Integrand size = 27, antiderivative size = 74 \[ \int \frac {\cos (c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {3 \log (1+\sin (c+d x))}{a^3 d}+\frac {\sin (c+d x)}{a^3 d}+\frac {1}{2 a d (a+a \sin (c+d x))^2}-\frac {3}{d \left (a^3+a^3 \sin (c+d x)\right )} \] Output:
-3*ln(1+sin(d*x+c))/a^3/d+sin(d*x+c)/a^3/d+1/2/a/d/(a+a*sin(d*x+c))^2-3/d/ (a^3+a^3*sin(d*x+c))
Time = 0.19 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.95 \[ \int \frac {\cos (c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {-12 \log (1+\sin (c+d x))+4 \sin (c+d x)+\frac {-9-10 \sin (c+d x)}{(1+\sin (c+d x))^2}+\frac {\sin ^2(c+d x)}{(1+\sin (c+d x))^2}}{4 a^3 d} \] Input:
Integrate[(Cos[c + d*x]*Sin[c + d*x]^3)/(a + a*Sin[c + d*x])^3,x]
Output:
(-12*Log[1 + Sin[c + d*x]] + 4*Sin[c + d*x] + (-9 - 10*Sin[c + d*x])/(1 + Sin[c + d*x])^2 + Sin[c + d*x]^2/(1 + Sin[c + d*x])^2)/(4*a^3*d)
Time = 0.30 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.89, 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.
\(\displaystyle \int \frac {\sin ^3(c+d x) \cos (c+d x)}{(a \sin (c+d x)+a)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^3 \cos (c+d x)}{(a \sin (c+d x)+a)^3}dx\) |
\(\Big \downarrow \) 3312 |
\(\displaystyle \frac {\int \frac {\sin ^3(c+d x)}{(\sin (c+d x) a+a)^3}d(a \sin (c+d x))}{a d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {a^3 \sin ^3(c+d x)}{(\sin (c+d x) a+a)^3}d(a \sin (c+d x))}{a^4 d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\int \left (-\frac {a^3}{(\sin (c+d x) a+a)^3}+\frac {3 a^2}{(\sin (c+d x) a+a)^2}-\frac {3 a}{\sin (c+d x) a+a}+1\right )d(a \sin (c+d x))}{a^4 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {a^3}{2 (a \sin (c+d x)+a)^2}-\frac {3 a^2}{a \sin (c+d x)+a}+a \sin (c+d x)-3 a \log (a \sin (c+d x)+a)}{a^4 d}\) |
Input:
Int[(Cos[c + d*x]*Sin[c + d*x]^3)/(a + a*Sin[c + d*x])^3,x]
Output:
(-3*a*Log[a + a*Sin[c + d*x]] + a*Sin[c + d*x] + a^3/(2*(a + a*Sin[c + d*x ])^2) - (3*a^2)/(a + a*Sin[c + d*x]))/(a^4*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 = 0.58 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.68
method | result | size |
derivativedivides | \(\frac {\sin \left (d x +c \right )-3 \ln \left (1+\sin \left (d x +c \right )\right )-\frac {3}{1+\sin \left (d x +c \right )}+\frac {1}{2 \left (1+\sin \left (d x +c \right )\right )^{2}}}{d \,a^{3}}\) | \(50\) |
default | \(\frac {\sin \left (d x +c \right )-3 \ln \left (1+\sin \left (d x +c \right )\right )-\frac {3}{1+\sin \left (d x +c \right )}+\frac {1}{2 \left (1+\sin \left (d x +c \right )\right )^{2}}}{d \,a^{3}}\) | \(50\) |
parallelrisch | \(\frac {\left (6 \cos \left (2 d x +2 c \right )-24 \sin \left (d x +c \right )-18\right ) \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )+\left (-12 \cos \left (2 d x +2 c \right )+48 \sin \left (d x +c \right )+36\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+9 \cos \left (2 d x +2 c \right )-15 \sin \left (d x +c \right )+\sin \left (3 d x +3 c \right )-9}{2 d \,a^{3} \left (-3+\cos \left (2 d x +2 c \right )-4 \sin \left (d x +c \right )\right )}\) | \(128\) |
