Integrand size = 21, antiderivative size = 194 \[ \int \frac {\sin ^5(c+d x)}{(a+b \sec (c+d x))^2} \, dx=-\frac {\left (a^4-6 a^2 b^2+5 b^4\right ) \cos (c+d x)}{a^6 d}-\frac {2 b \left (a^2-b^2\right ) \cos ^2(c+d x)}{a^5 d}+\frac {\left (2 a^2-3 b^2\right ) \cos ^3(c+d x)}{3 a^4 d}+\frac {b \cos ^4(c+d x)}{2 a^3 d}-\frac {\cos ^5(c+d x)}{5 a^2 d}+\frac {b^2 \left (a^2-b^2\right )^2}{a^7 d (b+a \cos (c+d x))}+\frac {2 b \left (a^4-4 a^2 b^2+3 b^4\right ) \log (b+a \cos (c+d x))}{a^7 d} \] Output:
-(a^4-6*a^2*b^2+5*b^4)*cos(d*x+c)/a^6/d-2*b*(a^2-b^2)*cos(d*x+c)^2/a^5/d+1 /3*(2*a^2-3*b^2)*cos(d*x+c)^3/a^4/d+1/2*b*cos(d*x+c)^4/a^3/d-1/5*cos(d*x+c )^5/a^2/d+b^2*(a^2-b^2)^2/a^7/d/(b+a*cos(d*x+c))+2*b*(a^4-4*a^2*b^2+3*b^4) *ln(b+a*cos(d*x+c))/a^7/d
Time = 0.94 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.44 \[ \int \frac {\sin ^5(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {-150 a^6+1740 a^4 b^2-2160 a^2 b^4+480 b^6-5 \left (25 a^6-168 a^4 b^2+144 a^2 b^4\right ) \cos (2 (c+d x))-115 a^5 b \cos (3 (c+d x))+120 a^3 b^3 \cos (3 (c+d x))+22 a^6 \cos (4 (c+d x))-30 a^4 b^2 \cos (4 (c+d x))+9 a^5 b \cos (5 (c+d x))-3 a^6 \cos (6 (c+d x))+960 a^4 b^2 \log (b+a \cos (c+d x))-3840 a^2 b^4 \log (b+a \cos (c+d x))+2880 b^6 \log (b+a \cos (c+d x))+120 a b \cos (c+d x) \left (-4 a^4+23 a^2 b^2-20 b^4+8 \left (a^4-4 a^2 b^2+3 b^4\right ) \log (b+a \cos (c+d x))\right )}{480 a^7 d (b+a \cos (c+d x))} \] Input:
Integrate[Sin[c + d*x]^5/(a + b*Sec[c + d*x])^2,x]
Output:
(-150*a^6 + 1740*a^4*b^2 - 2160*a^2*b^4 + 480*b^6 - 5*(25*a^6 - 168*a^4*b^ 2 + 144*a^2*b^4)*Cos[2*(c + d*x)] - 115*a^5*b*Cos[3*(c + d*x)] + 120*a^3*b ^3*Cos[3*(c + d*x)] + 22*a^6*Cos[4*(c + d*x)] - 30*a^4*b^2*Cos[4*(c + d*x) ] + 9*a^5*b*Cos[5*(c + d*x)] - 3*a^6*Cos[6*(c + d*x)] + 960*a^4*b^2*Log[b + a*Cos[c + d*x]] - 3840*a^2*b^4*Log[b + a*Cos[c + d*x]] + 2880*b^6*Log[b + a*Cos[c + d*x]] + 120*a*b*Cos[c + d*x]*(-4*a^4 + 23*a^2*b^2 - 20*b^4 + 8 *(a^4 - 4*a^2*b^2 + 3*b^4)*Log[b + a*Cos[c + d*x]]))/(480*a^7*d*(b + a*Cos [c + d*x]))
Time = 0.54 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.89, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 4360, 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.
