Integrand size = 21, antiderivative size = 127 \[ \int \frac {\cos ^5(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {2 \left (3 a^2-b^2\right ) \log (a+b \sin (c+d x))}{b^5 d}-\frac {3 a \sin (c+d x)}{b^4 d}+\frac {\sin ^2(c+d x)}{2 b^3 d}-\frac {\left (a^2-b^2\right )^2}{2 b^5 d (a+b \sin (c+d x))^2}+\frac {4 a \left (a^2-b^2\right )}{b^5 d (a+b \sin (c+d x))} \]
2*(3*a^2-b^2)*ln(a+b*sin(d*x+c))/b^5/d-3*a*sin(d*x+c)/b^4/d+1/2*sin(d*x+c) ^2/b^3/d-1/2*(a^2-b^2)^2/b^5/d/(a+b*sin(d*x+c))^2+4*a*(a^2-b^2)/b^5/d/(a+b *sin(d*x+c))
Time = 0.78 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.13 \[ \int \frac {\cos ^5(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {\frac {b^4 \cos ^4(c+d x)}{2 (a+b \sin (c+d x))^2}+2 a \left (2 a \log (a+b \sin (c+d x))-b \sin (c+d x)+\frac {(a-b) (a+b)}{a+b \sin (c+d x)}\right )+2 \left (-a^2+b^2\right ) \left (-\log (a+b \sin (c+d x))-\frac {3 a^2+b^2+4 a b \sin (c+d x)}{2 (a+b \sin (c+d x))^2}\right )}{b^5 d} \]
((b^4*Cos[c + d*x]^4)/(2*(a + b*Sin[c + d*x])^2) + 2*a*(2*a*Log[a + b*Sin[ c + d*x]] - b*Sin[c + d*x] + ((a - b)*(a + b))/(a + b*Sin[c + d*x])) + 2*( -a^2 + b^2)*(-Log[a + b*Sin[c + d*x]] - (3*a^2 + b^2 + 4*a*b*Sin[c + d*x]) /(2*(a + b*Sin[c + d*x])^2)))/(b^5*d)
Time = 0.30 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.85, 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.
| \(\displaystyle \int \frac {\cos ^5(c+d x)}{(a+b \sin (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^5}{(a+b \sin (c+d x))^3}dx\) |
| \(\Big \downarrow \) 3147 |
| \(\displaystyle \frac {\int \frac {\left (b^2-b^2 \sin ^2(c+d x)\right )^2}{(a+b \sin (c+d x))^3}d(b \sin (c+d x))}{b^5 d}\) |
| \(\Big \downarrow \) 476 |
| \(\displaystyle \frac {\int \left (\frac {\left (a^2-b^2\right )^2}{(a+b \sin (c+d x))^3}-3 a+b \sin (c+d x)+\frac {2 \left (3 a^2-b^2\right )}{a+b \sin (c+d x)}-\frac {4 \left (a^3-a b^2\right )}{(a+b \sin (c+d x))^2}\right )d(b \sin (c+d x))}{b^5 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {\left (a^2-b^2\right )^2}{2 (a+b \sin (c+d x))^2}+\frac {4 a \left (a^2-b^2\right )}{a+b \sin (c+d x)}+2 \left (3 a^2-b^2\right ) \log (a+b \sin (c+d x))-3 a b \sin (c+d x)+\frac {1}{2} b^2 \sin ^2(c+d x)}{b^5 d}\) |
(2*(3*a^2 - b^2)*Log[a + b*Sin[c + d*x]] - 3*a*b*Sin[c + d*x] + (b^2*Sin[c + d*x]^2)/2 - (a^2 - b^2)^2/(2*(a + b*Sin[c + d*x])^2) + (4*a*(a^2 - b^2) )/(a + b*Sin[c + d*x]))/(b^5*d)
3.5.50.3.1 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 = 1.61 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.92
| method | result | size |
| derivativedivides | \(\frac {-\frac {-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) b}{2}+3 a \sin \left (d x +c \right )}{b^{4}}+\frac {\left (6 a^{2}-2 b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{5}}-\frac {a^{4}-2 a^{2} b^{2}+b^{4}}{2 b^{5} \left (a +b \sin \left (d x +c \right )\right )^{2}}+\frac {4 a \left (a^{2}-b^{2}\right )}{b^{5} \left (a +b \sin \left (d x +c \right )\right )}}{d}\) | \(117\) |
| default | \(\frac {-\frac {-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) b}{2}+3 a \sin \left (d x +c \right )}{b^{4}}+\frac {\left (6 a^{2}-2 b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{5}}-\frac {a^{4}-2 a^{2} b^{2}+b^{4}}{2 b^{5} \left (a +b \sin \left (d x +c \right )\right )^{2}}+\frac {4 a \left (a^{2}-b^{2}\right )}{b^{5} \left (a +b \sin \left (d x +c \right )\right )}}{d}\) | \(117\) |