risch | \(\frac {3 i x}{a^{3}}-\frac {i {\mathrm e}^{i \left (d x +c \right )}}{2 d \,a^{3}}+\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{2 d \,a^{3}}+\frac {6 i c}{d \,a^{3}}-\frac {2 i \left (-3 \,{\mathrm e}^{i \left (d x +c \right )}+5 i {\mathrm e}^{2 i \left (d x +c \right )}+3 \,{\mathrm e}^{3 i \left (d x +c \right )}\right )}{d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{4}}-\frac {6 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{3}}\) | \(134\) |
norman | \(\frac {\frac {6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}+\frac {6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{d a}+\frac {110 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d a}+\frac {110 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d a}+\frac {156 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d a}+\frac {156 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d a}+\frac {24 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d a}+\frac {24 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d a}+\frac {192 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d a}+\frac {192 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d a}+\frac {56 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d a}+\frac {56 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{d a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {6 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{3}}+\frac {3 \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{d \,a^{3}}\) | \(303\) |
Input:
int(cos(d*x+c)*sin(d*x+c)^3/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/d/a^3*(sin(d*x+c)-3*ln(1+sin(d*x+c))-3/(1+sin(d*x+c))+1/2/(1+sin(d*x+c)) ^2)
Time = 0.08 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.28 \[ \int \frac {\cos (c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {4 \, \cos \left (d x + c\right )^{2} - 6 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) - 2\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (\cos \left (d x + c\right )^{2} + 1\right )} \sin \left (d x + c\right ) + 1}{2 \, {\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \sin \left (d x + c\right ) - 2 \, a^{3} d\right )}} \] Input:
integrate(cos(d*x+c)*sin(d*x+c)^3/(a+a*sin(d*x+c))^3,x, algorithm="fricas" )
Output:
1/2*(4*cos(d*x + c)^2 - 6*(cos(d*x + c)^2 - 2*sin(d*x + c) - 2)*log(sin(d* x + c) + 1) + 2*(cos(d*x + c)^2 + 1)*sin(d*x + c) + 1)/(a^3*d*cos(d*x + c) ^2 - 2*a^3*d*sin(d*x + c) - 2*a^3*d)
Leaf count of result is larger than twice the leaf count of optimal. 303 vs. \(2 (63) = 126\).
Time = 0.74 (sec) , antiderivative size = 303, normalized size of antiderivative = 4.09 \[ \int \frac {\cos (c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\begin {cases} - \frac {6 \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin ^{2}{\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} - \frac {12 \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin {\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} - \frac {6 \log {\left (\sin {\left (c + d x \right )} + 1 \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} + \frac {2 \sin ^{3}{\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} - \frac {12 \sin {\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} - \frac {9}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \sin ^{3}{\left (c \right )} \cos {\left (c \right )}}{\left (a \sin {\left (c \right )} + a\right )^{3}} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)*sin(d*x+c)**3/(a+a*sin(d*x+c))**3,x)
