\(\int \frac {\cos ^2(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx\) [1083]

Optimal result

Integrand size = 27, antiderivative size = 167 \[ \int \frac {\cos ^2(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {x}{b^3}+\frac {a \left (2 a^2-3 b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^3 \left (a^2-b^2\right )^{3/2} d}-\frac {a \cos ^3(c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}-\frac {\cos (c+d x) \left (2 \left (a^2-b^2\right )+a b \sin (c+d x)\right )}{2 b^2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))} \] Output:

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

Mathematica [A] (verified)

Time = 2.08 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.73 \[ \int \frac {\cos ^2(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {\frac {-8 (c+d x)+\frac {2 a \left (8 a^4-20 a^2 b^2+15 b^4\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac {a b \left (4 a^2-3 b^2\right ) \cos (c+d x)}{(a-b) (a+b) (a+b \sin (c+d x))^2}-\frac {3 b \left (4 a^4-7 a^2 b^2+2 b^4\right ) \cos (c+d x)}{(a-b)^2 (a+b)^2 (a+b \sin (c+d x))}}{b^3}-\frac {\frac {6 a b \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {\cos (c+d x) \left (a \left (2 a^2+b^2\right )+b \left (a^2+2 b^2\right ) \sin (c+d x)\right )}{(a+b \sin (c+d x))^2}}{(a-b)^2 (a+b)^2}}{8 d} \] Input:

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

Output:

((-8*(c + d*x) + (2*a*(8*a^4 - 20*a^2*b^2 + 15*b^4)*ArcTan[(b + a*Tan[(c + 
 d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(5/2) + (a*b*(4*a^2 - 3*b^2)*Cos[c 
 + d*x])/((a - b)*(a + b)*(a + b*Sin[c + d*x])^2) - (3*b*(4*a^4 - 7*a^2*b^ 
2 + 2*b^4)*Cos[c + d*x])/((a - b)^2*(a + b)^2*(a + b*Sin[c + d*x])))/b^3 - 
 ((6*a*b*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] 
 + (Cos[c + d*x]*(a*(2*a^2 + b^2) + b*(a^2 + 2*b^2)*Sin[c + d*x]))/(a + b* 
Sin[c + d*x])^2)/((a - b)^2*(a + b)^2))/(8*d)
 

Rubi [A] (verified)

Time = 0.79 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.14, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.407, Rules used = {3042, 3343, 3042, 3342, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3343

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3342

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3214

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 3342

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[g*(g*C 
os[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1)*((b*c*(m + p + 1) - a*d*p 
 + b*d*(m + 1)*Sin[e + f*x])/(b^2*f*(m + 1)*(m + p + 1))), x] + Simp[g^2*(( 
p - 1)/(b^2*(m + 1)*(m + p + 1)))   Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin 
[e + f*x])^(m + 1)*Simp[b*d*(m + 1) + (b*c*(m + p + 1) - a*d*p)*Sin[e + f*x 
], x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[a^2 - b^2, 0] && Lt 
Q[m, -1] && GtQ[p, 1] && NeQ[m + p + 1, 0] && IntegerQ[2*m]
 

rule 3343

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(b*c 
 - a*d))*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^(m + 1)/(f*g*(a^2 - 
 b^2)*(m + 1))), x] + Simp[1/((a^2 - b^2)*(m + 1))   Int[(g*Cos[e + f*x])^p 
*(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + p 
 + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ 
[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]
 

Maple [A] (verified)

