Integrand size = 21, antiderivative size = 118 \[ \int \frac {\cos ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{b^5 d}-\frac {a \left (a^2-2 b^2\right ) \sin (c+d x)}{b^4 d}+\frac {\left (a^2-2 b^2\right ) \sin ^2(c+d x)}{2 b^3 d}-\frac {a \sin ^3(c+d x)}{3 b^2 d}+\frac {\sin ^4(c+d x)}{4 b d} \]
(a^2-b^2)^2*ln(a+b*sin(d*x+c))/b^5/d-a*(a^2-2*b^2)*sin(d*x+c)/b^4/d+1/2*(a ^2-2*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
Time = 0.14 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.87 \[ \int \frac {\cos ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {3 b^4 \cos ^4(c+d x)+12 \left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))-12 a b \left (a^2-2 b^2\right ) \sin (c+d x)+6 b^2 \left (a^2-b^2\right ) \sin ^2(c+d x)-4 a b^3 \sin ^3(c+d x)}{12 b^5 d} \]
(3*b^4*Cos[c + d*x]^4 + 12*(a^2 - b^2)^2*Log[a + b*Sin[c + d*x]] - 12*a*b* (a^2 - 2*b^2)*Sin[c + d*x] + 6*b^2*(a^2 - b^2)*Sin[c + d*x]^2 - 4*a*b^3*Si n[c + d*x]^3)/(12*b^5*d)
Time = 0.29 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.89, 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.
\(\displaystyle \int \frac {\cos ^5(c+d x)}{a+b \sin (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^5}{a+b \sin (c+d x)}dx\) |
\(\Big \downarrow \) 3147 |
\(\displaystyle \frac {\int \frac {\left (b^2-b^2 \sin ^2(c+d x)\right )^2}{a+b \sin (c+d x)}d(b \sin (c+d x))}{b^5 d}\) |
\(\Big \downarrow \) 476 |
\(\displaystyle \frac {\int \left (-\left (\left (1-\frac {2 b^2}{a^2}\right ) a^3\right )-b^2 \sin ^2(c+d x) a+b^3 \sin ^3(c+d x)+b \left (a^2-2 b^2\right ) \sin (c+d x)+\frac {\left (a^2-b^2\right )^2}{a+b \sin (c+d x)}\right )d(b \sin (c+d x))}{b^5 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{2} b^2 \left (a^2-2 b^2\right ) \sin ^2(c+d x)-a b \left (a^2-2 b^2\right ) \sin (c+d x)+\left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))-\frac {1}{3} a b^3 \sin ^3(c+d x)+\frac {1}{4} b^4 \sin ^4(c+d x)}{b^5 d}\) |
((a^2 - b^2)^2*Log[a + b*Sin[c + d*x]] - a*b*(a^2 - 2*b^2)*Sin[c + d*x] + (b^2*(a^2 - 2*b^2)*Sin[c + d*x]^2)/2 - (a*b^3*Sin[c + d*x]^3)/3 + (b^4*Sin [c + d*x]^4)/4)/(b^5*d)
3.5.25.3.1 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 = 0.87 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.90
method | result | size |
derivativedivides | \(\frac {-\frac {-\frac {\left (\sin ^{4}\left (d x +c \right )\right ) b^{3}}{4}+\frac {a \left (\sin ^{3}\left (d x +c \right )\right ) b^{2}}{3}-\frac {\left (a^{2}-2 b^{2}\right ) \left (\sin ^{2}\left (d x +c \right )\right ) b}{2}+a \left (a^{2}-2 b^{2}\right ) \sin \left (d x +c \right )}{b^{4}}+\frac {\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{5}}}{d}\) | \(106\) |
default | \(\frac {-\frac {-\frac {\left (\sin ^{4}\left (d x +c \right )\right ) b^{3}}{4}+\frac {a \left (\sin ^{3}\left (d x +c \right )\right ) b^{2}}{3}-\frac {\left (a^{2}-2 b^{2}\right ) \left (\sin ^{2}\left (d x +c \right )\right ) b}{2}+a \left (a^{2}-2 b^{2}\right ) \sin \left (d x +c \right )}{b^{4}}+\frac {\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{5}}}{d}\) | \(106\) |
parallelrisch | \(\frac {96 \left (a -b \right )^{2} \left (a +b \right )^{2} \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )-96 \left (a -b \right )^{2} \left (a +b \right )^{2} \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-24 a^{2} b^{2}+36 b^{4}\right ) \cos \left (2 d x +2 c \right )+3 \cos \left (4 d x +4 c \right ) b^{4}+8 a \sin \left (3 d x +3 c \right ) b^{3}+\left (-96 a^{3} b +168 a \,b^{3}\right ) \sin \left (d x +c \right )+24 a^{2} b^{2}-39 b^{4}}{96 b^{5} d}\) | \(163\) |
risch | \(-\frac {i x}{b}-\frac {2 i c}{b d}+\frac {i a^{3} {\mathrm e}^{i \left (d x +c \right )}}{2 b^{4} d}-\frac {{\mathrm e}^{2 i \left (d x +c \right )} a^{2}}{8 b^{3} d}+\frac {3 \,{\mathrm e}^{2 i \left (d x +c \right )}}{16 b d}+\frac {2 i x \,a^{2}}{b^{3}}-\frac {i x \,a^{4}}{b^{5}}+\frac {7 i a \,{\mathrm e}^{-i \left (d x +c \right )}}{8 b^{2} d}-\frac {2 i a^{4} c}{b^{5} d}-\frac {{\mathrm e}^{-2 i \left (d x +c \right )} a^{2}}{8 b^{3} d}+\frac {3 \,{\mathrm e}^{-2 i \left (d x +c \right )}}{16 b d}-\frac {7 i a \,{\mathrm e}^{i \left (d x +c \right )}}{8 b^{2} d}-\frac {i a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{2 b^{4} d}+\frac {4 i a^{2} c}{b^{3} d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right ) a^{4}}{b^{5} d}-\frac {2 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right ) a^{2}}{b^{3} d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right )}{b d}+\frac {\cos \left (4 d x +4 c \right )}{32 b d}+\frac {a \sin \left (3 d x +3 c \right )}{12 b^{2} d}\) | \(366\) |
