\(\int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx\) [309]

Optimal result

Integrand size = 29, antiderivative size = 69 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {5 x}{2 a^2}-\frac {2 \cos (c+d x)}{a^2 d}+\frac {\cos (c+d x) \sin (c+d x)}{2 a^2 d}-\frac {2 \cos (c+d x)}{a^2 d (1+\sin (c+d x))} \]

[Out]

-5/2*x/a^2-2*cos(d*x+c)/a^2/d+1/2*cos(d*x+c)*sin(d*x+c)/a^2/d-2*cos(d*x+c)/a^2/d/(1+sin(d*x+c))

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2953, 3029, 2788, 2718, 2715, 8, 2727} \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {2 \cos (c+d x)}{a^2 d}+\frac {\sin (c+d x) \cos (c+d x)}{2 a^2 d}-\frac {2 \cos (c+d x)}{a^2 d (\sin (c+d x)+1)}-\frac {5 x}{2 a^2} \]

[In]

Int[(Cos[c + d*x]^2*Sin[c + d*x]^2)/(a + a*Sin[c + d*x])^2,x]

[Out]

(-5*x)/(2*a^2) - (2*Cos[c + d*x])/(a^2*d) + (Cos[c + d*x]*Sin[c + d*x])/(2*a^2*d) - (2*Cos[c + d*x])/(a^2*d*(1
+ Sin[c + d*x]))

Rule 8

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

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2727

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[-Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2788

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^(p_), x_Symbol] :> Dist[a^p, Int[Expan
dIntegrand[Sin[e + f*x]^p*((a + b*Sin[e + f*x])^(m - p/2)/(a - b*Sin[e + f*x])^(p/2)), x], x], x] /; FreeQ[{a,
b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p/2] && (LtQ[p, 0] || GtQ[m - p/2, 0])

Rule 2953

Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Dist[1/b^2, Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^(m + 1)*(a - b*Sin[e + f*x]), x], x] /;
FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && (ILtQ[m, 0] ||  !IGtQ[n, 0])

Rule 3029

Int[sin[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*
(x_)])^(n_.), x_Symbol] :> Dist[a^n*c^n, Int[Tan[e + f*x]^p*(a + b*Sin[e + f*x])^(m - n), x], x] /; FreeQ[{a,
b, c, d, e, f, m}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && EqQ[p + 2*n, 0] && IntegerQ[n]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {\sin ^2(c+d x) (a-a \sin (c+d x))}{a+a \sin (c+d x)} \, dx}{a^2} \\ & = \frac {\int (a-a \sin (c+d x))^2 \tan ^2(c+d x) \, dx}{a^4} \\ & = \frac {\int \left (-2+2 \sin (c+d x)-\sin ^2(c+d x)+\frac {2}{1+\sin (c+d x)}\right ) \, dx}{a^2} \\ & = -\frac {2 x}{a^2}-\frac {\int \sin ^2(c+d x) \, dx}{a^2}+\frac {2 \int \sin (c+d x) \, dx}{a^2}+\frac {2 \int \frac {1}{1+\sin (c+d x)} \, dx}{a^2} \\ & = -\frac {2 x}{a^2}-\frac {2 \cos (c+d x)}{a^2 d}+\frac {\cos (c+d x) \sin (c+d x)}{2 a^2 d}-\frac {2 \cos (c+d x)}{a^2 d (1+\sin (c+d x))}-\frac {\int 1 \, dx}{2 a^2} \\ & = -\frac {5 x}{2 a^2}-\frac {2 \cos (c+d x)}{a^2 d}+\frac {\cos (c+d x) \sin (c+d x)}{2 a^2 d}-\frac {2 \cos (c+d x)}{a^2 d (1+\sin (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {-10 (c+d x)-8 \cos (c+d x)+\frac {16 \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}+\sin (2 (c+d x))}{4 a^2 d} \]

[In]

Integrate[(Cos[c + d*x]^2*Sin[c + d*x]^2)/(a + a*Sin[c + d*x])^2,x]

[Out]

(-10*(c + d*x) - 8*Cos[c + d*x] + (16*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + Sin[2*(c + d*x
)])/(4*a^2*d)

