Integrand size = 21, antiderivative size = 102 \[ \int \frac {\sin ^5(c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {4 \cos (c+d x)}{a^3 d}+\frac {2 \cos ^2(c+d x)}{a^3 d}-\frac {4 \cos ^3(c+d x)}{3 a^3 d}+\frac {3 \cos ^4(c+d x)}{4 a^3 d}-\frac {\cos ^5(c+d x)}{5 a^3 d}+\frac {4 \log (1+\cos (c+d x))}{a^3 d} \] Output:
-4*cos(d*x+c)/a^3/d+2*cos(d*x+c)^2/a^3/d-4/3*cos(d*x+c)^3/a^3/d+3/4*cos(d* x+c)^4/a^3/d-1/5*cos(d*x+c)^5/a^3/d+4*ln(1+cos(d*x+c))/a^3/d
Time = 0.34 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.72 \[ \int \frac {\sin ^5(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {3857-4920 \cos (c+d x)+1320 \cos (2 (c+d x))-380 \cos (3 (c+d x))+90 \cos (4 (c+d x))-12 \cos (5 (c+d x))+7680 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{960 a^3 d} \] Input:
Integrate[Sin[c + d*x]^5/(a + a*Sec[c + d*x])^3,x]
Output:
(3857 - 4920*Cos[c + d*x] + 1320*Cos[2*(c + d*x)] - 380*Cos[3*(c + d*x)] + 90*Cos[4*(c + d*x)] - 12*Cos[5*(c + d*x)] + 7680*Log[Cos[(c + d*x)/2]])/( 960*a^3*d)
Time = 0.43 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.92, 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, 3315, 27, 99, 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.
\(\displaystyle \int \frac {\sin ^5(c+d x)}{(a \sec (c+d x)+a)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos \left (c+d x-\frac {\pi }{2}\right )^5}{\left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^3}dx\) |
\(\Big \downarrow \) 4360 |
\(\displaystyle \int -\frac {\sin ^5(c+d x) \cos ^3(c+d x)}{(a (-\cos (c+d x))-a)^3}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int -\frac {\cos ^3(c+d x) \sin ^5(c+d x)}{(\cos (c+d x) a+a)^3}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \int \frac {\sin ^5(c+d x) \cos ^3(c+d x)}{(a \cos (c+d x)+a)^3}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\sin \left (c+d x+\frac {\pi }{2}\right )^3 \cos \left (c+d x+\frac {\pi }{2}\right )^5}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\cos \left (\frac {1}{2} (2 c+\pi )+d x\right )^5 \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )^3}{\left (\sin \left (\frac {1}{2} (2 c+\pi )+d x\right ) a+a\right )^3}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle -\frac {\int \frac {\cos ^3(c+d x) (a-a \cos (c+d x))^2}{\cos (c+d x) a+a}d(a \cos (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {a^3 \cos ^3(c+d x) (a-a \cos (c+d x))^2}{\cos (c+d x) a+a}d(a \cos (c+d x))}{a^8 d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle -\frac {\int \left (-\frac {4 a^5}{\cos (c+d x) a+a}+\cos ^4(c+d x) a^4-3 \cos ^3(c+d x) a^4+4 \cos ^2(c+d x) a^4-4 \cos (c+d x) a^4+4 a^4\right )d(a \cos (c+d x))}{a^8 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\frac {1}{5} a^5 \cos ^5(c+d x)-\frac {3}{4} a^5 \cos ^4(c+d x)+\frac {4}{3} a^5 \cos ^3(c+d x)-2 a^5 \cos ^2(c+d x)+4 a^5 \cos (c+d x)-4 a^5 \log (a \cos (c+d x)+a)}{a^8 d}\) |
Input:
Int[Sin[c + d*x]^5/(a + a*Sec[c + d*x])^3,x]
Output:
-((4*a^5*Cos[c + d*x] - 2*a^5*Cos[c + d*x]^2 + (4*a^5*Cos[c + d*x]^3)/3 - (3*a^5*Cos[c + d*x]^4)/4 + (a^5*Cos[c + d*x]^5)/5 - 4*a^5*Log[a + a*Cos[c + d*x]])/(a^8*d))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
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 + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && Intege rQ[(p - 1)/2] && EqQ[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 = 0.48 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.67
method | result | size |
derivativedivides | \(\frac {-\frac {\cos \left (d x +c \right )^{5}}{5}+\frac {3 \cos \left (d x +c \right )^{4}}{4}-\frac {4 \cos \left (d x +c \right )^{3}}{3}+2 \cos \left (d x +c \right )^{2}-4 \cos \left (d x +c \right )+4 \ln \left (1+\cos \left (d x +c \right )\right )}{d \,a^{3}}\) | \(68\) |
default | \(\frac {-\frac {\cos \left (d x +c \right )^{5}}{5}+\frac {3 \cos \left (d x +c \right )^{4}}{4}-\frac {4 \cos \left (d x +c \right )^{3}}{3}+2 \cos \left (d x +c \right )^{2}-4 \cos \left (d x +c \right )+4 \ln \left (1+\cos \left (d x +c \right )\right )}{d \,a^{3}}\) | \(68\) |
parallelrisch | \(\frac {-1920 \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )-3361-6 \cos \left (5 d x +5 c \right )+45 \cos \left (4 d x +4 c \right )-190 \cos \left (3 d x +3 c \right )+660 \cos \left (2 d x +2 c \right )-2460 \cos \left (d x +c \right )}{480 d \,a^{3}}\) | \(77\) |
norman | \(\frac {-\frac {32 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d a}-\frac {166}{15 a d}-\frac {78 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d a}-\frac {154 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3 d a}-\frac {278 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{3 d a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5} a^{2}}-\frac {4 \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{d \,a^{3}}\) | \(128\) |
risch | \(-\frac {4 i x}{a^{3}}-\frac {41 \,{\mathrm e}^{i \left (d x +c \right )}}{16 d \,a^{3}}-\frac {41 \,{\mathrm e}^{-i \left (d x +c \right )}}{16 d \,a^{3}}-\frac {8 i c}{d \,a^{3}}+\frac {8 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \,a^{3}}-\frac {\cos \left (5 d x +5 c \right )}{80 a^{3} d}+\frac {3 \cos \left (4 d x +4 c \right )}{32 a^{3} d}-\frac {19 \cos \left (3 d x +3 c \right )}{48 a^{3} d}+\frac {11 \cos \left (2 d x +2 c \right )}{8 a^{3} d}\) | \(141\) |
Input:
int(sin(d*x+c)^5/(a+a*sec(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/d/a^3*(-1/5*cos(d*x+c)^5+3/4*cos(d*x+c)^4-4/3*cos(d*x+c)^3+2*cos(d*x+c)^ 2-4*cos(d*x+c)+4*ln(1+cos(d*x+c)))
Time = 0.13 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.69 \[ \int \frac {\sin ^5(c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {12 \, \cos \left (d x + c\right )^{5} - 45 \, \cos \left (d x + c\right )^{4} + 80 \, \cos \left (d x + c\right )^{3} - 120 \, \cos \left (d x + c\right )^{2} + 240 \, \cos \left (d x + c\right ) - 240 \, \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{60 \, a^{3} d} \] Input:
integrate(sin(d*x+c)^5/(a+a*sec(d*x+c))^3,x, algorithm="fricas")
Output:
-1/60*(12*cos(d*x + c)^5 - 45*cos(d*x + c)^4 + 80*cos(d*x + c)^3 - 120*cos (d*x + c)^2 + 240*cos(d*x + c) - 240*log(1/2*cos(d*x + c) + 1/2))/(a^3*d)
