\(\int \frac {1}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^2} \, dx\) [266]

Optimal result

Integrand size = 26, antiderivative size = 53 \[ \int \frac {1}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^2} \, dx=\frac {\sec (e+f x)}{3 a f \left (c^2-c^2 \sin (e+f x)\right )}+\frac {2 \tan (e+f x)}{3 a c^2 f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2815, 2751, 3852, 8} \[ \int \frac {1}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^2} \, dx=\frac {2 \tan (e+f x)}{3 a c^2 f}+\frac {\sec (e+f x)}{3 a f \left (c^2-c^2 \sin (e+f x)\right )} \]

[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 8

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

Rule 2751

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*C
os[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 2815

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 3852

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]

Rubi steps \begin{align*} \text {integral}& = \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 \text {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] (verified)

Time = 0.63 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.85 \[ \int \frac {1}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^2} \, dx=\frac {10 \cos (e+f x)+4 \cos (2 (e+f x))+8 \sin (e+f x)-5 \sin (2 (e+f x))}{12 a c^2 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )} \]

[In]

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

[Out]

(10*Cos[e + f*x] + 4*Cos[2*(e + f*x)] + 8*Sin[e + f*x] - 5*Sin[2*(e + f*x)])/(12*a*c^2*f*(Cos[(e + f*x)/2] - S
in[(e + f*x)/2])^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]))

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.77 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.02

method	result	size
risch	\(\frac {\frac {8 \,{\mathrm e}^{i \left (f x +e \right )}}{3}-\frac {4 i}{3}}{\left ({\mathrm e}^{i \left (f x +e \right )}-i\right )^{3} \left ({\mathrm e}^{i \left (f x +e \right )}+i\right ) a \,c^{2} f}\)	\(54\)
derivativedivides	\(\frac {-\frac {2}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {1}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {3}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}-\frac {1}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}}{a \,c^{2} f}\)	\(73\)
default	\(\frac {-\frac {2}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {1}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {3}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}-\frac {1}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}}{a \,c^{2} f}\)	\(73\)
parallelrisch	\(\frac {-2-6 \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+6 \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{3 f a \,c^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}\)	\(77\)
norman	\(\frac {\frac {2 \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a c f}-\frac {2}{3 a c f}-\frac {2 \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a c f}-\frac {2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{3 a c f}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) c \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}\)	\(107\)

[In]

int(1/(a+a*sin(f*x+e))/(c-c*sin(f*x+e))^2,x,method=_RETURNVERBOSE)

[Out]

4/3*(2*exp(I*(f*x+e))-I)/(exp(I*(f*x+e))-I)^3/(exp(I*(f*x+e))+I)/a/c^2/f

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.06 \[ \int \frac {1}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^2} \, dx=-\frac {2 \, \cos \left (f x + e\right )^{2} + 2 \, \sin \left (f x + e\right ) - 1}{3 \, {\left (a c^{2} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - a c^{2} f \cos \left (f x + e\right )\right )}} \]

[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))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 328 vs. \(2 (41) = 82\).

Time = 1.18 (sec) , antiderivative size = 328, normalized size of antiderivative = 6.19 \[ \int \frac {1}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^2} \, dx=\begin {cases} - \frac {6 \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{3 a c^{2} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 6 a c^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 6 a c^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 3 a c^{2} f} + \frac {6 \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{3 a c^{2} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 6 a c^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 6 a c^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 3 a c^{2} f} - \frac {2 \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{3 a c^{2} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 6 a c^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 6 a c^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 3 a c^{2} f} - \frac {2}{3 a c^{2} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 6 a c^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 6 a c^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 3 a c^{2} f} & \text {for}\: f \neq 0 \\\frac {x}{\left (a \sin {\left (e \right )} + a\right ) \left (- c \sin {\left (e \right )} + c\right )^{2}} & \text {otherwise} \end {cases} \]

[In]

integrate(1/(a+a*sin(f*x+e))/(c-c*sin(f*x+e))**2,x)

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 142 vs. \(2 (50) = 100\).

Time = 0.20 (sec) , antiderivative size = 142, normalized size of antiderivative = 2.68 \[ \int \frac {1}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^2} \, dx=\frac {2 \, {\left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {3 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + 1\right )}}{3 \, {\left (a c^{2} - \frac {2 \, a c^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {2 \, a c^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac {a c^{2} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}\right )} f} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.38 \[ \int \frac {1}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^2} \, dx=-\frac {\frac {3}{a c^{2} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}} + \frac {9 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 12 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 7}{a c^{2} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{3}}}{6 \, f} \]

[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

Mupad [B] (verification not implemented)

Time = 6.31 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.40 \[ \int \frac {1}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^2} \, dx=-\frac {2\,\left (3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3-3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+1\right )}{3\,a\,c^2\,f\,{\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )-1\right )}^3\,\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+1\right )} \]

[In]

int(1/((a + a*sin(e + f*x))*(c - c*sin(e + f*x))^2),x)

[Out]

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