\(\int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx\) [316]

Optimal result

Integrand size = 29, antiderivative size = 76 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {3 x}{a^3}+\frac {3 \cos (c+d x)}{a^3 d}-\frac {\cos ^3(c+d x)}{3 d (a+a \sin (c+d x))^3}+\frac {2 \cos ^3(c+d x)}{a d (a+a \sin (c+d x))^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2950, 2759, 2761, 8} \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {3 \cos (c+d x)}{a^3 d}+\frac {3 x}{a^3}+\frac {2 \cos ^3(c+d x)}{a d (a \sin (c+d x)+a)^2}-\frac {\cos ^3(c+d x)}{3 d (a \sin (c+d x)+a)^3} \]

[In]

Int[(Cos[c + d*x]^2*Sin[c + d*x]^2)/(a + a*Sin[c + d*x])^3,x]

[Out]

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

Rule 8

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

Rule 2759

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 2761

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[g*((g*Cos[e
+ f*x])^(p - 1)/(b*f*(p - 1))), x] + Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2), x], x] /; FreeQ[{a, b, e, f, g
}, x] && EqQ[a^2 - b^2, 0] && GtQ[p, 1] && IntegerQ[2*p]

Rule 2950

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*sin[(e_.) + (f_.)*(x_)]^2*((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*m)), x] - Dist[1/g^2, Int[(g*Cos
[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && E
qQ[m + p + 1, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {\cos ^3(c+d x)}{3 d (a+a \sin (c+d x))^3}-\int \frac {\cos ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx \\ & = -\frac {\cos ^3(c+d x)}{3 d (a+a \sin (c+d x))^3}+\frac {2 \cos ^3(c+d x)}{a d (a+a \sin (c+d x))^2}+\frac {3 \int \frac {\cos ^2(c+d x)}{a+a \sin (c+d x)} \, dx}{a^2} \\ & = \frac {3 \cos (c+d x)}{a^3 d}-\frac {\cos ^3(c+d x)}{3 d (a+a \sin (c+d x))^3}+\frac {2 \cos ^3(c+d x)}{a d (a+a \sin (c+d x))^2}+\frac {3 \int 1 \, dx}{a^3} \\ & = \frac {3 x}{a^3}+\frac {3 \cos (c+d x)}{a^3 d}-\frac {\cos ^3(c+d x)}{3 d (a+a \sin (c+d x))^3}+\frac {2 \cos ^3(c+d x)}{a d (a+a \sin (c+d x))^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.63 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.26 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {9 c+9 d x+3 \cos (c+d x)-\frac {2}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {2 \sin \left (\frac {1}{2} (c+d x)\right ) (11+13 \sin (c+d x))}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}}{3 a^3 d} \]

[In]

Integrate[(Cos[c + d*x]^2*Sin[c + d*x]^2)/(a + a*Sin[c + d*x])^3,x]

[Out]

(9*c + 9*d*x + 3*Cos[c + d*x] - 2/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 - (2*Sin[(c + d*x)/2]*(11 + 13*Sin[c
+ d*x]))/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3)/(3*a^3*d)

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.11

method	result	size
derivativedivides	\(\frac {\frac {8}{4+4 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+6 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {8}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {6}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{d \,a^{3}}\)	\(84\)
default	\(\frac {\frac {8}{4+4 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+6 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {8}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {6}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{d \,a^{3}}\)	\(84\)
risch	\(\frac {3 x}{a^{3}}+\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 d \,a^{3}}+\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 d \,a^{3}}+\frac {-\frac {26}{3}+16 i {\mathrm e}^{i \left (d x +c \right )}+10 \,{\mathrm e}^{2 i \left (d x +c \right )}}{d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{3}}\)	\(89\)
parallelrisch	\(\frac {18 d x \cos \left (3 d x +3 c \right )+54 d x \cos \left (d x +c \right )-26 \sin \left (3 d x +3 c \right )+28 \cos \left (3 d x +3 c \right )+72 \cos \left (2 d x +2 c \right )+84 \cos \left (d x +c \right )+6 \sin \left (d x +c \right )+3 \cos \left (4 d x +4 c \right )+37}{6 d \,a^{3} \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\)	\(114\)
norman	\(\frac {\frac {213 x \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {90 x \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {252 x \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {42 x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {252 x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {153 x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {213 x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {153 x \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {90 x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {42 x \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {28}{3 a d}+\frac {15 x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {3 x \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {3 x}{a}+\frac {30 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {302 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {122 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{3 d a}+\frac {1100 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {15 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a}+\frac {506 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {602 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {824 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {388 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {1252 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {922 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {238 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {6 \left (\tan ^{12}\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^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}\)	\(493\)

