\(\int \frac {\cos ^2(c+d x) \sin ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx\) [1074]

Optimal result

Integrand size = 29, antiderivative size = 188 \[ \int \frac {\cos ^2(c+d x) \sin ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {a \left (4 a^2-b^2\right ) x}{b^5}-\frac {2 a^2 \left (4 a^2-3 b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^5 \sqrt {a^2-b^2} d}+\frac {\left (12 a^2-b^2\right ) \cos (c+d x)}{3 b^4 d}-\frac {2 a \cos (c+d x) \sin (c+d x)}{b^3 d}+\frac {4 \cos (c+d x) \sin ^2(c+d x)}{3 b^2 d}-\frac {\cos (c+d x) \sin ^3(c+d x)}{b d (a+b \sin (c+d x))} \] Output:

a*(4*a^2-b^2)*x/b^5-2*a^2*(4*a^2-3*b^2)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a 
^2-b^2)^(1/2))/b^5/(a^2-b^2)^(1/2)/d+1/3*(12*a^2-b^2)*cos(d*x+c)/b^4/d-2*a 
*cos(d*x+c)*sin(d*x+c)/b^3/d+4/3*cos(d*x+c)*sin(d*x+c)^2/b^2/d-cos(d*x+c)* 
sin(d*x+c)^3/b/d/(a+b*sin(d*x+c))
 

Mathematica [A] (verified)

Time = 3.59 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.31 \[ \int \frac {\cos ^2(c+d x) \sin ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {-\frac {48 a^2 \left (4 a^2-3 b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {96 a^4 c-24 a^2 b^2 c+96 a^4 d x-24 a^2 b^2 d x+12 a b \left (8 a^2-b^2\right ) \cos (c+d x)+4 a b^3 \cos (3 (c+d x))+96 a^3 b c \sin (c+d x)-24 a b^3 c \sin (c+d x)+96 a^3 b d x \sin (c+d x)-24 a b^3 d x \sin (c+d x)+24 a^2 b^2 \sin (2 (c+d x))-2 b^4 \sin (2 (c+d x))-b^4 \sin (4 (c+d x))}{a+b \sin (c+d x)}}{24 b^5 d} \] Input:

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

Output:

((-48*a^2*(4*a^2 - 3*b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]] 
)/Sqrt[a^2 - b^2] + (96*a^4*c - 24*a^2*b^2*c + 96*a^4*d*x - 24*a^2*b^2*d*x 
 + 12*a*b*(8*a^2 - b^2)*Cos[c + d*x] + 4*a*b^3*Cos[3*(c + d*x)] + 96*a^3*b 
*c*Sin[c + d*x] - 24*a*b^3*c*Sin[c + d*x] + 96*a^3*b*d*x*Sin[c + d*x] - 24 
*a*b^3*d*x*Sin[c + d*x] + 24*a^2*b^2*Sin[2*(c + d*x)] - 2*b^4*Sin[2*(c + d 
*x)] - b^4*Sin[4*(c + d*x)])/(a + b*Sin[c + d*x]))/(24*b^5*d)
 

Rubi [A] (verified)

Time = 1.71 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.40, number of steps used = 21, number of rules used = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.690, Rules used = {3042, 3368, 3042, 3527, 25, 3042, 3529, 25, 3042, 3528, 27, 3042, 3502, 27, 3042, 3214, 3042, 3139, 1083, 217}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\sin ^3(c+d x) \cos ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (c+d x)^3 \cos (c+d x)^2}{(a+b \sin (c+d x))^2}dx\)

\(\Big \downarrow \) 3368

\(\displaystyle \int \frac {\sin ^3(c+d x) \left (1-\sin ^2(c+d x)\right )}{(a+b \sin (c+d x))^2}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (c+d x)^3 \left (1-\sin (c+d x)^2\right )}{(a+b \sin (c+d x))^2}dx\)

