Integrand size = 21, antiderivative size = 120 \[ \int \frac {\cos ^5(c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {4 a \left (a^2-b^2\right ) \log (a+b \sin (c+d x))}{b^5 d}+\frac {\left (3 a^2-2 b^2\right ) \sin (c+d x)}{b^4 d}-\frac {a \sin ^2(c+d x)}{b^3 d}+\frac {\sin ^3(c+d x)}{3 b^2 d}-\frac {\left (a^2-b^2\right )^2}{b^5 d (a+b \sin (c+d x))} \]
-4*a*(a^2-b^2)*ln(a+b*sin(d*x+c))/b^5/d+(3*a^2-2*b^2)*sin(d*x+c)/b^4/d-a*s in(d*x+c)^2/b^3/d+1/3*sin(d*x+c)^3/b^2/d-(a^2-b^2)^2/b^5/d/(a+b*sin(d*x+c) )
Time = 0.45 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.06 \[ \int \frac {\cos ^5(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\left (8 a^2 b-4 b^3\right ) \sin (c+d x)-2 a b^2 \sin ^2(c+d x)+\frac {b^4 \cos ^4(c+d x)-4 \left (a^2-b^2\right ) \left (a^2-b^2+3 a^2 \log (a+b \sin (c+d x))+3 a b \log (a+b \sin (c+d x)) \sin (c+d x)\right )}{a+b \sin (c+d x)}}{3 b^5 d} \]
((8*a^2*b - 4*b^3)*Sin[c + d*x] - 2*a*b^2*Sin[c + d*x]^2 + (b^4*Cos[c + d* x]^4 - 4*(a^2 - b^2)*(a^2 - b^2 + 3*a^2*Log[a + b*Sin[c + d*x]] + 3*a*b*Lo g[a + b*Sin[c + d*x]]*Sin[c + d*x]))/(a + b*Sin[c + d*x]))/(3*b^5*d)
Time = 0.30 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.87, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3147, 476, 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.
| \(\displaystyle \int \frac {\cos ^5(c+d x)}{(a+b \sin (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^5}{(a+b \sin (c+d x))^2}dx\) |
| \(\Big \downarrow \) 3147 |
| \(\displaystyle \frac {\int \frac {\left (b^2-b^2 \sin ^2(c+d x)\right )^2}{(a+b \sin (c+d x))^2}d(b \sin (c+d x))}{b^5 d}\) |
| \(\Big \downarrow \) 476 |
| \(\displaystyle \frac {\int \left (3 \left (1-\frac {2 b^2}{3 a^2}\right ) a^2-2 b \sin (c+d x) a+b^2 \sin ^2(c+d x)-\frac {4 \left (a^3-a b^2\right )}{a+b \sin (c+d x)}+\frac {\left (a^2-b^2\right )^2}{(a+b \sin (c+d x))^2}\right )d(b \sin (c+d x))}{b^5 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b \left (3 a^2-2 b^2\right ) \sin (c+d x)-\frac {\left (a^2-b^2\right )^2}{a+b \sin (c+d x)}-4 a \left (a^2-b^2\right ) \log (a+b \sin (c+d x))-a b^2 \sin ^2(c+d x)+\frac {1}{3} b^3 \sin ^3(c+d x)}{b^5 d}\) |
(-4*a*(a^2 - b^2)*Log[a + b*Sin[c + d*x]] + b*(3*a^2 - 2*b^2)*Sin[c + d*x] - a*b^2*Sin[c + d*x]^2 + (b^3*Sin[c + d*x]^3)/3 - (a^2 - b^2)^2/(a + b*Si n[c + d*x]))/(b^5*d)
3.5.38.3.1 Defintions of rubi rules used
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ ExpandIntegrand[(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[p, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^m*(b^2 - x^2)^((p - 1) /2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]
Time = 1.21 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.97
| method | result | size |
| derivativedivides | \(\frac {\frac {\frac {\left (\sin ^{3}\left (d x +c \right )\right ) b^{2}}{3}-\left (\sin ^{2}\left (d x +c \right )\right ) a b +3 a^{2} \sin \left (d x +c \right )-2 \sin \left (d x +c \right ) b^{2}}{b^{4}}-\frac {4 a \left (a^{2}-b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{5}}-\frac {a^{4}-2 a^{2} b^{2}+b^{4}}{b^{5} \left (a +b \sin \left (d x +c \right )\right )}}{d}\) | \(116\) |
