Integrand size = 21, antiderivative size = 152 \[ \int \frac {\sin ^5(c+d x)}{a+b \sec (c+d x)} \, dx=-\frac {\left (a^2-b^2\right )^2 \cos (c+d x)}{a^5 d}-\frac {b \left (2 a^2-b^2\right ) \cos ^2(c+d x)}{2 a^4 d}+\frac {\left (2 a^2-b^2\right ) \cos ^3(c+d x)}{3 a^3 d}+\frac {b \cos ^4(c+d x)}{4 a^2 d}-\frac {\cos ^5(c+d x)}{5 a d}+\frac {b \left (a^2-b^2\right )^2 \log (b+a \cos (c+d x))}{a^6 d} \]
-(a^2-b^2)^2*cos(d*x+c)/a^5/d-1/2*b*(2*a^2-b^2)*cos(d*x+c)^2/a^4/d+1/3*(2* a^2-b^2)*cos(d*x+c)^3/a^3/d+1/4*b*cos(d*x+c)^4/a^2/d-1/5*cos(d*x+c)^5/a/d+ b*(a^2-b^2)^2*ln(b+a*cos(d*x+c))/a^6/d
Time = 0.31 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.13 \[ \int \frac {\sin ^5(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {-60 a \left (5 a^4-14 a^2 b^2+8 b^4\right ) \cos (c+d x)-60 \left (3 a^4 b-2 a^2 b^3\right ) \cos (2 (c+d x))+50 a^5 \cos (3 (c+d x))-40 a^3 b^2 \cos (3 (c+d x))+15 a^4 b \cos (4 (c+d x))-6 a^5 \cos (5 (c+d x))+480 a^4 b \log (b+a \cos (c+d x))-960 a^2 b^3 \log (b+a \cos (c+d x))+480 b^5 \log (b+a \cos (c+d x))}{480 a^6 d} \]
(-60*a*(5*a^4 - 14*a^2*b^2 + 8*b^4)*Cos[c + d*x] - 60*(3*a^4*b - 2*a^2*b^3 )*Cos[2*(c + d*x)] + 50*a^5*Cos[3*(c + d*x)] - 40*a^3*b^2*Cos[3*(c + d*x)] + 15*a^4*b*Cos[4*(c + d*x)] - 6*a^5*Cos[5*(c + d*x)] + 480*a^4*b*Log[b + a*Cos[c + d*x]] - 960*a^2*b^3*Log[b + a*Cos[c + d*x]] + 480*b^5*Log[b + a* Cos[c + d*x]])/(480*a^6*d)
Time = 0.45 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.90, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 4360, 25, 25, 3042, 25, 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.
| \(\displaystyle \int \frac {\sin ^5(c+d x)}{a+b \sec (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos \left (c+d x-\frac {\pi }{2}\right )^5}{a-b \csc \left (c+d x-\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 4360 |
| \(\displaystyle \int -\frac {\sin ^5(c+d x) \cos (c+d x)}{-a \cos (c+d x)-b}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int -\frac {\cos (c+d x) \sin ^5(c+d x)}{b+a \cos (c+d x)}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \int \frac {\sin ^5(c+d x) \cos (c+d x)}{a \cos (c+d x)+b}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \cos \left (c+d x+\frac {\pi }{2}\right )^5}{a \sin \left (c+d x+\frac {\pi }{2}\right )+b}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {\cos \left (\frac {1}{2} (2 c+\pi )+d x\right )^5 \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )}{b+a \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle -\frac {\int \frac {\cos (c+d x) \left (a^2-a^2 \cos ^2(c+d x)\right )^2}{b+a \cos (c+d x)}d(a \cos (c+d x))}{a^5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {a \cos (c+d x) \left (a^2-a^2 \cos ^2(c+d x)\right )^2}{b+a \cos (c+d x)}d(a \cos (c+d x))}{a^6 d}\) |
| \(\Big \downarrow \) 522 |
| \(\displaystyle -\frac {\int \left (a^4 \cos ^4(c+d x)-a^3 b \cos ^3(c+d x)-a^2 \left (2 a^2-b^2\right ) \cos ^2(c+d x)-a b \left (b^2-2 a^2\right ) \cos (c+d x)+\left (a^2-b^2\right )^2-\frac {b \left (b^2-a^2\right )^2}{b+a \cos (c+d x)}\right )d(a \cos (c+d x))}{a^6 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\frac {1}{5} a^5 \cos ^5(c+d x)-\frac {1}{4} a^4 b \cos ^4(c+d x)+\frac {1}{2} a^2 b \left (2 a^2-b^2\right ) \cos ^2(c+d x)+a \left (a^2-b^2\right )^2 \cos (c+d x)-b \left (a^2-b^2\right )^2 \log (a \cos (c+d x)+b)-\frac {1}{3} a^3 \left (2 a^2-b^2\right ) \cos ^3(c+d x)}{a^6 d}\) |