| parallelrisch | \(\frac {96 \left (\frac {b^{2}}{2}-\frac {b^{2} \cos \left (2 d x +2 c \right )}{2}+2 \sin \left (d x +c \right ) a b +a^{2}\right ) \left (a^{2}-\frac {b^{2}}{3}\right ) \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )-96 \left (\frac {b^{2}}{2}-\frac {b^{2} \cos \left (2 d x +2 c \right )}{2}+2 \sin \left (d x +c \right ) a b +a^{2}\right ) \left (a^{2}-\frac {b^{2}}{3}\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (48 a^{2} b^{2}-20 b^{4}\right ) \cos \left (2 d x +2 c \right )-24 a \,b^{3} \sin \left (d x +c \right )+8 a \sin \left (3 d x +3 c \right ) b^{3}+\cos \left (4 d x +4 c \right ) b^{4}+48 a^{4}-64 a^{2} b^{2}+11 b^{4}}{8 b^{5} d \left (4 \sin \left (d x +c \right ) a b -b^{2} \cos \left (2 d x +2 c \right )+2 a^{2}+b^{2}\right )}\) | \(254\) |
| risch | \(-\frac {6 i x \,a^{2}}{b^{5}}+\frac {2 i x}{b^{3}}-\frac {{\mathrm e}^{2 i \left (d x +c \right )}}{8 b^{3} d}+\frac {3 i a \,{\mathrm e}^{i \left (d x +c \right )}}{2 b^{4} d}-\frac {3 i a \,{\mathrm e}^{-i \left (d x +c \right )}}{2 b^{4} d}-\frac {{\mathrm e}^{-2 i \left (d x +c \right )}}{8 b^{3} d}-\frac {12 i a^{2} c}{b^{5} d}+\frac {4 i c}{b^{3} d}+\frac {-8 i a^{3} b \,{\mathrm e}^{3 i \left (d x +c \right )}+8 i a \,b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+8 i a^{3} b \,{\mathrm e}^{i \left (d x +c \right )}-8 i a \,b^{3} {\mathrm e}^{i \left (d x +c \right )}+14 a^{4} {\mathrm e}^{2 i \left (d x +c \right )}-12 a^{2} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-2 b^{4} {\mathrm e}^{2 i \left (d x +c \right )}}{\left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+i b \right )^{2} d \,b^{5}}+\frac {6 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right ) a^{2}}{b^{5} d}-\frac {2 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right )}{b^{3} d}\) | \(339\) |
| norman | \(\frac {-\frac {\left (180 a^{4}-68 a^{2} b^{2}-10 b^{4}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2} b^{3} d}-\frac {\left (180 a^{4}-68 a^{2} b^{2}-10 b^{4}\right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2} b^{3} d}-\frac {\left (36 a^{4}-12 a^{2} b^{2}-2 b^{4}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2} b^{3} d}-\frac {\left (36 a^{4}-12 a^{2} b^{2}-2 b^{4}\right ) \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2} b^{3} d}-\frac {2 \left (90 a^{4}+2 a^{2} b^{2}-15 b^{4}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a \,b^{4} d}-\frac {2 \left (90 a^{4}+2 a^{2} b^{2}-15 b^{4}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a \,b^{4} d}-\frac {8 \left (30 a^{4}+2 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 {4 \left (18 a^{4}-2 a^{2} b^{2}-3 b^{4}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a \,b^{4} d}-\frac {4 \left (18 a^{4}-2 a^{2} b^{2}-3 b^{4}\right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a \,b^{4} d}-\frac {\left (360 a^{4}-144 a^{2} b^{2}-20 b^{4}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3} d \,a^{2}}-\frac {\left (360 a^{4}-144 a^{2} b^{2}-20 b^{4}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3} d \,a^{2}}-\frac {2 \left (6 a^{4}-2 a^{2} b^{2}-b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a \,b^{4} d}-\frac {2 \left (6 a^{4}-2 a^{2} b^{2}-b^{4}\right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a \,b^{4} 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 )}^{2}}-\frac {2 \left (3 a^{2}-b^{2}\right ) \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{5} d}+\frac {2 \left (3 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}\) | \(659\) |