\(\displaystyle \int \frac {\sin ^5(c+d x)}{(a+b \sec (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos \left (c+d x-\frac {\pi }{2}\right )^5}{\left (a-b \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 4360 |
\(\displaystyle \int \frac {\sin ^5(c+d x) \cos ^2(c+d x)}{(-a \cos (c+d x)-b)^2}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2 \cos \left (c+d x+\frac {\pi }{2}\right )^5}{\left (-a \sin \left (c+d x+\frac {\pi }{2}\right )-b\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\cos \left (\frac {1}{2} (2 c+\pi )+d x\right )^5 \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )^2}{\left (b+a \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^2}dx\) |
\(\Big \downarrow \) 3316 |
\(\displaystyle -\frac {\int \frac {\cos ^2(c+d x) \left (a^2-a^2 \cos ^2(c+d x)\right )^2}{(b+a \cos (c+d x))^2}d(a \cos (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {a^2 \cos ^2(c+d x) \left (a^2-a^2 \cos ^2(c+d x)\right )^2}{(b+a \cos (c+d x))^2}d(a \cos (c+d x))}{a^7 d}\) |
\(\Big \downarrow \) 522 |
\(\displaystyle -\frac {\int \left (\cos ^4(c+d x) a^4+\left (\frac {5 b^4-6 a^2 b^2}{a^4}+1\right ) a^4-2 b \cos ^3(c+d x) a^3-\left (2 a^2-3 b^2\right ) \cos ^2(c+d x) a^2-4 b \left (b^2-a^2\right ) \cos (c+d x) a-\frac {2 b \left (a^4-4 b^2 a^2+3 b^4\right )}{b+a \cos (c+d x)}+\frac {b^2 \left (a^2-b^2\right )^2}{(b+a \cos (c+d x))^2}\right )d(a \cos (c+d x))}{a^7 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\frac {1}{5} a^5 \cos ^5(c+d x)-\frac {1}{2} a^4 b \cos ^4(c+d x)+2 a^2 b \left (a^2-b^2\right ) \cos ^2(c+d x)-\frac {b^2 \left (a^2-b^2\right )^2}{a \cos (c+d x)+b}+a \left (a^4-6 a^2 b^2+5 b^4\right ) \cos (c+d x)-2 b \left (a^4-4 a^2 b^2+3 b^4\right ) \log (a \cos (c+d x)+b)-\frac {1}{3} a^3 \left (2 a^2-3 b^2\right ) \cos ^3(c+d x)}{a^7 d}\) |
Input:
Int[Sin[c + d*x]^5/(a + b*Sec[c + d*x])^2,x]
Output:
-((a*(a^4 - 6*a^2*b^2 + 5*b^4)*Cos[c + d*x] + 2*a^2*b*(a^2 - b^2)*Cos[c + d*x]^2 - (a^3*(2*a^2 - 3*b^2)*Cos[c + d*x]^3)/3 - (a^4*b*Cos[c + d*x]^4)/2 + (a^5*Cos[c + d*x]^5)/5 - (b^2*(a^2 - b^2)^2)/(b + a*Cos[c + d*x]) - 2*b *(a^4 - 4*a^2*b^2 + 3*b^4)*Log[b + a*Cos[c + d*x]])/(a^7*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]
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 = 1.22 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.02
method | result | size |
derivativedivides | \(\frac {-\frac {\frac {\cos \left (d x +c \right )^{5} a^{4}}{5}-\frac {b \cos \left (d x +c \right )^{4} a^{3}}{2}-\frac {2 a^{4} \cos \left (d x +c \right )^{3}}{3}+a^{2} b^{2} \cos \left (d x +c \right )^{3}+2 a^{3} b \cos \left (d x +c \right )^{2}-2 a \,b^{3} \cos \left (d x +c \right )^{2}+a^{4} \cos \left (d x +c \right )-6 \cos \left (d x +c \right ) a^{2} b^{2}+5 \cos \left (d x +c \right ) b^{4}}{a^{6}}+\frac {2 b \left (a^{4}-4 a^{2} b^{2}+3 b^{4}\right ) \ln \left (b +a \cos \left (d x +c \right )\right )}{a^{7}}+\frac {b^{2} \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}{a^{7} \left (b +a \cos \left (d x +c \right )\right )}}{d}\) | \(198\) |
