\(\int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx\) [1550]

Optimal result

Integrand size = 31, antiderivative size = 113 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx=\frac {\left (2 a A b-3 a^2 B+b^2 B\right ) \log (a+b \sin (c+d x))}{b^4 d}-\frac {(A b-2 a B) \sin (c+d x)}{b^3 d}-\frac {B \sin ^2(c+d x)}{2 b^2 d}+\frac {\left (a^2-b^2\right ) (A b-a B)}{b^4 d (a+b \sin (c+d x))} \] Output:

(2*A*a*b-3*B*a^2+B*b^2)*ln(a+b*sin(d*x+c))/b^4/d-(A*b-2*B*a)*sin(d*x+c)/b^ 
3/d-1/2*B*sin(d*x+c)^2/b^2/d+(a^2-b^2)*(A*b-B*a)/b^4/d/(a+b*sin(d*x+c))
 

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.98 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx=\frac {\frac {\left (-a^2+b^2\right ) B \log (a+b \sin (c+d x))}{b}+a B \sin (c+d x)-\frac {1}{2} b B \sin ^2(c+d x)+\left (A-\frac {a B}{b}\right ) \left (2 a \log (a+b \sin (c+d x))-b \sin (c+d x)+\frac {(a-b) (a+b)}{a+b \sin (c+d x)}\right )}{b^3 d} \] Input:

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

Output:

(((-a^2 + b^2)*B*Log[a + b*Sin[c + d*x]])/b + a*B*Sin[c + d*x] - (b*B*Sin[ 
c + d*x]^2)/2 + (A - (a*B)/b)*(2*a*Log[a + b*Sin[c + d*x]] - b*Sin[c + d*x 
] + ((a - b)*(a + b))/(a + b*Sin[c + d*x])))/(b^3*d)
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.89, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3042, 3316, 27, 652, 2009}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3316

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 652

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {\left (a^2-b^2\right ) (A b-a B)}{a+b \sin (c+d x)}+\left (-3 a^2 B+2 a A b+b^2 B\right ) \log (a+b \sin (c+d x))-b (A b-3 a B) \sin (c+d x)-\frac {1}{2} B (a+b \sin (c+d x))^2}{b^4 d}\)

Input:

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

Output:

((2*a*A*b - 3*a^2*B + b^2*B)*Log[a + b*Sin[c + d*x]] - b*(A*b - 3*a*B)*Sin 
[c + d*x] + ((a^2 - b^2)*(A*b - a*B))/(a + b*Sin[c + d*x]) - (B*(a + b*Sin 
[c + d*x])^2)/2)/(b^4*d)
 

Defintions of rubi rules used

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 652

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (c_.)*(x_ 
)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + c 
*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && IGtQ[p, 0]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 3316

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

Maple [A] (verified)

