3.266 \(\int \frac{1}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^2} \, dx\)

Optimal. Leaf size=53 \[ \frac{2 \tan (e+f x)}{3 a c^2 f}+\frac{\sec (e+f x)}{3 a f \left (c^2-c^2 \sin (e+f x)\right )} \]

[Out]

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

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Rubi [A]  time = 0.11045, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2736, 2672, 3767, 8} \[ \frac{2 \tan (e+f x)}{3 a c^2 f}+\frac{\sec (e+f x)}{3 a f \left (c^2-c^2 \sin (e+f x)\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Sin[e + f*x])*(c - c*Sin[e + f*x])^2),x]

[Out]

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

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 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

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

Rubi steps

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

Mathematica [A]  time = 0.425996, size = 87, normalized size = 1.64 \[ \frac{\sin (e+f x)+8 \sin (2 (e+f x))+\sin (3 (e+f x))+4 \cos (e+f x)-2 \cos (2 (e+f x))+4 \cos (3 (e+f x))-2}{24 a c^2 f (\sin (e+f x)-1)^2 (\sin (e+f x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + a*Sin[e + f*x])*(c - c*Sin[e + f*x])^2),x]

[Out]

(-2 + 4*Cos[e + f*x] - 2*Cos[2*(e + f*x)] + 4*Cos[3*(e + f*x)] + Sin[e + f*x] + 8*Sin[2*(e + f*x)] + Sin[3*(e
+ f*x)])/(24*a*c^2*f*(-1 + Sin[e + f*x])^2*(1 + Sin[e + f*x]))

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Maple [A]  time = 0.049, size = 73, normalized size = 1.4 \begin{align*} 2\,{\frac{1}{af{c}^{2}} \left ( -1/3\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-3}-1/2\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-2}-3/4\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-1}-1/4\, \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(1/(a+a*sin(f*x+e))/(c-c*sin(f*x+e))^2,x)

[Out]

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

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Maxima [B]  time = 1.41166, size = 192, normalized size = 3.62 \begin{align*} \frac{2 \,{\left (\frac{\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}} + \frac{3 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + 1\right )}}{3 \,{\left (a c^{2} - \frac{2 \, a c^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{2 \, a c^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac{a c^{2} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}\right )} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))/(c-c*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.24241, size = 142, normalized size = 2.68 \begin{align*} -\frac{2 \, \cos \left (f x + e\right )^{2} + 2 \, \sin \left (f x + e\right ) - 1}{3 \,{\left (a c^{2} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - a c^{2} f \cos \left (f x + e\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))/(c-c*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 8.5386, size = 328, normalized size = 6.19 \begin{align*} \begin{cases} \frac{\tan ^{4}{\left (\frac{e}{2} + \frac{f x}{2} \right )}}{3 a c^{2} f \tan ^{4}{\left (\frac{e}{2} + \frac{f x}{2} \right )} - 6 a c^{2} f \tan ^{3}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 6 a c^{2} f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} - 3 a c^{2} f} - \frac{8 \tan ^{3}{\left (\frac{e}{2} + \frac{f x}{2} \right )}}{3 a c^{2} f \tan ^{4}{\left (\frac{e}{2} + \frac{f x}{2} \right )} - 6 a c^{2} f \tan ^{3}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 6 a c^{2} f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} - 3 a c^{2} f} + \frac{6 \tan ^{2}{\left (\frac{e}{2} + \frac{f x}{2} \right )}}{3 a c^{2} f \tan ^{4}{\left (\frac{e}{2} + \frac{f x}{2} \right )} - 6 a c^{2} f \tan ^{3}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 6 a c^{2} f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} - 3 a c^{2} f} - \frac{3}{3 a c^{2} f \tan ^{4}{\left (\frac{e}{2} + \frac{f x}{2} \right )} - 6 a c^{2} f \tan ^{3}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 6 a c^{2} f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} - 3 a 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(1/(a+a*sin(f*x+e))/(c-c*sin(f*x+e))**2,x)

[Out]

Piecewise((tan(e/2 + f*x/2)**4/(3*a*c**2*f*tan(e/2 + f*x/2)**4 - 6*a*c**2*f*tan(e/2 + f*x/2)**3 + 6*a*c**2*f*t
an(e/2 + f*x/2) - 3*a*c**2*f) - 8*tan(e/2 + f*x/2)**3/(3*a*c**2*f*tan(e/2 + f*x/2)**4 - 6*a*c**2*f*tan(e/2 + f
*x/2)**3 + 6*a*c**2*f*tan(e/2 + f*x/2) - 3*a*c**2*f) + 6*tan(e/2 + f*x/2)**2/(3*a*c**2*f*tan(e/2 + f*x/2)**4 -
6*a*c**2*f*tan(e/2 + f*x/2)**3 + 6*a*c**2*f*tan(e/2 + f*x/2) - 3*a*c**2*f) - 3/(3*a*c**2*f*tan(e/2 + f*x/2)**
4 - 6*a*c**2*f*tan(e/2 + f*x/2)**3 + 6*a*c**2*f*tan(e/2 + f*x/2) - 3*a*c**2*f), Ne(f, 0)), (x/((a*sin(e) + a)*
(-c*sin(e) + c)**2), True))

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Giac [A]  time = 2.04361, size = 104, normalized size = 1.96 \begin{align*} -\frac{\frac{3}{a c^{2}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}} + \frac{9 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 12 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 7}{a c^{2}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}^{3}}}{6 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))/(c-c*sin(f*x+e))^2,x, algorithm="giac")

[Out]

-1/6*(3/(a*c^2*(tan(1/2*f*x + 1/2*e) + 1)) + (9*tan(1/2*f*x + 1/2*e)^2 - 12*tan(1/2*f*x + 1/2*e) + 7)/(a*c^2*(
tan(1/2*f*x + 1/2*e) - 1)^3))/f