Output:
Piecewise((-6*log(sin(c + d*x) + 1)*sin(c + d*x)**2/(2*a**3*d*sin(c + d*x) **2 + 4*a**3*d*sin(c + d*x) + 2*a**3*d) - 12*log(sin(c + d*x) + 1)*sin(c + d*x)/(2*a**3*d*sin(c + d*x)**2 + 4*a**3*d*sin(c + d*x) + 2*a**3*d) - 6*lo g(sin(c + d*x) + 1)/(2*a**3*d*sin(c + d*x)**2 + 4*a**3*d*sin(c + d*x) + 2* a**3*d) + 2*sin(c + d*x)**3/(2*a**3*d*sin(c + d*x)**2 + 4*a**3*d*sin(c + d *x) + 2*a**3*d) - 12*sin(c + d*x)/(2*a**3*d*sin(c + d*x)**2 + 4*a**3*d*sin (c + d*x) + 2*a**3*d) - 9/(2*a**3*d*sin(c + d*x)**2 + 4*a**3*d*sin(c + d*x ) + 2*a**3*d), Ne(d, 0)), (x*sin(c)**3*cos(c)/(a*sin(c) + a)**3, True))
Time = 0.03 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.96 \[ \int \frac {\cos (c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {6 \, \sin \left (d x + c\right ) + 5}{a^{3} \sin \left (d x + c\right )^{2} + 2 \, a^{3} \sin \left (d x + c\right ) + a^{3}} + \frac {6 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}} - \frac {2 \, \sin \left (d x + c\right )}{a^{3}}}{2 \, d} \] Input:
integrate(cos(d*x+c)*sin(d*x+c)^3/(a+a*sin(d*x+c))^3,x, algorithm="maxima" )
Output:
-1/2*((6*sin(d*x + c) + 5)/(a^3*sin(d*x + c)^2 + 2*a^3*sin(d*x + c) + a^3) + 6*log(sin(d*x + c) + 1)/a^3 - 2*sin(d*x + c)/a^3)/d
Time = 0.15 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.81 \[ \int \frac {\cos (c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {3 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3} d} + \frac {\sin \left (d x + c\right )}{a^{3} d} - \frac {6 \, \sin \left (d x + c\right ) + 5}{2 \, a^{3} d {\left (\sin \left (d x + c\right ) + 1\right )}^{2}} \] Input:
integrate(cos(d*x+c)*sin(d*x+c)^3/(a+a*sin(d*x+c))^3,x, algorithm="giac")
Output:
-3*log(abs(sin(d*x + c) + 1))/(a^3*d) + sin(d*x + c)/(a^3*d) - 1/2*(6*sin( d*x + c) + 5)/(a^3*d*(sin(d*x + c) + 1)^2)
Time = 0.05 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.80 \[ \int \frac {\cos (c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\sin \left (c+d\,x\right )}{a^3\,d}-\frac {3\,\ln \left (\sin \left (c+d\,x\right )+1\right )}{a^3\,d}-\frac {3\,\sin \left (c+d\,x\right )+\frac {5}{2}}{a^3\,d\,{\left (\sin \left (c+d\,x\right )+1\right )}^2} \] Input:
int((cos(c + d*x)*sin(c + d*x)^3)/(a + a*sin(c + d*x))^3,x)
Output:
sin(c + d*x)/(a^3*d) - (3*log(sin(c + d*x) + 1))/(a^3*d) - (3*sin(c + d*x) + 5/2)/(a^3*d*(sin(c + d*x) + 1)^2)
Time = 0.16 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.31 \[ \int \frac {\cos (c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {-6 \,\mathrm {log}\left (\sin \left (d x +c \right )+1\right ) \sin \left (d x +c \right )^{2}-12 \,\mathrm {log}\left (\sin \left (d x +c \right )+1\right ) \sin \left (d x +c \right )-6 \,\mathrm {log}\left (\sin \left (d x +c \right )+1\right )+2 \sin \left (d x +c \right )^{3}+6 \sin \left (d x +c \right )^{2}-3}{2 a^{3} d \left (\sin \left (d x +c \right )^{2}+2 \sin \left (d x +c \right )+1\right )} \] Input:
int(cos(d*x+c)*sin(d*x+c)^3/(a+a*sin(d*x+c))^3,x)
Output:
( - 6*log(sin(c + d*x) + 1)*sin(c + d*x)**2 - 12*log(sin(c + d*x) + 1)*sin (c + d*x) - 6*log(sin(c + d*x) + 1) + 2*sin(c + d*x)**3 + 6*sin(c + d*x)** 2 - 3)/(2*a**3*d*(sin(c + d*x)**2 + 2*sin(c + d*x) + 1))