\(\int \frac {\cos ^2(c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx\) [1282]

Optimal result

Integrand size = 29, antiderivative size = 191 \[ \int \frac {\cos ^2(c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\left (8 a^4-4 a^2 b^2-b^4\right ) x}{8 b^5}+\frac {2 a^3 \sqrt {a^2-b^2} \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^5 d}-\frac {a \left (3 a^2-b^2\right ) \cos (c+d x)}{3 b^4 d}+\frac {\left (4 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^3 d}-\frac {a \cos (c+d x) \sin ^2(c+d x)}{3 b^2 d}+\frac {\cos (c+d x) \sin ^3(c+d x)}{4 b d} \] Output:

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

Mathematica [A] (verified)

Time = 1.62 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.76 \[ \int \frac {\cos ^2(c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-12 \left (8 a^4-4 a^2 b^2-b^4\right ) (c+d x)+192 a^3 \sqrt {a^2-b^2} \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )+24 a b \left (-4 a^2+b^2\right ) \cos (c+d x)+8 a b^3 \cos (3 (c+d x))+24 a^2 b^2 \sin (2 (c+d x))-3 b^4 \sin (4 (c+d x))}{96 b^5 d} \] Input:

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

Output:

(-12*(8*a^4 - 4*a^2*b^2 - b^4)*(c + d*x) + 192*a^3*Sqrt[a^2 - b^2]*ArcTan[ 
(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]] + 24*a*b*(-4*a^2 + b^2)*Cos[c + 
d*x] + 8*a*b^3*Cos[3*(c + d*x)] + 24*a^2*b^2*Sin[2*(c + d*x)] - 3*b^4*Sin[ 
4*(c + d*x)])/(96*b^5*d)
 

Rubi [A] (verified)

Time = 1.43 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.16, 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, 3529, 25, 3042, 3528, 25, 3042, 3528, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3368

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3529

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3528

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3528

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3214

\(\displaystyle \frac {\sin ^3(c+d x) \cos (c+d x)}{4 b d}-\frac {\frac {4 a \sin ^2(c+d x) \cos (c+d x)}{3 b d}-\frac {\frac {3 \left (4 a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{2 b d}-\frac {\frac {3 \left (\frac {x \left (8 a^4-4 a^2 b^2-b^4\right )}{b}-\frac {8 a^3 \left (a^2-b^2\right ) \int \frac {1}{a+b \sin (c+d x)}dx}{b}\right )}{b}+\frac {8 a \left (3 a^2-b^2\right ) \cos (c+d x)}{b d}}{2 b}}{3 b}}{4 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sin ^3(c+d x) \cos (c+d x)}{4 b d}-\frac {\frac {4 a \sin ^2(c+d x) \cos (c+d x)}{3 b d}-\frac {\frac {3 \left (4 a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{2 b d}-\frac {\frac {3 \left (\frac {x \left (8 a^4-4 a^2 b^2-b^4\right )}{b}-\frac {8 a^3 \left (a^2-b^2\right ) \int \frac {1}{a+b \sin (c+d x)}dx}{b}\right )}{b}+\frac {8 a \left (3 a^2-b^2\right ) \cos (c+d x)}{b d}}{2 b}}{3 b}}{4 b}\)

\(\Big \downarrow \) 3139

\(\displaystyle \frac {\sin ^3(c+d x) \cos (c+d x)}{4 b d}-\frac {\frac {4 a \sin ^2(c+d x) \cos (c+d x)}{3 b d}-\frac {\frac {3 \left (4 a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{2 b d}-\frac {\frac {3 \left (\frac {x \left (8 a^4-4 a^2 b^2-b^4\right )}{b}-\frac {16 a^3 \left (a^2-b^2\right ) \int \frac {1}{a \tan ^2\left (\frac {1}{2} (c+d x)\right )+2 b \tan \left (\frac {1}{2} (c+d x)\right )+a}d\tan \left (\frac {1}{2} (c+d x)\right )}{b d}\right )}{b}+\frac {8 a \left (3 a^2-b^2\right ) \cos (c+d x)}{b d}}{2 b}}{3 b}}{4 b}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {\sin ^3(c+d x) \cos (c+d x)}{4 b d}-\frac {\frac {4 a \sin ^2(c+d x) \cos (c+d x)}{3 b d}-\frac {\frac {3 \left (4 a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{2 b d}-\frac {\frac {3 \left (\frac {32 a^3 \left (a^2-b^2\right ) \int \frac {1}{-\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )^2-4 \left (a^2-b^2\right )}d\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{b d}+\frac {x \left (8 a^4-4 a^2 b^2-b^4\right )}{b}\right )}{b}+\frac {8 a \left (3 a^2-b^2\right ) \cos (c+d x)}{b d}}{2 b}}{3 b}}{4 b}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\sin ^3(c+d x) \cos (c+d x)}{4 b d}-\frac {\frac {4 a \sin ^2(c+d x) \cos (c+d x)}{3 b d}-\frac {\frac {3 \left (4 a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{2 b d}-\frac {\frac {8 a \left (3 a^2-b^2\right ) \cos (c+d x)}{b d}+\frac {3 \left (\frac {x \left (8 a^4-4 a^2 b^2-b^4\right )}{b}-\frac {16 a^3 \sqrt {a^2-b^2} \arctan \left (\frac {2 a \tan \left (\frac {1}{2} (c+d x)\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{b d}\right )}{b}}{2 b}}{3 b}}{4 b}\)

