\(\int \frac {1}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^2} \, dx\) [275]

Optimal result

Integrand size = 26, antiderivative size = 38 \[ \int \frac {1}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^2} \, dx=\frac {\tan (e+f x)}{a^2 c^2 f}+\frac {\tan ^3(e+f x)}{3 a^2 c^2 f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2815, 3852} \[ \int \frac {1}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^2} \, dx=\frac {\tan ^3(e+f x)}{3 a^2 c^2 f}+\frac {\tan (e+f x)}{a^2 c^2 f} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.76 \[ \int \frac {1}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^2} \, dx=\frac {\tan (e+f x)+\frac {1}{3} \tan ^3(e+f x)}{a^2 c^2 f} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.12 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.79

method	result	size
default	\(-\frac {\left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (f x +e \right )\right )}{3}\right ) \tan \left (f x +e \right )}{a^{2} c^{2} f}\)	\(30\)
parallelrisch	\(\frac {2 \sin \left (3 f x +3 e \right )+6 \sin \left (f x +e \right )}{3 a^{2} c^{2} f \left (\cos \left (3 f x +3 e \right )+3 \cos \left (f x +e \right )\right )}\)	\(52\)
risch	\(\frac {4 i \left (3 \,{\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{3 \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )^{3} \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{3} f \,a^{2} c^{2}}\)	\(54\)
norman	\(\frac {-\frac {2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a c f}+\frac {4 \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 a c f}-\frac {2 \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a c f}}{a \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3} c \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}\)	\(99\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.97 \[ \int \frac {1}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^2} \, dx=\frac {{\left (2 \, \cos \left (f x + e\right )^{2} + 1\right )} \sin \left (f x + e\right )}{3 \, a^{2} c^{2} f \cos \left (f x + e\right )^{3}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 286 vs. \(2 (32) = 64\).

Time = 1.32 (sec) , antiderivative size = 286, normalized size of antiderivative = 7.53 \[ \int \frac {1}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^2} \, dx=\begin {cases} - \frac {6 \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{3 a^{2} c^{2} f \tan ^{6}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 9 a^{2} c^{2} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} c^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 3 a^{2} c^{2} f} + \frac {4 \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{3 a^{2} c^{2} f \tan ^{6}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 9 a^{2} c^{2} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} c^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 3 a^{2} c^{2} f} - \frac {6 \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{3 a^{2} c^{2} f \tan ^{6}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 9 a^{2} c^{2} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} c^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 3 a^{2} c^{2} f} & \text {for}\: f \neq 0 \\\frac {x}{\left (a \sin {\left (e \right )} + a\right )^{2} \left (- c \sin {\left (e \right )} + c\right )^{2}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-6*tan(e/2 + f*x/2)**5/(3*a**2*c**2*f*tan(e/2 + f*x/2)**6 - 9*a**2*c**2*f*tan(e/2 + f*x/2)**4 + 9*a
**2*c**2*f*tan(e/2 + f*x/2)**2 - 3*a**2*c**2*f) + 4*tan(e/2 + f*x/2)**3/(3*a**2*c**2*f*tan(e/2 + f*x/2)**6 - 9
*a**2*c**2*f*tan(e/2 + f*x/2)**4 + 9*a**2*c**2*f*tan(e/2 + f*x/2)**2 - 3*a**2*c**2*f) - 6*tan(e/2 + f*x/2)/(3*
a**2*c**2*f*tan(e/2 + f*x/2)**6 - 9*a**2*c**2*f*tan(e/2 + f*x/2)**4 + 9*a**2*c**2*f*tan(e/2 + f*x/2)**2 - 3*a*
*2*c**2*f), Ne(f, 0)), (x/((a*sin(e) + a)**2*(-c*sin(e) + c)**2), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.74 \[ \int \frac {1}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^2} \, dx=\frac {\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )}{3 \, a^{2} c^{2} f} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.74 \[ \int \frac {1}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^2} \, dx=\frac {\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )}{3 \, a^{2} c^{2} f} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 7.14 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.66 \[ \int \frac {1}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^2} \, dx=-\frac {2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+3\right )}{3\,a^2\,c^2\,f\,{\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )}^3} \]

[In]

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

[Out]

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