\(\Big \downarrow \) 3527

\(\displaystyle -\frac {\int -\frac {\sin ^2(c+d x) \left (3 \left (a^2-b^2\right )-4 \left (a^2-b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)}dx}{b \left (a^2-b^2\right )}-\frac {\sin ^3(c+d x) \cos (c+d x)}{b d (a+b \sin (c+d x))}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {\sin ^2(c+d x) \left (3 \left (a^2-b^2\right )-4 \left (a^2-b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)}dx}{b \left (a^2-b^2\right )}-\frac {\sin ^3(c+d x) \cos (c+d x)}{b d (a+b \sin (c+d x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\sin (c+d x)^2 \left (3 \left (a^2-b^2\right )-4 \left (a^2-b^2\right ) \sin (c+d x)^2\right )}{a+b \sin (c+d x)}dx}{b \left (a^2-b^2\right )}-\frac {\sin ^3(c+d x) \cos (c+d x)}{b d (a+b \sin (c+d x))}\)

\(\Big \downarrow \) 3529

\(\displaystyle \frac {\frac {\int -\frac {\sin (c+d x) \left (-12 a \left (a^2-b^2\right ) \sin ^2(c+d x)-b \left (a^2-b^2\right ) \sin (c+d x)+8 a \left (a^2-b^2\right )\right )}{a+b \sin (c+d x)}dx}{3 b}+\frac {4 \left (a^2-b^2\right ) \sin ^2(c+d x) \cos (c+d x)}{3 b d}}{b \left (a^2-b^2\right )}-\frac {\sin ^3(c+d x) \cos (c+d x)}{b d (a+b \sin (c+d x))}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {4 \left (a^2-b^2\right ) \sin ^2(c+d x) \cos (c+d x)}{3 b d}-\frac {\int \frac {\sin (c+d x) \left (-12 a \left (a^2-b^2\right ) \sin ^2(c+d x)-b \left (a^2-b^2\right ) \sin (c+d x)+8 a \left (a^2-b^2\right )\right )}{a+b \sin (c+d x)}dx}{3 b}}{b \left (a^2-b^2\right )}-\frac {\sin ^3(c+d x) \cos (c+d x)}{b d (a+b \sin (c+d x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {4 \left (a^2-b^2\right ) \sin ^2(c+d x) \cos (c+d x)}{3 b d}-\frac {\int \frac {\sin (c+d x) \left (-12 a \left (a^2-b^2\right ) \sin (c+d x)^2-b \left (a^2-b^2\right ) \sin (c+d x)+8 a \left (a^2-b^2\right )\right )}{a+b \sin (c+d x)}dx}{3 b}}{b \left (a^2-b^2\right )}-\frac {\sin ^3(c+d x) \cos (c+d x)}{b d (a+b \sin (c+d x))}\)