| default | \(\frac {\frac {\frac {\left (\sin ^{3}\left (d x +c \right )\right ) b^{2}}{3}-\left (\sin ^{2}\left (d x +c \right )\right ) a b +3 a^{2} \sin \left (d x +c \right )-2 \sin \left (d x +c \right ) b^{2}}{b^{4}}-\frac {4 a \left (a^{2}-b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{5}}-\frac {a^{4}-2 a^{2} b^{2}+b^{4}}{b^{5} \left (a +b \sin \left (d x +c \right )\right )}}{d}\) | \(116\) |
| parallelrisch | \(\frac {-96 a \left (a -b \right ) \left (a +b \right ) \left (a +b \sin \left (d x +c \right )\right ) \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )+96 a \left (a -b \right ) \left (a +b \right ) \left (a +b \sin \left (d x +c \right )\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-24 a^{2} b^{2}+20 b^{4}\right ) \cos \left (2 d x +2 c \right )-12 a \,b^{3} \sin \left (d x +c \right )+4 a \sin \left (3 d x +3 c \right ) b^{3}+\cos \left (4 d x +4 c \right ) b^{4}-96 a^{4}+120 a^{2} b^{2}-45 b^{4}}{24 b^{5} d \left (a +b \sin \left (d x +c \right )\right )}\) | \(185\) |
| risch | \(\frac {4 i a^{3} x}{b^{5}}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} a^{2}}{2 b^{4} d}+\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{4 b^{3} d}-\frac {7 i {\mathrm e}^{-i \left (d x +c \right )}}{8 b^{2} d}-\frac {4 i a x}{b^{3}}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} a^{2}}{2 b^{4} d}+\frac {8 i a^{3} c}{b^{5} d}+\frac {a \,{\mathrm e}^{-2 i \left (d x +c \right )}}{4 b^{3} d}+\frac {7 i {\mathrm e}^{i \left (d x +c \right )}}{8 b^{2} d}-\frac {8 i a c}{b^{3} d}-\frac {2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) {\mathrm e}^{i \left (d x +c \right )}}{b^{5} d \left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+i b \right )}-\frac {4 a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right )}{b^{5} d}+\frac {4 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right )}{b^{3} d}-\frac {\sin \left (3 d x +3 c \right )}{12 b^{2} d}\) | \(315\) |
| norman | \(\frac {\frac {4 \left (36 a^{2}-28 b^{2}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 b^{3} d}+\frac {\left (96 a^{2}-80 b^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 b^{3} d}+\frac {\left (96 a^{2}-80 b^{2}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 b^{3} d}+\frac {2 \left (4 a^{2}-4 b^{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3} d}+\frac {2 \left (4 a^{2}-4 b^{2}\right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3} d}+\frac {4 \left (20 a^{4}-24 a^{2} b^{2}+5 b^{4}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a \,b^{4} d}+\frac {4 \left (20 a^{4}-24 a^{2} b^{2}+5 b^{4}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a \,b^{4} d}+\frac {2 \left (60 a^{4}-68 a^{2} b^{2}+15 b^{4}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a \,b^{4} d}+\frac {2 \left (60 a^{4}-68 a^{2} b^{2}+15 b^{4}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a \,b^{4} d}+\frac {2 \left (4 a^{4}-4 a^{2} b^{2}+b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{b^{4} a d}+\frac {2 \left (4 a^{4}-4 a^{2} b^{2}+b^{4}\right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{4} a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5} \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}+\frac {4 a \left (a^{2}-b^{2}\right ) \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{5} d}-\frac {4 a \left (a^{2}-b^{2}\right ) \ln \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{b^{5} d}\) | \(516\) |
