3.232 \(\int \frac{a+a \sin (e+f x)}{(c-c \sin (e+f x))^2} \, dx\)

Optimal. Leaf size=30 \[ \frac{a c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^3} \]

[Out]

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

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Rubi [A]  time = 0.0729124, antiderivative size = 30, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {2736, 2671} \[ \frac{a c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [B]  time = 0.270363, size = 74, normalized size = 2.47 \[ -\frac{a \left (\cos \left (e+\frac{3 f x}{2}\right )-3 \cos \left (e+\frac{f x}{2}\right )\right )}{3 c^2 f \left (\cos \left (\frac{e}{2}\right )-\sin \left (\frac{e}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.067, size = 56, normalized size = 1.9 \begin{align*} 2\,{\frac{a}{f{c}^{2}} \left ( -2\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-2}-4/3\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-3}- \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))^2,x)

[Out]

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

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Maxima [B]  time = 1.24675, size = 293, normalized size = 9.77 \begin{align*} -\frac{2 \,{\left (\frac{a{\left (\frac{3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac{3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - 2\right )}}{c^{2} - \frac{3 \, c^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{3 \, c^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{c^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} - \frac{a{\left (\frac{3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}}{c^{2} - \frac{3 \, c^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{3 \, c^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{c^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}\right )}}{3 \, 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))^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.25734, size = 250, normalized size = 8.33 \begin{align*} \frac{a \cos \left (f x + e\right )^{2} - a \cos \left (f x + e\right ) -{\left (a \cos \left (f x + e\right ) + 2 \, a\right )} \sin \left (f x + e\right ) - 2 \, a}{3 \,{\left (c^{2} f \cos \left (f x + e\right )^{2} - c^{2} f \cos \left (f x + e\right ) - 2 \, c^{2} f +{\left (c^{2} f \cos \left (f x + e\right ) + 2 \, c^{2} 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))^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 6.83427, size = 158, normalized size = 5.27 \begin{align*} \begin{cases} - \frac{6 a \tan ^{2}{\left (\frac{e}{2} + \frac{f x}{2} \right )}}{3 c^{2} f \tan ^{3}{\left (\frac{e}{2} + \frac{f x}{2} \right )} - 9 c^{2} f \tan ^{2}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 9 c^{2} f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} - 3 c^{2} f} - \frac{2 a}{3 c^{2} f \tan ^{3}{\left (\frac{e}{2} + \frac{f x}{2} \right )} - 9 c^{2} f \tan ^{2}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 9 c^{2} f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} - 3 c^{2} f} & \text{for}\: f \neq 0 \\\frac{x \left (a \sin{\left (e \right )} + a\right )}{\left (- c \sin{\left (e \right )} + c\right )^{2}} & \text{otherwise} \end{cases} \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))**2,x)

[Out]

Piecewise((-6*a*tan(e/2 + f*x/2)**2/(3*c**2*f*tan(e/2 + f*x/2)**3 - 9*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*ta
n(e/2 + f*x/2) - 3*c**2*f) - 2*a/(3*c**2*f*tan(e/2 + f*x/2)**3 - 9*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e
/2 + f*x/2) - 3*c**2*f), Ne(f, 0)), (x*(a*sin(e) + a)/(-c*sin(e) + c)**2, True))

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Giac [A]  time = 1.93706, size = 53, normalized size = 1.77 \begin{align*} -\frac{2 \,{\left (3 \, a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + a\right )}}{3 \, c^{2} f{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}^{3}} \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))^2,x, algorithm="giac")

[Out]

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