Integrand size = 25, antiderivative size = 86 \[ \int \cos ^4(c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \log (\sin (c+d x))}{d}+\frac {a \sin (c+d x)}{d}-\frac {a \sin ^2(c+d x)}{d}-\frac {2 a \sin ^3(c+d x)}{3 d}+\frac {a \sin ^4(c+d x)}{4 d}+\frac {a \sin ^5(c+d x)}{5 d} \]
a*ln(sin(d*x+c))/d+a*sin(d*x+c)/d-a*sin(d*x+c)^2/d-2/3*a*sin(d*x+c)^3/d+1/ 4*a*sin(d*x+c)^4/d+1/5*a*sin(d*x+c)^5/d
Time = 0.03 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00 \[ \int \cos ^4(c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \log (\sin (c+d x))}{d}+\frac {a \sin (c+d x)}{d}-\frac {a \sin ^2(c+d x)}{d}-\frac {2 a \sin ^3(c+d x)}{3 d}+\frac {a \sin ^4(c+d x)}{4 d}+\frac {a \sin ^5(c+d x)}{5 d} \]
(a*Log[Sin[c + d*x]])/d + (a*Sin[c + d*x])/d - (a*Sin[c + d*x]^2)/d - (2*a *Sin[c + d*x]^3)/(3*d) + (a*Sin[c + d*x]^4)/(4*d) + (a*Sin[c + d*x]^5)/(5* d)
Time = 0.26 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 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.
\(\displaystyle \int \cos ^4(c+d x) \cot (c+d x) (a \sin (c+d x)+a) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^5 (a \sin (c+d x)+a)}{\sin (c+d x)}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \csc (c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^3d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\csc (c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^3}{a}d(a \sin (c+d x))}{a^4 d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left (\sin ^4(c+d x) a^4+\sin ^3(c+d x) a^4-2 \sin ^2(c+d x) a^4+\csc (c+d x) a^4-2 \sin (c+d x) a^4+a^4\right )d(a \sin (c+d x))}{a^4 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{5} a^5 \sin ^5(c+d x)+\frac {1}{4} a^5 \sin ^4(c+d x)-\frac {2}{3} a^5 \sin ^3(c+d x)-a^5 \sin ^2(c+d x)+a^5 \sin (c+d x)+a^5 \log (a \sin (c+d x))}{a^4 d}\) |
(a^5*Log[a*Sin[c + d*x]] + a^5*Sin[c + d*x] - a^5*Sin[c + d*x]^2 - (2*a^5* Sin[c + d*x]^3)/3 + (a^5*Sin[c + d*x]^4)/4 + (a^5*Sin[c + d*x]^5)/5)/(a^4* d)
3.5.99.3.1 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[((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]
Time = 0.32 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.76
method | result | size |
derivativedivides | \(\frac {\frac {a \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+a \left (\frac {\left (\cos ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\cos ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\) | \(65\) |
default | \(\frac {\frac {a \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+a \left (\frac {\left (\cos ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\cos ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\) | \(65\) |
parallelrisch | \(\frac {a \left (480 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-480 \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-195+50 \sin \left (3 d x +3 c \right )+300 \sin \left (d x +c \right )+6 \sin \left (5 d x +5 c \right )+180 \cos \left (2 d x +2 c \right )+15 \cos \left (4 d x +4 c \right )\right )}{480 d}\) | \(87\) |
risch | \(-i a x +\frac {3 a \,{\mathrm e}^{2 i \left (d x +c \right )}}{16 d}+\frac {3 a \,{\mathrm e}^{-2 i \left (d x +c \right )}}{16 d}-\frac {2 i a c}{d}+\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}+\frac {5 a \sin \left (d x +c \right )}{8 d}+\frac {a \sin \left (5 d x +5 c \right )}{80 d}+\frac {a \cos \left (4 d x +4 c \right )}{32 d}+\frac {5 a \sin \left (3 d x +3 c \right )}{48 d}\) | \(119\) |
norman | \(\frac {\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {8 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {116 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {8 a \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 a \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {8 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {8 a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\) | \(205\) |
1/d*(1/5*a*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+a*(1/4*cos(d*x+c )^4+1/2*cos(d*x+c)^2+ln(sin(d*x+c))))
Time = 0.29 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.86 \[ \int \cos ^4(c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=\frac {15 \, a \cos \left (d x + c\right )^{4} + 30 \, a \cos \left (d x + c\right )^{2} + 60 \, a \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 4 \, {\left (3 \, a \cos \left (d x + c\right )^{4} + 4 \, a \cos \left (d x + c\right )^{2} + 8 \, a\right )} \sin \left (d x + c\right )}{60 \, d} \]
1/60*(15*a*cos(d*x + c)^4 + 30*a*cos(d*x + c)^2 + 60*a*log(1/2*sin(d*x + c )) + 4*(3*a*cos(d*x + c)^4 + 4*a*cos(d*x + c)^2 + 8*a)*sin(d*x + c))/d
\[ \int \cos ^4(c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=a \left (\int \cos ^{5}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx + \int \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx\right ) \]
a*(Integral(cos(c + d*x)**5*csc(c + d*x), x) + Integral(sin(c + d*x)*cos(c + d*x)**5*csc(c + d*x), x))
Time = 0.21 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.80 \[ \int \cos ^4(c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=\frac {12 \, a \sin \left (d x + c\right )^{5} + 15 \, a \sin \left (d x + c\right )^{4} - 40 \, a \sin \left (d x + c\right )^{3} - 60 \, a \sin \left (d x + c\right )^{2} + 60 \, a \log \left (\sin \left (d x + c\right )\right ) + 60 \, a \sin \left (d x + c\right )}{60 \, d} \]
1/60*(12*a*sin(d*x + c)^5 + 15*a*sin(d*x + c)^4 - 40*a*sin(d*x + c)^3 - 60 *a*sin(d*x + c)^2 + 60*a*log(sin(d*x + c)) + 60*a*sin(d*x + c))/d
Time = 0.41 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.81 \[ \int \cos ^4(c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=\frac {12 \, a \sin \left (d x + c\right )^{5} + 15 \, a \sin \left (d x + c\right )^{4} - 40 \, a \sin \left (d x + c\right )^{3} - 60 \, a \sin \left (d x + c\right )^{2} + 60 \, a \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 60 \, a \sin \left (d x + c\right )}{60 \, d} \]
1/60*(12*a*sin(d*x + c)^5 + 15*a*sin(d*x + c)^4 - 40*a*sin(d*x + c)^3 - 60 *a*sin(d*x + c)^2 + 60*a*log(abs(sin(d*x + c))) + 60*a*sin(d*x + c))/d
Time = 9.86 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.47 \[ \int \cos ^4(c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=\frac {a\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {a\,\ln \left (\frac {1}{{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}\right )}{d}+\frac {a\,{\cos \left (c+d\,x\right )}^2}{2\,d}+\frac {a\,{\cos \left (c+d\,x\right )}^4}{4\,d}+\frac {8\,a\,\sin \left (c+d\,x\right )}{15\,d}+\frac {4\,a\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{15\,d}+\frac {a\,{\cos \left (c+d\,x\right )}^4\,\sin \left (c+d\,x\right )}{5\,d} \]