f*x + e)^3 - 15*a*cos(f*x + e)^2 - 35*a*cos(f*x + e) - 70*a)*sin(f*x + e) -
70*a)/(c^5*f*cos(f*x + e)^5 + 5*c^5*f*cos(f*x + e)^4 - 8*c^5*f*cos(f*x + e)^3 - 20*c^5*f*cos(f*x + e)^2 + 8*c
^5*f*cos(f*x + e) + 16*c^5*f - (c^5*f*cos(f*x + e)^4 - 4*c^5*f*cos(f*x + e)^3 - 12*c^5*f*cos(f*x + e)^2 + 8*c^
5*f*cos(f*x + e) + 16*c^5*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((a+a*sin(f*x+e))/(c-c*sin(f*x+e))**5,x)

[Out]

Timed out

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Giac [A]  time = 1.68009, size = 194, normalized size = 1.54 \begin{align*} -\frac{2 \,{\left (315 \, a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{8} - 945 \, a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{7} + 2625 \, a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} - 3465 \, a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 3843 \, a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 2247 \, a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 1143 \, a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 207 \, a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 58 \, a\right )}}{315 \, c^{5} 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((a+a*sin(f*x+e))/(c-c*sin(f*x+e))^5,x, algorithm="giac")

[Out]

-2/315*(315*a*tan(1/2*f*x + 1/2*e)^8 - 945*a*tan(1/2*f*x + 1/2*e)^7 + 2625*a*tan(1/2*f*x + 1/2*e)^6 - 3465*a*t
an(1/2*f*x + 1/2*e)^5 + 3843*a*tan(1/2*f*x + 1/2*e)^4 - 2247*a*tan(1/2*f*x + 1/2*e)^3 + 1143*a*tan(1/2*f*x + 1
/2*e)^2 - 207*a*tan(1/2*f*x + 1/2*e) + 58*a)/(c^5*f*(tan(1/2*f*x + 1/2*e) - 1)^9)