Integrand size = 27, antiderivative size = 87 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {4 \log (1+\sin (c+d x))}{a^2 d}+\frac {3 \sin (c+d x)}{a^2 d}-\frac {\sin ^2(c+d x)}{a^2 d}+\frac {\sin ^3(c+d x)}{3 a^2 d}-\frac {1}{d \left (a^2+a^2 \sin (c+d x)\right )} \] Output:
-4*ln(1+sin(d*x+c))/a^2/d+3*sin(d*x+c)/a^2/d-sin(d*x+c)^2/a^2/d+1/3*sin(d* x+c)^3/a^2/d-1/d/(a^2+a^2*sin(d*x+c))
Time = 0.23 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.97 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {-3 (1+4 \log (1+\sin (c+d x)))+(9-12 \log (1+\sin (c+d x))) \sin (c+d x)+6 \sin ^2(c+d x)-2 \sin ^3(c+d x)+\sin ^4(c+d x)}{3 a^2 d (1+\sin (c+d x))} \] Input:
Integrate[(Cos[c + d*x]*Sin[c + d*x]^4)/(a + a*Sin[c + d*x])^2,x]
Output:
(-3*(1 + 4*Log[1 + Sin[c + d*x]]) + (9 - 12*Log[1 + Sin[c + d*x]])*Sin[c + d*x] + 6*Sin[c + d*x]^2 - 2*Sin[c + d*x]^3 + Sin[c + d*x]^4)/(3*a^2*d*(1 + Sin[c + d*x]))
Time = 0.31 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.92, 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.
\(\displaystyle \int \frac {\sin ^4(c+d x) \cos (c+d x)}{(a \sin (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^4 \cos (c+d x)}{(a \sin (c+d x)+a)^2}dx\) |
\(\Big \downarrow \) 3312 |
\(\displaystyle \frac {\int \frac {\sin ^4(c+d x)}{(\sin (c+d x) a+a)^2}d(a \sin (c+d x))}{a d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {a^4 \sin ^4(c+d x)}{(\sin (c+d x) a+a)^2}d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\int \left (\frac {a^4}{(\sin (c+d x) a+a)^2}-\frac {4 a^3}{\sin (c+d x) a+a}+\sin ^2(c+d x) a^2-2 \sin (c+d x) a^2+3 a^2\right )d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {a^4}{a \sin (c+d x)+a}+\frac {1}{3} a^3 \sin ^3(c+d x)-a^3 \sin ^2(c+d x)+3 a^3 \sin (c+d x)-4 a^3 \log (a \sin (c+d x)+a)}{a^5 d}\) |
Input:
Int[(Cos[c + d*x]*Sin[c + d*x]^4)/(a + a*Sin[c + d*x])^2,x]
Output:
(-4*a^3*Log[a + a*Sin[c + d*x]] + 3*a^3*Sin[c + d*x] - a^3*Sin[c + d*x]^2 + (a^3*Sin[c + d*x]^3)/3 - a^4/(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 = 0.81 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.69
method | result | size |
derivativedivides | \(\frac {\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )^{2}+3 \sin \left (d x +c \right )-4 \ln \left (1+\sin \left (d x +c \right )\right )-\frac {1}{1+\sin \left (d x +c \right )}}{d \,a^{2}}\) | \(60\) |
default | \(\frac {\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )^{2}+3 \sin \left (d x +c \right )-4 \ln \left (1+\sin \left (d x +c \right )\right )-\frac {1}{1+\sin \left (d x +c \right )}}{d \,a^{2}}\) | \(60\) |
parallelrisch | \(\frac {\left (96 \sin \left (d x +c \right )+96\right ) \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )+\left (-192 \sin \left (d x +c \right )-192\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-28 \cos \left (2 d x +2 c \right )+\cos \left (4 d x +4 c \right )+84 \sin \left (d x +c \right )+4 \sin \left (3 d x +3 c \right )+27}{24 \left (1+\sin \left (d x +c \right )\right ) d \,a^{2}}\) | \(106\) |