Input:

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

Output:

(Cos[c + d*x]*Sin[c + d*x]^3)/(4*b*d) - ((4*a*Cos[c + d*x]*Sin[c + d*x]^2) 
/(3*b*d) - (-1/2*((3*(((8*a^4 - 4*a^2*b^2 - b^4)*x)/b - (16*a^3*Sqrt[a^2 - 
 b^2]*ArcTan[(2*b + 2*a*Tan[(c + d*x)/2])/(2*Sqrt[a^2 - b^2])])/(b*d)))/b 
+ (8*a*(3*a^2 - b^2)*Cos[c + d*x])/(b*d))/b + (3*(4*a^2 - b^2)*Cos[c + d*x 
]*Sin[c + d*x])/(2*b*d))/(3*b))/(4*b)
 

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 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 [C] (verified)

Result contains complex when optimal does not.

Time = 0.84 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.39

method	result	size
risch	\(-\frac {x \,a^{4}}{b^{5}}+\frac {x \,a^{2}}{2 b^{3}}+\frac {x}{8 b}-\frac {a^{3} {\mathrm e}^{i \left (d x +c \right )}}{2 b^{4} d}+\frac {a \,{\mathrm e}^{i \left (d x +c \right )}}{8 b^{2} d}-\frac {a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{2 b^{4} d}+\frac {a \,{\mathrm e}^{-i \left (d x +c \right )}}{8 b^{2} d}+\frac {i \sqrt {a^{2}-b^{2}}\, a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}+a \right )}{b}\right )}{d \,b^{5}}-\frac {i \sqrt {a^{2}-b^{2}}\, a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i \left (\sqrt {a^{2}-b^{2}}-a \right )}{b}\right )}{d \,b^{5}}-\frac {\sin \left (4 d x +4 c \right )}{32 b d}+\frac {a \cos \left (3 d x +3 c \right )}{12 b^{2} d}+\frac {a^{2} \sin \left (2 d x +2 c \right )}{4 b^{3} d}\)	\(265\)
derivativedivides	\(\frac {-\frac {2 \left (\frac {\left (\frac {1}{2} a^{2} b^{2}-\frac {1}{8} b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}+\left (a^{3} b -a \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (\frac {1}{2} a^{2} b^{2}+\frac {7}{8} b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (3 a^{3} b -a \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-\frac {1}{2} a^{2} b^{2}-\frac {7}{8} b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (3 a^{3} b -\frac {1}{3} a \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (-\frac {1}{2} a^{2} b^{2}+\frac {1}{8} b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a^{3} b -\frac {a \,b^{3}}{3}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}+\frac {\left (8 a^{4}-4 a^{2} b^{2}-b^{4}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}\right )}{b^{5}}+\frac {2 a^{3} \sqrt {a^{2}-b^{2}}\, \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{b^{5}}}{d}\)	\(296\)
default	\(\frac {-\frac {2 \left (\frac {\left (\frac {1}{2} a^{2} b^{2}-\frac {1}{8} b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}+\left (a^{3} b -a \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (\frac {1}{2} a^{2} b^{2}+\frac {7}{8} b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (3 a^{3} b -a \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-\frac {1}{2} a^{2} b^{2}-\frac {7}{8} b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (3 a^{3} b -\frac {1}{3} a \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (-\frac {1}{2} a^{2} b^{2}+\frac {1}{8} b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a^{3} b -\frac {a \,b^{3}}{3}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}+\frac {\left (8 a^{4}-4 a^{2} b^{2}-b^{4}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}\right )}{b^{5}}+\frac {2 a^{3} \sqrt {a^{2}-b^{2}}\, \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{b^{5}}}{d}\)	\(296\)

Input:

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

Output:

