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

Optimal. Leaf size=72 \[ -\frac{2 a^2 \cos (e+f x)}{f \left (c^2-c^2 \sin (e+f x)\right )}+\frac{a^2 x}{c^2}+\frac{2 a^2 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.136745, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {2736, 2680, 8} \[ -\frac{2 a^2 \cos (e+f x)}{f \left (c^2-c^2 \sin (e+f x)\right )}+\frac{a^2 x}{c^2}+\frac{2 a^2 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2680

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^2} \, dx &=\left (a^2 c^2\right ) \int \frac{\cos ^4(e+f x)}{(c-c \sin (e+f x))^4} \, dx\\ &=\frac{2 a^2 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^3}-a^2 \int \frac{\cos ^2(e+f x)}{(c-c \sin (e+f x))^2} \, dx\\ &=\frac{2 a^2 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^3}-\frac{2 a^2 \cos (e+f x)}{f \left (c^2-c^2 \sin (e+f x)\right )}+\frac{a^2 \int 1 \, dx}{c^2}\\ &=\frac{a^2 x}{c^2}+\frac{2 a^2 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^3}-\frac{2 a^2 \cos (e+f x)}{f \left (c^2-c^2 \sin (e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.613442, size = 121, normalized size = 1.68 \[ -\frac{a^2 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (-3 (3 e+3 f x+8) \cos \left (\frac{1}{2} (e+f x)\right )+(3 e+3 f x+16) \cos \left (\frac{3}{2} (e+f x)\right )+6 \sin \left (\frac{1}{2} (e+f x)\right ) (2 (e+f x+2)+(e+f x) \cos (e+f x))\right )}{6 c^2 f (\sin (e+f x)-1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(-3*(8 + 3*e + 3*f*x)*Cos[(e + f*x)/2] + (16 + 3*e + 3*f*x)*Cos[(3
*(e + f*x))/2] + 6*(2*(2 + e + f*x) + (e + f*x)*Cos[e + f*x])*Sin[(e + f*x)/2]))/(6*c^2*f*(-1 + Sin[e + f*x])^
2)

________________________________________________________________________________________

Maple [A]  time = 0.075, size = 71, normalized size = 1. \begin{align*} -{\frac{16\,{a}^{2}}{3\,f{c}^{2}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) ^{-3}}-8\,{\frac{{a}^{2}}{f{c}^{2} \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{2}}}+2\,{\frac{{a}^{2}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{f{c}^{2}}} \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))^2,x)

[Out]

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

________________________________________________________________________________________

Maxima [B]  time = 1.77773, size = 491, normalized size = 6.82 \begin{align*} \frac{2 \,{\left (a^{2}{\left (\frac{\frac{9 \, \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}} - 4}{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{3 \, \arctan \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c^{2}}\right )} - \frac{a^{2}{\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{2 \, a^{2}{\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))^2/(c-c*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

2/3*(a^2*((9*sin(f*x + e)/(cos(f*x + e) + 1) - 3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 4)/(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
) + 3*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/c^2) - a^2*(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) + 2*a^2*(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

________________________________________________________________________________________

Fricas [B]  time = 1.35024, size = 362, normalized size = 5.03 \begin{align*} -\frac{6 \, a^{2} f x -{\left (3 \, a^{2} f x + 8 \, a^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, a^{2} +{\left (3 \, a^{2} f x - 4 \, a^{2}\right )} \cos \left (f x + e\right ) -{\left (6 \, a^{2} f x - 4 \, a^{2} +{\left (3 \, a^{2} f x - 8 \, a^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{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))^2/(c-c*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

-1/3*(6*a^2*f*x - (3*a^2*f*x + 8*a^2)*cos(f*x + e)^2 + 4*a^2 + (3*a^2*f*x - 4*a^2)*cos(f*x + e) - (6*a^2*f*x -
4*a^2 + (3*a^2*f*x - 8*a^2)*cos(f*x + e))*sin(f*x + e))/(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))

________________________________________________________________________________________

Sympy [A]  time = 16.866, size = 486, normalized size = 6.75 \begin{align*} \begin{cases} \frac{3 a^{2} f x \tan ^{3}{\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{9 a^{2} f x \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{9 a^{2} f x \tan{\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{3 a^{2} f x}{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{8 a^{2} \tan ^{3}{\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{24 a^{2} \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} & \text{for}\: f \neq 0 \\\frac{x \left (a \sin{\left (e \right )} + a\right )^{2}}{\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))**2/(c-c*sin(f*x+e))**2,x)

[Out]

Piecewise((3*a**2*f*x*tan(e/2 + f*x/2)**3/(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) - 9*a**2*f*x*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*tan(e/2 + f*x/2) - 3*c**2*f) + 9*a**2*f*x*tan(e/2 + f*x/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*tan(e/2 + f*x/2) - 3*c**2*f) - 3*a**2*f*x/(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) + 8*a**2*tan(e/2 + f*x/2)**3
/(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) - 24*a**
2*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*tan(e/2 + f*x/2)
- 3*c**2*f), Ne(f, 0)), (x*(a*sin(e) + a)**2/(-c*sin(e) + c)**2, True))

________________________________________________________________________________________

Giac [A]  time = 2.00909, size = 81, normalized size = 1.12 \begin{align*} \frac{\frac{3 \,{\left (f x + e\right )} a^{2}}{c^{2}} - \frac{8 \,{\left (3 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - a^{2}\right )}}{c^{2}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}^{3}}}{3 \, f} \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))^2,x, algorithm="giac")

[Out]

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