risch | \(\frac {4 i x}{a^{2}}-\frac {13 i {\mathrm e}^{i \left (d x +c \right )}}{8 d \,a^{2}}+\frac {13 i {\mathrm e}^{-i \left (d x +c \right )}}{8 d \,a^{2}}+\frac {8 i c}{d \,a^{2}}-\frac {2 i {\mathrm e}^{i \left (d x +c \right )}}{d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{2}}-\frac {8 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{2}}-\frac {\sin \left (3 d x +3 c \right )}{12 d \,a^{2}}+\frac {\cos \left (2 d x +2 c \right )}{2 d \,a^{2}}\) | \(142\) |
norman | \(\frac {\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}+\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{d a}+\frac {72 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d a}+\frac {72 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d a}+\frac {16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d a}+\frac {16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d a}+\frac {128 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d a}+\frac {128 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{3 d a}+\frac {368 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{3 d a}+\frac {368 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d a}+\frac {304 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 d a}+\frac {304 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{3 d a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5} a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {8 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{2}}+\frac {4 \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{d \,a^{2}}\) | \(303\) |
Input:
int(cos(d*x+c)*sin(d*x+c)^4/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d/a^2*(1/3*sin(d*x+c)^3-sin(d*x+c)^2+3*sin(d*x+c)-4*ln(1+sin(d*x+c))-1/( 1+sin(d*x+c)))
Time = 0.08 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.93 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {2 \, \cos \left (d x + c\right )^{4} - 16 \, \cos \left (d x + c\right )^{2} - 24 \, {\left (\sin \left (d x + c\right ) + 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (4 \, \cos \left (d x + c\right )^{2} + 17\right )} \sin \left (d x + c\right ) + 11}{6 \, {\left (a^{2} d \sin \left (d x + c\right ) + a^{2} d\right )}} \] Input:
integrate(cos(d*x+c)*sin(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="fricas" )
Output:
1/6*(2*cos(d*x + c)^4 - 16*cos(d*x + c)^2 - 24*(sin(d*x + c) + 1)*log(sin( d*x + c) + 1) + (4*cos(d*x + c)^2 + 17)*sin(d*x + c) + 11)/(a^2*d*sin(d*x + c) + a^2*d)
Leaf count of result is larger than twice the leaf count of optimal. 201 vs. \(2 (75) = 150\).
Time = 0.87 (sec) , antiderivative size = 201, normalized size of antiderivative = 2.31 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\begin {cases} - \frac {12 \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin {\left (c + d x \right )}}{3 a^{2} d \sin {\left (c + d x \right )} + 3 a^{2} d} - \frac {12 \log {\left (\sin {\left (c + d x \right )} + 1 \right )}}{3 a^{2} d \sin {\left (c + d x \right )} + 3 a^{2} d} + \frac {\sin ^{4}{\left (c + d x \right )}}{3 a^{2} d \sin {\left (c + d x \right )} + 3 a^{2} d} - \frac {2 \sin ^{3}{\left (c + d x \right )}}{3 a^{2} d \sin {\left (c + d x \right )} + 3 a^{2} d} + \frac {6 \sin ^{2}{\left (c + d x \right )}}{3 a^{2} d \sin {\left (c + d x \right )} + 3 a^{2} d} - \frac {12}{3 a^{2} d \sin {\left (c + d x \right )} + 3 a^{2} d} & \text {for}\: d \neq 0 \\\frac {x \sin ^{4}{\left (c \right )} \cos {\left (c \right )}}{\left (a \sin {\left (c \right )} + a\right )^{2}} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)*sin(d*x+c)**4/(a+a*sin(d*x+c))**2,x)
