Integrand size = 29, antiderivative size = 149 \[ \int \frac {\cos ^3(c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a^3 \left (a^2-b^2\right ) \log (a+b \sin (c+d x))}{b^6 d}-\frac {a^2 \left (a^2-b^2\right ) \sin (c+d x)}{b^5 d}+\frac {a \left (a^2-b^2\right ) \sin ^2(c+d x)}{2 b^4 d}-\frac {\left (a^2-b^2\right ) \sin ^3(c+d x)}{3 b^3 d}+\frac {a \sin ^4(c+d x)}{4 b^2 d}-\frac {\sin ^5(c+d x)}{5 b d} \] Output:
a^3*(a^2-b^2)*ln(a+b*sin(d*x+c))/b^6/d-a^2*(a^2-b^2)*sin(d*x+c)/b^5/d+1/2* a*(a^2-b^2)*sin(d*x+c)^2/b^4/d-1/3*(a^2-b^2)*sin(d*x+c)^3/b^3/d+1/4*a*sin( d*x+c)^4/b^2/d-1/5*sin(d*x+c)^5/b/d
Time = 0.30 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.85 \[ \int \frac {\cos ^3(c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {60 a^3 (a-b) (a+b) \log (a+b \sin (c+d x))}{b^6}-\frac {60 a^2 (a-b) (a+b) \sin (c+d x)}{b^5}+\frac {30 a (a-b) (a+b) \sin ^2(c+d x)}{b^4}-\frac {20 (a-b) (a+b) \sin ^3(c+d x)}{b^3}+\frac {15 a \sin ^4(c+d x)}{b^2}-\frac {12 \sin ^5(c+d x)}{b}}{60 d} \] Input:
Integrate[(Cos[c + d*x]^3*Sin[c + d*x]^3)/(a + b*Sin[c + d*x]),x]
Output:
((60*a^3*(a - b)*(a + b)*Log[a + b*Sin[c + d*x]])/b^6 - (60*a^2*(a - b)*(a + b)*Sin[c + d*x])/b^5 + (30*a*(a - b)*(a + b)*Sin[c + d*x]^2)/b^4 - (20* (a - b)*(a + b)*Sin[c + d*x]^3)/b^3 + (15*a*Sin[c + d*x]^4)/b^2 - (12*Sin[ c + d*x]^5)/b)/(60*d)
Time = 0.38 (sec) , antiderivative size = 133, 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.
| \(\displaystyle \int \frac {\sin ^3(c+d x) \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^3 \cos (c+d x)^3}{a+b \sin (c+d x)}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {\int \frac {\sin ^3(c+d x) \left (b^2-b^2 \sin ^2(c+d x)\right )}{a+b \sin (c+d x)}d(b \sin (c+d x))}{b^3 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {b^3 \sin ^3(c+d x) \left (b^2-b^2 \sin ^2(c+d x)\right )}{a+b \sin (c+d x)}d(b \sin (c+d x))}{b^6 d}\) |
| \(\Big \downarrow \) 522 |
| \(\displaystyle \frac {\int \left (-\left (\left (1-\frac {b^2}{a^2}\right ) a^4\right )+b^3 \sin ^3(c+d x) a+b \left (a^2-b^2\right ) \sin (c+d x) a-b^4 \sin ^4(c+d x)-b^2 \left (a^2-b^2\right ) \sin ^2(c+d x)+\frac {a^5-a^3 b^2}{a+b \sin (c+d x)}\right )d(b \sin (c+d x))}{b^6 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {1}{2} a b^2 \left (a^2-b^2\right ) \sin ^2(c+d x)-a^2 b \left (a^2-b^2\right ) \sin (c+d x)-\frac {1}{3} b^3 \left (a^2-b^2\right ) \sin ^3(c+d x)+a^3 \left (a^2-b^2\right ) \log (a+b \sin (c+d x))+\frac {1}{4} a b^4 \sin ^4(c+d x)-\frac {1}{5} b^5 \sin ^5(c+d x)}{b^6 d}\) |
Input:
Int[(Cos[c + d*x]^3*Sin[c + d*x]^3)/(a + b*Sin[c + d*x]),x]
Output:
(a^3*(a^2 - b^2)*Log[a + b*Sin[c + d*x]] - a^2*b*(a^2 - b^2)*Sin[c + d*x] + (a*b^2*(a^2 - b^2)*Sin[c + d*x]^2)/2 - (b^3*(a^2 - b^2)*Sin[c + d*x]^3)/ 3 + (a*b^4*Sin[c + d*x]^4)/4 - (b^5*Sin[c + d*x]^5)/5)/(b^6*d)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
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]
Time = 0.96 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.99
| method | result | size |