default | \(\frac {-\frac {\frac {\cos \left (d x +c \right )^{5} a^{4}}{5}-\frac {b \cos \left (d x +c \right )^{4} a^{3}}{2}-\frac {2 a^{4} \cos \left (d x +c \right )^{3}}{3}+a^{2} b^{2} \cos \left (d x +c \right )^{3}+2 a^{3} b \cos \left (d x +c \right )^{2}-2 a \,b^{3} \cos \left (d x +c \right )^{2}+a^{4} \cos \left (d x +c \right )-6 \cos \left (d x +c \right ) a^{2} b^{2}+5 \cos \left (d x +c \right ) b^{4}}{a^{6}}+\frac {2 b \left (a^{4}-4 a^{2} b^{2}+3 b^{4}\right ) \ln \left (b +a \cos \left (d x +c \right )\right )}{a^{7}}+\frac {b^{2} \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}{a^{7} \left (b +a \cos \left (d x +c \right )\right )}}{d}\) | \(198\) |
parallelrisch | \(\frac {960 b^{2} \left (a -b \right ) \left (a +b \right ) \left (a^{2}-3 b^{2}\right ) \left (b +a \cos \left (d x +c \right )\right ) \ln \left (-2 a +\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (a -b \right )\right )-960 b^{2} \left (a -b \right ) \left (a +b \right ) \left (a^{2}-3 b^{2}\right ) \left (b +a \cos \left (d x +c \right )\right ) \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )+\left (-125 a^{6} b +840 a^{4} b^{3}-720 a^{2} b^{5}\right ) \cos \left (2 d x +2 c \right )+\left (-115 a^{5} b^{2}+120 a^{3} b^{4}\right ) \cos \left (3 d x +3 c \right )+\left (22 a^{6} b -30 a^{4} b^{3}\right ) \cos \left (4 d x +4 c \right )+9 \cos \left (5 d x +5 c \right ) a^{5} b^{2}-3 b \cos \left (6 d x +6 c \right ) a^{6}+\left (-256 a^{7}-918 a^{5} b^{2}+3912 a^{3} b^{4}-2880 a \,b^{6}\right ) \cos \left (d x +c \right )-406 a^{6} b +1302 a^{4} b^{3}-1008 a^{2} b^{5}}{480 a^{7} d b \left (b +a \cos \left (d x +c \right )\right )}\) | \(297\) |
risch | \(\frac {21 \,{\mathrm e}^{-i \left (d x +c \right )} b^{2}}{8 a^{4} d}-\frac {5 \,{\mathrm e}^{-i \left (d x +c \right )} b^{4}}{2 a^{6} d}-\frac {3 b \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 a^{3} d}+\frac {b^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{2 a^{5} d}+\frac {2 b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{a^{3} d}-\frac {8 b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{a^{5} d}+\frac {6 b^{5} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{a^{7} d}-\frac {5 \,{\mathrm e}^{-i \left (d x +c \right )}}{16 d \,a^{2}}-\frac {\cos \left (3 d x +3 c \right ) b^{2}}{4 d \,a^{4}}-\frac {2 i x b}{a^{3}}+\frac {2 b^{2} \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) {\mathrm e}^{i \left (d x +c \right )}}{a^{7} d \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}-\frac {3 b \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 a^{3} d}+\frac {b^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{2 a^{5} d}+\frac {21 \,{\mathrm e}^{i \left (d x +c \right )} b^{2}}{8 a^{4} d}-\frac {5 \,{\mathrm e}^{i \left (d x +c \right )} b^{4}}{2 a^{6} d}+\frac {8 i x \,b^{3}}{a^{5}}-\frac {6 i x \,b^{5}}{a^{7}}-\frac {5 \,{\mathrm e}^{i \left (d x +c \right )}}{16 d \,a^{2}}+\frac {5 \cos \left (3 d x +3 c \right )}{48 a^{2} d}-\frac {\cos \left (5 d x +5 c \right )}{80 a^{2} d}-\frac {4 i b c}{a^{3} d}+\frac {b \cos \left (4 d x +4 c \right )}{16 d \,a^{3}}+\frac {16 i b^{3} c}{a^{5} d}-\frac {12 i b^{5} c}{a^{7} d}\) | \(503\) |