-x/b^5*a^4+1/2*x/b^3*a^2+1/8*x/b-1/2*a^3/b^4/d*exp(I*(d*x+c))+1/8*a/b^2/d* 
exp(I*(d*x+c))-1/2*a^3/b^4/d*exp(-I*(d*x+c))+1/8*a/b^2/d*exp(-I*(d*x+c))+I 
*(a^2-b^2)^(1/2)*a^3/d/b^5*ln(exp(I*(d*x+c))+I*((a^2-b^2)^(1/2)+a)/b)-I*(a 
^2-b^2)^(1/2)*a^3/d/b^5*ln(exp(I*(d*x+c))-I*((a^2-b^2)^(1/2)-a)/b)-1/32/b/ 
d*sin(4*d*x+4*c)+1/12/b^2/d*a*cos(3*d*x+3*c)+1/4/b^3*a^2/d*sin(2*d*x+2*c)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 380, normalized size of antiderivative = 1.99 \[ \int \frac {\cos ^2(c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\left [\frac {8 \, a b^{3} \cos \left (d x + c\right )^{3} - 24 \, a^{3} b \cos \left (d x + c\right ) + 12 \, \sqrt {-a^{2} + b^{2}} a^{3} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) - 3 \, {\left (8 \, a^{4} - 4 \, a^{2} b^{2} - b^{4}\right )} d x - 3 \, {\left (2 \, b^{4} \cos \left (d x + c\right )^{3} - {\left (4 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, b^{5} d}, \frac {8 \, a b^{3} \cos \left (d x + c\right )^{3} - 24 \, a^{3} b \cos \left (d x + c\right ) - 24 \, \sqrt {a^{2} - b^{2}} a^{3} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) - 3 \, {\left (8 \, a^{4} - 4 \, a^{2} b^{2} - b^{4}\right )} d x - 3 \, {\left (2 \, b^{4} \cos \left (d x + c\right )^{3} - {\left (4 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, b^{5} d}\right ] \] Input:

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

Output:

[1/24*(8*a*b^3*cos(d*x + c)^3 - 24*a^3*b*cos(d*x + c) + 12*sqrt(-a^2 + b^2 
)*a^3*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)) - 3*(8*a^4 - 4*a^2*b^2 - 
 b^4)*d*x - 3*(2*b^4*cos(d*x + c)^3 - (4*a^2*b^2 + b^4)*cos(d*x + c))*sin( 
d*x + c))/(b^5*d), 1/24*(8*a*b^3*cos(d*x + c)^3 - 24*a^3*b*cos(d*x + c) - 
24*sqrt(a^2 - b^2)*a^3*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d 
*x + c))) - 3*(8*a^4 - 4*a^2*b^2 - b^4)*d*x - 3*(2*b^4*cos(d*x + c)^3 - (4 
*a^2*b^2 + b^4)*cos(d*x + c))*sin(d*x + c))/(b^5*d)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

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

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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 366 vs. \(2 (176) = 352\).

Time = 0.18 (sec) , antiderivative size = 366, normalized size of antiderivative = 1.92 \[ \int \frac {\cos ^2(c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {3 \, {\left (8 \, a^{4} - 4 \, a^{2} b^{2} - b^{4}\right )} {\left (d x + c\right )}}{b^{5}} - \frac {48 \, {\left (a^{5} - a^{3} b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} b^{5}} + \frac {2 \, {\left (12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 3 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 24 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 21 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 72 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 24 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 21 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 72 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 8 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, a^{3} - 8 \, a b^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4} b^{4}}}{24 \, d} \] Input:

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

Output:

-1/24*(3*(8*a^4 - 4*a^2*b^2 - b^4)*(d*x + c)/b^5 - 48*(a^5 - a^3*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) + 2*(12*a^2*b*tan(1/2*d*x + 1/2*c 
)^7 - 3*b^3*tan(1/2*d*x + 1/2*c)^7 + 24*a^3*tan(1/2*d*x + 1/2*c)^6 - 24*a* 
b^2*tan(1/2*d*x + 1/2*c)^6 + 12*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 21*b^3*tan( 
1/2*d*x + 1/2*c)^5 + 72*a^3*tan(1/2*d*x + 1/2*c)^4 - 24*a*b^2*tan(1/2*d*x 
+ 1/2*c)^4 - 12*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 21*b^3*tan(1/2*d*x + 1/2*c) 
^3 + 72*a^3*tan(1/2*d*x + 1/2*c)^2 - 8*a*b^2*tan(1/2*d*x + 1/2*c)^2 - 12*a 
^2*b*tan(1/2*d*x + 1/2*c) + 3*b^3*tan(1/2*d*x + 1/2*c) + 24*a^3 - 8*a*b^2) 
/((tan(1/2*d*x + 1/2*c)^2 + 1)^4*b^4))/d
 

Mupad [B] (verification not implemented)

Time = 20.84 (sec) , antiderivative size = 2616, normalized size of antiderivative = 13.70 \[ \int \frac {\cos ^2(c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