| derivativedivides | \(\frac {-\frac {\frac {\sin \left (d x +c \right )^{5} b^{4}}{5}-\frac {a \sin \left (d x +c \right )^{4} b^{3}}{4}+\frac {a^{2} b^{2} \sin \left (d x +c \right )^{3}}{3}-\frac {b^{4} \sin \left (d x +c \right )^{3}}{3}-\frac {a^{3} b \sin \left (d x +c \right )^{2}}{2}+\frac {a \,b^{3} \sin \left (d x +c \right )^{2}}{2}+a^{4} \sin \left (d x +c \right )-\sin \left (d x +c \right ) a^{2} b^{2}}{b^{5}}+\frac {a^{3} \left (a^{2}-b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{6}}}{d}\) | \(147\) |
| default | \(\frac {-\frac {\frac {\sin \left (d x +c \right )^{5} b^{4}}{5}-\frac {a \sin \left (d x +c \right )^{4} b^{3}}{4}+\frac {a^{2} b^{2} \sin \left (d x +c \right )^{3}}{3}-\frac {b^{4} \sin \left (d x +c \right )^{3}}{3}-\frac {a^{3} b \sin \left (d x +c \right )^{2}}{2}+\frac {a \,b^{3} \sin \left (d x +c \right )^{2}}{2}+a^{4} \sin \left (d x +c \right )-\sin \left (d x +c \right ) a^{2} b^{2}}{b^{5}}+\frac {a^{3} \left (a^{2}-b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{6}}}{d}\) | \(147\) |
| parallelrisch | \(\frac {480 \left (a^{5}-a^{3} b^{2}\right ) \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )+480 \left (-a^{5}+a^{3} b^{2}\right ) \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )+60 \left (-2 a^{3} b^{2}+a \,b^{4}\right ) \cos \left (2 d x +2 c \right )+10 \left (4 a^{2} b^{3}-b^{5}\right ) \sin \left (3 d x +3 c \right )+15 b^{4} a \cos \left (4 d x +4 c \right )-6 b^{5} \sin \left (5 d x +5 c \right )+60 \left (-8 a^{4} b +6 a^{2} b^{3}+b^{5}\right ) \sin \left (d x +c \right )+120 a^{3} b^{2}-75 a \,b^{4}}{480 d \,b^{6}}\) | \(197\) |
| risch | \(-\frac {i x \,a^{5}}{b^{6}}+\frac {i a^{3} x}{b^{4}}-\frac {a^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d \,b^{4}}+\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{16 b^{2} d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} a^{4}}{2 b^{5} d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} a^{2}}{8 b^{3} d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{16 b d}-\frac {2 i a^{5} c}{b^{6} d}-\frac {i {\mathrm e}^{i \left (d x +c \right )}}{16 b d}+\frac {2 i a^{3} c}{b^{4} d}-\frac {a^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d \,b^{4}}+\frac {a \,{\mathrm e}^{-2 i \left (d x +c \right )}}{16 b^{2} d}+\frac {i {\mathrm e}^{i \left (d x +c \right )} a^{4}}{2 b^{5} d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} a^{2}}{8 b^{3} d}+\frac {a^{5} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{b^{6} d}-\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{b^{4} d}-\frac {\sin \left (5 d x +5 c \right )}{80 b d}+\frac {a \cos \left (4 d x +4 c \right )}{32 b^{2} d}+\frac {\sin \left (3 d x +3 c \right ) a^{2}}{12 b^{3} d}-\frac {\sin \left (3 d x +3 c \right )}{48 b d}\) | \(393\) |
| norman | \(\frac {\frac {\left (8 a^{3}-4 a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d \,b^{4}}+\frac {\left (8 a^{3}-4 a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d \,b^{4}}-\frac {4 \left (25 a^{4}-15 a^{2} b^{2}-2 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5 b^{5} d}-\frac {4 \left (25 a^{4}-15 a^{2} b^{2}-2 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{5 b^{5} d}+\frac {4 \left (3 a^{3}-a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d \,b^{4}}+\frac {2 \left (a^{3}-a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d \,b^{4}}+\frac {2 \left (a^{3}-a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{d \,b^{4}}-\frac {2 a^{2} \left (a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,b^{5}}-\frac {2 a^{2} \left (a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d \,b^{5}}-\frac {2 \left (a^{2}-b^{2}\right ) \left (15 a^{2}+4 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 b^{5} d}-\frac {2 \left (a^{2}-b^{2}\right ) \left (15 a^{2}+4 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 b^{5} d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{6}}+\frac {a^{3} \left (a^{2}-b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{d \,b^{6}}-\frac {a^{3} \left (a^{2}-b^{2}\right ) \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{d \,b^{6}}\) | \(463\) |
Input:
int(cos(d*x+c)^3*sin(d*x+c)^3/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(-1/b^5*(1/5*sin(d*x+c)^5*b^4-1/4*a*sin(d*x+c)^4*b^3+1/3*a^2*b^2*sin(d *x+c)^3-1/3*b^4*sin(d*x+c)^3-1/2*a^3*b*sin(d*x+c)^2+1/2*a*b^3*sin(d*x+c)^2 +a^4*sin(d*x+c)-sin(d*x+c)*a^2*b^2)+a^3*(a^2-b^2)/b^6*ln(a+b*sin(d*x+c)))
Time = 0.10 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.85 \[ \int \frac {\cos ^3(c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {15 \, a b^{4} \cos \left (d x + c\right )^{4} - 30 \, a^{3} b^{2} \cos \left (d x + c\right )^{2} + 60 \, {\left (a^{5} - a^{3} b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) - 4 \, {\left (3 \, b^{5} \cos \left (d x + c\right )^{4} + 15 \, a^{4} b - 10 \, a^{2} b^{3} - 2 \, b^{5} - {\left (5 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{60 \, b^{6} d} \] Input:
integrate(cos(d*x+c)^3*sin(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="fricas" )
Output:
1/60*(15*a*b^4*cos(d*x + c)^4 - 30*a^3*b^2*cos(d*x + c)^2 + 60*(a^5 - a^3* b^2)*log(b*sin(d*x + c) + a) - 4*(3*b^5*cos(d*x + c)^4 + 15*a^4*b - 10*a^2 *b^3 - 2*b^5 - (5*a^2*b^3 + b^5)*cos(d*x + c)^2)*sin(d*x + c))/(b^6*d)
Timed out. \[ \int \frac {\cos ^3(c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**3*sin(d*x+c)**3/(a+b*sin(d*x+c)),x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.88 \[ \int \frac {\cos ^3(c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {12 \, b^{4} \sin \left (d x + c\right )^{5} - 15 \, a b^{3} \sin \left (d x + c\right )^{4} + 20 \, {\left (a^{2} b^{2} - b^{4}\right )} \sin \left (d x + c\right )^{3} - 30 \, {\left (a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )^{2} + 60 \, {\left (a^{4} - a^{2} b^{2}\right )} \sin \left (d x + c\right )}{b^{5}} - \frac {60 \, {\left (a^{5} - a^{3} b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{6}}}{60 \, d} \] Input:
integrate(cos(d*x+c)^3*sin(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="maxima" )
Output:
-1/60*((12*b^4*sin(d*x + c)^5 - 15*a*b^3*sin(d*x + c)^4 + 20*(a^2*b^2 - b^ 4)*sin(d*x + c)^3 - 30*(a^3*b - a*b^3)*sin(d*x + c)^2 + 60*(a^4 - a^2*b^2) *sin(d*x + c))/b^5 - 60*(a^5 - a^3*b^2)*log(b*sin(d*x + c) + a)/b^6)/d