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.94

method	result	size
parallelrisch	\(\frac {-20 d x \cos \left (d x +c \right )+\sin \left (3 d x +3 c \right )-8 \cos \left (2 d x +2 c \right )-32 \cos \left (d x +c \right )+17 \sin \left (d x +c \right )-24}{8 d \,a^{2} \cos \left (d x +c \right )}\)	\(65\)
risch	\(-\frac {5 x}{2 a^{2}}-\frac {{\mathrm e}^{i \left (d x +c \right )}}{d \,a^{2}}-\frac {{\mathrm e}^{-i \left (d x +c \right )}}{d \,a^{2}}-\frac {4}{d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}+\frac {\sin \left (2 d x +2 c \right )}{4 d \,a^{2}}\)	\(81\)
derivativedivides	\(\frac {-\frac {4 \left (\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4}+1\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-5 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {4}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{d \,a^{2}}\)	\(91\)
default	\(\frac {-\frac {4 \left (\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4}+1\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-5 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {4}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{d \,a^{2}}\)	\(91\)
norman	\(\frac {-\frac {8}{a d}-\frac {19 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}-\frac {5 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {41 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {15 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {68 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {36 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {5 x}{2 a}-\frac {15 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}-\frac {35 x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}-\frac {65 x \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}-\frac {45 x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {55 x \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {55 x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {45 x \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {65 x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}-\frac {35 x \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}-\frac {15 x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}-\frac {5 x \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}-\frac {84 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {60 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {94 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {82 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}\)	\(421\)

[In]

int(cos(d*x+c)^2*sin(d*x+c)^2/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/8/d/a^2*(-20*d*x*cos(d*x+c)+sin(3*d*x+3*c)-8*cos(2*d*x+2*c)-32*cos(d*x+c)+17*sin(d*x+c)-24)/cos(d*x+c)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.45 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\cos \left (d x + c\right )^{3} + 5 \, d x + {\left (5 \, d x + 7\right )} \cos \left (d x + c\right ) + 4 \, \cos \left (d x + c\right )^{2} + {\left (5 \, d x - \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) - 4\right )} \sin \left (d x + c\right ) + 4}{2 \, {\left (a^{2} d \cos \left (d x + c\right ) + a^{2} d \sin \left (d x + c\right ) + a^{2} d\right )}} \]

[In]

integrate(cos(d*x+c)^2*sin(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/2*(cos(d*x + c)^3 + 5*d*x + (5*d*x + 7)*cos(d*x + c) + 4*cos(d*x + c)^2 + (5*d*x - cos(d*x + c)^2 + 3*cos(d
*x + c) - 4)*sin(d*x + c) + 4)/(a^2*d*cos(d*x + c) + a^2*d*sin(d*x + c) + a^2*d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1248 vs. \(2 (63) = 126\).

Time = 7.17 (sec) , antiderivative size = 1248, normalized size of antiderivative = 18.09 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\text {Too large to display} \]

[In]

integrate(cos(d*x+c)**2*sin(d*x+c)**2/(a+a*sin(d*x+c))**2,x)

[Out]