-((a*(a^2 - b^2)^2*Cos[c + d*x] + (a^2*b*(2*a^2 - b^2)*Cos[c + d*x]^2)/2 - (a^3*(2*a^2 - b^2)*Cos[c + d*x]^3)/3 - (a^4*b*Cos[c + d*x]^4)/4 + (a^5*Co s[c + d*x]^5)/5 - b*(a^2 - b^2)^2*Log[b + a*Cos[c + d*x]])/(a^6*d))
3.2.97.3.1 Defintions of rubi rules used
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]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Cos[e + f*x])^p*((b + a*Sin[e + f*x])^m/Si n[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]
Time = 0.83 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.05
| method | result | size |
| derivativedivides | \(\frac {-\frac {\frac {\cos \left (d x +c \right )^{5} a^{4}}{5}-\frac {b \cos \left (d x +c \right )^{4} a^{3}}{4}-\frac {2 \cos \left (d x +c \right )^{3} a^{4}}{3}+\frac {\cos \left (d x +c \right )^{3} a^{2} b^{2}}{3}+\cos \left (d x +c \right )^{2} a^{3} b -\frac {\cos \left (d x +c \right )^{2} a \,b^{3}}{2}+\cos \left (d x +c \right ) a^{4}-2 \cos \left (d x +c \right ) a^{2} b^{2}+\cos \left (d x +c \right ) b^{4}}{a^{5}}+\frac {b \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \ln \left (b +a \cos \left (d x +c \right )\right )}{a^{6}}}{d}\) | \(160\) |
| default | \(\frac {-\frac {\frac {\cos \left (d x +c \right )^{5} a^{4}}{5}-\frac {b \cos \left (d x +c \right )^{4} a^{3}}{4}-\frac {2 \cos \left (d x +c \right )^{3} a^{4}}{3}+\frac {\cos \left (d x +c \right )^{3} a^{2} b^{2}}{3}+\cos \left (d x +c \right )^{2} a^{3} b -\frac {\cos \left (d x +c \right )^{2} a \,b^{3}}{2}+\cos \left (d x +c \right ) a^{4}-2 \cos \left (d x +c \right ) a^{2} b^{2}+\cos \left (d x +c \right ) b^{4}}{a^{5}}+\frac {b \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \ln \left (b +a \cos \left (d x +c \right )\right )}{a^{6}}}{d}\) | \(160\) |
| parallelrisch | \(\frac {480 b \left (a -b \right )^{2} \left (a +b \right )^{2} \ln \left (-2 a +\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (a -b \right )\right )-480 b \left (a -b \right )^{2} \left (a +b \right )^{2} \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )-6 \left (\left (30 a^{3} b -20 a \,b^{3}\right ) \cos \left (2 d x +2 c \right )+\left (-\frac {25}{3} a^{4}+\frac {20}{3} a^{2} b^{2}\right ) \cos \left (3 d x +3 c \right )-\frac {5 b \cos \left (4 d x +4 c \right ) a^{3}}{2}+\cos \left (5 d x +5 c \right ) a^{4}+\left (50 a^{4}-140 a^{2} b^{2}+80 b^{4}\right ) \cos \left (d x +c \right )+\frac {128 a^{4}}{3}-\frac {55 a^{3} b}{2}-\frac {400 a^{2} b^{2}}{3}+20 a \,b^{3}+80 b^{4}\right ) a}{480 d \,a^{6}}\) | \(209\) |
