Integrand size = 27, antiderivative size = 70 \[ \int \frac {\cos (c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {3 \log (1+\sin (c+d x))}{a^2 d}-\frac {2 \sin (c+d x)}{a^2 d}+\frac {\sin ^2(c+d x)}{2 a^2 d}+\frac {1}{d \left (a^2+a^2 \sin (c+d x)\right )} \] Output:
3*ln(1+sin(d*x+c))/a^2/d-2*sin(d*x+c)/a^2/d+1/2*sin(d*x+c)^2/a^2/d+1/d/(a^ 2+a^2*sin(d*x+c))
Time = 0.13 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.01 \[ \int \frac {\cos (c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {2+6 \log (1+\sin (c+d x))+(-4+6 \log (1+\sin (c+d x))) \sin (c+d x)-3 \sin ^2(c+d x)+\sin ^3(c+d x)}{2 a^2 d (1+\sin (c+d x))} \] Input:
Integrate[(Cos[c + d*x]*Sin[c + d*x]^3)/(a + a*Sin[c + d*x])^2,x]
Output:
(2 + 6*Log[1 + Sin[c + d*x]] + (-4 + 6*Log[1 + Sin[c + d*x]])*Sin[c + d*x] - 3*Sin[c + d*x]^2 + Sin[c + d*x]^3)/(2*a^2*d*(1 + Sin[c + d*x]))
Time = 0.30 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.94, 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.
\(\displaystyle \int \frac {\sin ^3(c+d x) \cos (c+d x)}{(a \sin (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^3 \cos (c+d x)}{(a \sin (c+d x)+a)^2}dx\) |
\(\Big \downarrow \) 3312 |
\(\displaystyle \frac {\int \frac {\sin ^3(c+d x)}{(\sin (c+d x) a+a)^2}d(a \sin (c+d x))}{a d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {a^3 \sin ^3(c+d x)}{(\sin (c+d x) a+a)^2}d(a \sin (c+d x))}{a^4 d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\int \left (-\frac {a^3}{(\sin (c+d x) a+a)^2}+\frac {3 a^2}{\sin (c+d x) a+a}+\sin (c+d x) a-2 a\right )d(a \sin (c+d x))}{a^4 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {a^3}{a \sin (c+d x)+a}+\frac {1}{2} a^2 \sin ^2(c+d x)-2 a^2 \sin (c+d x)+3 a^2 \log (a \sin (c+d x)+a)}{a^4 d}\) |
Input:
Int[(Cos[c + d*x]*Sin[c + d*x]^3)/(a + a*Sin[c + d*x])^2,x]
Output:
(3*a^2*Log[a + a*Sin[c + d*x]] - 2*a^2*Sin[c + d*x] + (a^2*Sin[c + d*x]^2) /2 + a^3/(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 = 0.49 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.69
method | result | size |
derivativedivides | \(\frac {\frac {\sin \left (d x +c \right )^{2}}{2}-2 \sin \left (d x +c \right )+\frac {1}{1+\sin \left (d x +c \right )}+3 \ln \left (1+\sin \left (d x +c \right )\right )}{d \,a^{2}}\) | \(48\) |
default | \(\frac {\frac {\sin \left (d x +c \right )^{2}}{2}-2 \sin \left (d x +c \right )+\frac {1}{1+\sin \left (d x +c \right )}+3 \ln \left (1+\sin \left (d x +c \right )\right )}{d \,a^{2}}\) | \(48\) |
parallelrisch | \(\frac {\left (-24 \sin \left (d x +c \right )-24\right ) \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )+\left (48 \sin \left (d x +c \right )+48\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+6 \cos \left (2 d x +2 c \right )-21 \sin \left (d x +c \right )-\sin \left (3 d x +3 c \right )-6}{8 \left (1+\sin \left (d x +c \right )\right ) d \,a^{2}}\) | \(97\) |
risch | \(-\frac {3 i x}{a^{2}}+\frac {i {\mathrm e}^{i \left (d x +c \right )}}{d \,a^{2}}-\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{d \,a^{2}}-\frac {6 i c}{d \,a^{2}}+\frac {2 i {\mathrm e}^{i \left (d x +c \right )}}{d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{2}}+\frac {6 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{2}}-\frac {\cos \left (2 d x +2 c \right )}{4 d \,a^{2}}\) | \(125\) |
norman | \(\frac {-\frac {12 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d a}-\frac {12 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d a}-\frac {6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}-\frac {26 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d a}-\frac {26 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d a}-\frac {6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{d a}-\frac {38 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d a}-\frac {38 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d a}-\frac {46 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d a}-\frac {46 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {6 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{2}}-\frac {3 \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{d \,a^{2}}\) | \(265\) |