Piecewise((-5*d*x*tan(c/2 + d*x/2)**5/(2*a**2*d*tan(c/2 + d*x/2)**5 + 2*a**2*d*tan(c/2 + d*x/2)**4 + 4*a**2*d*
tan(c/2 + d*x/2)**3 + 4*a**2*d*tan(c/2 + d*x/2)**2 + 2*a**2*d*tan(c/2 + d*x/2) + 2*a**2*d) - 5*d*x*tan(c/2 + d
*x/2)**4/(2*a**2*d*tan(c/2 + d*x/2)**5 + 2*a**2*d*tan(c/2 + d*x/2)**4 + 4*a**2*d*tan(c/2 + d*x/2)**3 + 4*a**2*
d*tan(c/2 + d*x/2)**2 + 2*a**2*d*tan(c/2 + d*x/2) + 2*a**2*d) - 10*d*x*tan(c/2 + d*x/2)**3/(2*a**2*d*tan(c/2 +
d*x/2)**5 + 2*a**2*d*tan(c/2 + d*x/2)**4 + 4*a**2*d*tan(c/2 + d*x/2)**3 + 4*a**2*d*tan(c/2 + d*x/2)**2 + 2*a*
*2*d*tan(c/2 + d*x/2) + 2*a**2*d) - 10*d*x*tan(c/2 + d*x/2)**2/(2*a**2*d*tan(c/2 + d*x/2)**5 + 2*a**2*d*tan(c/
2 + d*x/2)**4 + 4*a**2*d*tan(c/2 + d*x/2)**3 + 4*a**2*d*tan(c/2 + d*x/2)**2 + 2*a**2*d*tan(c/2 + d*x/2) + 2*a*
*2*d) - 5*d*x*tan(c/2 + d*x/2)/(2*a**2*d*tan(c/2 + d*x/2)**5 + 2*a**2*d*tan(c/2 + d*x/2)**4 + 4*a**2*d*tan(c/2
+ d*x/2)**3 + 4*a**2*d*tan(c/2 + d*x/2)**2 + 2*a**2*d*tan(c/2 + d*x/2) + 2*a**2*d) - 5*d*x/(2*a**2*d*tan(c/2
+ d*x/2)**5 + 2*a**2*d*tan(c/2 + d*x/2)**4 + 4*a**2*d*tan(c/2 + d*x/2)**3 + 4*a**2*d*tan(c/2 + d*x/2)**2 + 2*a
**2*d*tan(c/2 + d*x/2) + 2*a**2*d) - 10*tan(c/2 + d*x/2)**4/(2*a**2*d*tan(c/2 + d*x/2)**5 + 2*a**2*d*tan(c/2 +
d*x/2)**4 + 4*a**2*d*tan(c/2 + d*x/2)**3 + 4*a**2*d*tan(c/2 + d*x/2)**2 + 2*a**2*d*tan(c/2 + d*x/2) + 2*a**2*
d) - 10*tan(c/2 + d*x/2)**3/(2*a**2*d*tan(c/2 + d*x/2)**5 + 2*a**2*d*tan(c/2 + d*x/2)**4 + 4*a**2*d*tan(c/2 +
d*x/2)**3 + 4*a**2*d*tan(c/2 + d*x/2)**2 + 2*a**2*d*tan(c/2 + d*x/2) + 2*a**2*d) - 22*tan(c/2 + d*x/2)**2/(2*a
**2*d*tan(c/2 + d*x/2)**5 + 2*a**2*d*tan(c/2 + d*x/2)**4 + 4*a**2*d*tan(c/2 + d*x/2)**3 + 4*a**2*d*tan(c/2 + d
*x/2)**2 + 2*a**2*d*tan(c/2 + d*x/2) + 2*a**2*d) - 6*tan(c/2 + d*x/2)/(2*a**2*d*tan(c/2 + d*x/2)**5 + 2*a**2*d
*tan(c/2 + d*x/2)**4 + 4*a**2*d*tan(c/2 + d*x/2)**3 + 4*a**2*d*tan(c/2 + d*x/2)**2 + 2*a**2*d*tan(c/2 + d*x/2)
+ 2*a**2*d) - 16/(2*a**2*d*tan(c/2 + d*x/2)**5 + 2*a**2*d*tan(c/2 + d*x/2)**4 + 4*a**2*d*tan(c/2 + d*x/2)**3
+ 4*a**2*d*tan(c/2 + d*x/2)**2 + 2*a**2*d*tan(c/2 + d*x/2) + 2*a**2*d), Ne(d, 0)), (x*sin(c)**2*cos(c)**2/(a*s
in(c) + a)**2, True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 226 vs. \(2 (65) = 130\).

Time = 0.28 (sec) , antiderivative size = 226, normalized size of antiderivative = 3.28 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {11 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {5 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 8}{a^{2} + \frac {a^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {2 \, a^{2} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{2} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}} + \frac {5 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{d} \]

[In]

integrate(cos(d*x+c)^2*sin(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-((3*sin(d*x + c)/(cos(d*x + c) + 1) + 11*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 5*sin(d*x + c)^3/(cos(d*x + c)
+ 1)^3 + 5*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 8)/(a^2 + a^2*sin(d*x + c)/(cos(d*x + c) + 1) + 2*a^2*sin(d*
x + c)^2/(cos(d*x + c) + 1)^2 + 2*a^2*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + a^2*sin(d*x + c)^4/(cos(d*x + c) +
1)^4 + a^2*sin(d*x + c)^5/(cos(d*x + c) + 1)^5) + 5*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.32 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {5 \, {\left (d x + c\right )}}{a^{2}} + \frac {2 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{2}} + \frac {8}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}}}{2 \, d} \]

[In]

integrate(cos(d*x+c)^2*sin(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 12.18 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.38 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {5\,x}{2\,a^2}-\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+11\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+8}{a^2\,d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^2} \]

[In]

int((cos(c + d*x)^2*sin(c + d*x)^2)/(a + a*sin(c + d*x))^2,x)

[Out]

- (5*x)/(2*a^2) - (3*tan(c/2 + (d*x)/2) + 11*tan(c/2 + (d*x)/2)^2 + 5*tan(c/2 + (d*x)/2)^3 + 5*tan(c/2 + (d*x)
/2)^4 + 8)/(a^2*d*(tan(c/2 + (d*x)/2) + 1)*(tan(c/2 + (d*x)/2)^2 + 1)^2)