Integrand size = 27, antiderivative size = 95 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {4 \log (1+\sin (c+d x))}{a^4 d}+\frac {\sin (c+d x)}{a^4 d}-\frac {1}{3 a d (a+a \sin (c+d x))^3}+\frac {2}{d \left (a^2+a^2 \sin (c+d x)\right )^2}-\frac {6}{d \left (a^4+a^4 \sin (c+d x)\right )} \] Output:
-4*ln(1+sin(d*x+c))/a^4/d+sin(d*x+c)/a^4/d-1/3/a/d/(a+a*sin(d*x+c))^3+2/d/ (a^2+a^2*sin(d*x+c))^2-6/d/(a^4+a^4*sin(d*x+c))
Time = 0.41 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.15 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {-13-12 \log (1+\sin (c+d x))-9 (3+4 \log (1+\sin (c+d x))) \sin (c+d x)-9 (1+4 \log (1+\sin (c+d x))) \sin ^2(c+d x)+(9-12 \log (1+\sin (c+d x))) \sin ^3(c+d x)+3 \sin ^4(c+d x)}{3 a^4 d (1+\sin (c+d x))^3} \] Input:
Integrate[(Cos[c + d*x]*Sin[c + d*x]^4)/(a + a*Sin[c + d*x])^4,x]
Output:
(-13 - 12*Log[1 + Sin[c + d*x]] - 9*(3 + 4*Log[1 + Sin[c + d*x]])*Sin[c + d*x] - 9*(1 + 4*Log[1 + Sin[c + d*x]])*Sin[c + d*x]^2 + (9 - 12*Log[1 + Si n[c + d*x]])*Sin[c + d*x]^3 + 3*Sin[c + d*x]^4)/(3*a^4*d*(1 + Sin[c + d*x] )^3)
Time = 0.31 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.87, 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.
| \(\displaystyle \int \frac {\sin ^4(c+d x) \cos (c+d x)}{(a \sin (c+d x)+a)^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^4 \cos (c+d x)}{(a \sin (c+d x)+a)^4}dx\) |
| \(\Big \downarrow \) 3312 |
| \(\displaystyle \frac {\int \frac {\sin ^4(c+d x)}{(\sin (c+d x) a+a)^4}d(a \sin (c+d x))}{a d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a^4 \sin ^4(c+d x)}{(\sin (c+d x) a+a)^4}d(a \sin (c+d x))}{a^5 d}\) |
| \(\Big \downarrow \) 49 |
| \(\displaystyle \frac {\int \left (\frac {a^4}{(\sin (c+d x) a+a)^4}-\frac {4 a^3}{(\sin (c+d x) a+a)^3}+\frac {6 a^2}{(\sin (c+d x) a+a)^2}-\frac {4 a}{\sin (c+d x) a+a}+1\right )d(a \sin (c+d x))}{a^5 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {a^4}{3 (a \sin (c+d x)+a)^3}+\frac {2 a^3}{(a \sin (c+d x)+a)^2}-\frac {6 a^2}{a \sin (c+d x)+a}+a \sin (c+d x)-4 a \log (a \sin (c+d x)+a)}{a^5 d}\) |
Input:
Int[(Cos[c + d*x]*Sin[c + d*x]^4)/(a + a*Sin[c + d*x])^4,x]
Output:
(-4*a*Log[a + a*Sin[c + d*x]] + a*Sin[c + d*x] - a^4/(3*(a + a*Sin[c + d*x ])^3) + (2*a^3)/(a + a*Sin[c + d*x])^2 - (6*a^2)/(a + a*Sin[c + d*x]))/(a^ 5*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_.), 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 = 0.92 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.65
| method | result | size |
| derivativedivides | \(\frac {\sin \left (d x +c \right )+\frac {2}{\left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {1}{3 \left (1+\sin \left (d x +c \right )\right )^{3}}-4 \ln \left (1+\sin \left (d x +c \right )\right )-\frac {6}{1+\sin \left (d x +c \right )}}{d \,a^{4}}\) | \(62\) |
| default | \(\frac {\sin \left (d x +c \right )+\frac {2}{\left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {1}{3 \left (1+\sin \left (d x +c \right )\right )^{3}}-4 \ln \left (1+\sin \left (d x +c \right )\right )-\frac {6}{1+\sin \left (d x +c \right )}}{d \,a^{4}}\) | \(62\) |
| risch | \(\frac {4 i x}{a^{4}}-\frac {i {\mathrm e}^{i \left (d x +c \right )}}{2 d \,a^{4}}+\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{2 d \,a^{4}}+\frac {8 i c}{d \,a^{4}}-\frac {4 i \left (30 i {\mathrm e}^{4 i \left (d x +c \right )}+9 \,{\mathrm e}^{5 i \left (d x +c \right )}-30 i {\mathrm e}^{2 i \left (d x +c \right )}-44 \,{\mathrm e}^{3 i \left (d x +c \right )}+9 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{3 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{6}}-\frac {8 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{4}}\) | \(157\) |
