Integrand size = 27, antiderivative size = 116 \[ \int \frac {\cos (c+d x) \sin ^5(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {10 \log (1+\sin (c+d x))}{a^4 d}-\frac {4 \sin (c+d x)}{a^4 d}+\frac {\sin ^2(c+d x)}{2 a^4 d}+\frac {1}{3 a d (a+a \sin (c+d x))^3}-\frac {5}{2 d \left (a^2+a^2 \sin (c+d x)\right )^2}+\frac {10}{d \left (a^4+a^4 \sin (c+d x)\right )} \]
10*ln(1+sin(d*x+c))/a^4/d-4*sin(d*x+c)/a^4/d+1/2*sin(d*x+c)^2/a^4/d+1/3/a/ d/(a+a*sin(d*x+c))^3-5/2/d/(a^2+a^2*sin(d*x+c))^2+10/d/(a^4+a^4*sin(d*x+c) )
Time = 0.62 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.03 \[ \int \frac {\cos (c+d x) \sin ^5(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {47+60 \log (1+\sin (c+d x))+9 (9+20 \log (1+\sin (c+d x))) \sin (c+d x)+9 (-1+20 \log (1+\sin (c+d x))) \sin ^2(c+d x)+(-63+60 \log (1+\sin (c+d x))) \sin ^3(c+d x)-15 \sin ^4(c+d x)+3 \sin ^5(c+d x)}{6 a^4 d (1+\sin (c+d x))^3} \]
(47 + 60*Log[1 + Sin[c + d*x]] + 9*(9 + 20*Log[1 + Sin[c + d*x]])*Sin[c + d*x] + 9*(-1 + 20*Log[1 + Sin[c + d*x]])*Sin[c + d*x]^2 + (-63 + 60*Log[1 + Sin[c + d*x]])*Sin[c + d*x]^3 - 15*Sin[c + d*x]^4 + 3*Sin[c + d*x]^5)/(6 *a^4*d*(1 + Sin[c + d*x])^3)
Time = 0.30 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.91, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3042, 3312, 27, 49, 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.
\(\displaystyle \int \frac {\sin ^5(c+d x) \cos (c+d x)}{(a \sin (c+d x)+a)^4} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^5 \cos (c+d x)}{(a \sin (c+d x)+a)^4}dx\) |
\(\Big \downarrow \) 3312 |
\(\displaystyle \frac {\int \frac {\sin ^5(c+d x)}{(\sin (c+d x) a+a)^4}d(a \sin (c+d x))}{a d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {a^5 \sin ^5(c+d x)}{(\sin (c+d x) a+a)^4}d(a \sin (c+d x))}{a^6 d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\int \left (-\frac {a^5}{(\sin (c+d x) a+a)^4}+\frac {5 a^4}{(\sin (c+d x) a+a)^3}-\frac {10 a^3}{(\sin (c+d x) a+a)^2}+\frac {10 a^2}{\sin (c+d x) a+a}+\sin (c+d x) a-4 a\right )d(a \sin (c+d x))}{a^6 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {a^5}{3 (a \sin (c+d x)+a)^3}-\frac {5 a^4}{2 (a \sin (c+d x)+a)^2}+\frac {10 a^3}{a \sin (c+d x)+a}+\frac {1}{2} a^2 \sin ^2(c+d x)-4 a^2 \sin (c+d x)+10 a^2 \log (a \sin (c+d x)+a)}{a^6 d}\) |
(10*a^2*Log[a + a*Sin[c + d*x]] - 4*a^2*Sin[c + d*x] + (a^2*Sin[c + d*x]^2 )/2 + a^5/(3*(a + a*Sin[c + d*x])^3) - (5*a^4)/(2*(a + a*Sin[c + d*x])^2) + (10*a^3)/(a + a*Sin[c + d*x]))/(a^6*d)
3.3.49.3.1 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_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*(( c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b*f) Su bst[Int[(a + x)^m*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]
Time = 0.54 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.64
method | result | size |
derivativedivides | \(\frac {\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-4 \sin \left (d x +c \right )+\frac {10}{1+\sin \left (d x +c \right )}+10 \ln \left (1+\sin \left (d x +c \right )\right )+\frac {1}{3 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {5}{2 \left (1+\sin \left (d x +c \right )\right )^{2}}}{d \,a^{4}}\) | \(74\) |
