Integrand size = 27, antiderivative size = 85 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\log (1+\sin (c+d x))}{a d}-\frac {\sin (c+d x)}{a d}+\frac {\sin ^2(c+d x)}{2 a d}-\frac {\sin ^3(c+d x)}{3 a d}+\frac {\sin ^4(c+d x)}{4 a d} \] Output:
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 = 0.09 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.71 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {12 \log (1+\sin (c+d x))-12 \sin (c+d x)+6 \sin ^2(c+d x)-4 \sin ^3(c+d x)+3 \sin ^4(c+d x)}{12 a d} \] Input:
Integrate[(Cos[c + d*x]*Sin[c + d*x]^4)/(a + a*Sin[c + d*x]),x]
Output:
(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 = 0.29 (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 |
\(\displaystyle \frac {\frac {1}{4} a^4 \sin ^4(c+d x)-\frac {1}{3} a^4 \sin ^3(c+d x)+\frac {1}{2} a^4 \sin ^2(c+d x)-a^4 \sin (c+d x)+a^4 \log (a \sin (c+d x)+a)}{a^5 d}\) |
Input:
Int[(Cos[c + d*x]*Sin[c + d*x]^4)/(a + a*Sin[c + d*x]),x]
Output:
(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)
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 = 0.57 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.66
method | result | size |
derivativedivides | \(\frac {\frac {\sin \left (d x +c \right )^{4}}{4}-\frac {\sin \left (d x +c \right )^{3}}{3}+\frac {\sin \left (d x +c \right )^{2}}{2}-\sin \left (d x +c \right )+\ln \left (1+\sin \left (d x +c \right )\right )}{d a}\) | \(56\) |
default | \(\frac {\frac {\sin \left (d x +c \right )^{4}}{4}-\frac {\sin \left (d x +c \right )^{3}}{3}+\frac {\sin \left (d x +c \right )^{2}}{2}-\sin \left (d x +c \right )+\ln \left (1+\sin \left (d x +c \right )\right )}{d a}\) | \(56\) |
parallelrisch | \(\frac {-96 \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )+192 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+33-36 \cos \left (2 d x +2 c \right )+3 \cos \left (4 d x +4 c \right )-120 \sin \left (d x +c \right )+8 \sin \left (3 d x +3 c \right )}{96 d a}\) | \(80\) |
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 | \(\frac {\frac {2}{a d}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d a}+\frac {10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d a}+\frac {10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d a}+\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d a}+\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{3 d a}+\frac {38 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 d a}+\frac {38 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{3 d a}+\frac {58 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{3 d a}+\frac {58 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a d}-\frac {\ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{a d}\) | \(253\) |
Input:
int(cos(d*x+c)*sin(d*x+c)^4/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d/a*(1/4*sin(d*x+c)^4-1/3*sin(d*x+c)^3+1/2*sin(d*x+c)^2-sin(d*x+c)+ln(1+ sin(d*x+c)))
Time = 0.08 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.68 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {3 \, \cos \left (d x + c\right )^{4} - 12 \, \cos \left (d x + c\right )^{2} + 4 \, {\left (\cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right ) + 12 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{12 \, a d} \] Input:
integrate(cos(d*x+c)*sin(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="fricas")
Output:
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 = 0.56 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.94 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\begin {cases} \frac {\log {\left (\sin {\left (c + d x \right )} + 1 \right )}}{a d} + \frac {\sin ^{4}{\left (c + d x \right )}}{4 a d} - \frac {\sin ^{3}{\left (c + d x \right )}}{3 a d} - \frac {\sin {\left (c + d x \right )}}{a d} - \frac {\cos ^{2}{\left (c + d x \right )}}{2 a d} & \text {for}\: d \neq 0 \\\frac {x \sin ^{4}{\left (c \right )} \cos {\left (c \right )}}{a \sin {\left (c \right )} + a} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)*sin(d*x+c)**4/(a+a*sin(d*x+c)),x)
Output:
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)/(a*d) - cos(c + d*x)**2/(2*a*d), Ne(d, 0)) , (x*sin(c)**4*cos(c)/(a*sin(c) + a), True))
Time = 0.03 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.74 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {3 \, \sin \left (d x + c\right )^{4} - 4 \, \sin \left (d x + c\right )^{3} + 6 \, \sin \left (d x + c\right )^{2} - 12 \, \sin \left (d x + c\right )}{a} + \frac {12 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a}}{12 \, d} \] Input:
integrate(cos(d*x+c)*sin(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="maxima")
Output:
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 = 0.15 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.05 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a d} + \frac {3 \, a^{3} d^{3} \sin \left (d x + c\right )^{4} - 4 \, a^{3} d^{3} \sin \left (d x + c\right )^{3} + 6 \, a^{3} d^{3} \sin \left (d x + c\right )^{2} - 12 \, a^{3} d^{3} \sin \left (d x + c\right )}{12 \, a^{4} d^{4}} \] Input:
integrate(cos(d*x+c)*sin(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
log(abs(sin(d*x + c) + 1))/(a*d) + 1/12*(3*a^3*d^3*sin(d*x + c)^4 - 4*a^3* d^3*sin(d*x + c)^3 + 6*a^3*d^3*sin(d*x + c)^2 - 12*a^3*d^3*sin(d*x + c))/( a^4*d^4)
Time = 17.98 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.80 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )}{a}-\frac {\sin \left (c+d\,x\right )}{a}+\frac {{\sin \left (c+d\,x\right )}^2}{2\,a}-\frac {{\sin \left (c+d\,x\right )}^3}{3\,a}+\frac {{\sin \left (c+d\,x\right )}^4}{4\,a}}{d} \] Input:
int((cos(c + d*x)*sin(c + d*x)^4)/(a + a*sin(c + d*x)),x)
Output:
(log(sin(c + d*x) + 1)/a - sin(c + d*x)/a + sin(c + d*x)^2/(2*a) - sin(c + d*x)^3/(3*a) + sin(c + d*x)^4/(4*a))/d
Time = 0.17 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.68 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {-6 \cos \left (d x +c \right )^{2}+12 \,\mathrm {log}\left (\sin \left (d x +c \right )+1\right )+3 \sin \left (d x +c \right )^{4}-4 \sin \left (d x +c \right )^{3}-12 \sin \left (d x +c \right )}{12 a d} \] Input:
int(cos(d*x+c)*sin(d*x+c)^4/(a+a*sin(d*x+c)),x)
Output:
( - 6*cos(c + d*x)**2 + 12*log(sin(c + d*x) + 1) + 3*sin(c + d*x)**4 - 4*s in(c + d*x)**3 - 12*sin(c + d*x))/(12*a*d)