| parallelrisch | \(\frac {\left (144 \cos \left (2 d x +2 c \right )-360 \sin \left (d x +c \right )+24 \sin \left (3 d x +3 c \right )-240\right ) \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )+\left (-288 \cos \left (2 d x +2 c \right )+720 \sin \left (d x +c \right )-48 \sin \left (3 d x +3 c \right )+480\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+132 \cos \left (2 d x +2 c \right )-3 \cos \left (4 d x +4 c \right )-228 \sin \left (d x +c \right )+44 \sin \left (3 d x +3 c \right )-129}{6 d \,a^{4} \left (\sin \left (3 d x +3 c \right )-15 \sin \left (d x +c \right )-10+6 \cos \left (2 d x +2 c \right )\right )}\) | \(174\) |
| norman | \(\frac {\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}+\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{16}}{d a}+\frac {48 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d a}+\frac {48 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{d a}+\frac {464 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d a}+\frac {464 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}}{3 d a}+\frac {1112 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{3 d a}+\frac {1112 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{3 d a}+\frac {5288 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{3 d a}+\frac {5288 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 d a}+\frac {2152 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 d a}+\frac {2152 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{3 d a}+\frac {3376 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{3 d a}+\frac {3376 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3 d a}+\frac {4592 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d a}+\frac {4592 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{3 d a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5} a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}-\frac {8 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{4}}+\frac {4 \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{d \,a^{4}}\) | \(379\) |
Input:
int(cos(d*x+c)*sin(d*x+c)^4/(a+a*sin(d*x+c))^4,x,method=_RETURNVERBOSE)
Output:
1/d/a^4*(sin(d*x+c)+2/(1+sin(d*x+c))^2-1/3/(1+sin(d*x+c))^3-4*ln(1+sin(d*x +c))-6/(1+sin(d*x+c)))
Time = 0.08 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.39 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {3 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{2} + 12 \, {\left (3 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right ) - 4\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 9 \, {\left (\cos \left (d x + c\right )^{2} + 2\right )} \sin \left (d x + c\right ) - 19}{3 \, {\left (3 \, a^{4} d \cos \left (d x + c\right )^{2} - 4 \, a^{4} d + {\left (a^{4} d \cos \left (d x + c\right )^{2} - 4 \, a^{4} d\right )} \sin \left (d x + c\right )\right )}} \] Input:
integrate(cos(d*x+c)*sin(d*x+c)^4/(a+a*sin(d*x+c))^4,x, algorithm="fricas" )
Output:
-1/3*(3*cos(d*x + c)^4 + 3*cos(d*x + c)^2 + 12*(3*cos(d*x + c)^2 + (cos(d* x + c)^2 - 4)*sin(d*x + c) - 4)*log(sin(d*x + c) + 1) - 9*(cos(d*x + c)^2 + 2)*sin(d*x + c) - 19)/(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. 527 vs. \(2 (82) = 164\).