norman | \(\frac {\frac {\left (32 a^{5}+32 a^{4} b +96 a^{3} b^{2}-360 a^{2} b^{3}-144 a \,b^{4}+360 b^{5}\right ) \left (a +b \right )}{60 a^{6} b d}-\frac {\left (32 a^{6}-128 a^{5} b +120 a^{3} b^{3}-408 a^{2} b^{4}-72 a \,b^{5}+360 b^{6}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{12 a^{6} b d}+\frac {\left (32 a^{6}+128 a^{5} b -120 a^{3} b^{3}-408 a^{2} b^{4}+72 a \,b^{5}+360 b^{6}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{12 a^{6} b d}-\frac {\left (32 a^{6}-64 a^{5} b +128 a^{4} b^{2}+264 a^{3} b^{3}-504 a^{2} b^{4}-216 a \,b^{5}+360 b^{6}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{60 a^{6} d b}-\frac {\left (64 a^{6}-160 a^{5} b +168 a^{4} b^{2}+432 a^{3} b^{3}-888 a^{2} b^{4}-288 a \,b^{5}+720 b^{6}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{30 a^{6} b d}+\frac {\left (64 a^{6}+160 a^{5} b +168 a^{4} b^{2}-432 a^{3} b^{3}-888 a^{2} b^{4}+288 a \,b^{5}+720 b^{6}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{30 a^{6} d b}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}-\frac {2 b \left (a^{4}-4 a^{2} b^{2}+3 b^{4}\right ) \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{a^{7} d}+\frac {2 b \left (a^{4}-4 a^{2} b^{2}+3 b^{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}{a^{7} d}\) | \(538\) |
Input:
int(sin(d*x+c)^5/(a+b*sec(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(-1/a^6*(1/5*cos(d*x+c)^5*a^4-1/2*b*cos(d*x+c)^4*a^3-2/3*a^4*cos(d*x+c )^3+a^2*b^2*cos(d*x+c)^3+2*a^3*b*cos(d*x+c)^2-2*a*b^3*cos(d*x+c)^2+a^4*cos (d*x+c)-6*cos(d*x+c)*a^2*b^2+5*cos(d*x+c)*b^4)+2/a^7*b*(a^4-4*a^2*b^2+3*b^ 4)*ln(b+a*cos(d*x+c))+b^2*(a^4-2*a^2*b^2+b^4)/a^7/(b+a*cos(d*x+c)))
Time = 0.12 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.24 \[ \int \frac {\sin ^5(c+d x)}{(a+b \sec (c+d x))^2} \, dx=-\frac {48 \, a^{6} \cos \left (d x + c\right )^{6} - 72 \, a^{5} b \cos \left (d x + c\right )^{5} - 435 \, a^{4} b^{2} + 720 \, a^{2} b^{4} - 240 \, b^{6} - 40 \, {\left (4 \, a^{6} - 3 \, a^{4} b^{2}\right )} \cos \left (d x + c\right )^{4} + 80 \, {\left (4 \, a^{5} b - 3 \, a^{3} b^{3}\right )} \cos \left (d x + c\right )^{3} + 240 \, {\left (a^{6} - 4 \, a^{4} b^{2} + 3 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (3 \, a^{5} b - 80 \, a^{3} b^{3} + 80 \, a b^{5}\right )} \cos \left (d x + c\right ) - 480 \, {\left (a^{4} b^{2} - 4 \, a^{2} b^{4} + 3 \, b^{6} + {\left (a^{5} b - 4 \, a^{3} b^{3} + 3 \, a b^{5}\right )} \cos \left (d x + c\right )\right )} \log \left (a \cos \left (d x + c\right ) + b\right )}{240 \, {\left (a^{8} d \cos \left (d x + c\right ) + a^{7} b d\right )}} \] Input:
integrate(sin(d*x+c)^5/(a+b*sec(d*x+c))^2,x, algorithm="fricas")
Output:
-1/240*(48*a^6*cos(d*x + c)^6 - 72*a^5*b*cos(d*x + c)^5 - 435*a^4*b^2 + 72 0*a^2*b^4 - 240*b^6 - 40*(4*a^6 - 3*a^4*b^2)*cos(d*x + c)^4 + 80*(4*a^5*b - 3*a^3*b^3)*cos(d*x + c)^3 + 240*(a^6 - 4*a^4*b^2 + 3*a^2*b^4)*cos(d*x + c)^2 + 15*(3*a^5*b - 80*a^3*b^3 + 80*a*b^5)*cos(d*x + c) - 480*(a^4*b^2 - 4*a^2*b^4 + 3*b^6 + (a^5*b - 4*a^3*b^3 + 3*a*b^5)*cos(d*x + c))*log(a*cos( d*x + c) + b))/(a^8*d*cos(d*x + c) + a^7*b*d)
Timed out. \[ \int \frac {\sin ^5(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\text {Timed out} \] Input:
integrate(sin(d*x+c)**5/(a+b*sec(d*x+c))**2,x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.95 \[ \int \frac {\sin ^5(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {\frac {30 \, {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )}}{a^{8} \cos \left (d x + c\right ) + a^{7} b} - \frac {6 \, a^{4} \cos \left (d x + c\right )^{5} - 15 \, a^{3} b \cos \left (d x + c\right )^{4} - 10 \, {\left (2 \, a^{4} - 3 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{3} + 60 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{2} + 30 \, {\left (a^{4} - 6 \, a^{2} b^{2} + 5 \, b^{4}\right )} \cos \left (d x + c\right )}{a^{6}} + \frac {60 \, {\left (a^{4} b - 4 \, a^{2} b^{3} + 3 \, b^{5}\right )} \log \left (a \cos \left (d x + c\right ) + b\right )}{a^{7}}}{30 \, d} \] Input:
integrate(sin(d*x+c)^5/(a+b*sec(d*x+c))^2,x, algorithm="maxima")
Output:
1/30*(30*(a^4*b^2 - 2*a^2*b^4 + b^6)/(a^8*cos(d*x + c) + a^7*b) - (6*a^4*c os(d*x + c)^5 - 15*a^3*b*cos(d*x + c)^4 - 10*(2*a^4 - 3*a^2*b^2)*cos(d*x + c)^3 + 60*(a^3*b - a*b^3)*cos(d*x + c)^2 + 30*(a^4 - 6*a^2*b^2 + 5*b^4)*c os(d*x + c))/a^6 + 60*(a^4*b - 4*a^2*b^3 + 3*b^5)*log(a*cos(d*x + c) + b)/ a^7)/d
Time = 0.14 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.23 \[ \int \frac {\sin ^5(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {2 \, {\left (a^{4} b - 4 \, a^{2} b^{3} + 3 \, b^{5}\right )} \log \left ({\left | a \cos \left (d x + c\right ) + b \right |}\right )}{a^{7} d} + \frac {a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}}{{\left (a \cos \left (d x + c\right ) + b\right )} a^{7} d} - \frac {6 \, a^{8} d^{4} \cos \left (d x + c\right )^{5} - 15 \, a^{7} b d^{4} \cos \left (d x + c\right )^{4} - 20 \, a^{8} d^{4} \cos \left (d x + c\right )^{3} + 30 \, a^{6} b^{2} d^{4} \cos \left (d x + c\right )^{3} + 60 \, a^{7} b d^{4} \cos \left (d x + c\right )^{2} - 60 \, a^{5} b^{3} d^{4} \cos \left (d x + c\right )^{2} + 30 \, a^{8} d^{4} \cos \left (d x + c\right ) - 180 \, a^{6} b^{2} d^{4} \cos \left (d x + c\right ) + 150 \, a^{4} b^{4} d^{4} \cos \left (d x + c\right )}{30 \, a^{10} d^{5}} \] Input:
integrate(sin(d*x+c)^5/(a+b*sec(d*x+c))^2,x, algorithm="giac")
Output:
2*(a^4*b - 4*a^2*b^3 + 3*b^5)*log(abs(a*cos(d*x + c) + b))/(a^7*d) + (a^4* b^2 - 2*a^2*b^4 + b^6)/((a*cos(d*x + c) + b)*a^7*d) - 1/30*(6*a^8*d^4*cos( d*x + c)^5 - 15*a^7*b*d^4*cos(d*x + c)^4 - 20*a^8*d^4*cos(d*x + c)^3 + 30* a^6*b^2*d^4*cos(d*x + c)^3 + 60*a^7*b*d^4*cos(d*x + c)^2 - 60*a^5*b^3*d^4* cos(d*x + c)^2 + 30*a^8*d^4*cos(d*x + c) - 180*a^6*b^2*d^4*cos(d*x + c) + 150*a^4*b^4*d^4*cos(d*x + c))/(a^10*d^5)