default | \(\frac {\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-4 \sin \left (d x +c \right )+\frac {10}{1+\sin \left (d x +c \right )}+10 \ln \left (1+\sin \left (d x +c \right )\right )+\frac {1}{3 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {5}{2 \left (1+\sin \left (d x +c \right )\right )^{2}}}{d \,a^{4}}\) | \(74\) |
parallelrisch | \(\frac {\left (-1440 \cos \left (2 d x +2 c \right )+3600 \sin \left (d x +c \right )-240 \sin \left (3 d x +3 c \right )+2400\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (2880 \cos \left (2 d x +2 c \right )-7200 \sin \left (d x +c \right )+480 \sin \left (3 d x +3 c \right )-4800\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-1320 \cos \left (2 d x +2 c \right )+30 \cos \left (4 d x +4 c \right )+2250 \sin \left (d x +c \right )-425 \sin \left (3 d x +3 c \right )-3 \sin \left (5 d x +5 c \right )+1290}{24 d \,a^{4} \left (-10+6 \cos \left (2 d x +2 c \right )+\sin \left (3 d x +3 c \right )-15 \sin \left (d x +c \right )\right )}\) | \(185\) |
risch | \(-\frac {10 i x}{a^{4}}-\frac {{\mathrm e}^{2 i \left (d x +c \right )}}{8 d \,a^{4}}+\frac {2 i {\mathrm e}^{i \left (d x +c \right )}}{d \,a^{4}}-\frac {2 i {\mathrm e}^{-i \left (d x +c \right )}}{d \,a^{4}}-\frac {{\mathrm e}^{-2 i \left (d x +c \right )}}{8 d \,a^{4}}-\frac {20 i c}{d \,a^{4}}+\frac {2 i \left (-154 \,{\mathrm e}^{3 i \left (d x +c \right )}-105 i {\mathrm e}^{2 i \left (d x +c \right )}+105 i {\mathrm e}^{4 i \left (d x +c \right )}+30 \,{\mathrm e}^{5 i \left (d x +c \right )}+30 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{3 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{6}}+\frac {20 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{4}}\) | \(191\) |
1/d/a^4*(1/2*sin(d*x+c)^2-4*sin(d*x+c)+10/(1+sin(d*x+c))+10*ln(1+sin(d*x+c ))+1/3/(1+sin(d*x+c))^3-5/2/(1+sin(d*x+c))^2)
Time = 0.29 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.24 \[ \int \frac {\cos (c+d x) \sin ^5(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {30 \, \cos \left (d x + c\right )^{4} - 87 \, \cos \left (d x + c\right )^{2} + 120 \, {\left (3 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right ) - 4\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (2 \, \cos \left (d x + c\right )^{4} + 39 \, \cos \left (d x + c\right )^{2} + 10\right )} \sin \left (d x + c\right ) - 34}{12 \, {\left (3 \, a^{4} d \cos \left (d x + c\right )^{2} - 4 \, a^{4} d + {\left (a^{4} d \cos \left (d x + c\right )^{2} - 4 \, a^{4} d\right )} \sin \left (d x + c\right )\right )}} \]
1/12*(30*cos(d*x + c)^4 - 87*cos(d*x + c)^2 + 120*(3*cos(d*x + c)^2 + (cos (d*x + c)^2 - 4)*sin(d*x + c) - 4)*log(sin(d*x + c) + 1) - 3*(2*cos(d*x + c)^4 + 39*cos(d*x + c)^2 + 10)*sin(d*x + c) - 34)/(3*a^4*d*cos(d*x + c)^2 - 4*a^4*d + (a^4*d*cos(d*x + c)^2 - 4*a^4*d)*sin(d*x + c))
Leaf count of result is larger than twice the leaf count of optimal. 588 vs. \(2 (100) = 200\).