Time = 1.27 (sec) , antiderivative size = 527, normalized size of antiderivative = 5.55 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\begin {cases} - \frac {12 \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin ^{3}{\left (c + d x \right )}}{3 a^{4} d \sin ^{3}{\left (c + d x \right )} + 9 a^{4} d \sin ^{2}{\left (c + d x \right )} + 9 a^{4} d \sin {\left (c + d x \right )} + 3 a^{4} d} - \frac {36 \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin ^{2}{\left (c + d x \right )}}{3 a^{4} d \sin ^{3}{\left (c + d x \right )} + 9 a^{4} d \sin ^{2}{\left (c + d x \right )} + 9 a^{4} d \sin {\left (c + d x \right )} + 3 a^{4} d} - \frac {36 \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin {\left (c + d x \right )}}{3 a^{4} d \sin ^{3}{\left (c + d x \right )} + 9 a^{4} d \sin ^{2}{\left (c + d x \right )} + 9 a^{4} d \sin {\left (c + d x \right )} + 3 a^{4} d} - \frac {12 \log {\left (\sin {\left (c + d x \right )} + 1 \right )}}{3 a^{4} d \sin ^{3}{\left (c + d x \right )} + 9 a^{4} d \sin ^{2}{\left (c + d x \right )} + 9 a^{4} d \sin {\left (c + d x \right )} + 3 a^{4} d} + \frac {3 \sin ^{4}{\left (c + d x \right )}}{3 a^{4} d \sin ^{3}{\left (c + d x \right )} + 9 a^{4} d \sin ^{2}{\left (c + d x \right )} + 9 a^{4} d \sin {\left (c + d x \right )} + 3 a^{4} d} - \frac {36 \sin ^{2}{\left (c + d x \right )}}{3 a^{4} d \sin ^{3}{\left (c + d x \right )} + 9 a^{4} d \sin ^{2}{\left (c + d x \right )} + 9 a^{4} d \sin {\left (c + d x \right )} + 3 a^{4} d} - \frac {54 \sin {\left (c + d x \right )}}{3 a^{4} d \sin ^{3}{\left (c + d x \right )} + 9 a^{4} d \sin ^{2}{\left (c + d x \right )} + 9 a^{4} d \sin {\left (c + d x \right )} + 3 a^{4} d} - \frac {22}{3 a^{4} d \sin ^{3}{\left (c + d x \right )} + 9 a^{4} d \sin ^{2}{\left (c + d x \right )} + 9 a^{4} d \sin {\left (c + d x \right )} + 3 a^{4} d} & \text {for}\: d \neq 0 \\\frac {x \sin ^{4}{\left (c \right )} \cos {\left (c \right )}}{\left (a \sin {\left (c \right )} + a\right )^{4}} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)*sin(d*x+c)**4/(a+a*sin(d*x+c))**4,x)
Output:
Piecewise((-12*log(sin(c + d*x) + 1)*sin(c + d*x)**3/(3*a**4*d*sin(c + d*x )**3 + 9*a**4*d*sin(c + d*x)**2 + 9*a**4*d*sin(c + d*x) + 3*a**4*d) - 36*l og(sin(c + d*x) + 1)*sin(c + d*x)**2/(3*a**4*d*sin(c + d*x)**3 + 9*a**4*d* sin(c + d*x)**2 + 9*a**4*d*sin(c + d*x) + 3*a**4*d) - 36*log(sin(c + d*x) + 1)*sin(c + d*x)/(3*a**4*d*sin(c + d*x)**3 + 9*a**4*d*sin(c + d*x)**2 + 9 *a**4*d*sin(c + d*x) + 3*a**4*d) - 12*log(sin(c + d*x) + 1)/(3*a**4*d*sin( c + d*x)**3 + 9*a**4*d*sin(c + d*x)**2 + 9*a**4*d*sin(c + d*x) + 3*a**4*d) + 3*sin(c + d*x)**4/(3*a**4*d*sin(c + d*x)**3 + 9*a**4*d*sin(c + d*x)**2 + 9*a**4*d*sin(c + d*x) + 3*a**4*d) - 36*sin(c + d*x)**2/(3*a**4*d*sin(c + d*x)**3 + 9*a**4*d*sin(c + d*x)**2 + 9*a**4*d*sin(c + d*x) + 3*a**4*d) - 54*sin(c + d*x)/(3*a**4*d*sin(c + d*x)**3 + 9*a**4*d*sin(c + d*x)**2 + 9*a **4*d*sin(c + d*x) + 3*a**4*d) - 22/(3*a**4*d*sin(c + d*x)**3 + 9*a**4*d*s in(c + d*x)**2 + 9*a**4*d*sin(c + d*x) + 3*a**4*d), Ne(d, 0)), (x*sin(c)** 4*cos(c)/(a*sin(c) + a)**4, True))