norman | \(\frac {\frac {\left (2 a^{2}-4 b^{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3} d}+\frac {\left (2 a^{2}-4 b^{2}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3} d}+\frac {2 \left (3 a^{2}-4 b^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3} d}+\frac {2 \left (3 a^{2}-4 b^{2}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3} d}-\frac {2 a \left (a^{2}-2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{b^{4} d}-\frac {2 a \left (a^{2}-2 b^{2}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{4} d}-\frac {8 a \left (3 a^{2}-5 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 b^{4} d}-\frac {8 a \left (3 a^{2}-5 b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 b^{4} d}-\frac {4 a \left (9 a^{2}-14 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 b^{4} d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \ln \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{b^{5} d}-\frac {\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{5} d}\) | \(373\) |
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^2-2*b^2)* sin(d*x+c)^2*b+a*(a^2-2*b^2)*sin(d*x+c))+(a^4-2*a^2*b^2+b^4)/b^5*ln(a+b*si n(d*x+c)))
Time = 0.29 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.91 \[ \int \frac {\cos ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {3 \, b^{4} \cos \left (d x + c\right )^{4} - 6 \, {\left (a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{2} + 12 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + 4 \, {\left (a b^{3} \cos \left (d x + c\right )^{2} - 3 \, a^{3} b + 5 \, a b^{3}\right )} \sin \left (d x + c\right )}{12 \, b^{5} d} \]
1/12*(3*b^4*cos(d*x + c)^4 - 6*(a^2*b^2 - b^4)*cos(d*x + c)^2 + 12*(a^4 - 2*a^2*b^2 + b^4)*log(b*sin(d*x + c) + a) + 4*(a*b^3*cos(d*x + c)^2 - 3*a^3 *b + 5*a*b^3)*sin(d*x + c))/(b^5*d)
Timed out. \[ \int \frac {\cos ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \]
Time = 0.18 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.92 \[ \int \frac {\cos ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {3 \, b^{3} \sin \left (d x + c\right )^{4} - 4 \, a b^{2} \sin \left (d x + c\right )^{3} + 6 \, {\left (a^{2} b - 2 \, b^{3}\right )} \sin \left (d x + c\right )^{2} - 12 \, {\left (a^{3} - 2 \, a b^{2}\right )} \sin \left (d x + c\right )}{b^{4}} + \frac {12 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{5}}}{12 \, d} \]
1/12*((3*b^3*sin(d*x + c)^4 - 4*a*b^2*sin(d*x + c)^3 + 6*(a^2*b - 2*b^3)*s in(d*x + c)^2 - 12*(a^3 - 2*a*b^2)*sin(d*x + c))/b^4 + 12*(a^4 - 2*a^2*b^2 + b^4)*log(b*sin(d*x + c) + a)/b^5)/d
Time = 0.31 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.02 \[ \int \frac {\cos ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {3 \, b^{3} \sin \left (d x + c\right )^{4} - 4 \, a b^{2} \sin \left (d x + c\right )^{3} + 6 \, a^{2} b \sin \left (d x + c\right )^{2} - 12 \, b^{3} \sin \left (d x + c\right )^{2} - 12 \, a^{3} \sin \left (d x + c\right ) + 24 \, a b^{2} \sin \left (d x + c\right )}{b^{4}} + \frac {12 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{5}}}{12 \, d} \]
1/12*((3*b^3*sin(d*x + c)^4 - 4*a*b^2*sin(d*x + c)^3 + 6*a^2*b*sin(d*x + c )^2 - 12*b^3*sin(d*x + c)^2 - 12*a^3*sin(d*x + c) + 24*a*b^2*sin(d*x + c)) /b^4 + 12*(a^4 - 2*a^2*b^2 + b^4)*log(abs(b*sin(d*x + c) + a))/b^5)/d
Time = 4.44 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.92 \[ \int \frac {\cos ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {{\sin \left (c+d\,x\right )}^4}{4\,b}-{\sin \left (c+d\,x\right )}^2\,\left (\frac {1}{b}-\frac {a^2}{2\,b^3}\right )+\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}{b^5}-\frac {a\,{\sin \left (c+d\,x\right )}^3}{3\,b^2}+\frac {a\,\sin \left (c+d\,x\right )\,\left (\frac {2}{b}-\frac {a^2}{b^3}\right )}{b}}{d} \]