Time = 1.85 (sec) , antiderivative size = 588, normalized size of antiderivative = 5.07 \[ \int \frac {\cos (c+d x) \sin ^5(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\begin {cases} \frac {60 \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin ^{3}{\left (c + d x \right )}}{6 a^{4} d \sin ^{3}{\left (c + d x \right )} + 18 a^{4} d \sin ^{2}{\left (c + d x \right )} + 18 a^{4} d \sin {\left (c + d x \right )} + 6 a^{4} d} + \frac {180 \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin ^{2}{\left (c + d x \right )}}{6 a^{4} d \sin ^{3}{\left (c + d x \right )} + 18 a^{4} d \sin ^{2}{\left (c + d x \right )} + 18 a^{4} d \sin {\left (c + d x \right )} + 6 a^{4} d} + \frac {180 \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin {\left (c + d x \right )}}{6 a^{4} d \sin ^{3}{\left (c + d x \right )} + 18 a^{4} d \sin ^{2}{\left (c + d x \right )} + 18 a^{4} d \sin {\left (c + d x \right )} + 6 a^{4} d} + \frac {60 \log {\left (\sin {\left (c + d x \right )} + 1 \right )}}{6 a^{4} d \sin ^{3}{\left (c + d x \right )} + 18 a^{4} d \sin ^{2}{\left (c + d x \right )} + 18 a^{4} d \sin {\left (c + d x \right )} + 6 a^{4} d} + \frac {3 \sin ^{5}{\left (c + d x \right )}}{6 a^{4} d \sin ^{3}{\left (c + d x \right )} + 18 a^{4} d \sin ^{2}{\left (c + d x \right )} + 18 a^{4} d \sin {\left (c + d x \right )} + 6 a^{4} d} - \frac {15 \sin ^{4}{\left (c + d x \right )}}{6 a^{4} d \sin ^{3}{\left (c + d x \right )} + 18 a^{4} d \sin ^{2}{\left (c + d x \right )} + 18 a^{4} d \sin {\left (c + d x \right )} + 6 a^{4} d} + \frac {180 \sin ^{2}{\left (c + d x \right )}}{6 a^{4} d \sin ^{3}{\left (c + d x \right )} + 18 a^{4} d \sin ^{2}{\left (c + d x \right )} + 18 a^{4} d \sin {\left (c + d x \right )} + 6 a^{4} d} + \frac {270 \sin {\left (c + d x \right )}}{6 a^{4} d \sin ^{3}{\left (c + d x \right )} + 18 a^{4} d \sin ^{2}{\left (c + d x \right )} + 18 a^{4} d \sin {\left (c + d x \right )} + 6 a^{4} d} + \frac {110}{6 a^{4} d \sin ^{3}{\left (c + d x \right )} + 18 a^{4} d \sin ^{2}{\left (c + d x \right )} + 18 a^{4} d \sin {\left (c + d x \right )} + 6 a^{4} d} & \text {for}\: d \neq 0 \\\frac {x \sin ^{5}{\left (c \right )} \cos {\left (c \right )}}{\left (a \sin {\left (c \right )} + a\right )^{4}} & \text {otherwise} \end {cases} \]
Piecewise((60*log(sin(c + d*x) + 1)*sin(c + d*x)**3/(6*a**4*d*sin(c + d*x) **3 + 18*a**4*d*sin(c + d*x)**2 + 18*a**4*d*sin(c + d*x) + 6*a**4*d) + 180 *log(sin(c + d*x) + 1)*sin(c + d*x)**2/(6*a**4*d*sin(c + d*x)**3 + 18*a**4 *d*sin(c + d*x)**2 + 18*a**4*d*sin(c + d*x) + 6*a**4*d) + 180*log(sin(c + d*x) + 1)*sin(c + d*x)/(6*a**4*d*sin(c + d*x)**3 + 18*a**4*d*sin(c + d*x)* *2 + 18*a**4*d*sin(c + d*x) + 6*a**4*d) + 