\(\Big \downarrow \) 3528

\(\displaystyle \frac {\frac {4 \left (a^2-b^2\right ) \sin ^2(c+d x) \cos (c+d x)}{3 b d}-\frac {\frac {\int -\frac {2 \left (6 \left (a^2-b^2\right ) a^2-2 b \left (a^2-b^2\right ) \sin (c+d x) a-\left (a^2-b^2\right ) \left (12 a^2-b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)}dx}{2 b}+\frac {6 a \left (a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{b d}}{3 b}}{b \left (a^2-b^2\right )}-\frac {\sin ^3(c+d x) \cos (c+d x)}{b d (a+b \sin (c+d x))}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {4 \left (a^2-b^2\right ) \sin ^2(c+d x) \cos (c+d x)}{3 b d}-\frac {\frac {6 a \left (a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{b d}-\frac {\int \frac {6 \left (a^2-b^2\right ) a^2-2 b \left (a^2-b^2\right ) \sin (c+d x) a-\left (a^2-b^2\right ) \left (12 a^2-b^2\right ) \sin ^2(c+d x)}{a+b \sin (c+d x)}dx}{b}}{3 b}}{b \left (a^2-b^2\right )}-\frac {\sin ^3(c+d x) \cos (c+d x)}{b d (a+b \sin (c+d x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {4 \left (a^2-b^2\right ) \sin ^2(c+d x) \cos (c+d x)}{3 b d}-\frac {\frac {6 a \left (a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{b d}-\frac {\int \frac {6 \left (a^2-b^2\right ) a^2-2 b \left (a^2-b^2\right ) \sin (c+d x) a-\left (a^2-b^2\right ) \left (12 a^2-b^2\right ) \sin (c+d x)^2}{a+b \sin (c+d x)}dx}{b}}{3 b}}{b \left (a^2-b^2\right )}-\frac {\sin ^3(c+d x) \cos (c+d x)}{b d (a+b \sin (c+d x))}\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {\frac {4 \left (a^2-b^2\right ) \sin ^2(c+d x) \cos (c+d x)}{3 b d}-\frac {\frac {6 a \left (a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{b d}-\frac {\frac {\int \frac {3 \left (2 b \left (a^2-b^2\right ) a^2+\left (a^2-b^2\right ) \left (4 a^2-b^2\right ) \sin (c+d x) a\right )}{a+b \sin (c+d x)}dx}{b}+\frac {\left (a^2-b^2\right ) \left (12 a^2-b^2\right ) \cos (c+d x)}{b d}}{b}}{3 b}}{b \left (a^2-b^2\right )}-\frac {\sin ^3(c+d x) \cos (c+d x)}{b d (a+b \sin (c+d x))}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {4 \left (a^2-b^2\right ) \sin ^2(c+d x) \cos (c+d x)}{3 b d}-\frac {\frac {6 a \left (a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{b d}-\frac {\frac {3 \int \frac {2 b \left (a^2-b^2\right ) a^2+\left (a^2-b^2\right ) \left (4 a^2-b^2\right ) \sin (c+d x) a}{a+b \sin (c+d x)}dx}{b}+\frac {\left (a^2-b^2\right ) \left (12 a^2-b^2\right ) \cos (c+d x)}{b d}}{b}}{3 b}}{b \left (a^2-b^2\right )}-\frac {\sin ^3(c+d x) \cos (c+d x)}{b d (a+b \sin (c+d x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {4 \left (a^2-b^2\right ) \sin ^2(c+d x) \cos (c+d x)}{3 b d}-\frac {\frac {6 a \left (a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{b d}-\frac {\frac {3 \int \frac {2 b \left (a^2-b^2\right ) a^2+\left (a^2-b^2\right ) \left (4 a^2-b^2\right ) \sin (c+d x) a}{a+b \sin (c+d x)}dx}{b}+\frac {\left (a^2-b^2\right ) \left (12 a^2-b^2\right ) \cos (c+d x)}{b d}}{b}}{3 b}}{b \left (a^2-b^2\right )}-\frac {\sin ^3(c+d x) \cos (c+d x)}{b d (a+b \sin (c+d x))}\)

\(\Big \downarrow \) 3214

\(\displaystyle \frac {\frac {4 \left (a^2-b^2\right ) \sin ^2(c+d x) \cos (c+d x)}{3 b d}-\frac {\frac {6 a \left (a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{b d}-\frac {\frac {3 \left (\frac {a x \left (a^2-b^2\right ) \left (4 a^2-b^2\right )}{b}-\frac {a^2 \left (4 a^2-3 b^2\right ) \left (a^2-b^2\right ) \int \frac {1}{a+b \sin (c+d x)}dx}{b}\right )}{b}+\frac {\left (a^2-b^2\right ) \left (12 a^2-b^2\right ) \cos (c+d x)}{b d}}{b}}{3 b}}{b \left (a^2-b^2\right )}-\frac {\sin ^3(c+d x) \cos (c+d x)}{b d (a+b \sin (c+d x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {4 \left (a^2-b^2\right ) \sin ^2(c+d x) \cos (c+d x)}{3 b d}-\frac {\frac {6 a \left (a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{b d}-\frac {\frac {3 \left (\frac {a x \left (a^2-b^2\right ) \left (4 a^2-b^2\right )}{b}-\frac {a^2 \left (4 a^2-3 b^2\right ) \left (a^2-b^2\right ) \int \frac {1}{a+b \sin (c+d x)}dx}{b}\right )}{b}+\frac {\left (a^2-b^2\right ) \left (12 a^2-b^2\right ) \cos (c+d x)}{b d}}{b}}{3 b}}{b \left (a^2-b^2\right )}-\frac {\sin ^3(c+d x) \cos (c+d x)}{b d (a+b \sin (c+d x))}\)