Output:
Piecewise((-12*log(sin(c + d*x) + 1)*sin(c + d*x)/(3*a**2*d*sin(c + d*x) + 3*a**2*d) - 12*log(sin(c + d*x) + 1)/(3*a**2*d*sin(c + d*x) + 3*a**2*d) + sin(c + d*x)**4/(3*a**2*d*sin(c + d*x) + 3*a**2*d) - 2*sin(c + d*x)**3/(3 *a**2*d*sin(c + d*x) + 3*a**2*d) + 6*sin(c + d*x)**2/(3*a**2*d*sin(c + d*x ) + 3*a**2*d) - 12/(3*a**2*d*sin(c + d*x) + 3*a**2*d), Ne(d, 0)), (x*sin(c )**4*cos(c)/(a*sin(c) + a)**2, True))
Time = 0.03 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.80 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {3}{a^{2} \sin \left (d x + c\right ) + a^{2}} - \frac {\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )^{2} + 9 \, \sin \left (d x + c\right )}{a^{2}} + \frac {12 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}}}{3 \, d} \] Input:
integrate(cos(d*x+c)*sin(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="maxima" )
Output:
-1/3*(3/(a^2*sin(d*x + c) + a^2) - (sin(d*x + c)^3 - 3*sin(d*x + c)^2 + 9* sin(d*x + c))/a^2 + 12*log(sin(d*x + c) + 1)/a^2)/d
Time = 0.19 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.05 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {4 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2} d} - \frac {1}{a^{2} d {\left (\sin \left (d x + c\right ) + 1\right )}} + \frac {a^{4} d^{2} \sin \left (d x + c\right )^{3} - 3 \, a^{4} d^{2} \sin \left (d x + c\right )^{2} + 9 \, a^{4} d^{2} \sin \left (d x + c\right )}{3 \, a^{6} d^{3}} \] Input:
integrate(cos(d*x+c)*sin(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="giac")
Output:
-4*log(abs(sin(d*x + c) + 1))/(a^2*d) - 1/(a^2*d*(sin(d*x + c) + 1)) + 1/3 *(a^4*d^2*sin(d*x + c)^3 - 3*a^4*d^2*sin(d*x + c)^2 + 9*a^4*d^2*sin(d*x + c))/(a^6*d^3)
Time = 0.03 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.83 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {1}{a^2\,\sin \left (c+d\,x\right )+a^2}+\frac {4\,\ln \left (\sin \left (c+d\,x\right )+1\right )}{a^2}-\frac {3\,\sin \left (c+d\,x\right )}{a^2}+\frac {{\sin \left (c+d\,x\right )}^2}{a^2}-\frac {{\sin \left (c+d\,x\right )}^3}{3\,a^2}}{d} \] Input:
int((cos(c + d*x)*sin(c + d*x)^4)/(a + a*sin(c + d*x))^2,x)
Output:
-(1/(a^2*sin(c + d*x) + a^2) + (4*log(sin(c + d*x) + 1))/a^2 - (3*sin(c + d*x))/a^2 + sin(c + d*x)^2/a^2 - sin(c + d*x)^3/(3*a^2))/d
Time = 0.16 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.14 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {-6 \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )-6 \cos \left (d x +c \right )^{2}-12 \,\mathrm {log}\left (\sin \left (d x +c \right )+1\right ) \sin \left (d x +c \right )-12 \,\mathrm {log}\left (\sin \left (d x +c \right )+1\right )+\sin \left (d x +c \right )^{4}-8 \sin \left (d x +c \right )^{3}+12 \sin \left (d x +c \right )}{3 a^{2} d \left (\sin \left (d x +c \right )+1\right )} \] Input:
int(cos(d*x+c)*sin(d*x+c)^4/(a+a*sin(d*x+c))^2,x)
Output:
( - 6*cos(c + d*x)**2*sin(c + d*x) - 6*cos(c + d*x)**2 - 12*log(sin(c + d* x) + 1)*sin(c + d*x) - 12*log(sin(c + d*x) + 1) + sin(c + d*x)**4 - 8*sin( c + d*x)**3 + 12*sin(c + d*x))/(3*a**2*d*(sin(c + d*x) + 1))