Integrand size = 21, antiderivative size = 55 \[ \int \frac {\sin ^5(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\cos ^3(c+d x)}{3 a d}-\frac {\cos ^5(c+d x)}{5 a d}+\frac {\sin ^4(c+d x)}{4 a d} \]
Time = 0.17 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.76 \[ \int \frac {\sin ^5(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {2 (13+21 \cos (c+d x)+6 \cos (2 (c+d x))) \sin ^6\left (\frac {1}{2} (c+d x)\right )}{15 a d} \]
Time = 0.49 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.93, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 4360, 25, 25, 3042, 25, 3314, 25, 3042, 3044, 15, 3045, 244, 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.
\(\displaystyle \int \frac {\sin ^5(c+d x)}{a \sec (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos \left (c+d x-\frac {\pi }{2}\right )^5}{a-a \csc \left (c+d x-\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 4360 |
\(\displaystyle \int -\frac {\sin ^5(c+d x) \cos (c+d x)}{a (-\cos (c+d x))-a}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int -\frac {\cos (c+d x) \sin ^5(c+d x)}{\cos (c+d x) a+a}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \int \frac {\sin ^5(c+d x) \cos (c+d x)}{a \cos (c+d x)+a}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \cos \left (c+d x+\frac {\pi }{2}\right )^5}{a \sin \left (c+d x+\frac {\pi }{2}\right )+a}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\cos \left (\frac {1}{2} (2 c+\pi )+d x\right )^5 \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )}{\sin \left (\frac {1}{2} (2 c+\pi )+d x\right ) a+a}dx\) |
\(\Big \downarrow \) 3314 |
\(\displaystyle \frac {\int -\cos ^2(c+d x) \sin ^3(c+d x)dx}{a}-\frac {\int -\cos (c+d x) \sin ^3(c+d x)dx}{a}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \cos (c+d x) \sin ^3(c+d x)dx}{a}-\frac {\int \cos ^2(c+d x) \sin ^3(c+d x)dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \cos (c+d x) \sin (c+d x)^3dx}{a}-\frac {\int \cos (c+d x)^2 \sin (c+d x)^3dx}{a}\) |
\(\Big \downarrow \) 3044 |
\(\displaystyle \frac {\int \sin ^3(c+d x)d\sin (c+d x)}{a d}-\frac {\int \cos (c+d x)^2 \sin (c+d x)^3dx}{a}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {\sin ^4(c+d x)}{4 a d}-\frac {\int \cos (c+d x)^2 \sin (c+d x)^3dx}{a}\) |
\(\Big \downarrow \) 3045 |
\(\displaystyle \frac {\int \cos ^2(c+d x) \left (1-\cos ^2(c+d x)\right )d\cos (c+d x)}{a d}+\frac {\sin ^4(c+d x)}{4 a d}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {\int \left (\cos ^2(c+d x)-\cos ^4(c+d x)\right )d\cos (c+d x)}{a d}+\frac {\sin ^4(c+d x)}{4 a d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sin ^4(c+d x)}{4 a d}+\frac {\frac {1}{3} \cos ^3(c+d x)-\frac {1}{5} \cos ^5(c+d x)}{a d}\) |
3.1.59.3.1 Defintions of rubi rules used
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_ Symbol] :> Simp[1/(a*f) Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a *Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] && !(I ntegerQ[(m - 1)/2] && LtQ[0, m, n])
Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_ Symbol] :> Simp[-(a*f)^(-1) Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] && !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])
Int[(cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/(( a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[1/a Int[Cos[e + f *x]^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Simp[1/(b*d) Int[Cos[e + f*x]^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] & & IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && IntegerQ[n] && (LtQ[0, n, (p + 1)/2] || (LeQ[p, -n] && LtQ[-n, 2*p - 3]) || (GtQ[n, 0] && LeQ[n, -p]))
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 = 0.53 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.89
method | result | size |
derivativedivides | \(\frac {-\frac {\cos \left (d x +c \right )^{5}}{5}+\frac {\cos \left (d x +c \right )^{4}}{4}+\frac {\cos \left (d x +c \right )^{3}}{3}-\frac {\cos \left (d x +c \right )^{2}}{2}}{d a}\) | \(49\) |
default | \(\frac {-\frac {\cos \left (d x +c \right )^{5}}{5}+\frac {\cos \left (d x +c \right )^{4}}{4}+\frac {\cos \left (d x +c \right )^{3}}{3}-\frac {\cos \left (d x +c \right )^{2}}{2}}{d a}\) | \(49\) |
parallelrisch | \(\frac {-60 \cos \left (2 d x +2 c \right )+109+15 \cos \left (4 d x +4 c \right )+60 \cos \left (d x +c \right )+10 \cos \left (3 d x +3 c \right )-6 \cos \left (5 d x +5 c \right )}{480 d a}\) | \(63\) |
norman | \(\frac {\frac {4}{15 a d}+\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d a}+\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3 d a}+\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{3 d a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5}}\) | \(83\) |
risch | \(\frac {\cos \left (d x +c \right )}{8 a d}-\frac {\cos \left (5 d x +5 c \right )}{80 a d}+\frac {\cos \left (4 d x +4 c \right )}{32 a d}+\frac {\cos \left (3 d x +3 c \right )}{48 a d}-\frac {\cos \left (2 d x +2 c \right )}{8 a d}\) | \(84\) |
Time = 0.27 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.89 \[ \int \frac {\sin ^5(c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {12 \, \cos \left (d x + c\right )^{5} - 15 \, \cos \left (d x + c\right )^{4} - 20 \, \cos \left (d x + c\right )^{3} + 30 \, \cos \left (d x + c\right )^{2}}{60 \, a d} \]
Timed out. \[ \int \frac {\sin ^5(c+d x)}{a+a \sec (c+d x)} \, dx=\text {Timed out} \]
Time = 0.20 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.89 \[ \int \frac {\sin ^5(c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {12 \, \cos \left (d x + c\right )^{5} - 15 \, \cos \left (d x + c\right )^{4} - 20 \, \cos \left (d x + c\right )^{3} + 30 \, \cos \left (d x + c\right )^{2}}{60 \, a d} \]
Time = 0.32 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.76 \[ \int \frac {\sin ^5(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {4 \, {\left (\frac {5 \, {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {10 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {30 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - 1\right )}}{15 \, a d {\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1\right )}^{5}} \]
4/15*(5*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 10*(cos(d*x + c) - 1)^2/(c os(d*x + c) + 1)^2 + 30*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 - 1)/(a* d*((cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 1)^5)
Time = 0.07 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.05 \[ \int \frac {\sin ^5(c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {\frac {{\cos \left (c+d\,x\right )}^2}{2\,a}-\frac {{\cos \left (c+d\,x\right )}^3}{3\,a}-\frac {{\cos \left (c+d\,x\right )}^4}{4\,a}+\frac {{\cos \left (c+d\,x\right )}^5}{5\,a}}{d} \]