Time = 0.17 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.17 \[ \int \frac {\cos ^3(c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {{\left (a^{5} - a^{3} b^{2}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{6} d} - \frac {12 \, b^{4} d^{4} \sin \left (d x + c\right )^{5} - 15 \, a b^{3} d^{4} \sin \left (d x + c\right )^{4} + 20 \, a^{2} b^{2} d^{4} \sin \left (d x + c\right )^{3} - 20 \, b^{4} d^{4} \sin \left (d x + c\right )^{3} - 30 \, a^{3} b d^{4} \sin \left (d x + c\right )^{2} + 30 \, a b^{3} d^{4} \sin \left (d x + c\right )^{2} + 60 \, a^{4} d^{4} \sin \left (d x + c\right ) - 60 \, a^{2} b^{2} d^{4} \sin \left (d x + c\right )}{60 \, b^{5} d^{5}} \] Input:
integrate(cos(d*x+c)^3*sin(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="giac")
Output:
(a^5 - a^3*b^2)*log(abs(b*sin(d*x + c) + a))/(b^6*d) - 1/60*(12*b^4*d^4*si n(d*x + c)^5 - 15*a*b^3*d^4*sin(d*x + c)^4 + 20*a^2*b^2*d^4*sin(d*x + c)^3 - 20*b^4*d^4*sin(d*x + c)^3 - 30*a^3*b*d^4*sin(d*x + c)^2 + 30*a*b^3*d^4* sin(d*x + c)^2 + 60*a^4*d^4*sin(d*x + c) - 60*a^2*b^2*d^4*sin(d*x + c))/(b ^5*d^5)
Time = 0.05 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.89 \[ \int \frac {\cos ^3(c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {{\sin \left (c+d\,x\right )}^3\,\left (\frac {1}{3\,b}-\frac {a^2}{3\,b^3}\right )-\frac {{\sin \left (c+d\,x\right )}^5}{5\,b}+\frac {a\,{\sin \left (c+d\,x\right )}^4}{4\,b^2}+\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (a^5-a^3\,b^2\right )}{b^6}-\frac {a\,{\sin \left (c+d\,x\right )}^2\,\left (\frac {1}{b}-\frac {a^2}{b^3}\right )}{2\,b}+\frac {a^2\,\sin \left (c+d\,x\right )\,\left (\frac {1}{b}-\frac {a^2}{b^3}\right )}{b^2}}{d} \] Input:
int((cos(c + d*x)^3*sin(c + d*x)^3)/(a + b*sin(c + d*x)),x)
Output:
(sin(c + d*x)^3*(1/(3*b) - a^2/(3*b^3)) - sin(c + d*x)^5/(5*b) + (a*sin(c + d*x)^4)/(4*b^2) + (log(a + b*sin(c + d*x))*(a^5 - a^3*b^2))/b^6 - (a*sin (c + d*x)^2*(1/b - a^2/b^3))/(2*b) + (a^2*sin(c + d*x)*(1/b - a^2/b^3))/b^ 2)/d
Time = 0.19 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.64 \[ \int \frac {\cos ^3(c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) a^{5}+60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) a^{3} b^{2}+60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) a^{5}-60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) a^{3} b^{2}-12 \sin \left (d x +c \right )^{5} b^{5}+15 \sin \left (d x +c \right )^{4} a \,b^{4}-20 \sin \left (d x +c \right )^{3} a^{2} b^{3}+20 \sin \left (d x +c \right )^{3} b^{5}+30 \sin \left (d x +c \right )^{2} a^{3} b^{2}-30 \sin \left (d x +c \right )^{2} a \,b^{4}-60 \sin \left (d x +c \right ) a^{4} b +60 \sin \left (d x +c \right ) a^{2} b^{3}-24 a^{3} b^{2}+24 a \,b^{4}}{60 b^{6} d} \] Input:
int(cos(d*x+c)^3*sin(d*x+c)^3/(a+b*sin(d*x+c)),x)
Output:
( - 60*log(tan((c + d*x)/2)**2 + 1)*a**5 + 60*log(tan((c + d*x)/2)**2 + 1) *a**3*b**2 + 60*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*a**5 - 60*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*a**3*b**2 - 12 *sin(c + d*x)**5*b**5 + 15*sin(c + d*x)**4*a*b**4 - 20*sin(c + d*x)**3*a** 2*b**3 + 20*sin(c + d*x)**3*b**5 + 30*sin(c + d*x)**2*a**3*b**2 - 30*sin(c + d*x)**2*a*b**4 - 60*sin(c + d*x)*a**4*b + 60*sin(c + d*x)*a**2*b**3 - 2 4*a**3*b**2 + 24*a*b**4)/(60*b**6*d)