3.308 \(\int \frac{\cos ^2(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx\)
Optimal. Leaf size=83 \[ -\frac{\cos ^3(c+d x)}{3 a^2 d}+\frac{3 \cos (c+d x)}{a^2 d}-\frac{\sin (c+d x) \cos (c+d x)}{a^2 d}+\frac{2 \cos (c+d x)}{a^2 d (\sin (c+d x)+1)}+\frac{3 x}{a^2} \]
[Out]
(3*x)/a^2 + (3*Cos[c + d*x])/(a^2*d) - Cos[c + d*x]^3/(3*a^2*d) - (Cos[c + d*x]*Sin[c + d*x])/(a^2*d) + (2*Cos
[c + d*x])/(a^2*d*(1 + Sin[c + d*x]))
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Rubi [A] time = 0.221578, antiderivative size = 83, normalized size of antiderivative = 1.,
number of steps used = 9, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241, Rules used =
{2874, 2966, 2638, 2635, 8, 2633, 2648} \[ -\frac{\cos ^3(c+d x)}{3 a^2 d}+\frac{3 \cos (c+d x)}{a^2 d}-\frac{\sin (c+d x) \cos (c+d x)}{a^2 d}+\frac{2 \cos (c+d x)}{a^2 d (\sin (c+d x)+1)}+\frac{3 x}{a^2} \]
Antiderivative was successfully verified.
[In]
Int[(Cos[c + d*x]^2*Sin[c + d*x]^3)/(a + a*Sin[c + d*x])^2,x]
[Out]
(3*x)/a^2 + (3*Cos[c + d*x])/(a^2*d) - Cos[c + d*x]^3/(3*a^2*d) - (Cos[c + d*x]*Sin[c + d*x])/(a^2*d) + (2*Cos
[c + d*x])/(a^2*d*(1 + Sin[c + d*x]))
Rule 2874
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 2966
Int[sin[(e_.) + (f_.)*(x_)]^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Int[ExpandTrig[sin[e + f*x]^n*(a + b*sin[e + f*x])^m*(A + B*sin[e + f*x]), x], x] /; Fr
eeQ[{a, b, e, f, A, B}, x] && EqQ[A*b + a*B, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && IntegerQ[n]
Rule 2638
Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]
Rule 2635
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] && Integer
Q[2*n]
Rule 8
Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
Rule 2633
Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]
Rule 2648
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]
Rubi steps
\begin{align*} \int \frac{\cos ^2(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac{\int \frac{\sin ^3(c+d x) (a-a \sin (c+d x))}{a+a \sin (c+d x)} \, dx}{a^2}\\ &=\frac{\int \left (2-2 \sin (c+d x)+2 \sin ^2(c+d x)-\sin ^3(c+d x)-\frac{2}{1+\sin (c+d x)}\right ) \, dx}{a^2}\\ &=\frac{2 x}{a^2}-\frac{\int \sin ^3(c+d x) \, dx}{a^2}-\frac{2 \int \sin (c+d x) \, dx}{a^2}+\frac{2 \int \sin ^2(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)}{a^2 d}+\frac{2 \cos (c+d x)}{a^2 d (1+\sin (c+d x))}+\frac{\int 1 \, dx}{a^2}+\frac{\operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{a^2 d}\\ &=\frac{3 x}{a^2}+\frac{3 \cos (c+d x)}{a^2 d}-\frac{\cos ^3(c+d x)}{3 a^2 d}-\frac{\cos (c+d x) \sin (c+d x)}{a^2 d}+\frac{2 \cos (c+d x)}{a^2 d (1+\sin (c+d x))}\\ \end{align*}
Mathematica [A] time = 0.92207, size = 165, normalized size = 1.99 \[ -\frac{-72 d x \sin \left (c+\frac{d x}{2}\right )-27 \sin \left (2 c+\frac{3 d x}{2}\right )+5 \sin \left (2 c+\frac{5 d x}{2}\right )+\sin \left (4 c+\frac{7 d x}{2}\right )-31 \cos \left (c+\frac{d x}{2}\right )-27 \cos \left (c+\frac{3 d x}{2}\right )-5 \cos \left (3 c+\frac{5 d x}{2}\right )+\cos \left (3 c+\frac{7 d x}{2}\right )+131 \sin \left (\frac{d x}{2}\right )-72 d x \cos \left (\frac{d x}{2}\right )}{24 a^2 d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )} \]
Antiderivative was successfully verified.