[In]

int(cos(d*x+c)^2*sin(d*x+c)^2/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.89 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {{\left (9 \, d x - 16\right )} \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right )^{3} - 18 \, d x - {\left (9 \, d x + 17\right )} \cos \left (d x + c\right ) - {\left (18 \, d x + {\left (9 \, d x + 19\right )} \cos \left (d x + c\right ) + 3 \, \cos \left (d x + c\right )^{2} + 2\right )} \sin \left (d x + c\right ) + 2}{3 \, {\left (a^{3} d \cos \left (d x + c\right )^{2} - a^{3} d \cos \left (d x + c\right ) - 2 \, a^{3} d - {\left (a^{3} d \cos \left (d x + c\right ) + 2 \, a^{3} d\right )} \sin \left (d x + c\right )\right )}} \]

[In]

integrate(cos(d*x+c)^2*sin(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/3*((9*d*x - 16)*cos(d*x + c)^2 + 3*cos(d*x + c)^3 - 18*d*x - (9*d*x + 17)*cos(d*x + c) - (18*d*x + (9*d*x +
19)*cos(d*x + c) + 3*cos(d*x + c)^2 + 2)*sin(d*x + c) + 2)/(a^3*d*cos(d*x + c)^2 - a^3*d*cos(d*x + c) - 2*a^3*
d - (a^3*d*cos(d*x + c) + 2*a^3*d)*sin(d*x + c))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1246 vs. \(2 (68) = 136\).

Time = 13.71 (sec) , antiderivative size = 1246, normalized size of antiderivative = 16.39 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\text {Too large to display} \]

[In]

integrate(cos(d*x+c)**2*sin(d*x+c)**2/(a+a*sin(d*x+c))**3,x)

[Out]

Piecewise((9*d*x*tan(c/2 + d*x/2)**5/(3*a**3*d*tan(c/2 + d*x/2)**5 + 9*a**3*d*tan(c/2 + d*x/2)**4 + 12*a**3*d*
tan(c/2 + d*x/2)**3 + 12*a**3*d*tan(c/2 + d*x/2)**2 + 9*a**3*d*tan(c/2 + d*x/2) + 3*a**3*d) + 27*d*x*tan(c/2 +
d*x/2)**4/(3*a**3*d*tan(c/2 + d*x/2)**5 + 9*a**3*d*tan(c/2 + d*x/2)**4 + 12*a**3*d*tan(c/2 + d*x/2)**3 + 12*a
**3*d*tan(c/2 + d*x/2)**2 + 9*a**3*d*tan(c/2 + d*x/2) + 3*a**3*d) + 36*d*x*tan(c/2 + d*x/2)**3/(3*a**3*d*tan(c
/2 + d*x/2)**5 + 9*a**3*d*tan(c/2 + d*x/2)**4 + 12*a**3*d*tan(c/2 + d*x/2)**3 + 12*a**3*d*tan(c/2 + d*x/2)**2
+ 9*a**3*d*tan(c/2 + d*x/2) + 3*a**3*d) + 36*d*x*tan(c/2 + d*x/2)**2/(3*a**3*d*tan(c/2 + d*x/2)**5 + 9*a**3*d*
tan(c/2 + d*x/2)**4 + 12*a**3*d*tan(c/2 + d*x/2)**3 + 12*a**3*d*tan(c/2 + d*x/2)**2 + 9*a**3*d*tan(c/2 + d*x/2
) + 3*a**3*d) + 27*d*x*tan(c/2 + d*x/2)/(3*a**3*d*tan(c/2 + d*x/2)**5 + 9*a**3*d*tan(c/2 + d*x/2)**4 + 12*a**3
*d*tan(c/2 + d*x/2)**3 + 12*a**3*d*tan(c/2 + d*x/2)**2 + 9*a**3*d*tan(c/2 + d*x/2) + 3*a**3*d) + 9*d*x/(3*a**3
*d*tan(c/2 + d*x/2)**5 + 9*a**3*d*tan(c/2 + d*x/2)**4 + 12*a**3*d*tan(c/2 + d*x/2)**3 + 12*a**3*d*tan(c/2 + d*
x/2)**2 + 9*a**3*d*tan(c/2 + d*x/2) + 3*a**3*d) + 18*tan(c/2 + d*x/2)**4/(3*a**3*d*tan(c/2 + d*x/2)**5 + 9*a**
3*d*tan(c/2 + d*x/2)**4 + 12*a**3*d*tan(c/2 + d*x/2)**3 + 12*a**3*d*tan(c/2 + d*x/2)**2 + 9*a**3*d*tan(c/2 + d
*x/2) + 3*a**3*d) + 54*tan(c/2 + d*x/2)**3/(3*a**3*d*tan(c/2 + d*x/2)**5 + 9*a**3*d*tan(c/2 + d*x/2)**4 + 12*a
**3*d*tan(c/2 + d*x/2)**3 + 12*a**3*d*tan(c/2 + d*x/2)**2 + 9*a**3*d*tan(c/2 + d*x/2) + 3*a**3*d) + 58*tan(c/2
+ d*x/2)**2/(3*a**3*d*tan(c/2 + d*x/2)**5 + 9*a**3*d*tan(c/2 + d*x/2)**4 + 12*a**3*d*tan(c/2 + d*x/2)**3 + 12
*a**3*d*tan(c/2 + d*x/2)**2 + 9*a**3*d*tan(c/2 + d*x/2) + 3*a**3*d) + 66*tan(c/2 + d*x/2)/(3*a**3*d*tan(c/2 +
d*x/2)**5 + 9*a**3*d*tan(c/2 + d*x/2)**4 + 12*a**3*d*tan(c/2 + d*x/2)**3 + 12*a**3*d*tan(c/2 + d*x/2)**2 + 9*a
**3*d*tan(c/2 + d*x/2) + 3*a**3*d) + 28/(3*a**3*d*tan(c/2 + d*x/2)**5 + 9*a**3*d*tan(c/2 + d*x/2)**4 + 12*a**3
*d*tan(c/2 + d*x/2)**3 + 12*a**3*d*tan(c/2 + d*x/2)**2 + 9*a**3*d*tan(c/2 + d*x/2) + 3*a**3*d), Ne(d, 0)), (x*
sin(c)**2*cos(c)**2/(a*sin(c) + a)**3, True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 228 vs. \(2 (74) = 148\).