| norman | \(\frac {\frac {\left (2 a^{3} b +2 a^{2} b^{2}-2 a \,b^{3}-2 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d \,a^{5}}+\frac {-16 a^{4}+50 a^{2} b^{2}-30 b^{4}}{15 d \,a^{5}}+\frac {2 \left (5 a^{3} b +6 a^{2} b^{2}-3 a \,b^{3}-4 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d \,a^{5}}+\frac {\left (-16 a^{4}+6 a^{3} b +44 a^{2} b^{2}-6 a \,b^{3}-24 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3 d \,a^{5}}+\frac {2 \left (-16 a^{4}+15 a^{3} b +32 a^{2} b^{2}-9 a \,b^{3}-18 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{3 d \,a^{5}}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5}}+\frac {\left (a +b \right ) b \left (a^{3}-a^{2} b -a \,b^{2}+b^{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}{d \,a^{6}}-\frac {\left (a +b \right ) b \left (a^{3}-a^{2} b -a \,b^{2}+b^{3}\right ) \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{d \,a^{6}}\) | \(345\) |
| risch | \(-\frac {\cos \left (5 d x +5 c \right )}{80 d a}-\frac {5 \,{\mathrm e}^{-i \left (d x +c \right )}}{16 a d}-\frac {2 i b^{5} c}{a^{6} d}+\frac {4 i b^{3} c}{a^{4} d}-\frac {i x \,b^{5}}{a^{6}}+\frac {5 \cos \left (3 d x +3 c \right )}{48 a d}-\frac {2 i b c}{a^{2} d}-\frac {i x b}{a^{2}}-\frac {5 \,{\mathrm e}^{i \left (d x +c \right )}}{16 d a}+\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{a^{2} d}-\frac {2 b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{a^{4} d}+\frac {b^{5} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{a^{6} d}-\frac {3 b \,{\mathrm e}^{-2 i \left (d x +c \right )}}{16 a^{2} d}+\frac {b^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 a^{4} d}+\frac {7 \,{\mathrm e}^{i \left (d x +c \right )} b^{2}}{8 a^{3} d}-\frac {{\mathrm e}^{i \left (d x +c \right )} b^{4}}{2 a^{5} d}+\frac {7 \,{\mathrm e}^{-i \left (d x +c \right )} b^{2}}{8 a^{3} d}-\frac {{\mathrm e}^{-i \left (d x +c \right )} b^{4}}{2 a^{5} d}+\frac {b \cos \left (4 d x +4 c \right )}{32 d \,a^{2}}-\frac {\cos \left (3 d x +3 c \right ) b^{2}}{12 a^{3} d}+\frac {2 i x \,b^{3}}{a^{4}}-\frac {3 b \,{\mathrm e}^{2 i \left (d x +c \right )}}{16 a^{2} d}+\frac {b^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{8 a^{4} d}\) | \(439\) |
1/d*(-1/a^5*(1/5*cos(d*x+c)^5*a^4-1/4*b*cos(d*x+c)^4*a^3-2/3*cos(d*x+c)^3* a^4+1/3*cos(d*x+c)^3*a^2*b^2+cos(d*x+c)^2*a^3*b-1/2*cos(d*x+c)^2*a*b^3+cos (d*x+c)*a^4-2*cos(d*x+c)*a^2*b^2+cos(d*x+c)*b^4)+b*(a^4-2*a^2*b^2+b^4)/a^6 *ln(b+a*cos(d*x+c)))
Time = 0.28 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.92 \[ \int \frac {\sin ^5(c+d x)}{a+b \sec (c+d x)} \, dx=-\frac {12 \, a^{5} \cos \left (d x + c\right )^{5} - 15 \, a^{4} b \cos \left (d x + c\right )^{4} - 20 \, {\left (2 \, a^{5} - a^{3} b^{2}\right )} \cos \left (d x + c\right )^{3} + 30 \, {\left (2 \, a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2} + 60 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right ) - 60 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \log \left (a \cos \left (d x + c\right ) + b\right )}{60 \, a^{6} d} \]
-1/60*(12*a^5*cos(d*x + c)^5 - 15*a^4*b*cos(d*x + c)^4 - 20*(2*a^5 - a^3*b ^2)*cos(d*x + c)^3 + 30*(2*a^4*b - a^2*b^3)*cos(d*x + c)^2 + 60*(a^5 - 2*a ^3*b^2 + a*b^4)*cos(d*x + c) - 60*(a^4*b - 2*a^2*b^3 + b^5)*log(a*cos(d*x + c) + b))/(a^6*d)
Timed out. \[ \int \frac {\sin ^5(c+d x)}{a+b \sec (c+d x)} \, dx=\text {Timed out} \]