[In]
Integrate[(Cos[c + d*x]^2*Sin[c + d*x]^3)/(a + a*Sin[c + d*x])^2,x]
[Out]
-(-72*d*x*Cos[(d*x)/2] - 31*Cos[c + (d*x)/2] - 27*Cos[c + (3*d*x)/2] - 5*Cos[3*c + (5*d*x)/2] + Cos[3*c + (7*d
*x)/2] + 131*Sin[(d*x)/2] - 72*d*x*Sin[c + (d*x)/2] - 27*Sin[2*c + (3*d*x)/2] + 5*Sin[2*c + (5*d*x)/2] + Sin[4
*c + (7*d*x)/2])/(24*a^2*d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))
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Maple [B] time = 0.104, size = 198, normalized size = 2.4 \begin{align*} 2\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}}{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+4\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}}{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+12\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}-2\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+{\frac{16}{3\,d{a}^{2}} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-3}}+6\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{2}}}+4\,{\frac{1}{d{a}^{2} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)^2*sin(d*x+c)^3/(a+a*sin(d*x+c))^2,x)
[Out]
2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^5+4/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^
4+12/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^2-2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*
c)+16/3/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3+6/d/a^2*arctan(tan(1/2*d*x+1/2*c))+4/d/a^2/(tan(1/2*d*x+1/2*c)+1)
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Maxima [B] time = 1.72474, size = 421, normalized size = 5.07 \begin{align*} \frac{2 \,{\left (\frac{\frac{5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{33 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{18 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{24 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{9 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{9 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + 14}{a^{2} + \frac{a^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{3 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{3 \, a^{2} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{3 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{3 \, a^{2} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac{a^{2} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}} + \frac{9 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )}}{3 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^2*sin(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="maxima")
[Out]
2/3*((5*sin(d*x + c)/(cos(d*x + c) + 1) + 33*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 18*sin(d*x + c)^3/(cos(d*x
+ c) + 1)^3 + 24*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 9*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 9*sin(d*x + c)^
6/(cos(d*x + c) + 1)^6 + 14)/(a^2 + a^2*sin(d*x + c)/(cos(d*x + c) + 1) + 3*a^2*sin(d*x + c)^2/(cos(d*x + c) +
1)^2 + 3*a^2*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 3*a^2*sin(d*x