Time = 0.28 (sec) , antiderivative size = 228, normalized size of antiderivative = 3.00 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {2 \, {\left (\frac {\frac {33 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {29 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {27 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {9 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 14}{a^{3} + \frac {3 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {4 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {4 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}} + \frac {9 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )}}{3 \, d} \]

[In]

integrate(cos(d*x+c)^2*sin(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

2/3*((33*sin(d*x + c)/(cos(d*x + c) + 1) + 29*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 27*sin(d*x + c)^3/(cos(d*x
+ c) + 1)^3 + 9*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 14)/(a^3 + 3*a^3*sin(d*x + c)/(cos(d*x + c) + 1) + 4*a^
3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 4*a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*a^3*sin(d*x + c)^4/(cos(
d*x + c) + 1)^4 + a^3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5) + 9*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3)/d

Giac [A] (verification not implemented)

none

Time = 0.45 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.05 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {9 \, {\left (d x + c\right )}}{a^{3}} + \frac {6}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a^{3}} + \frac {2 \, {\left (9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 24 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 11\right )}}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{3}}}{3 \, d} \]

[In]

integrate(cos(d*x+c)^2*sin(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/3*(9*(d*x + c)/a^3 + 6/((tan(1/2*d*x + 1/2*c)^2 + 1)*a^3) + 2*(9*tan(1/2*d*x + 1/2*c)^2 + 24*tan(1/2*d*x + 1
/2*c) + 11)/(a^3*(tan(1/2*d*x + 1/2*c) + 1)^3))/d

Mupad [B] (verification not implemented)

Time = 12.47 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.24 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {3\,x}{a^3}+\frac {6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+18\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {58\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+22\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {28}{3}}{a^3\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}^3\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

[In]

int((cos(c + d*x)^2*sin(c + d*x)^2)/(a + a*sin(c + d*x))^3,x)

[Out]

(3*x)/a^3 + (22*tan(c/2 + (d*x)/2) + (58*tan(c/2 + (d*x)/2)^2)/3 + 18*tan(c/2 + (d*x)/2)^3 + 6*tan(c/2 + (d*x)
/2)^4 + 28/3)/(a^3*d*(tan(c/2 + (d*x)/2) + 1)^3*(tan(c/2 + (d*x)/2)^2 + 1))