1/d*(1/b^4*(1/3*sin(d*x+c)^3*b^2-sin(d*x+c)^2*a*b+3*a^2*sin(d*x+c)-2*sin(d *x+c)*b^2)-4*a/b^5*(a^2-b^2)*ln(a+b*sin(d*x+c))-1/b^5*(a^4-2*a^2*b^2+b^4)/ (a+b*sin(d*x+c)))
Time = 0.30 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.30 \[ \int \frac {\cos ^5(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {2 \, b^{4} \cos \left (d x + c\right )^{4} - 6 \, a^{4} + 27 \, a^{2} b^{2} - 16 \, b^{4} - 4 \, {\left (3 \, a^{2} b^{2} - 2 \, b^{4}\right )} \cos \left (d x + c\right )^{2} - 24 \, {\left (a^{4} - a^{2} b^{2} + {\left (a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + {\left (4 \, a b^{3} \cos \left (d x + c\right )^{2} + 18 \, a^{3} b - 13 \, a b^{3}\right )} \sin \left (d x + c\right )}{6 \, {\left (b^{6} d \sin \left (d x + c\right ) + a b^{5} d\right )}} \]
1/6*(2*b^4*cos(d*x + c)^4 - 6*a^4 + 27*a^2*b^2 - 16*b^4 - 4*(3*a^2*b^2 - 2 *b^4)*cos(d*x + c)^2 - 24*(a^4 - a^2*b^2 + (a^3*b - a*b^3)*sin(d*x + c))*l og(b*sin(d*x + c) + a) + (4*a*b^3*cos(d*x + c)^2 + 18*a^3*b - 13*a*b^3)*si n(d*x + c))/(b^6*d*sin(d*x + c) + a*b^5*d)
Timed out. \[ \int \frac {\cos ^5(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Timed out} \]
Time = 0.19 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.97 \[ \int \frac {\cos ^5(c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {\frac {3 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )}}{b^{6} \sin \left (d x + c\right ) + a b^{5}} - \frac {b^{2} \sin \left (d x + c\right )^{3} - 3 \, a b \sin \left (d x + c\right )^{2} + 3 \, {\left (3 \, a^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )}{b^{4}} + \frac {12 \, {\left (a^{3} - a b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{5}}}{3 \, d} \]
-1/3*(3*(a^4 - 2*a^2*b^2 + b^4)/(b^6*sin(d*x + c) + a*b^5) - (b^2*sin(d*x + c)^3 - 3*a*b*sin(d*x + c)^2 + 3*(3*a^2 - 2*b^2)*sin(d*x + c))/b^4 + 12*( a^3 - a*b^2)*log(b*sin(d*x + c) + a)/b^5)/d
Time = 0.33 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.25 \[ \int \frac {\cos ^5(c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {\frac {12 \, {\left (a^{3} - a b^{2}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{5}} - \frac {b^{4} \sin \left (d x + c\right )^{3} - 3 \, a b^{3} \sin \left (d x + c\right )^{2} + 9 \, a^{2} b^{2} \sin \left (d x + c\right ) - 6 \, b^{4} \sin \left (d x + c\right )}{b^{6}} - \frac {3 \, {\left (4 \, a^{3} b \sin \left (d x + c\right ) - 4 \, a b^{3} \sin \left (d x + c\right ) + 3 \, a^{4} - 2 \, a^{2} b^{2} - b^{4}\right )}}{{\left (b \sin \left (d x + c\right ) + a\right )} b^{5}}}{3 \, d} \]
-1/3*(12*(a^3 - a*b^2)*log(abs(b*sin(d*x + c) + a))/b^5 - (b^4*sin(d*x + c )^3 - 3*a*b^3*sin(d*x + c)^2 + 9*a^2*b^2*sin(d*x + c) - 6*b^4*sin(d*x + c) )/b^6 - 3*(4*a^3*b*sin(d*x + c) - 4*a*b^3*sin(d*x + c) + 3*a^4 - 2*a^2*b^2 - b^4)/((b*sin(d*x + c) + a)*b^5))/d
Time = 0.09 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.98 \[ \int \frac {\cos ^5(c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {\sin \left (c+d\,x\right )\,\left (\frac {2}{b^2}-\frac {3\,a^2}{b^4}\right )-\frac {{\sin \left (c+d\,x\right )}^3}{3\,b^2}+\frac {a\,{\sin \left (c+d\,x\right )}^2}{b^3}-\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (4\,a\,b^2-4\,a^3\right )}{b^5}+\frac {a^4-2\,a^2\,b^2+b^4}{b\,\left (\sin \left (c+d\,x\right )\,b^5+a\,b^4\right )}}{d} \]