1/d*(-1/b^4*(-1/2*sin(d*x+c)^2*b+3*a*sin(d*x+c))+(6*a^2-2*b^2)/b^5*ln(a+b* sin(d*x+c))-1/2/b^5*(a^4-2*a^2*b^2+b^4)/(a+b*sin(d*x+c))^2+4*a/b^5*(a^2-b^ 2)/(a+b*sin(d*x+c)))
Time = 0.29 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.67 \[ \int \frac {\cos ^5(c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {2 \, b^{4} \cos \left (d x + c\right )^{4} + 14 \, a^{4} - 35 \, a^{2} b^{2} - b^{4} + {\left (22 \, a^{2} b^{2} - 3 \, b^{4}\right )} \cos \left (d x + c\right )^{2} + 8 \, {\left (3 \, a^{4} + 2 \, a^{2} b^{2} - b^{4} - {\left (3 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + 2 \, {\left (4 \, a b^{3} \cos \left (d x + c\right )^{2} + 2 \, a^{3} b - 13 \, a b^{3}\right )} \sin \left (d x + c\right )}{4 \, {\left (b^{7} d \cos \left (d x + c\right )^{2} - 2 \, a b^{6} d \sin \left (d x + c\right ) - {\left (a^{2} b^{5} + b^{7}\right )} d\right )}} \]
-1/4*(2*b^4*cos(d*x + c)^4 + 14*a^4 - 35*a^2*b^2 - b^4 + (22*a^2*b^2 - 3*b ^4)*cos(d*x + c)^2 + 8*(3*a^4 + 2*a^2*b^2 - b^4 - (3*a^2*b^2 - b^4)*cos(d* x + c)^2 + 2*(3*a^3*b - a*b^3)*sin(d*x + c))*log(b*sin(d*x + c) + a) + 2*( 4*a*b^3*cos(d*x + c)^2 + 2*a^3*b - 13*a*b^3)*sin(d*x + c))/(b^7*d*cos(d*x + c)^2 - 2*a*b^6*d*sin(d*x + c) - (a^2*b^5 + b^7)*d)
Timed out. \[ \int \frac {\cos ^5(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Timed out} \]
Time = 0.19 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.03 \[ \int \frac {\cos ^5(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {\frac {7 \, a^{4} - 6 \, a^{2} b^{2} - b^{4} + 8 \, {\left (a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )}{b^{7} \sin \left (d x + c\right )^{2} + 2 \, a b^{6} \sin \left (d x + c\right ) + a^{2} b^{5}} + \frac {b \sin \left (d x + c\right )^{2} - 6 \, a \sin \left (d x + c\right )}{b^{4}} + \frac {4 \, {\left (3 \, a^{2} - b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{5}}}{2 \, d} \]
1/2*((7*a^4 - 6*a^2*b^2 - b^4 + 8*(a^3*b - a*b^3)*sin(d*x + c))/(b^7*sin(d *x + c)^2 + 2*a*b^6*sin(d*x + c) + a^2*b^5) + (b*sin(d*x + c)^2 - 6*a*sin( d*x + c))/b^4 + 4*(3*a^2 - b^2)*log(b*sin(d*x + c) + a)/b^5)/d
Time = 0.34 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.12 \[ \int \frac {\cos ^5(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {\frac {4 \, {\left (3 \, a^{2} - b^{2}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{5}} + \frac {b^{3} \sin \left (d x + c\right )^{2} - 6 \, a b^{2} \sin \left (d x + c\right )}{b^{6}} - \frac {18 \, a^{2} b^{2} \sin \left (d x + c\right )^{2} - 6 \, b^{4} \sin \left (d x + c\right )^{2} + 28 \, a^{3} b \sin \left (d x + c\right ) - 4 \, a b^{3} \sin \left (d x + c\right ) + 11 \, a^{4} + b^{4}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{2} b^{5}}}{2 \, d} \]
1/2*(4*(3*a^2 - b^2)*log(abs(b*sin(d*x + c) + a))/b^5 + (b^3*sin(d*x + c)^ 2 - 6*a*b^2*sin(d*x + c))/b^6 - (18*a^2*b^2*sin(d*x + c)^2 - 6*b^4*sin(d*x + c)^2 + 28*a^3*b*sin(d*x + c) - 4*a*b^3*sin(d*x + c) + 11*a^4 + b^4)/((b *sin(d*x + c) + a)^2*b^5))/d
Time = 4.74 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.12 \[ \int \frac {\cos ^5(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {{\sin \left (c+d\,x\right )}^2}{2\,b^3\,d}-\frac {\frac {-7\,a^4+6\,a^2\,b^2+b^4}{2\,b}+\sin \left (c+d\,x\right )\,\left (4\,a\,b^2-4\,a^3\right )}{d\,\left (a^2\,b^4+2\,a\,b^5\,\sin \left (c+d\,x\right )+b^6\,{\sin \left (c+d\,x\right )}^2\right )}-\frac {3\,a\,\sin \left (c+d\,x\right )}{b^4\,d}+\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (6\,a^2-2\,b^2\right )}{b^5\,d} \]