60*log(sin(c + d*x) + 1)/(6*a**4 *d*sin(c + d*x)**3 + 18*a**4*d*sin(c + d*x)**2 + 18*a**4*d*sin(c + d*x) + 6*a**4*d) + 3*sin(c + d*x)**5/(6*a**4*d*sin(c + d*x)**3 + 18*a**4*d*sin(c + d*x)**2 + 18*a**4*d*sin(c + d*x) + 6*a**4*d) - 15*sin(c + d*x)**4/(6*a** 4*d*sin(c + d*x)**3 + 18*a**4*d*sin(c + d*x)**2 + 18*a**4*d*sin(c + d*x) + 6*a**4*d) + 180*sin(c + d*x)**2/(6*a**4*d*sin(c + d*x)**3 + 18*a**4*d*sin (c + d*x)**2 + 18*a**4*d*sin(c + d*x) + 6*a**4*d) + 270*sin(c + d*x)/(6*a* *4*d*sin(c + d*x)**3 + 18*a**4*d*sin(c + d*x)**2 + 18*a**4*d*sin(c + d*x) + 6*a**4*d) + 110/(6*a**4*d*sin(c + d*x)**3 + 18*a**4*d*sin(c + d*x)**2 + 18*a**4*d*sin(c + d*x) + 6*a**4*d), Ne(d, 0)), (x*sin(c)**5*cos(c)/(a*sin( c) + a)**4, True))
Time = 0.20 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.91 \[ \int \frac {\cos (c+d x) \sin ^5(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\frac {60 \, \sin \left (d x + c\right )^{2} + 105 \, \sin \left (d x + c\right ) + 47}{a^{4} \sin \left (d x + c\right )^{3} + 3 \, a^{4} \sin \left (d x + c\right )^{2} + 3 \, a^{4} \sin \left (d x + c\right ) + a^{4}} + \frac {3 \, {\left (\sin \left (d x + c\right )^{2} - 8 \, \sin \left (d x + c\right )\right )}}{a^{4}} + \frac {60 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{4}}}{6 \, d} \]
1/6*((60*sin(d*x + c)^2 + 105*sin(d*x + c) + 47)/(a^4*sin(d*x + c)^3 + 3*a ^4*sin(d*x + c)^2 + 3*a^4*sin(d*x + c) + a^4) + 3*(sin(d*x + c)^2 - 8*sin( d*x + c))/a^4 + 60*log(sin(d*x + c) + 1)/a^4)/d
Time = 0.36 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.72 \[ \int \frac {\cos (c+d x) \sin ^5(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\frac {60 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{4}} + \frac {60 \, \sin \left (d x + c\right )^{2} + 105 \, \sin \left (d x + c\right ) + 47}{a^{4} {\left (\sin \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, {\left (a^{4} \sin \left (d x + c\right )^{2} - 8 \, a^{4} \sin \left (d x + c\right )\right )}}{a^{8}}}{6 \, d} \]
1/6*(60*log(abs(sin(d*x + c) + 1))/a^4 + (60*sin(d*x + c)^2 + 105*sin(d*x + c) + 47)/(a^4*(sin(d*x + c) + 1)^3) + 3*(a^4*sin(d*x + c)^2 - 8*a^4*sin( d*x + c))/a^8)/d
Time = 10.37 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.98 \[ \int \frac {\cos (c+d x) \sin ^5(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {10\,\ln \left (\sin \left (c+d\,x\right )+1\right )}{a^4\,d}-\frac {4\,\sin \left (c+d\,x\right )}{a^4\,d}+\frac {10\,{\sin \left (c+d\,x\right )}^2+\frac {35\,\sin \left (c+d\,x\right )}{2}+\frac {47}{6}}{d\,\left (a^4\,{\sin \left (c+d\,x\right )}^3+3\,a^4\,{\sin \left (c+d\,x\right )}^2+3\,a^4\,\sin \left (c+d\,x\right )+a^4\right )}+\frac {{\sin \left (c+d\,x\right )}^2}{2\,a^4\,d} \]