(7*tan(c/2 + (d*x)/2)^3)/(4*d*(b + 4*b*tan(c/2 + (d*x)/2)^2 + 6*b*tan(c/2 
+ (d*x)/2)^4 + 4*b*tan(c/2 + (d*x)/2)^6 + b*tan(c/2 + (d*x)/2)^8)) - (7*ta 
n(c/2 + (d*x)/2)^5)/(4*d*(b + 4*b*tan(c/2 + (d*x)/2)^2 + 6*b*tan(c/2 + (d* 
x)/2)^4 + 4*b*tan(c/2 + (d*x)/2)^6 + b*tan(c/2 + (d*x)/2)^8)) + tan(c/2 + 
(d*x)/2)^7/(4*d*(b + 4*b*tan(c/2 + (d*x)/2)^2 + 6*b*tan(c/2 + (d*x)/2)^4 + 
 4*b*tan(c/2 + (d*x)/2)^6 + b*tan(c/2 + (d*x)/2)^8)) + (2*a)/(3*d*(4*b^2*t 
an(c/2 + (d*x)/2)^2 + 6*b^2*tan(c/2 + (d*x)/2)^4 + 4*b^2*tan(c/2 + (d*x)/2 
)^6 + b^2*tan(c/2 + (d*x)/2)^8 + b^2)) - (2*a^3)/(d*(4*b^4*tan(c/2 + (d*x) 
/2)^2 + 6*b^4*tan(c/2 + (d*x)/2)^4 + 4*b^4*tan(c/2 + (d*x)/2)^6 + b^4*tan( 
c/2 + (d*x)/2)^8 + b^4)) + atan((11*a^3*tan(c/2 + (d*x)/2))/(8*((a*b^2)/8 
+ (11*a^3)/8 + (3*a^5)/(2*b^2) - (11*a^7)/b^4 + (8*a^9)/b^6)) + (3*a^5*tan 
(c/2 + (d*x)/2))/(2*((a*b^4)/8 + (3*a^5)/2 + (11*a^3*b^2)/8 - (11*a^7)/b^2 
 + (8*a^9)/b^4)) - (11*a^7*tan(c/2 + (d*x)/2))/((a*b^6)/8 - 11*a^7 + (11*a 
^3*b^4)/8 + (3*a^5*b^2)/2 + (8*a^9)/b^2) + (8*a^9*tan(c/2 + (d*x)/2))/((a* 
b^8)/8 + 8*a^9 + (11*a^3*b^6)/8 + (3*a^5*b^4)/2 - 11*a^7*b^2) + (a*b*tan(c 
/2 + (d*x)/2))/(8*((a*b)/8 + (11*a^3)/(8*b) + (3*a^5)/(2*b^3) - (11*a^7)/b 
^5 + (8*a^9)/b^7)))/(4*b*d) - tan(c/2 + (d*x)/2)/(4*d*(b + 4*b*tan(c/2 + ( 
d*x)/2)^2 + 6*b*tan(c/2 + (d*x)/2)^4 + 4*b*tan(c/2 + (d*x)/2)^6 + b*tan(c/ 
2 + (d*x)/2)^8)) + (2*a*tan(c/2 + (d*x)/2)^2)/(3*d*(4*b^2*tan(c/2 + (d*x)/ 
2)^2 + 6*b^2*tan(c/2 + (d*x)/2)^4 + 4*b^2*tan(c/2 + (d*x)/2)^6 + b^2*ta...
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.09 \[ \int \frac {\cos ^2(c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {48 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) a^{3}+6 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} b^{4}-8 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a \,b^{3}+12 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b^{2}-3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b^{4}-24 \cos \left (d x +c \right ) a^{3} b +8 \cos \left (d x +c \right ) a \,b^{3}-24 a^{4} c -24 a^{4} d x -12 a^{3} b +12 a^{2} b^{2} c +12 a^{2} b^{2} d x -4 a \,b^{3}+3 b^{4} c +3 b^{4} d x}{24 b^{5} d} \] Input:

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

Output:

(48*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*a** 
3 + 6*cos(c + d*x)*sin(c + d*x)**3*b**4 - 8*cos(c + d*x)*sin(c + d*x)**2*a 
*b**3 + 12*cos(c + d*x)*sin(c + d*x)*a**2*b**2 - 3*cos(c + d*x)*sin(c + d* 
x)*b**4 - 24*cos(c + d*x)*a**3*b + 8*cos(c + d*x)*a*b**3 - 24*a**4*c - 24* 
a**4*d*x - 12*a**3*b + 12*a**2*b**2*c + 12*a**2*b**2*d*x - 4*a*b**3 + 3*b* 
*4*c + 3*b**4*d*x)/(24*b**5*d)