Time = 0.72 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.51

method	result	size
derivativedivides	\(\frac {-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3}}+\frac {\frac {2 \left (-\frac {a^{2} b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a^{2}-b^{2}\right )}-\frac {b \left (2 a^{4}+3 a^{2} b^{2}-2 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 a \left (a^{2}-b^{2}\right )}-\frac {b^{2} \left (7 a^{2}-4 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a^{2}-b^{2}\right )}-\frac {a b \left (2 a^{2}-b^{2}\right )}{2 \left (a^{2}-b^{2}\right )}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )^{2}}+\frac {a \left (2 a^{2}-3 b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a^{2}-b^{2}\right )^{\frac {3}{2}}}}{b^{3}}}{d}\)	\(252\)
default	\(\frac {-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3}}+\frac {\frac {2 \left (-\frac {a^{2} b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a^{2}-b^{2}\right )}-\frac {b \left (2 a^{4}+3 a^{2} b^{2}-2 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 a \left (a^{2}-b^{2}\right )}-\frac {b^{2} \left (7 a^{2}-4 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a^{2}-b^{2}\right )}-\frac {a b \left (2 a^{2}-b^{2}\right )}{2 \left (a^{2}-b^{2}\right )}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )^{2}}+\frac {a \left (2 a^{2}-3 b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a^{2}-b^{2}\right )^{\frac {3}{2}}}}{b^{3}}}{d}\)	\(252\)
risch	\(-\frac {x}{b^{3}}+\frac {i \left (-4 i a^{3} b \,{\mathrm e}^{3 i \left (d x +c \right )}+3 i a \,b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+8 i a^{3} b \,{\mathrm e}^{i \left (d x +c \right )}-5 i a \,b^{3} {\mathrm e}^{i \left (d x +c \right )}+6 a^{4} {\mathrm e}^{2 i \left (d x +c \right )}-b^{2} a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-2 b^{4} {\mathrm e}^{2 i \left (d x +c \right )}-3 a^{2} b^{2}+2 b^{4}\right )}{\left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+i b +2 a \,{\mathrm e}^{i \left (d x +c \right )}\right )^{2} \left (a^{2}-b^{2}\right ) d \,b^{3}}-\frac {i a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{3}}+\frac {3 i a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d b}+\frac {i a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{3}}-\frac {3 i a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d b}\)	\(519\)

Input:

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

Output:

1/d*(-2/b^3*arctan(tan(1/2*d*x+1/2*c))+2/b^3*((-1/2*a^2*b^2/(a^2-b^2)*tan( 
1/2*d*x+1/2*c)^3-1/2*b*(2*a^4+3*a^2*b^2-2*b^4)/a/(a^2-b^2)*tan(1/2*d*x+1/2 
*c)^2-1/2*b^2*(7*a^2-4*b^2)/(a^2-b^2)*tan(1/2*d*x+1/2*c)-1/2*a*b*(2*a^2-b^ 
2)/(a^2-b^2))/(tan(1/2*d*x+1/2*c)^2*a+2*b*tan(1/2*d*x+1/2*c)+a)^2+1/2*a*(2 
*a^2-3*b^2)/(a^2-b^2)^(3/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b 
^2)^(1/2))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 355 vs. \(2 (157) = 314\).

Time = 0.12 (sec) , antiderivative size = 793, normalized size of antiderivative = 4.75 \[ \int \frac {\cos ^2(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

