3.243 \(\int \frac{(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^3} \, dx\)

Optimal. Leaf size=34 \[ \frac{a^2 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^5} \]

[Out]

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

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Rubi [A]  time = 0.0938423, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {2736, 2671} \[ \frac{a^2 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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))^2}{(c-c \sin (e+f x))^3} \, dx &=\left (a^2 c^2\right ) \int \frac{\cos ^4(e+f x)}{(c-c \sin (e+f x))^5} \, dx\\ &=\frac{a^2 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^5}\\ \end{align*}

Mathematica [B]  time = 0.401749, size = 81, normalized size = 2.38 \[ \frac{a^2 \left (-10 \sin \left (\frac{1}{2} (e+f x)\right )-5 \sin \left (\frac{3}{2} (e+f x)\right )+\sin \left (\frac{5}{2} (e+f x)\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )}{10 c^3 f (\sin (e+f x)-1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(-10*Sin[(e + f*x)/2] - 5*Sin[(3*(e + f*x))/2] + Sin[(5*(e + f*x))/
2]))/(10*c^3*f*(-1 + Sin[e + f*x])^3)

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Maple [B]  time = 0.088, size = 88, normalized size = 2.6 \begin{align*} 2\,{\frac{{a}^{2}}{f{c}^{3}} \left ( -{\frac{16}{5\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{5}}}-8\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-4}-8\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-3}- \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-1}-4\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-2} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^3,x)

[Out]

2/f*a^2/c^3*(-16/5/(tan(1/2*f*x+1/2*e)-1)^5-8/(tan(1/2*f*x+1/2*e)-1)^4-8/(tan(1/2*f*x+1/2*e)-1)^3-1/(tan(1/2*f
*x+1/2*e)-1)-4/(tan(1/2*f*x+1/2*e)-1)^2)

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Maxima [B]  time = 1.2695, size = 752, normalized size = 22.12 \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))^2/(c-c*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

-2/15*(a^2*(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/(c
os(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) - 6*a^2*(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)
/(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) + 2*a^2*(5*sin(f*x + e)
/(cos(f*x + e) + 1) - 10*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 1)/(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))/f

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Fricas [B]  time = 1.28247, size = 398, normalized size = 11.71 \begin{align*} \frac{a^{2} \cos \left (f x + e\right )^{3} + 3 \, a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} \cos \left (f x + e\right ) - 4 \, a^{2} +{\left (a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} \cos \left (f x + e\right ) - 4 \, a^{2}\right )} \sin \left (f x + e\right )}{5 \,{\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))^2/(c-c*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

1/5*(a^2*cos(f*x + e)^3 + 3*a^2*cos(f*x + e)^2 - 2*a^2*cos(f*x + e) - 4*a^2 + (a^2*cos(f*x + e)^2 - 2*a^2*cos(
f*x + e) - 4*a^2)*sin(f*x + e))/(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*cos(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 = 49.3753, size = 364, normalized size = 10.71 \begin{align*} \begin{cases} - \frac{2 a^{2} \tan ^{5}{\left (\frac{e}{2} + \frac{f x}{2} \right )}}{5 c^{3} f \tan ^{5}{\left (\frac{e}{2} + \frac{f x}{2} \right )} - 25 c^{3} f \tan ^{4}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 50 c^{3} f \tan ^{3}{\left (\frac{e}{2} + \frac{f x}{2} \right )} - 50 c^{3} f \tan ^{2}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 25 c^{3} f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} - 5 c^{3} f} - \frac{20 a^{2} \tan ^{3}{\left (\frac{e}{2} + \frac{f x}{2} \right )}}{5 c^{3} f \tan ^{5}{\left (\frac{e}{2} + \frac{f x}{2} \right )} - 25 c^{3} f \tan ^{4}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 50 c^{3} f \tan ^{3}{\left (\frac{e}{2} + \frac{f x}{2} \right )} - 50 c^{3} f \tan ^{2}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 25 c^{3} f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} - 5 c^{3} f} - \frac{10 a^{2} \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )}}{5 c^{3} f \tan ^{5}{\left (\frac{e}{2} + \frac{f x}{2} \right )} - 25 c^{3} f \tan ^{4}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 50 c^{3} f \tan ^{3}{\left (\frac{e}{2} + \frac{f x}{2} \right )} - 50 c^{3} f \tan ^{2}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 25 c^{3} f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} - 5 c^{3} f} & \text{for}\: f \neq 0 \\\frac{x \left (a \sin{\left (e \right )} + a\right )^{2}}{\left (- c \sin{\left (e \right )} + c\right )^{3}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))**2/(c-c*sin(f*x+e))**3,x)

[Out]

Piecewise((-2*a**2*tan(e/2 + f*x/2)**5/(5*c**3*f*tan(e/2 + f*x/2)**5 - 25*c**3*f*tan(e/2 + f*x/2)**4 + 50*c**3
*f*tan(e/2 + f*x/2)**3 - 50*c**3*f*tan(e/2 + f*x/2)**2 + 25*c**3*f*tan(e/2 + f*x/2) - 5*c**3*f) - 20*a**2*tan(
e/2 + f*x/2)**3/(5*c**3*f*tan(e/2 + f*x/2)**5 - 25*c**3*f*tan(e/2 + f*x/2)**4 + 50*c**3*f*tan(e/2 + f*x/2)**3
- 50*c**3*f*tan(e/2 + f*x/2)**2 + 25*c**3*f*tan(e/2 + f*x/2) - 5*c**3*f) - 10*a**2*tan(e/2 + f*x/2)/(5*c**3*f*
tan(e/2 + f*x/2)**5 - 25*c**3*f*tan(e/2 + f*x/2)**4 + 50*c**3*f*tan(e/2 + f*x/2)**3 - 50*c**3*f*tan(e/2 + f*x/
2)**2 + 25*c**3*f*tan(e/2 + f*x/2) - 5*c**3*f), Ne(f, 0)), (x*(a*sin(e) + a)**2/(-c*sin(e) + c)**3, True))

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Giac [A]  time = 2.17096, size = 81, normalized size = 2.38 \begin{align*} -\frac{2 \,{\left (5 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 10 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + a^{2}\right )}}{5 \, 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))^2/(c-c*sin(f*x+e))^3,x, algorithm="giac")

[Out]

-2/5*(5*a^2*tan(1/2*f*x + 1/2*e)^4 + 10*a^2*tan(1/2*f*x + 1/2*e)^2 + a^2)/(c^3*f*(tan(1/2*f*x + 1/2*e) - 1)^5)