+ c)^5/(cos(d*x + c) + 1)^5 + a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + a^2*sin(d*x + c)^7/(cos(d*x + c) + 1)^
7) + 9*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d
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Fricas [A] time = 1.66617, size = 313, normalized size = 3.77 \begin{align*} -\frac{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{3} - 9 \, d x - 3 \,{\left (3 \, d x + 4\right )} \cos \left (d x + c\right ) - 9 \, \cos \left (d x + c\right )^{2} +{\left (\cos \left (d x + c\right )^{3} - 9 \, d x + 3 \, \cos \left (d x + c\right )^{2} - 6 \, \cos \left (d x + c\right ) + 6\right )} \sin \left (d x + c\right ) - 6}{3 \,{\left (a^{2} d \cos \left (d x + c\right ) + a^{2} d \sin \left (d x + c\right ) + a^{2} d\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^2*sin(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="fricas")
[Out]
-1/3*(cos(d*x + c)^4 - 2*cos(d*x + c)^3 - 9*d*x - 3*(3*d*x + 4)*cos(d*x + c) - 9*cos(d*x + c)^2 + (cos(d*x + c
)^3 - 9*d*x + 3*cos(d*x + c)^2 - 6*cos(d*x + c) + 6)*sin(d*x + c) - 6)/(a^2*d*cos(d*x + c) + a^2*d*sin(d*x + c
) + a^2*d)
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Sympy [A] time = 48.7996, size = 2263, normalized size = 27.27 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)**2*sin(d*x+c)**3/(a+a*sin(d*x+c))**2,x)
[Out]
Piecewise((9*d*x*tan(c/2 + d*x/2)**7/(3*a**2*d*tan(c/2 + d*x/2)**7 + 3*a**2*d*tan(c/2 + d*x/2)**6 + 9*a**2*d*t
an(c/2 + d*x/2)**5 + 9*a**2*d*tan(c/2 + d*x/2)**4 + 9*a**2*d*tan(c/2 + d*x/2)**3 + 9*a**2*d*tan(c/2 + d*x/2)**
2 + 3*a**2*d*tan(c/2 + d*x/2) + 3*a**2*d) + 9*d*x*tan(c/2 + d*x/2)**6/(3*a**2*d*tan(c/2 + d*x/2)**7 + 3*a**2*d
*tan(c/2 + d*x/2)**6 + 9*a**2*d*tan(c/2 + d*x/2)**5 + 9*a**2*d*tan(c/2 + d*x/2)**4 + 9*a**2*d*tan(c/2 + d*x/2)
**3 + 9*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d*tan(c/2 + d*x/2) + 3*a**2*d) + 27*d*x*tan(c/2 + d*x/2)**5/(3*a**
2*d*tan(c/2 + d*x/2)**7 + 3*a**2*d*tan(c/2 + d*x/2)**6 + 9*a**2*d*tan(c/2 + d*x/2)**5 + 9*a**2*d*tan(c/2 + d*x
/2)**4 + 9*a**2*d*tan(c/2 + d*x/2)**3 + 9*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d*tan(c/2 + d*x/2) + 3*a**2*d) +
27*d*x*tan(c/2 + d*x/2)**4/(3*a**2*d*tan(c/2 + d*x/2)**7 + 3*a**2*d*tan(c/2 + d*x/2)**6 + 9*a**2*d*tan(c/2 +
d*x/2)**5 + 9*a**2*d*tan(c/2 + d*x/2)**4 + 9*a**2*d*tan(c/2 + d*x/2)**3 + 9*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**
2*d*tan(c/2 + d*x/2) + 3*a**2*d) + 27*d*x*tan(c/2 + d*x/2)**3/(3*a**2*d*tan(c/2 + d*x/2)**7 + 3*a**2*d*tan(c/2
+ d*x/2)**6 + 9*a**2*d*tan(c/2 + d*x/2)**5 + 9*a**2*d*tan(c/2 + d*x/2)**4 + 9*a**2*d*tan(c/2 + d*x/2)**3 + 9*
a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d*tan(c/2 + d*x/2) + 3*a**2*d) + 27*d*x*tan(c/2 + d*x/2)**2/(3*a**2*d*tan(
c/2 + d*x/2)**7 + 3*a**2*d*tan(c/2 + d*x/2)**6 + 9*a**2*d*tan(c/2 + d*x/2)**5 + 9*a**2*d*tan(c/2 + d*x/2)**4 +
9*a**2*d*tan(c/2 + d*x/2)**3 + 9*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d*tan(c/2 + d*x/2) + 3*a**2*d) + 9*d*x*t
an(c/2 + d*x/2)/(3*a**2*d*tan(c/2 + d*x/2)**7 + 3*a**2*d*tan(c/2 + d*x/2)**6 + 9*a**2*d*tan(c/2 + d*x/2)**5 +