Time = 0.20 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.93 \[ \int \frac {\sin ^5(c+d x)}{a+b \sec (c+d x)} \, dx=-\frac {\frac {12 \, a^{4} \cos \left (d x + c\right )^{5} - 15 \, a^{3} b \cos \left (d x + c\right )^{4} - 20 \, {\left (2 \, a^{4} - a^{2} b^{2}\right )} \cos \left (d x + c\right )^{3} + 30 \, {\left (2 \, a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{2} + 60 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )}{a^{5}} - \frac {60 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \log \left (a \cos \left (d x + c\right ) + b\right )}{a^{6}}}{60 \, d} \]
-1/60*((12*a^4*cos(d*x + c)^5 - 15*a^3*b*cos(d*x + c)^4 - 20*(2*a^4 - a^2* b^2)*cos(d*x + c)^3 + 30*(2*a^3*b - a*b^3)*cos(d*x + c)^2 + 60*(a^4 - 2*a^ 2*b^2 + b^4)*cos(d*x + c))/a^5 - 60*(a^4*b - 2*a^2*b^3 + b^5)*log(a*cos(d* x + c) + b)/a^6)/d
Leaf count of result is larger than twice the leaf count of optimal. 867 vs. \(2 (144) = 288\).
Time = 0.33 (sec) , antiderivative size = 867, normalized size of antiderivative = 5.70 \[ \int \frac {\sin ^5(c+d x)}{a+b \sec (c+d x)} \, dx=\text {Too large to display} \]
1/60*(60*(a^5*b - a^4*b^2 - 2*a^3*b^3 + 2*a^2*b^4 + a*b^5 - b^6)*log(abs(a + b + a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - b*(cos(d*x + c) - 1)/(cos (d*x + c) + 1)))/(a^7 - a^6*b) - 60*(a^4*b - 2*a^2*b^3 + b^5)*log(abs(-(co s(d*x + c) - 1)/(cos(d*x + c) + 1) + 1))/a^6 + (64*a^5 - 137*a^4*b - 200*a ^3*b^2 + 274*a^2*b^3 + 120*a*b^4 - 137*b^5 - 320*a^5*(cos(d*x + c) - 1)/(c os(d*x + c) + 1) + 805*a^4*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 880*a ^3*b^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 1490*a^2*b^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 480*a*b^4*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 685*b^5*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 640*a^5*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 1970*a^4*b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 1280*a^3*b^2*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 3100*a^2 *b^3*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 720*a*b^4*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 1370*b^5*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 1970*a^4*b*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 720*a^3*b^2 *(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 - 3100*a^2*b^3*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 - 480*a*b^4*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 1370*b^5*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 - 805*a^4*b*(co s(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 - 120*a^3*b^2*(cos(d*x + c) - 1)^4/ (cos(d*x + c) + 1)^4 + 1490*a^2*b^3*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1 )^4 + 120*a*b^4*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 - 685*b^5*(co...
Time = 13.42 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.99 \[ \int \frac {\sin ^5(c+d x)}{a+b \sec (c+d x)} \, dx=-\frac {\cos \left (c+d\,x\right )\,\left (\frac {1}{a}-\frac {b^2\,\left (\frac {2}{a}-\frac {b^2}{a^3}\right )}{a^2}\right )-{\cos \left (c+d\,x\right )}^3\,\left (\frac {2}{3\,a}-\frac {b^2}{3\,a^3}\right )+\frac {{\cos \left (c+d\,x\right )}^5}{5\,a}-\frac {b\,{\cos \left (c+d\,x\right )}^4}{4\,a^2}-\frac {\ln \left (b+a\,\cos \left (c+d\,x\right )\right )\,\left (a^4\,b-2\,a^2\,b^3+b^5\right )}{a^6}+\frac {b\,{\cos \left (c+d\,x\right )}^2\,\left (\frac {2}{a}-\frac {b^2}{a^3}\right )}{2\,a}}{d} \]