[-1/4*(4*(a^4*b^2 - 2*a^2*b^4 + b^6)*d*x*cos(d*x + c)^2 - 4*(a^6 - a^4*b^2 
 - a^2*b^4 + b^6)*d*x - (2*a^5 - a^3*b^2 - 3*a*b^4 - (2*a^3*b^2 - 3*a*b^4) 
*cos(d*x + c)^2 + 2*(2*a^4*b - 3*a^2*b^3)*sin(d*x + c))*sqrt(-a^2 + b^2)*l 
og(((2*a^2 - b^2)*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2 + 2*(a*c 
os(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)) - 2*(2*a^5*b - 3*a^3*b^3 + a*b^5 
)*cos(d*x + c) - 2*(4*(a^5*b - 2*a^3*b^3 + a*b^5)*d*x + (3*a^4*b^2 - 5*a^2 
*b^4 + 2*b^6)*cos(d*x + c))*sin(d*x + c))/((a^4*b^5 - 2*a^2*b^7 + b^9)*d*c 
os(d*x + c)^2 - 2*(a^5*b^4 - 2*a^3*b^6 + a*b^8)*d*sin(d*x + c) - (a^6*b^3 
- a^4*b^5 - a^2*b^7 + b^9)*d), -1/2*(2*(a^4*b^2 - 2*a^2*b^4 + b^6)*d*x*cos 
(d*x + c)^2 - 2*(a^6 - a^4*b^2 - a^2*b^4 + b^6)*d*x - (2*a^5 - a^3*b^2 - 3 
*a*b^4 - (2*a^3*b^2 - 3*a*b^4)*cos(d*x + c)^2 + 2*(2*a^4*b - 3*a^2*b^3)*si 
n(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)* 
cos(d*x + c))) - (2*a^5*b - 3*a^3*b^3 + a*b^5)*cos(d*x + c) - (4*(a^5*b - 
2*a^3*b^3 + a*b^5)*d*x + (3*a^4*b^2 - 5*a^2*b^4 + 2*b^6)*cos(d*x + c))*sin 
(d*x + c))/((a^4*b^5 - 2*a^2*b^7 + b^9)*d*cos(d*x + c)^2 - 2*(a^5*b^4 - 2* 
a^3*b^6 + a*b^8)*d*sin(d*x + c) - (a^6*b^3 - a^4*b^5 - a^2*b^7 + b^9)*d)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate(cos(d*x+c)^2*sin(d*x+c)/(a+b*sin(d*x+c))^3,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 [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.53 \[ \int \frac {\cos ^2(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {\frac {{\left (2 \, a^{3} - 3 \, a b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{2} b^{3} - b^{5}\right )} \sqrt {a^{2} - b^{2}}} - \frac {a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 7 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, a^{4} - a^{2} b^{2}}{{\left (a^{3} b^{2} - a b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )}^{2}} - \frac {d x + c}{b^{3}}}{d} \] Input:

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

Output:

((2*a^3 - 3*a*b^2)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan((a*ta 
n(1/2*d*x + 1/2*c) + b)/sqrt(a^2 - b^2)))/((a^2*b^3 - b^5)*sqrt(a^2 - b^2) 
) - (a^3*b*tan(1/2*d*x + 1/2*c)^3 + 2*a^4*tan(1/2*d*x + 1/2*c)^2 + 3*a^2*b 
^2*tan(1/2*d*x + 1/2*c)^2 - 2*b^4*tan(1/2*d*x + 1/2*c)^2 + 7*a^3*b*tan(1/2 
*d*x + 1/2*c) - 4*a*b^3*tan(1/2*d*x + 1/2*c) + 2*a^4 - a^2*b^2)/((a^3*b^2 
- a*b^4)*(a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)^2) - (d 
*x + c)/b^3)/d
 

Mupad [B] (verification not implemented)

Time = 37.88 (sec) , antiderivative size = 2709, normalized size of antiderivative = 16.22 \[ \int \frac {\cos ^2(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Too large to display} \] Input:

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

Output:

(2*atan((64*a*b^5*tan(c/2 + (d*x)/2))/((176*a^3*b^12)/(b^9 - 2*a^2*b^7 + a 
^4*b^5) - (160*a^5*b^10)/(b^9 - 2*a^2*b^7 + a^4*b^5) + (48*a^7*b^8)/(b^9 - 
 2*a^2*b^7 + a^4*b^5) - (64*a*b^14)/(b^9 - 2*a^2*b^7 + a^4*b^5)) - (48*a^3 
*b^3*tan(c/2 + (d*x)/2))/((176*a^3*b^12)/(b^9 - 2*a^2*b^7 + a^4*b^5) - (16 
0*a^5*b^10)/(b^9 - 2*a^2*b^7 + a^4*b^5) + (48*a^7*b^8)/(b^9 - 2*a^2*b^7 + 
a^4*b^5) - (64*a*b^14)/(b^9 - 2*a^2*b^7 + a^4*b^5))))/(b^3*d) - ((tan(c/2 
+ (d*x)/2)*(7*a^2 - 4*b^2))/(b*(a^2 - b^2)) - (a*b^2 - 2*a^3)/(b^2*(a^2 - 
b^2)) + (a^2*tan(c/2 + (d*x)/2)^3)/(b*(a^2 - b^2)) + (tan(c/2 + (d*x)/2)^2 
*(a^2 + 2*b^2)*(2*a^2 - b^2))/(a*b^2*(a^2 - b^2)))/(d*(tan(c/2 + (d*x)/2)^ 
2*(2*a^2 + 4*b^2) + a^2*tan(c/2 + (d*x)/2)^4 + a^2 + 4*a*b*tan(c/2 + (d*x) 
/2)^3 + 4*a*b*tan(c/2 + (d*x)/2))) - (a*atan(((a*(2*a^2 - 3*b^2)*(-(a + b) 
^3*(a - b)^3)^(1/2)*((8*(4*a^2*b^6 - 8*a^4*b^4 + 4*a^6*b^2))/(b^9 - 2*a^2* 
b^7 + a^4*b^5) + (8*tan(c/2 + (d*x)/2)*(8*a*b^8 - 29*a^3*b^6 + 28*a^5*b^4 
- 8*a^7*b^2))/(b^10 - 2*a^2*b^8 + a^4*b^6) - (a*(2*a^2 - 3*b^2)*(-(a + b)^ 
3*(a - b)^3)^(1/2)*((8*(4*a*b^10 - 6*a^3*b^8 + 2*a^5*b^6))/(b^9 - 2*a^2*b^ 
7 + a^4*b^5) + (8*tan(c/2 + (d*x)/2)*(12*a^2*b^10 - 20*a^4*b^8 + 8*a^6*b^6 
))/(b^10 - 2*a^2*b^8 + a^4*b^6) - (a*(2*a^2 - 3*b^2)*(-(a + b)^3*(a - b)^3 
)^(1/2)*((8*(4*a^2*b^12 - 8*a^4*b^10 + 4*a^6*b^8))/(b^9 - 2*a^2*b^7 + a^4* 
b^5) + (8*tan(c/2 + (d*x)/2)*(12*a*b^14 - 32*a^3*b^12 + 28*a^5*b^10 - 8*a^ 
7*b^8))/(b^10 - 2*a^2*b^8 + a^4*b^6)))/(2*(b^9 - 3*a^2*b^7 + 3*a^4*b^5 ...
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 619, normalized size of antiderivative = 3.71 \[ \int \frac {\cos ^2(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {4 \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 )^{2} a^{3} b^{2}-6 \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 )^{2} a \,b^{4}+8 \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 -12 \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}+4 \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}-6 \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}-3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{4} b^{2}+5 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b^{4}-2 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b^{6}-2 \cos \left (d x +c \right ) a^{5} b +3 \cos \left (d x +c \right ) a^{3} b^{3}-\cos \left (d x +c \right ) a \,b^{5}-2 \sin \left (d x +c \right )^{2} a^{4} b^{2} d x +4 \sin \left (d x +c \right )^{2} a^{2} b^{4} d x -2 \sin \left (d x +c \right )^{2} b^{6} d x -4 \sin \left (d x +c \right ) a^{5} b d x +8 \sin \left (d x +c \right ) a^{3} b^{3} d x -4 \sin \left (d x +c \right ) a \,b^{5} d x -2 a^{6} d x +4 a^{4} b^{2} d x -2 a^{2} b^{4} d x}{2 b^{3} d \left (\sin \left (d x +c \right )^{2} a^{4} b^{2}-2 \sin \left (d x +c \right )^{2} a^{2} b^{4}+\sin \left (d x +c \right )^{2} b^{6}+2 \sin \left (d x +c \right ) a^{5} b -4 \sin \left (d x +c \right ) a^{3} b^{3}+2 \sin \left (d x +c \right ) a \,b^{5}+a^{6}-2 a^{4} b^{2}+a^{2} b^{4}\right )} \] Input:

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

Output:

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