9*a**2*d*tan(c/2 + d*x/2)**4 + 9*a**2*d*tan(c/2 + d*x/2)**3 + 9*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d*tan(c/2
+ d*x/2) + 3*a**2*d) + 9*d*x/(3*a**2*d*tan(c/2 + d*x/2)**7 + 3*a**2*d*tan(c/2 + d*x/2)**6 + 9*a**2*d*tan(c/2 +
d*x/2)**5 + 9*a**2*d*tan(c/2 + d*x/2)**4 + 9*a**2*d*tan(c/2 + d*x/2)**3 + 9*a**2*d*tan(c/2 + d*x/2)**2 + 3*a*
*2*d*tan(c/2 + d*x/2) + 3*a**2*d) + 18*tan(c/2 + d*x/2)**6/(3*a**2*d*tan(c/2 + d*x/2)**7 + 3*a**2*d*tan(c/2 +
d*x/2)**6 + 9*a**2*d*tan(c/2 + d*x/2)**5 + 9*a**2*d*tan(c/2 + d*x/2)**4 + 9*a**2*d*tan(c/2 + d*x/2)**3 + 9*a**
2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d*tan(c/2 + d*x/2) + 3*a**2*d) + 18*tan(c/2 + d*x/2)**5/(3*a**2*d*tan(c/2 + d
*x/2)**7 + 3*a**2*d*tan(c/2 + d*x/2)**6 + 9*a**2*d*tan(c/2 + d*x/2)**5 + 9*a**2*d*tan(c/2 + d*x/2)**4 + 9*a**2
*d*tan(c/2 + d*x/2)**3 + 9*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d*tan(c/2 + d*x/2) + 3*a**2*d) + 48*tan(c/2 + d
*x/2)**4/(3*a**2*d*tan(c/2 + d*x/2)**7 + 3*a**2*d*tan(c/2 + d*x/2)**6 + 9*a**2*d*tan(c/2 + d*x/2)**5 + 9*a**2*
d*tan(c/2 + d*x/2)**4 + 9*a**2*d*tan(c/2 + d*x/2)**3 + 9*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d*tan(c/2 + d*x/2
) + 3*a**2*d) + 36*tan(c/2 + d*x/2)**3/(3*a**2*d*tan(c/2 + d*x/2)**7 + 3*a**2*d*tan(c/2 + d*x/2)**6 + 9*a**2*d
*tan(c/2 + d*x/2)**5 + 9*a**2*d*tan(c/2 + d*x/2)**4 + 9*a**2*d*tan(c/2 + d*x/2)**3 + 9*a**2*d*tan(c/2 + d*x/2)
**2 + 3*a**2*d*tan(c/2 + d*x/2) + 3*a**2*d) + 66*tan(c/2 + d*x/2)**2/(3*a**2*d*tan(c/2 + d*x/2)**7 + 3*a**2*d*
tan(c/2 + d*x/2)**6 + 9*a**2*d*tan(c/2 + d*x/2)**5 + 9*a**2*d*tan(c/2 + d*x/2)**4 + 9*a**2*d*tan(c/2 + d*x/2)*
*3 + 9*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d*tan(c/2 + d*x/2) + 3*a**2*d) + 10*tan(c/2 + d*x/2)/(3*a**2*d*tan(
c/2 + d*x/2)**7 + 3*a**2*d*tan(c/2 + d*x/2)**6 + 9*a**2*d*tan(c/2 + d*x/2)**5 + 9*a**2*d*tan(c/2 + d*x/2)**4 +
9*a**2*d*tan(c/2 + d*x/2)**3 + 9*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d*tan(c/2 + d*x/2) + 3*a**2*d) + 28/(3*a
**2*d*tan(c/2 + d*x/2)**7 + 3*a**2*d*tan(c/2 + d*x/2)**6 + 9*a**2*d*tan(c/2 + d*x/2)**5 + 9*a**2*d*tan(c/2 + d
*x/2)**4 + 9*a**2*d*tan(c/2 + d*x/2)**3 + 9*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d*tan(c/2 + d*x/2) + 3*a**2*d)
, Ne(d, 0)), (x*sin(c)**3*cos(c)**2/(a*sin(c) + a)**2, True))
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Giac [A] time = 1.28067, size = 143, normalized size = 1.72 \begin{align*} \frac{\frac{9 \,{\left (d x + c\right )}}{a^{2}} + \frac{12}{a^{2}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}} + \frac{2 \,{\left (3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 18 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 8\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{3} a^{2}}}{3 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^2*sin(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="giac")
[Out]
1/3*(9*(d*x + c)/a^2 + 12/(a^2*(tan(1/2*d*x + 1/2*c) + 1)) + 2*(3*tan(1/2*d*x + 1/2*c)^5 + 6*tan(1/2*d*x + 1/2
*c)^4 + 18*tan(1/2*d*x + 1/2*c)^2 - 3*tan(1/2*d*x + 1/2*c) + 8)/((tan(1/2*d*x + 1/2*c)^2 + 1)^3*a^2))/d