3.233 \(\int \frac{a+a \sin (e+f x)}{(c-c \sin (e+f x))^3} \, dx\)

Optimal. Leaf size=60 \[ \frac{a \cos ^3(e+f x)}{15 f (c-c \sin (e+f x))^3}+\frac{a c \cos ^3(e+f x)}{5 f (c-c \sin (e+f x))^4} \]

[Out]

(a*c*Cos[e + f*x]^3)/(5*f*(c - c*Sin[e + f*x])^4) + (a*Cos[e + f*x]^3)/(15*f*(c - c*Sin[e + f*x])^3)

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Rubi [A]  time = 0.116896, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {2736, 2672, 2671} \[ \frac{a \cos ^3(e+f x)}{15 f (c-c \sin (e+f x))^3}+\frac{a c \cos ^3(e+f x)}{5 f (c-c \sin (e+f x))^4} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])/(c - c*Sin[e + f*x])^3,x]

[Out]

(a*c*Cos[e + f*x]^3)/(5*f*(c - c*Sin[e + f*x])^4) + (a*Cos[e + f*x]^3)/(15*f*(c - c*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))^3} \, dx &=(a c) \int \frac{\cos ^2(e+f x)}{(c-c \sin (e+f x))^4} \, dx\\ &=\frac{a c \cos ^3(e+f x)}{5 f (c-c \sin (e+f x))^4}+\frac{1}{5} a \int \frac{\cos ^2(e+f x)}{(c-c \sin (e+f x))^3} \, dx\\ &=\frac{a c \cos ^3(e+f x)}{5 f (c-c \sin (e+f x))^4}+\frac{a \cos ^3(e+f x)}{15 f (c-c \sin (e+f x))^3}\\ \end{align*}

Mathematica [A]  time = 0.32822, size = 96, normalized size = 1.6 \[ \frac{a \left (\sin \left (2 e+\frac{5 f x}{2}\right )+15 \cos \left (e+\frac{f x}{2}\right )-5 \cos \left (e+\frac{3 f x}{2}\right )+5 \sin \left (\frac{f x}{2}\right )\right )}{30 c^3 f \left (\cos \left (\frac{e}{2}\right )-\sin \left (\frac{e}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])/(c - c*Sin[e + f*x])^3,x]

[Out]

(a*(15*Cos[e + (f*x)/2] - 5*Cos[e + (3*f*x)/2] + 5*Sin[(f*x)/2] + Sin[2*e + (5*f*x)/2]))/(30*c^3*f*(Cos[e/2] -
Sin[e/2])*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5)

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Maple [A]  time = 0.076, size = 86, normalized size = 1.4 \begin{align*} 2\,{\frac{a}{f{c}^{3}} \left ( -4\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-4}-14/3\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-3}-3\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-2}-8/5\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-5}- \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-1} \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))^3,x)

[Out]

2/f*a/c^3*(-4/(tan(1/2*f*x+1/2*e)-1)^4-14/3/(tan(1/2*f*x+1/2*e)-1)^3-3/(tan(1/2*f*x+1/2*e)-1)^2-8/5/(tan(1/2*f
*x+1/2*e)-1)^5-1/(tan(1/2*f*x+1/2*e)-1))

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Maxima [B]  time = 1.25825, size = 525, normalized size = 8.75 \begin{align*} -\frac{2 \,{\left (\frac{a{\left (\frac{20 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac{40 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{30 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac{15 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 7\right )}}{c^{3} - \frac{5 \, c^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{10 \, c^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{10 \, c^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{5 \, c^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac{c^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}} - \frac{3 \, a{\left (\frac{5 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac{5 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{5 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - 1\right )}}{c^{3} - \frac{5 \, c^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{10 \, c^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{10 \, c^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{5 \, c^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac{c^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}}\right )}}{15 \, f} \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))^3,x, algorithm="maxima")

[Out]

-2/15*(a*(20*sin(f*x + e)/(cos(f*x + e) + 1) - 40*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 30*sin(f*x + e)^3/(cos
(f*x + e) + 1)^3 - 15*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 7)/(c^3 - 5*c^3*sin(f*x + e)/(cos(f*x + e) + 1) +
10*c^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 10*c^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*c^3*sin(f*x + e)^4
/(cos(f*x + e) + 1)^4 - c^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) - 3*a*(5*sin(f*x + e)/(cos(f*x + e) + 1) - 5*
sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 5*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 1)/(c^3 - 5*c^3*sin(f*x + e)/(co
s(f*x + e) + 1) + 10*c^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 10*c^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*
c^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - c^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5))/f

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Fricas [B]  time = 1.30051, size = 382, normalized size = 6.37 \begin{align*} -\frac{a \cos \left (f x + e\right )^{3} - 2 \, a \cos \left (f x + e\right )^{2} + 3 \, a \cos \left (f x + e\right ) +{\left (a \cos \left (f x + e\right )^{2} + 3 \, a \cos \left (f x + e\right ) + 6 \, a\right )} \sin \left (f x + e\right ) + 6 \, a}{15 \,{\left (c^{3} f \cos \left (f x + e\right )^{3} + 3 \, c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f -{\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} 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))^3,x, algorithm="fricas")

[Out]

-1/15*(a*cos(f*x + e)^3 - 2*a*cos(f*x + e)^2 + 3*a*cos(f*x + e) + (a*cos(f*x + e)^2 + 3*a*cos(f*x + e) + 6*a)*
sin(f*x + e) + 6*a)/(c^3*f*cos(f*x + e)^3 + 3*c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) - 4*c^3*f - (c^3*f*c
os(f*x + e)^2 - 2*c^3*f*cos(f*x + e) - 4*c^3*f)*sin(f*x + e))

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Sympy [A]  time = 15.0005, size = 573, normalized size = 9.55 \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))**3,x)

[Out]

Piecewise((-2*a*tan(e/2 + f*x/2)**5/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*
f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) - 20*a*tan(e/
2 + f*x/2)**4/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3
- 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) + 10*a*tan(e/2 + f*x/2)**3/(15*c**3
*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 +
f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) - 30*a*tan(e/2 + f*x/2)**2/(15*c**3*f*tan(e/2 + f*x/2)**5
- 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f
*tan(e/2 + f*x/2) - 15*c**3*f) - 6*a/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3
*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f), Ne(f, 0)),
(x*(a*sin(e) + a)/(-c*sin(e) + c)**3, True))

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Giac [A]  time = 1.63448, size = 113, normalized size = 1.88 \begin{align*} -\frac{2 \,{\left (15 \, a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 15 \, a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 25 \, a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 5 \, a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 4 \, a\right )}}{15 \, c^{3} f{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}^{5}} \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))^3,x, algorithm="giac")

[Out]

-2/15*(15*a*tan(1/2*f*x + 1/2*e)^4 - 15*a*tan(1/2*f*x + 1/2*e)^3 + 25*a*tan(1/2*f*x + 1/2*e)^2 - 5*a*tan(1/2*f
*x + 1/2*e) + 4*a)/(c^3*f*(tan(1/2*f*x + 1/2*e) - 1)^5)