\(\int \frac {\cos ^3(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx\) [1292]

Optimal result

Integrand size = 29, antiderivative size = 119 \[ \int \frac {\cos ^3(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {a^2 \left (a^2-b^2\right ) \log (a+b \sin (c+d x))}{b^5 d}+\frac {a \left (a^2-b^2\right ) \sin (c+d x)}{b^4 d}-\frac {\left (a^2-b^2\right ) \sin ^2(c+d x)}{2 b^3 d}+\frac {a \sin ^3(c+d x)}{3 b^2 d}-\frac {\sin ^4(c+d x)}{4 b d} \] Output:

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

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.87 \[ \int \frac {\cos ^3(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {12 a^2 \left (-a^2+b^2\right ) \log (a+b \sin (c+d x))+12 a b \left (a^2-b^2\right ) \sin (c+d x)+6 b^2 \left (-a^2+b^2\right ) \sin ^2(c+d x)+4 a b^3 \sin ^3(c+d x)-3 b^4 \sin ^4(c+d x)}{12 b^5 d} \] Input:

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

Output:

(12*a^2*(-a^2 + b^2)*Log[a + b*Sin[c + d*x]] + 12*a*b*(a^2 - b^2)*Sin[c + 
d*x] + 6*b^2*(-a^2 + b^2)*Sin[c + d*x]^2 + 4*a*b^3*Sin[c + d*x]^3 - 3*b^4* 
Sin[c + d*x]^4)/(12*b^5*d)
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.89, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 3316, 27, 522, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3316

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

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {b^2 \sin ^2(c+d x) \left (b^2-b^2 \sin ^2(c+d x)\right )}{a+b \sin (c+d x)}d(b \sin (c+d x))}{b^5 d}\)

\(\Big \downarrow \) 522

\(\displaystyle \frac {\int \left (\left (1-\frac {b^2}{a^2}\right ) a^3+\frac {\left (b^2-a^2\right ) a^2}{a+b \sin (c+d x)}+b^2 \sin ^2(c+d x) a-b^3 \sin ^3(c+d x)-b \left (a^2-b^2\right ) \sin (c+d x)\right )d(b \sin (c+d x))}{b^5 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {1}{2} b^2 \left (a^2-b^2\right ) \sin ^2(c+d x)+a b \left (a^2-b^2\right ) \sin (c+d x)-a^2 \left (a^2-b^2\right ) \log (a+b \sin (c+d x))+\frac {1}{3} a b^3 \sin ^3(c+d x)-\frac {1}{4} b^4 \sin ^4(c+d x)}{b^5 d}\)

Input:

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

Output:

(-(a^2*(a^2 - b^2)*Log[a + b*Sin[c + d*x]]) + a*b*(a^2 - b^2)*Sin[c + d*x] 
 - (b^2*(a^2 - b^2)*Sin[c + d*x]^2)/2 + (a*b^3*Sin[c + d*x]^3)/3 - (b^4*Si 
n[c + d*x]^4)/4)/(b^5*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 522

Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. 
), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], 
x] /; FreeQ[{a, b, c, d, e, 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 = 0.73 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.87

