Integrand size = 27, antiderivative size = 85 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\log (1+\sin (c+d x))}{a d}-\frac {\sin (c+d x)}{a d}+\frac {\sin ^2(c+d x)}{2 a d}-\frac {\sin ^3(c+d x)}{3 a d}+\frac {\sin ^4(c+d x)}{4 a d} \]
ln(1+sin(d*x+c))/a/d-sin(d*x+c)/a/d+1/2*sin(d*x+c)^2/a/d-1/3*sin(d*x+c)^3/ a/d+1/4*sin(d*x+c)^4/a/d
Time = 0.10 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.71 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {12 \log (1+\sin (c+d x))-12 \sin (c+d x)+6 \sin ^2(c+d x)-4 \sin ^3(c+d x)+3 \sin ^4(c+d x)}{12 a d} \]
(12*Log[1 + Sin[c + d*x]] - 12*Sin[c + d*x] + 6*Sin[c + d*x]^2 - 4*Sin[c + d*x]^3 + 3*Sin[c + d*x]^4)/(12*a*d)
Time = 0.27 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.93, 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.
\(\displaystyle \int \frac {\sin ^4(c+d x) \cos (c+d x)}{a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^4 \cos (c+d x)}{a \sin (c+d x)+a}dx\) |
\(\Big \downarrow \) 3312 |
\(\displaystyle \frac {\int \frac {\sin ^4(c+d x)}{\sin (c+d x) a+a}d(a \sin (c+d x))}{a d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {a^4 \sin ^4(c+d x)}{\sin (c+d x) a+a}d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\int \left (\frac {a^4}{\sin (c+d x) a+a}+\sin ^3(c+d x) a^3-\sin ^2(c+d x) a^3+\sin (c+d x) a^3-a^3\right )d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{4} a^4 \sin ^4(c+d x)-\frac {1}{3} a^4 \sin ^3(c+d x)+\frac {1}{2} a^4 \sin ^2(c+d x)-a^4 \sin (c+d x)+a^4 \log (a \sin (c+d x)+a)}{a^5 d}\) |
(a^4*Log[a + a*Sin[c + d*x]] - a^4*Sin[c + d*x] + (a^4*Sin[c + d*x]^2)/2 - (a^4*Sin[c + d*x]^3)/3 + (a^4*Sin[c + d*x]^4)/4)/(a^5*d)
3.3.24.3.1 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 = 0.22 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.66
method | result | size |
derivativedivides | \(\frac {\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\sin \left (d x +c \right )+\ln \left (1+\sin \left (d x +c \right )\right )}{d a}\) | \(56\) |
default | \(\frac {\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\sin \left (d x +c \right )+\ln \left (1+\sin \left (d x +c \right )\right )}{d a}\) | \(56\) |
parallelrisch | \(\frac {192 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-96 \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+33-36 \cos \left (2 d x +2 c \right )+3 \cos \left (4 d x +4 c \right )-120 \sin \left (d x +c \right )+8 \sin \left (3 d x +3 c \right )}{96 d a}\) | \(80\) |
risch | \(-\frac {i x}{a}+\frac {5 i {\mathrm e}^{i \left (d x +c \right )}}{8 d a}-\frac {5 i {\mathrm e}^{-i \left (d x +c \right )}}{8 d a}-\frac {2 i c}{a d}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a d}+\frac {\cos \left (4 d x +4 c \right )}{32 a d}+\frac {\sin \left (3 d x +3 c \right )}{12 d a}-\frac {3 \cos \left (2 d x +2 c \right )}{8 a d}\) | \(127\) |
norman | \(\frac {\frac {2}{a d}+\frac {2 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {10 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {10 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {4 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {4 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {38 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {38 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {58 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {58 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a d}-\frac {\ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}\) | \(253\) |
Time = 0.27 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.68 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {3 \, \cos \left (d x + c\right )^{4} - 12 \, \cos \left (d x + c\right )^{2} + 4 \, {\left (\cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right ) + 12 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{12 \, a d} \]
1/12*(3*cos(d*x + c)^4 - 12*cos(d*x + c)^2 + 4*(cos(d*x + c)^2 - 4)*sin(d* x + c) + 12*log(sin(d*x + c) + 1))/(a*d)
Time = 0.49 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.94 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\begin {cases} \frac {\log {\left (\sin {\left (c + d x \right )} + 1 \right )}}{a d} + \frac {\sin ^{4}{\left (c + d x \right )}}{4 a d} - \frac {\sin ^{3}{\left (c + d x \right )}}{3 a d} + \frac {\sin ^{2}{\left (c + d x \right )}}{2 a d} - \frac {\sin {\left (c + d x \right )}}{a d} & \text {for}\: d \neq 0 \\\frac {x \sin ^{4}{\left (c \right )} \cos {\left (c \right )}}{a \sin {\left (c \right )} + a} & \text {otherwise} \end {cases} \]
Piecewise((log(sin(c + d*x) + 1)/(a*d) + sin(c + d*x)**4/(4*a*d) - sin(c + d*x)**3/(3*a*d) + sin(c + d*x)**2/(2*a*d) - sin(c + d*x)/(a*d), Ne(d, 0)) , (x*sin(c)**4*cos(c)/(a*sin(c) + a), True))
Time = 0.19 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.74 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {3 \, \sin \left (d x + c\right )^{4} - 4 \, \sin \left (d x + c\right )^{3} + 6 \, \sin \left (d x + c\right )^{2} - 12 \, \sin \left (d x + c\right )}{a} + \frac {12 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a}}{12 \, d} \]
1/12*((3*sin(d*x + c)^4 - 4*sin(d*x + c)^3 + 6*sin(d*x + c)^2 - 12*sin(d*x + c))/a + 12*log(sin(d*x + c) + 1)/a)/d
Time = 0.32 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.89 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {12 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} + \frac {3 \, a^{3} \sin \left (d x + c\right )^{4} - 4 \, a^{3} \sin \left (d x + c\right )^{3} + 6 \, a^{3} \sin \left (d x + c\right )^{2} - 12 \, a^{3} \sin \left (d x + c\right )}{a^{4}}}{12 \, d} \]
1/12*(12*log(abs(sin(d*x + c) + 1))/a + (3*a^3*sin(d*x + c)^4 - 4*a^3*sin( d*x + c)^3 + 6*a^3*sin(d*x + c)^2 - 12*a^3*sin(d*x + c))/a^4)/d
Time = 9.55 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.80 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )}{a}-\frac {\sin \left (c+d\,x\right )}{a}+\frac {{\sin \left (c+d\,x\right )}^2}{2\,a}-\frac {{\sin \left (c+d\,x\right )}^3}{3\,a}+\frac {{\sin \left (c+d\,x\right )}^4}{4\,a}}{d} \]