Time = 2.51 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.04

method	result	size
derivativedivides	\(-\frac {\frac {\frac {B \sin \left (d x +c \right )^{2} b}{2}+A \sin \left (d x +c \right ) b -2 a B \sin \left (d x +c \right )}{b^{3}}+\frac {\left (-2 A a b +3 a^{2} B -B \,b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{4}}-\frac {A \,a^{2} b -A \,b^{3}-a^{3} B +B a \,b^{2}}{b^{4} \left (a +b \sin \left (d x +c \right )\right )}}{d}\)	\(118\)
default	\(-\frac {\frac {\frac {B \sin \left (d x +c \right )^{2} b}{2}+A \sin \left (d x +c \right ) b -2 a B \sin \left (d x +c \right )}{b^{3}}+\frac {\left (-2 A a b +3 a^{2} B -B \,b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{4}}-\frac {A \,a^{2} b -A \,b^{3}-a^{3} B +B a \,b^{2}}{b^{4} \left (a +b \sin \left (d x +c \right )\right )}}{d}\)	\(118\)
parallelrisch	\(\frac {16 \left (A a b -\frac {3}{2} a^{2} B +\frac {1}{2} B \,b^{2}\right ) \left (a +b \sin \left (d x +c \right )\right ) a \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )-16 \left (A a b -\frac {3}{2} a^{2} B +\frac {1}{2} B \,b^{2}\right ) \left (a +b \sin \left (d x +c \right )\right ) a \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )+\left (4 A a \,b^{3}-6 B \,a^{2} b^{2}\right ) \cos \left (2 d x +2 c \right )+B a \,b^{3} \sin \left (3 d x +3 c \right )+\left (-16 A \,a^{2} b^{2}+8 A \,b^{4}+24 B \,a^{3} b -11 B a \,b^{3}\right ) \sin \left (d x +c \right )-4 A a \,b^{3}+6 B \,a^{2} b^{2}}{8 a d \,b^{4} \left (a +b \sin \left (d x +c \right )\right )}\)	\(219\)
risch	\(\frac {i {\mathrm e}^{-i \left (d x +c \right )} B a}{b^{3} d}+\frac {i {\mathrm e}^{i \left (d x +c \right )} A}{2 b^{2} d}+\frac {6 i B \,a^{2} c}{b^{4} d}+\frac {{\mathrm e}^{2 i \left (d x +c \right )} B}{8 b^{2} d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} B a}{b^{3} d}-\frac {i B x}{b^{2}}-\frac {2 i x A a}{b^{3}}+\frac {3 i x B \,a^{2}}{b^{4}}+\frac {{\mathrm e}^{-2 i \left (d x +c \right )} B}{8 b^{2} d}-\frac {2 i B c}{b^{2} d}-\frac {4 i A a c}{b^{3} d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} A}{2 b^{2} d}-\frac {2 \left (-A \,a^{2} b +A \,b^{3}+a^{3} B -B a \,b^{2}\right ) {\mathrm e}^{i \left (d x +c \right )}}{b^{4} d \left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+i b +2 a \,{\mathrm e}^{i \left (d x +c \right )}\right )}+\frac {2 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right ) A a}{b^{3} d}-\frac {3 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right ) B \,a^{2}}{b^{4} d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right ) B}{b^{2} d}\)	\(369\)
norman	\(\frac {-\frac {2 \left (6 A b -9 B a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{b^{2} d}-\frac {2 \left (6 A b -9 B a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{b^{2} d}-\frac {\left (4 A b -6 B a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{b^{2} d}-\frac {\left (4 A b -6 B a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{b^{2} d}-\frac {4 \left (6 A \,a^{2} b -3 A \,b^{3}-9 a^{3} B +5 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{b^{3} d a}-\frac {4 \left (4 A \,a^{2} b -2 A \,b^{3}-6 a^{3} B +3 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{b^{3} d a}-\frac {4 \left (4 A \,a^{2} b -2 A \,b^{3}-6 a^{3} B +3 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{b^{3} d a}-\frac {2 \left (2 A \,a^{2} b -A \,b^{3}-3 a^{3} B +B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a \,b^{3} d}-\frac {2 \left (2 A \,a^{2} b -A \,b^{3}-3 a^{3} B +B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{a \,b^{3} d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}+\frac {\left (2 A a b -3 a^{2} B +B \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{b^{4} d}-\frac {\left (2 A a b -3 a^{2} B +B \,b^{2}\right ) \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{b^{4} d}\)	\(492\)

Input:

int(cos(d*x+c)^3*(A+B*sin(d*x+c))/(a+b*sin(d*x+c))^2,x,method=_RETURNVERBO 
SE)
 

Output:

-1/d*(1/b^3*(1/2*B*sin(d*x+c)^2*b+A*sin(d*x+c)*b-2*a*B*sin(d*x+c))+(-2*A*a 
*b+3*B*a^2-B*b^2)/b^4*ln(a+b*sin(d*x+c))-1/b^4*(A*a^2*b-A*b^3-B*a^3+B*a*b^ 
2)/(a+b*sin(d*x+c)))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.58 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx=-\frac {4 \, B a^{3} - 4 \, A a^{2} b - 11 \, B a b^{2} + 8 \, A b^{3} + 2 \, {\left (3 \, B a b^{2} - 2 \, A b^{3}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (3 \, B a^{3} - 2 \, A a^{2} b - B a b^{2} + {\left (3 \, B a^{2} b - 2 \, A a b^{2} - B b^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) - {\left (2 \, B b^{3} \cos \left (d x + c\right )^{2} + 8 \, B a^{2} b - 4 \, A a b^{2} - B b^{3}\right )} \sin \left (d x + c\right )}{4 \, {\left (b^{5} d \sin \left (d x + c\right ) + a b^{4} d\right )}} \] Input:

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

Output:

-1/4*(4*B*a^3 - 4*A*a^2*b - 11*B*a*b^2 + 8*A*b^3 + 2*(3*B*a*b^2 - 2*A*b^3) 
*cos(d*x + c)^2 + 4*(3*B*a^3 - 2*A*a^2*b - B*a*b^2 + (3*B*a^2*b - 2*A*a*b^ 
2 - B*b^3)*sin(d*x + c))*log(b*sin(d*x + c) + a) - (2*B*b^3*cos(d*x + c)^2 
 + 8*B*a^2*b - 4*A*a*b^2 - B*b^3)*sin(d*x + c))/(b^5*d*sin(d*x + c) + a*b^ 
4*d)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.04 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx=-\frac {\frac {2 \, {\left (B a^{3} - A a^{2} b - B a b^{2} + A b^{3}\right )}}{b^{5} \sin \left (d x + c\right ) + a b^{4}} + \frac {B b \sin \left (d x + c\right )^{2} - 2 \, {\left (2 \, B a - A b\right )} \sin \left (d x + c\right )}{b^{3}} + \frac {2 \, {\left (3 \, B a^{2} - 2 \, A a b - B b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{4}}}{2 \, d} \] Input:

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

Output:

-1/2*(2*(B*a^3 - A*a^2*b - B*a*b^2 + A*b^3)/(b^5*sin(d*x + c) + a*b^4) + ( 
B*b*sin(d*x + c)^2 - 2*(2*B*a - A*b)*sin(d*x + c))/b^3 + 2*(3*B*a^2 - 2*A* 
a*b - B*b^2)*log(b*sin(d*x + c) + a)/b^4)/d
 

Giac [A] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.17 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx=-\frac {{\left (3 \, B a^{2} - 2 \, A a b - B b^{2}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{4} d} - \frac {B b^{2} d \sin \left (d x + c\right )^{2} - 4 \, B a b d \sin \left (d x + c\right ) + 2 \, A b^{2} d \sin \left (d x + c\right )}{2 \, b^{4} d^{2}} - \frac {B a^{3} - A a^{2} b - B a b^{2} + A b^{3}}{{\left (b \sin \left (d x + c\right ) + a\right )} b^{4} d} \] Input:

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

Output:

-(3*B*a^2 - 2*A*a*b - B*b^2)*log(abs(b*sin(d*x + c) + a))/(b^4*d) - 1/2*(B 
*b^2*d*sin(d*x + c)^2 - 4*B*a*b*d*sin(d*x + c) + 2*A*b^2*d*sin(d*x + c))/( 
b^4*d^2) - (B*a^3 - A*a^2*b - B*a*b^2 + A*b^3)/((b*sin(d*x + c) + a)*b^4*d 
)
 

Mupad [B] (verification not implemented)

Time = 36.39 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.13 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx=\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (-3\,B\,a^2+2\,A\,a\,b+B\,b^2\right )}{b^4\,d}-\frac {B\,a^3-A\,a^2\,b-B\,a\,b^2+A\,b^3}{b\,d\,\left (\sin \left (c+d\,x\right )\,b^4+a\,b^3\right )}-\frac {\sin \left (c+d\,x\right )\,\left (\frac {A}{b^2}-\frac {2\,B\,a}{b^3}\right )}{d}-\frac {B\,{\sin \left (c+d\,x\right )}^2}{2\,b^2\,d} \] Input:

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

Output:

(log(a + b*sin(c + d*x))*(B*b^2 - 3*B*a^2 + 2*A*a*b))/(b^4*d) - (A*b^3 + B 
*a^3 - A*a^2*b - B*a*b^2)/(b*d*(a*b^3 + b^4*sin(c + d*x))) - (sin(c + d*x) 
*(A/b^2 - (2*B*a)/b^3))/d - (B*sin(c + d*x)^2)/(2*b^2*d)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.25 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx=\frac {2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) a^{2}-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) b^{2}-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) a^{2}+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) b^{2}-\sin \left (d x +c \right )^{2} b^{2}+2 \sin \left (d x +c \right ) a b +2 b^{2}}{2 b^{3} d} \] Input:

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

Output:

(2*log(tan((c + d*x)/2)**2 + 1)*a**2 - 2*log(tan((c + d*x)/2)**2 + 1)*b**2 
 - 2*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*a**2 + 2*log(ta 
n((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*b**2 - sin(c + d*x)**2*b** 
2 + 2*sin(c + d*x)*a*b + 2*b**2)/(2*b**3*d)