Time = 0.03 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.99 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {\frac {18 \, \sin \left (d x + c\right )^{2} + 30 \, \sin \left (d x + c\right ) + 13}{a^{4} \sin \left (d x + c\right )^{3} + 3 \, a^{4} \sin \left (d x + c\right )^{2} + 3 \, a^{4} \sin \left (d x + c\right ) + a^{4}} + \frac {12 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{4}} - \frac {3 \, \sin \left (d x + c\right )}{a^{4}}}{3 \, d} \] Input:
integrate(cos(d*x+c)*sin(d*x+c)^4/(a+a*sin(d*x+c))^4,x, algorithm="maxima" )
Output:
-1/3*((18*sin(d*x + c)^2 + 30*sin(d*x + c) + 13)/(a^4*sin(d*x + c)^3 + 3*a ^4*sin(d*x + c)^2 + 3*a^4*sin(d*x + c) + a^4) + 12*log(sin(d*x + c) + 1)/a ^4 - 3*sin(d*x + c)/a^4)/d
Time = 0.17 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.74 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {4 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{4} d} + \frac {\sin \left (d x + c\right )}{a^{4} d} - \frac {18 \, \sin \left (d x + c\right )^{2} + 30 \, \sin \left (d x + c\right ) + 13}{3 \, a^{4} d {\left (\sin \left (d x + c\right ) + 1\right )}^{3}} \] Input:
integrate(cos(d*x+c)*sin(d*x+c)^4/(a+a*sin(d*x+c))^4,x, algorithm="giac")
Output:
-4*log(abs(sin(d*x + c) + 1))/(a^4*d) + sin(d*x + c)/(a^4*d) - 1/3*(18*sin (d*x + c)^2 + 30*sin(d*x + c) + 13)/(a^4*d*(sin(d*x + c) + 1)^3)
Time = 0.08 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.73 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\sin \left (c+d\,x\right )}{a^4\,d}-\frac {4\,\ln \left (\sin \left (c+d\,x\right )+1\right )}{a^4\,d}-\frac {6\,{\sin \left (c+d\,x\right )}^2+10\,\sin \left (c+d\,x\right )+\frac {13}{3}}{a^4\,d\,{\left (\sin \left (c+d\,x\right )+1\right )}^3} \] Input:
int((cos(c + d*x)*sin(c + d*x)^4)/(a + a*sin(c + d*x))^4,x)
Output:
sin(c + d*x)/(a^4*d) - (4*log(sin(c + d*x) + 1))/(a^4*d) - (10*sin(c + d*x ) + 6*sin(c + d*x)^2 + 13/3)/(a^4*d*(sin(c + d*x) + 1)^3)
Time = 0.16 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.41 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {-12 \,\mathrm {log}\left (\sin \left (d x +c \right )+1\right ) \sin \left (d x +c \right )^{3}-36 \,\mathrm {log}\left (\sin \left (d x +c \right )+1\right ) \sin \left (d x +c \right )^{2}-36 \,\mathrm {log}\left (\sin \left (d x +c \right )+1\right ) \sin \left (d x +c \right )-12 \,\mathrm {log}\left (\sin \left (d x +c \right )+1\right )+3 \sin \left (d x +c \right )^{4}+12 \sin \left (d x +c \right )^{3}-18 \sin \left (d x +c \right )-10}{3 a^{4} d \left (\sin \left (d x +c \right )^{3}+3 \sin \left (d x +c \right )^{2}+3 \sin \left (d x +c \right )+1\right )} \] Input:
int(cos(d*x+c)*sin(d*x+c)^4/(a+a*sin(d*x+c))^4,x)
Output:
( - 12*log(sin(c + d*x) + 1)*sin(c + d*x)**3 - 36*log(sin(c + d*x) + 1)*si n(c + d*x)**2 - 36*log(sin(c + d*x) + 1)*sin(c + d*x) - 12*log(sin(c + d*x ) + 1) + 3*sin(c + d*x)**4 + 12*sin(c + d*x)**3 - 18*sin(c + d*x) - 10)/(3 *a**4*d*(sin(c + d*x)**3 + 3*sin(c + d*x)**2 + 3*sin(c + d*x) + 1))