\(\Big \downarrow \) 3139

\(\displaystyle \frac {\frac {4 \left (a^2-b^2\right ) \sin ^2(c+d x) \cos (c+d x)}{3 b d}-\frac {\frac {6 a \left (a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{b d}-\frac {\frac {3 \left (\frac {a x \left (a^2-b^2\right ) \left (4 a^2-b^2\right )}{b}-\frac {2 a^2 \left (4 a^2-3 b^2\right ) \left (a^2-b^2\right ) \int \frac {1}{a \tan ^2\left (\frac {1}{2} (c+d x)\right )+2 b \tan \left (\frac {1}{2} (c+d x)\right )+a}d\tan \left (\frac {1}{2} (c+d x)\right )}{b d}\right )}{b}+\frac {\left (a^2-b^2\right ) \left (12 a^2-b^2\right ) \cos (c+d x)}{b d}}{b}}{3 b}}{b \left (a^2-b^2\right )}-\frac {\sin ^3(c+d x) \cos (c+d x)}{b d (a+b \sin (c+d x))}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {\frac {4 \left (a^2-b^2\right ) \sin ^2(c+d x) \cos (c+d x)}{3 b d}-\frac {\frac {6 a \left (a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{b d}-\frac {\frac {3 \left (\frac {4 a^2 \left (4 a^2-3 b^2\right ) \left (a^2-b^2\right ) \int \frac {1}{-\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )^2-4 \left (a^2-b^2\right )}d\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{b d}+\frac {a x \left (a^2-b^2\right ) \left (4 a^2-b^2\right )}{b}\right )}{b}+\frac {\left (a^2-b^2\right ) \left (12 a^2-b^2\right ) \cos (c+d x)}{b d}}{b}}{3 b}}{b \left (a^2-b^2\right )}-\frac {\sin ^3(c+d x) \cos (c+d x)}{b d (a+b \sin (c+d x))}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\frac {4 \left (a^2-b^2\right ) \sin ^2(c+d x) \cos (c+d x)}{3 b d}-\frac {\frac {6 a \left (a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{b d}-\frac {\frac {3 \left (\frac {a x \left (a^2-b^2\right ) \left (4 a^2-b^2\right )}{b}-\frac {2 a^2 \left (4 a^2-3 b^2\right ) \sqrt {a^2-b^2} \arctan \left (\frac {2 a \tan \left (\frac {1}{2} (c+d x)\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{b d}\right )}{b}+\frac {\left (a^2-b^2\right ) \left (12 a^2-b^2\right ) \cos (c+d x)}{b d}}{b}}{3 b}}{b \left (a^2-b^2\right )}-\frac {\sin ^3(c+d x) \cos (c+d x)}{b d (a+b \sin (c+d x))}\)

Input:

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

Output:

-((Cos[c + d*x]*Sin[c + d*x]^3)/(b*d*(a + b*Sin[c + d*x]))) + ((4*(a^2 - b 
^2)*Cos[c + d*x]*Sin[c + d*x]^2)/(3*b*d) - (-(((3*((a*(a^2 - b^2)*(4*a^2 - 
 b^2)*x)/b - (2*a^2*(4*a^2 - 3*b^2)*Sqrt[a^2 - b^2]*ArcTan[(2*b + 2*a*Tan[ 
(c + d*x)/2])/(2*Sqrt[a^2 - b^2])])/(b*d)))/b + ((a^2 - b^2)*(12*a^2 - b^2 
)*Cos[c + d*x])/(b*d))/b) + (6*a*(a^2 - b^2)*Cos[c + d*x]*Sin[c + d*x])/(b 
*d))/(3*b))/(b*(a^2 - b^2))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 1083