Timed out. \[ \int \frac {\sin ^5(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\text {Timed out} \] Input:
integrate(sin(d*x+c)**5/(a+a*sec(d*x+c))**3,x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.72 \[ \int \frac {\sin ^5(c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {\frac {12 \, \cos \left (d x + c\right )^{5} - 45 \, \cos \left (d x + c\right )^{4} + 80 \, \cos \left (d x + c\right )^{3} - 120 \, \cos \left (d x + c\right )^{2} + 240 \, \cos \left (d x + c\right )}{a^{3}} - \frac {240 \, \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{3}}}{60 \, d} \] Input:
integrate(sin(d*x+c)^5/(a+a*sec(d*x+c))^3,x, algorithm="maxima")
Output:
-1/60*((12*cos(d*x + c)^5 - 45*cos(d*x + c)^4 + 80*cos(d*x + c)^3 - 120*co s(d*x + c)^2 + 240*cos(d*x + c))/a^3 - 240*log(cos(d*x + c) + 1)/a^3)/d
Time = 0.13 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.04 \[ \int \frac {\sin ^5(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {4 \, \log \left ({\left | \cos \left (d x + c\right ) + 1 \right |}\right )}{a^{3} d} - \frac {12 \, a^{12} d^{4} \cos \left (d x + c\right )^{5} - 45 \, a^{12} d^{4} \cos \left (d x + c\right )^{4} + 80 \, a^{12} d^{4} \cos \left (d x + c\right )^{3} - 120 \, a^{12} d^{4} \cos \left (d x + c\right )^{2} + 240 \, a^{12} d^{4} \cos \left (d x + c\right )}{60 \, a^{15} d^{5}} \] Input:
integrate(sin(d*x+c)^5/(a+a*sec(d*x+c))^3,x, algorithm="giac")
Output:
4*log(abs(cos(d*x + c) + 1))/(a^3*d) - 1/60*(12*a^12*d^4*cos(d*x + c)^5 - 45*a^12*d^4*cos(d*x + c)^4 + 80*a^12*d^4*cos(d*x + c)^3 - 120*a^12*d^4*cos (d*x + c)^2 + 240*a^12*d^4*cos(d*x + c))/(a^15*d^5)
Time = 11.24 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.80 \[ \int \frac {\sin ^5(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\frac {4\,\ln \left (\cos \left (c+d\,x\right )+1\right )}{a^3}-\frac {4\,\cos \left (c+d\,x\right )}{a^3}+\frac {2\,{\cos \left (c+d\,x\right )}^2}{a^3}-\frac {4\,{\cos \left (c+d\,x\right )}^3}{3\,a^3}+\frac {3\,{\cos \left (c+d\,x\right )}^4}{4\,a^3}-\frac {{\cos \left (c+d\,x\right )}^5}{5\,a^3}}{d} \] Input:
int(sin(c + d*x)^5/(a + a/cos(c + d*x))^3,x)
Output:
((4*log(cos(c + d*x) + 1))/a^3 - (4*cos(c + d*x))/a^3 + (2*cos(c + d*x)^2) /a^3 - (4*cos(c + d*x)^3)/(3*a^3) + (3*cos(c + d*x)^4)/(4*a^3) - cos(c + d *x)^5/(5*a^3))/d
Time = 0.17 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.84 \[ \int \frac {\sin ^5(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {-12 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}+104 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}-332 \cos \left (d x +c \right )-240 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )+45 \sin \left (d x +c \right )^{4}-210 \sin \left (d x +c \right )^{2}+332}{60 a^{3} d} \] Input:
int(sin(d*x+c)^5/(a+a*sec(d*x+c))^3,x)
Output:
( - 12*cos(c + d*x)*sin(c + d*x)**4 + 104*cos(c + d*x)*sin(c + d*x)**2 - 3 32*cos(c + d*x) - 240*log(tan((c + d*x)/2)**2 + 1) + 45*sin(c + d*x)**4 - 210*sin(c + d*x)**2 + 332)/(60*a**3*d)