Time = 0.13 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.30 \[ \int \frac {\sin ^5(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {{\cos \left (c+d\,x\right )}^3\,\left (\frac {2}{3\,a^2}-\frac {b^2}{a^4}\right )}{d}-\frac {{\cos \left (c+d\,x\right )}^2\,\left (\frac {b^3}{a^5}+\frac {b\,\left (\frac {2}{a^2}-\frac {3\,b^2}{a^4}\right )}{a}\right )}{d}-\frac {\cos \left (c+d\,x\right )\,\left (\frac {1}{a^2}+\frac {b^2\,\left (\frac {2}{a^2}-\frac {3\,b^2}{a^4}\right )}{a^2}-\frac {2\,b\,\left (\frac {2\,b^3}{a^5}+\frac {2\,b\,\left (\frac {2}{a^2}-\frac {3\,b^2}{a^4}\right )}{a}\right )}{a}\right )}{d}-\frac {{\cos \left (c+d\,x\right )}^5}{5\,a^2\,d}+\frac {b\,{\cos \left (c+d\,x\right )}^4}{2\,a^3\,d}+\frac {\ln \left (b+a\,\cos \left (c+d\,x\right )\right )\,\left (2\,a^4\,b-8\,a^2\,b^3+6\,b^5\right )}{a^7\,d}+\frac {a^4\,b^2-2\,a^2\,b^4+b^6}{a\,d\,\left (\cos \left (c+d\,x\right )\,a^7+b\,a^6\right )} \] Input:
int(sin(c + d*x)^5/(a + b/cos(c + d*x))^2,x)
Output:
(cos(c + d*x)^3*(2/(3*a^2) - b^2/a^4))/d - (cos(c + d*x)^2*(b^3/a^5 + (b*( 2/a^2 - (3*b^2)/a^4))/a))/d - (cos(c + d*x)*(1/a^2 + (b^2*(2/a^2 - (3*b^2) /a^4))/a^2 - (2*b*((2*b^3)/a^5 + (2*b*(2/a^2 - (3*b^2)/a^4))/a))/a))/d - c os(c + d*x)^5/(5*a^2*d) + (b*cos(c + d*x)^4)/(2*a^3*d) + (log(b + a*cos(c + d*x))*(2*a^4*b + 6*b^5 - 8*a^2*b^3))/(a^7*d) + (b^6 - 2*a^2*b^4 + a^4*b^ 2)/(a*d*(a^7*cos(c + d*x) + a^6*b))
Time = 0.20 (sec) , antiderivative size = 700, normalized size of antiderivative = 3.61 \[ \int \frac {\sin ^5(c+d x)}{(a+b \sec (c+d x))^2} \, dx =\text {Too large to display} \] Input:
int(sin(d*x+c)^5/(a+b*sec(d*x+c))^2,x)
Output:
( - 60*cos(c + d*x)*log(tan((c + d*x)/2)**2 + 1)*a**5*b + 240*cos(c + d*x) *log(tan((c + d*x)/2)**2 + 1)*a**3*b**3 - 180*cos(c + d*x)*log(tan((c + d* x)/2)**2 + 1)*a*b**5 + 60*cos(c + d*x)*log(tan((c + d*x)/2)**2*a - tan((c + d*x)/2)**2*b - a - b)*a**5*b - 240*cos(c + d*x)*log(tan((c + d*x)/2)**2* a - tan((c + d*x)/2)**2*b - a - b)*a**3*b**3 + 180*cos(c + d*x)*log(tan((c + d*x)/2)**2*a - tan((c + d*x)/2)**2*b - a - b)*a*b**5 + 9*cos(c + d*x)*s in(c + d*x)**4*a**5*b + 22*cos(c + d*x)*sin(c + d*x)**2*a**5*b - 30*cos(c + d*x)*sin(c + d*x)**2*a**3*b**3 + 16*cos(c + d*x)*a**6 - 16*cos(c + d*x)* a**5*b - 180*cos(c + d*x)*a**4*b**2 + 180*cos(c + d*x)*a**3*b**3 + 180*cos (c + d*x)*a**2*b**4 - 180*cos(c + d*x)*a*b**5 - 60*log(tan((c + d*x)/2)**2 + 1)*a**4*b**2 + 240*log(tan((c + d*x)/2)**2 + 1)*a**2*b**4 - 180*log(tan ((c + d*x)/2)**2 + 1)*b**6 + 60*log(tan((c + d*x)/2)**2*a - tan((c + d*x)/ 2)**2*b - a - b)*a**4*b**2 - 240*log(tan((c + d*x)/2)**2*a - tan((c + d*x) /2)**2*b - a - b)*a**2*b**4 + 180*log(tan((c + d*x)/2)**2*a - tan((c + d*x )/2)**2*b - a - b)*b**6 + 6*sin(c + d*x)**6*a**6 + 2*sin(c + d*x)**4*a**6 - 15*sin(c + d*x)**4*a**4*b**2 + 8*sin(c + d*x)**2*a**6 - 90*sin(c + d*x)* *2*a**4*b**2 + 90*sin(c + d*x)**2*a**2*b**4 - 16*a**6 + 16*a**5*b + 180*a* *4*b**2 - 180*a**3*b**3 - 180*a**2*b**4 + 180*a*b**5)/(30*a**7*d*(cos(c + d*x)*a + b))