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3139

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre 
eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d)   Subst[Int[1/(a + 2*b*e*x + a 
*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ 
[a^2 - b^2, 0]
 

rule 3214

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. 
)*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d   Int[1/(c + d 
*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
 

rule 3368

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

rule 3502

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m 
+ 2))   Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m 
 + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] 
 &&  !LtQ[m, -1]
 

rule 3527

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> 
Simp[(-(c^2*C + A*d^2))*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + 
 f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^ 
2))   Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A* 
d*(b*d*m + a*c*(n + 1)) + c*C*(b*c*m + a*d*(n + 1)) - (A*d*(a*d*(n + 2) - b 
*c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*( 
A*d^2*(m + n + 2) + C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x]^2, x], x], 
x] /; FreeQ[{a, b, c, d, e, f, A, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - 
b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]
 

rule 3528

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) 
+ (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_ 
.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x 
])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Simp[1/(d*(m + 
n + 2))   Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A* 
d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2) - C*(a 
*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + 
 n + 2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n} 
, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[ 
m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
 

rule 3529

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) 
+ (f_.)*(x_)])^(n_.)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] : 
> Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 
1)/(d*f*(m + n + 2))), x] + Simp[1/(d*(m + n + 2))   Int[(a + b*Sin[e + f*x 
])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*( 
n + 1)) + (A*b*d*(m + n + 2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + C* 
(a*d*m - b*c*(m + 1))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f 
, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 
0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 
0])))
 

Maple [A] (verified)

Time = 1.56 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.32

method	result	size
derivativedivides	\(\frac {\frac {\frac {4 \left (\frac {a \,b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2}+\left (\frac {3}{2} a^{2} b -\frac {1}{2} b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+3 a^{2} b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {a \,b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}+\frac {3 a^{2} b}{2}-\frac {b^{3}}{6}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+2 a \left (4 a^{2}-b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{5}}-\frac {4 a^{2} \left (\frac {-\frac {b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}-\frac {a b}{2}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+\frac {\left (4 a^{2}-3 b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}}\right )}{b^{5}}}{d}\)	\(249\)
default	\(\frac {\frac {\frac {4 \left (\frac {a \,b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2}+\left (\frac {3}{2} a^{2} b -\frac {1}{2} b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+3 a^{2} b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {a \,b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}+\frac {3 a^{2} b}{2}-\frac {b^{3}}{6}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+2 a \left (4 a^{2}-b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{5}}-\frac {4 a^{2} \left (\frac {-\frac {b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}-\frac {a b}{2}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+\frac {\left (4 a^{2}-3 b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}}\right )}{b^{5}}}{d}\)	\(249\)
risch	\(\frac {4 a^{3} x}{b^{5}}-\frac {a x}{b^{3}}+\frac {i a \,{\mathrm e}^{2 i \left (d x +c \right )}}{4 b^{3} d}+\frac {3 \,{\mathrm e}^{i \left (d x +c \right )} a^{2}}{2 b^{4} d}-\frac {{\mathrm e}^{i \left (d x +c \right )}}{8 b^{2} d}+\frac {3 \,{\mathrm e}^{-i \left (d x +c \right )} a^{2}}{2 b^{4} d}-\frac {{\mathrm e}^{-i \left (d x +c \right )}}{8 b^{2} d}-\frac {i a \,{\mathrm e}^{-2 i \left (d x +c \right )}}{4 b^{3} d}+\frac {2 i a^{3} \left (-i a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}{b^{5} d \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}-b +2 i a \,{\mathrm e}^{i \left (d x +c \right )}\right )}+\frac {4 i a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, d \,b^{5}}-\frac {3 i a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, d \,b^{3}}-\frac {4 i a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, d \,b^{5}}+\frac {3 i a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, d \,b^{3}}-\frac {\cos \left (3 d x +3 c \right )}{12 d \,b^{2}}\)	\(496\)

Input:

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

Output:

1/d*(4/b^5*((1/2*a*b^2*tan(1/2*d*x+1/2*c)^5+(3/2*a^2*b-1/2*b^3)*tan(1/2*d* 
x+1/2*c)^4+3*a^2*b*tan(1/2*d*x+1/2*c)^2-1/2*a*b^2*tan(1/2*d*x+1/2*c)+3/2*a 
^2*b-1/6*b^3)/(1+tan(1/2*d*x+1/2*c)^2)^3+1/2*a*(4*a^2-b^2)*arctan(tan(1/2* 
d*x+1/2*c)))-4*a^2/b^5*((-1/2*b^2*tan(1/2*d*x+1/2*c)-1/2*a*b)/(tan(1/2*d*x 
+1/2*c)^2*a+2*b*tan(1/2*d*x+1/2*c)+a)+1/2*(4*a^2-3*b^2)/(a^2-b^2)^(1/2)*ar 
ctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))))
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 643, normalized size of antiderivative = 3.42 \[ \int \frac {\cos ^2(c+d x) \sin ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\left [\frac {4 \, {\left (a^{3} b^{3} - a b^{5}\right )} \cos \left (d x + c\right )^{3} + 6 \, {\left (4 \, a^{6} - 5 \, a^{4} b^{2} + a^{2} b^{4}\right )} d x + 3 \, {\left (4 \, a^{5} - 3 \, a^{3} b^{2} + {\left (4 \, a^{4} b - 3 \, a^{2} b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) + 6 \, {\left (4 \, a^{5} b - 5 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right ) - 2 \, {\left ({\left (a^{2} b^{4} - b^{6}\right )} \cos \left (d x + c\right )^{3} - 3 \, {\left (4 \, a^{5} b - 5 \, a^{3} b^{3} + a b^{5}\right )} d x - 6 \, {\left (a^{4} b^{2} - a^{2} b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left ({\left (a^{2} b^{6} - b^{8}\right )} d \sin \left (d x + c\right ) + {\left (a^{3} b^{5} - a b^{7}\right )} d\right )}}, \frac {2 \, {\left (a^{3} b^{3} - a b^{5}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (4 \, a^{6} - 5 \, a^{4} b^{2} + a^{2} b^{4}\right )} d x + 3 \, {\left (4 \, a^{5} - 3 \, a^{3} b^{2} + {\left (4 \, a^{4} b - 3 \, a^{2} b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) + 3 \, {\left (4 \, a^{5} b - 5 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right ) - {\left ({\left (a^{2} b^{4} - b^{6}\right )} \cos \left (d x + c\right )^{3} - 3 \, {\left (4 \, a^{5} b - 5 \, a^{3} b^{3} + a b^{5}\right )} d x - 6 \, {\left (a^{4} b^{2} - a^{2} b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{3 \, {\left ({\left (a^{2} b^{6} - b^{8}\right )} d \sin \left (d x + c\right ) + {\left (a^{3} b^{5} - a b^{7}\right )} d\right )}}\right ] \] Input:

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

Output:

[1/6*(4*(a^3*b^3 - a*b^5)*cos(d*x + c)^3 + 6*(4*a^6 - 5*a^4*b^2 + a^2*b^4) 
*d*x + 3*(4*a^5 - 3*a^3*b^2 + (4*a^4*b - 3*a^2*b^3)*sin(d*x + c))*sqrt(-a^ 
2 + b^2)*log(((2*a^2 - b^2)*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^ 
2 + 2*(a*cos(d*x + c)*sin(d*x + c) + b*cos(d*x + c))*sqrt(-a^2 + b^2))/(b^ 
2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2)) + 6*(4*a^5*b - 5*a^3*b 
^3 + a*b^5)*cos(d*x + c) - 2*((a^2*b^4 - b^6)*cos(d*x + c)^3 - 3*(4*a^5*b 
- 5*a^3*b^3 + a*b^5)*d*x - 6*(a^4*b^2 - a^2*b^4)*cos(d*x + c))*sin(d*x + c 
))/((a^2*b^6 - b^8)*d*sin(d*x + c) + (a^3*b^5 - a*b^7)*d), 1/3*(2*(a^3*b^3 
 - a*b^5)*cos(d*x + c)^3 + 3*(4*a^6 - 5*a^4*b^2 + a^2*b^4)*d*x + 3*(4*a^5 
- 3*a^3*b^2 + (4*a^4*b - 3*a^2*b^3)*sin(d*x + c))*sqrt(a^2 - b^2)*arctan(- 
(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) + 3*(4*a^5*b - 5*a^3* 
b^3 + a*b^5)*cos(d*x + c) - ((a^2*b^4 - b^6)*cos(d*x + c)^3 - 3*(4*a^5*b - 
 5*a^3*b^3 + a*b^5)*d*x - 6*(a^4*b^2 - a^2*b^4)*cos(d*x + c))*sin(d*x + c) 
)/((a^2*b^6 - b^8)*d*sin(d*x + c) + (a^3*b^5 - a*b^7)*d)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^2(c+d x) \sin ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cos ^2(c+d x) \sin ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f 
or more de
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.39 \[ \int \frac {\cos ^2(c+d x) \sin ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\frac {3 \, {\left (4 \, a^{3} - a b^{2}\right )} {\left (d x + c\right )}}{b^{5}} - \frac {6 \, {\left (4 \, a^{4} - 3 \, a^{2} b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} b^{5}} + \frac {6 \, {\left (a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3}\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )} b^{4}} + \frac {2 \, {\left (3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 3 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 18 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, a^{2} - b^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} b^{4}}}{3 \, d} \] Input:

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

Output:

1/3*(3*(4*a^3 - a*b^2)*(d*x + c)/b^5 - 6*(4*a^4 - 3*a^2*b^2)*(pi*floor(1/2 
*(d*x + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 
 - b^2)))/(sqrt(a^2 - b^2)*b^5) + 6*(a^2*b*tan(1/2*d*x + 1/2*c) + a^3)/((a 
*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)*b^4) + 2*(3*a*b*ta 
n(1/2*d*x + 1/2*c)^5 + 9*a^2*tan(1/2*d*x + 1/2*c)^4 - 3*b^2*tan(1/2*d*x + 
1/2*c)^4 + 18*a^2*tan(1/2*d*x + 1/2*c)^2 - 3*a*b*tan(1/2*d*x + 1/2*c) + 9* 
a^2 - b^2)/((tan(1/2*d*x + 1/2*c)^2 + 1)^3*b^4))/d
 

Mupad [B] (verification not implemented)

Time = 38.55 (sec) , antiderivative size = 1688, normalized size of antiderivative = 8.98 \[ \int \frac {\cos ^2(c+d x) \sin ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

((28*a^2*tan(c/2 + (d*x)/2)^3)/b^3 - (2*(a*b^2 - 12*a^3))/(3*b^4) + (4*a^2 
*tan(c/2 + (d*x)/2)^7)/b^3 + (2*tan(c/2 + (d*x)/2)^6*(a*b^2 + 4*a^3))/b^4 
- (2*tan(c/2 + (d*x)/2)^4*(a*b^2 - 12*a^3))/b^4 - (2*tan(c/2 + (d*x)/2)^2* 
(7*a*b^2 - 36*a^3))/(3*b^4) + (4*tan(c/2 + (d*x)/2)*(9*a^2 - b^2))/(3*b^3) 
 + (4*tan(c/2 + (d*x)/2)^5*(5*a^2 - b^2))/b^3)/(d*(a + 2*b*tan(c/2 + (d*x) 
/2) + 4*a*tan(c/2 + (d*x)/2)^2 + 6*a*tan(c/2 + (d*x)/2)^4 + 4*a*tan(c/2 + 
(d*x)/2)^6 + a*tan(c/2 + (d*x)/2)^8 + 6*b*tan(c/2 + (d*x)/2)^3 + 6*b*tan(c 
/2 + (d*x)/2)^5 + 2*b*tan(c/2 + (d*x)/2)^7)) + (atan((64*a^4*tan(c/2 + (d* 
x)/2))/(64*a^4 - (256*a^6)/b^2) + (256*a^6*tan(c/2 + (d*x)/2))/(256*a^6 - 
64*a^4*b^2))*(a*b^2*1i - a^3*4i)*2i)/(b^5*d) + (a^2*atan(((a^2*(-(a + b)*( 
a - b))^(1/2)*(4*a^2 - 3*b^2)*((32*(a^4*b^8 - 8*a^6*b^6 + 16*a^8*b^4))/b^1 
1 + (32*tan(c/2 + (d*x)/2)*(2*a^3*b^10 - 26*a^5*b^8 + 64*a^7*b^6 - 32*a^9* 
b^4))/b^12 + (a^2*(-(a + b)*(a - b))^(1/2)*(4*a^2 - 3*b^2)*((32*(a^2*b^12 
- 2*a^4*b^10))/b^11 + (32*tan(c/2 + (d*x)/2)*(6*a^3*b^12 - 8*a^5*b^10))/b^ 
12 + (a^2*(-(a + b)*(a - b))^(1/2)*(4*a^2 - 3*b^2)*(32*a^2*b^3 + (32*tan(c 
/2 + (d*x)/2)*(3*a*b^16 - 2*a^3*b^14))/b^12))/(b^7 - a^2*b^5)))/(b^7 - a^2 
*b^5))*1i)/(b^7 - a^2*b^5) + (a^2*(-(a + b)*(a - b))^(1/2)*(4*a^2 - 3*b^2) 
*((32*(a^4*b^8 - 8*a^6*b^6 + 16*a^8*b^4))/b^11 + (32*tan(c/2 + (d*x)/2)*(2 
*a^3*b^10 - 26*a^5*b^8 + 64*a^7*b^6 - 32*a^9*b^4))/b^12 - (a^2*(-(a + b)*( 
a - b))^(1/2)*(4*a^2 - 3*b^2)*((32*(a^2*b^12 - 2*a^4*b^10))/b^11 + (32*...