method	result	size
derivativedivides	\(\frac {\frac {-\frac {\sin \left (d x +c \right )^{4} b^{3}}{4}+\frac {a \sin \left (d x +c \right )^{3} b^{2}}{3}-\frac {\left (a^{2}-b^{2}\right ) \sin \left (d x +c \right )^{2} b}{2}+\sin \left (d x +c \right ) a \left (a^{2}-b^{2}\right )}{b^{4}}-\frac {a^{2} \left (a^{2}-b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{5}}}{d}\)	\(103\)
default	\(\frac {\frac {-\frac {\sin \left (d x +c \right )^{4} b^{3}}{4}+\frac {a \sin \left (d x +c \right )^{3} b^{2}}{3}-\frac {\left (a^{2}-b^{2}\right ) \sin \left (d x +c \right )^{2} b}{2}+\sin \left (d x +c \right ) a \left (a^{2}-b^{2}\right )}{b^{4}}-\frac {a^{2} \left (a^{2}-b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{5}}}{d}\)	\(103\)
parallelrisch	\(\frac {96 a^{4} \left (-\ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )+\ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )\right )-96 a^{2} b^{2} \left (-\ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )+\ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )\right )-24 a^{2} b^{2}+15 b^{4}+96 a^{3} b \sin \left (d x +c \right )+24 a^{2} b^{2} \cos \left (2 d x +2 c \right )-72 a \,b^{3} \sin \left (d x +c \right )-8 a \sin \left (3 d x +3 c \right ) b^{3}-12 b^{4} \cos \left (2 d x +2 c \right )-3 \cos \left (4 d x +4 c \right ) b^{4}}{96 b^{5} d}\)	\(204\)
risch	\(\frac {i x \,a^{4}}{b^{5}}+\frac {3 i a \,{\mathrm e}^{i \left (d x +c \right )}}{8 b^{2} d}+\frac {{\mathrm e}^{2 i \left (d x +c \right )} a^{2}}{8 b^{3} d}-\frac {{\mathrm e}^{2 i \left (d x +c \right )}}{16 b d}-\frac {2 i a^{2} c}{d \,b^{3}}-\frac {i a^{3} {\mathrm e}^{i \left (d x +c \right )}}{2 b^{4} d}-\frac {3 i a \,{\mathrm e}^{-i \left (d x +c \right )}}{8 b^{2} d}+\frac {i a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{2 b^{4} d}+\frac {{\mathrm e}^{-2 i \left (d x +c \right )} a^{2}}{8 b^{3} d}-\frac {{\mathrm e}^{-2 i \left (d x +c \right )}}{16 b d}-\frac {i a^{2} x}{b^{3}}+\frac {2 i a^{4} c}{d \,b^{5}}-\frac {a^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{d \,b^{5}}+\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{d \,b^{3}}-\frac {\cos \left (4 d x +4 c \right )}{32 b d}-\frac {a \sin \left (3 d x +3 c \right )}{12 b^{2} d}\)	\(314\)
norman	\(\frac {-\frac {2 \left (3 a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d \,b^{3}}-\frac {2 \left (3 a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d \,b^{3}}-\frac {\left (2 a^{2}-2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d \,b^{3}}-\frac {\left (2 a^{2}-2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d \,b^{3}}+\frac {2 a \left (a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{b^{4} d}+\frac {2 a \left (a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{b^{4} d}+\frac {8 a \left (3 a^{2}-2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 b^{4} d}+\frac {8 a \left (3 a^{2}-2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 b^{4} d}+\frac {4 a \left (9 a^{2}-5 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 b^{4} d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5}}+\frac {a^{2} \left (a^{2}-b^{2}\right ) \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{d \,b^{5}}-\frac {a^{2} \left (a^{2}-b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{d \,b^{5}}\)	\(369\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.82 \[ \int \frac {\cos ^3(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {3 \, b^{4} \cos \left (d x + c\right )^{4} - 6 \, a^{2} b^{2} \cos \left (d x + c\right )^{2} + 12 \, {\left (a^{4} - a^{2} b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + 4 \, {\left (a b^{3} \cos \left (d x + c\right )^{2} - 3 \, a^{3} b + 2 \, a b^{3}\right )} \sin \left (d x + c\right )}{12 \, b^{5} d} \] Input:

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

Output:

-1/12*(3*b^4*cos(d*x + c)^4 - 6*a^2*b^2*cos(d*x + c)^2 + 12*(a^4 - a^2*b^2 
)*log(b*sin(d*x + c) + a) + 4*(a*b^3*cos(d*x + c)^2 - 3*a^3*b + 2*a*b^3)*s 
in(d*x + c))/(b^5*d)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.88 \[ \int \frac {\cos ^3(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {3 \, b^{3} \sin \left (d x + c\right )^{4} - 4 \, a b^{2} \sin \left (d x + c\right )^{3} + 6 \, {\left (a^{2} b - b^{3}\right )} \sin \left (d x + c\right )^{2} - 12 \, {\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )}{b^{4}} + \frac {12 \, {\left (a^{4} - a^{2} b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{5}}}{12 \, d} \] Input:

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

Output:

-1/12*((3*b^3*sin(d*x + c)^4 - 4*a*b^2*sin(d*x + c)^3 + 6*(a^2*b - b^3)*si 
n(d*x + c)^2 - 12*(a^3 - a*b^2)*sin(d*x + c))/b^4 + 12*(a^4 - a^2*b^2)*log 
(b*sin(d*x + c) + a)/b^5)/d
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.15 \[ \int \frac {\cos ^3(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {{\left (a^{4} - a^{2} b^{2}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{5} d} - \frac {3 \, b^{3} d^{3} \sin \left (d x + c\right )^{4} - 4 \, a b^{2} d^{3} \sin \left (d x + c\right )^{3} + 6 \, a^{2} b d^{3} \sin \left (d x + c\right )^{2} - 6 \, b^{3} d^{3} \sin \left (d x + c\right )^{2} - 12 \, a^{3} d^{3} \sin \left (d x + c\right ) + 12 \, a b^{2} d^{3} \sin \left (d x + c\right )}{12 \, b^{4} d^{4}} \] Input:

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

Output:

-(a^4 - a^2*b^2)*log(abs(b*sin(d*x + c) + a))/(b^5*d) - 1/12*(3*b^3*d^3*si 
n(d*x + c)^4 - 4*a*b^2*d^3*sin(d*x + c)^3 + 6*a^2*b*d^3*sin(d*x + c)^2 - 6 
*b^3*d^3*sin(d*x + c)^2 - 12*a^3*d^3*sin(d*x + c) + 12*a*b^2*d^3*sin(d*x + 
 c))/(b^4*d^4)
 

Mupad [B] (verification not implemented)

Time = 19.65 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.90 \[ \int \frac {\cos ^3(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {{\sin \left (c+d\,x\right )}^4}{4\,b}-{\sin \left (c+d\,x\right )}^2\,\left (\frac {1}{2\,b}-\frac {a^2}{2\,b^3}\right )-\frac {a\,{\sin \left (c+d\,x\right )}^3}{3\,b^2}+\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (a^4-a^2\,b^2\right )}{b^5}+\frac {a\,\sin \left (c+d\,x\right )\,\left (\frac {1}{b}-\frac {a^2}{b^3}\right )}{b}}{d} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.78 \[ \int \frac {\cos ^3(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) a^{4}-12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) a^{2} b^{2}-12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) a^{4}+12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) a^{2} b^{2}-3 \sin \left (d x +c \right )^{4} b^{4}+4 \sin \left (d x +c \right )^{3} a \,b^{3}-6 \sin \left (d x +c \right )^{2} a^{2} b^{2}+6 \sin \left (d x +c \right )^{2} b^{4}+12 \sin \left (d x +c \right ) a^{3} b -12 \sin \left (d x +c \right ) a \,b^{3}+6 a^{2} b^{2}-6 b^{4}}{12 b^{5} d} \] Input:

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

Output:

(12*log(tan((c + d*x)/2)**2 + 1)*a**4 - 12*log(tan((c + d*x)/2)**2 + 1)*a* 
*2*b**2 - 12*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*a**4 + 
12*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*a**2*b**2 - 3*sin 
(c + d*x)**4*b**4 + 4*sin(c + d*x)**3*a*b**3 - 6*sin(c + d*x)**2*a**2*b**2 
 + 6*sin(c + d*x)**2*b**4 + 12*sin(c + d*x)*a**3*b - 12*sin(c + d*x)*a*b** 
3 + 6*a**2*b**2 - 6*b**4)/(12*b**5*d)