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 533, normalized size of antiderivative = 2.84 \[ \int \frac {\cos ^2(c+d x) \sin ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {-24 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (d x +c \right ) a^{4} b +18 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (d x +c \right ) a^{2} b^{3}+12 \sin \left (d x +c \right ) a^{5} b d x -15 \sin \left (d x +c \right ) a^{3} b^{3} d x +3 \sin \left (d x +c \right ) a \,b^{5} d x -24 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) a^{5}+12 \cos \left (d x +c \right ) a^{5} b -13 \cos \left (d x +c \right ) a^{3} b^{3}+6 \sin \left (d x +c \right ) a^{4} b^{2}-7 \sin \left (d x +c \right ) a^{2} b^{4}+12 a^{6} d x +\cos \left (d x +c \right ) a \,b^{5}+18 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) a^{3} b^{2}-2 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{3} b^{3}+2 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a \,b^{5}+6 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{4} b^{2}-7 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b^{4}-15 a^{4} b^{2} d x +3 a^{2} b^{4} d x +6 a^{5} b -7 a^{3} b^{3}-\cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} b^{6}+\cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a^{2} b^{4}+\cos \left (d x +c \right ) \sin \left (d x +c \right ) b^{6}+\sin \left (d x +c \right ) b^{6}+a \,b^{5}}{3 b^{5} d \left (\sin \left (d x +c \right ) a^{2} b -\sin \left (d x +c \right ) b^{3}+a^{3}-a \,b^{2}\right )} \] Input:

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

Output:

( - 24*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))* 
sin(c + d*x)*a**4*b + 18*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/s 
qrt(a**2 - b**2))*sin(c + d*x)*a**2*b**3 - 24*sqrt(a**2 - b**2)*atan((tan( 
(c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*a**5 + 18*sqrt(a**2 - b**2)*atan((t 
an((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*a**3*b**2 + cos(c + d*x)*sin(c + 
 d*x)**3*a**2*b**4 - cos(c + d*x)*sin(c + d*x)**3*b**6 - 2*cos(c + d*x)*si 
n(c + d*x)**2*a**3*b**3 + 2*cos(c + d*x)*sin(c + d*x)**2*a*b**5 + 6*cos(c 
+ d*x)*sin(c + d*x)*a**4*b**2 - 7*cos(c + d*x)*sin(c + d*x)*a**2*b**4 + co 
s(c + d*x)*sin(c + d*x)*b**6 + 12*cos(c + d*x)*a**5*b - 13*cos(c + d*x)*a* 
*3*b**3 + cos(c + d*x)*a*b**5 + 12*sin(c + d*x)*a**5*b*d*x + 6*sin(c + d*x 
)*a**4*b**2 - 15*sin(c + d*x)*a**3*b**3*d*x - 7*sin(c + d*x)*a**2*b**4 + 3 
*sin(c + d*x)*a*b**5*d*x + sin(c + d*x)*b**6 + 12*a**6*d*x + 6*a**5*b - 15 
*a**4*b**2*d*x - 7*a**3*b**3 + 3*a**2*b**4*d*x + a*b**5)/(3*b**5*d*(sin(c 
+ d*x)*a**2*b - sin(c + d*x)*b**3 + a**3 - a*b**2))