Input:
int(cos(d*x+c)*sin(d*x+c)^3/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d/a^2*(1/2*sin(d*x+c)^2-2*sin(d*x+c)+1/(1+sin(d*x+c))+3*ln(1+sin(d*x+c)) )
Time = 0.10 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.03 \[ \int \frac {\cos (c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {6 \, \cos \left (d x + c\right )^{2} + 12 \, {\left (\sin \left (d x + c\right ) + 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, \cos \left (d x + c\right )^{2} + 7\right )} \sin \left (d x + c\right ) - 3}{4 \, {\left (a^{2} d \sin \left (d x + c\right ) + a^{2} d\right )}} \] Input:
integrate(cos(d*x+c)*sin(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="fricas" )
Output:
1/4*(6*cos(d*x + c)^2 + 12*(sin(d*x + c) + 1)*log(sin(d*x + c) + 1) - (2*c os(d*x + c)^2 + 7)*sin(d*x + c) - 3)/(a^2*d*sin(d*x + c) + a^2*d)
Leaf count of result is larger than twice the leaf count of optimal. 170 vs. \(2 (61) = 122\).
Time = 0.64 (sec) , antiderivative size = 170, normalized size of antiderivative = 2.43 \[ \int \frac {\cos (c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\begin {cases} \frac {6 \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin {\left (c + d x \right )}}{2 a^{2} d \sin {\left (c + d x \right )} + 2 a^{2} d} + \frac {6 \log {\left (\sin {\left (c + d x \right )} + 1 \right )}}{2 a^{2} d \sin {\left (c + d x \right )} + 2 a^{2} d} + \frac {\sin ^{3}{\left (c + d x \right )}}{2 a^{2} d \sin {\left (c + d x \right )} + 2 a^{2} d} - \frac {3 \sin ^{2}{\left (c + d x \right )}}{2 a^{2} d \sin {\left (c + d x \right )} + 2 a^{2} d} + \frac {6}{2 a^{2} d \sin {\left (c + d x \right )} + 2 a^{2} d} & \text {for}\: d \neq 0 \\\frac {x \sin ^{3}{\left (c \right )} \cos {\left (c \right )}}{\left (a \sin {\left (c \right )} + a\right )^{2}} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)*sin(d*x+c)**3/(a+a*sin(d*x+c))**2,x)
Output:
Piecewise((6*log(sin(c + d*x) + 1)*sin(c + d*x)/(2*a**2*d*sin(c + d*x) + 2 *a**2*d) + 6*log(sin(c + d*x) + 1)/(2*a**2*d*sin(c + d*x) + 2*a**2*d) + si n(c + d*x)**3/(2*a**2*d*sin(c + d*x) + 2*a**2*d) - 3*sin(c + d*x)**2/(2*a* *2*d*sin(c + d*x) + 2*a**2*d) + 6/(2*a**2*d*sin(c + d*x) + 2*a**2*d), Ne(d , 0)), (x*sin(c)**3*cos(c)/(a*sin(c) + a)**2, True))
Time = 0.03 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.84 \[ \int \frac {\cos (c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {2}{a^{2} \sin \left (d x + c\right ) + a^{2}} + \frac {\sin \left (d x + c\right )^{2} - 4 \, \sin \left (d x + c\right )}{a^{2}} + \frac {6 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}}}{2 \, d} \] Input:
integrate(cos(d*x+c)*sin(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="maxima" )
Output:
1/2*(2/(a^2*sin(d*x + c) + a^2) + (sin(d*x + c)^2 - 4*sin(d*x + c))/a^2 + 6*log(sin(d*x + c) + 1)/a^2)/d
Time = 0.19 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00 \[ \int \frac {\cos (c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {3 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2} d} + \frac {1}{a^{2} d {\left (\sin \left (d x + c\right ) + 1\right )}} + \frac {a^{2} d \sin \left (d x + c\right )^{2} - 4 \, a^{2} d \sin \left (d x + c\right )}{2 \, a^{4} d^{2}} \] Input:
integrate(cos(d*x+c)*sin(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="giac")
Output:
3*log(abs(sin(d*x + c) + 1))/(a^2*d) + 1/(a^2*d*(sin(d*x + c) + 1)) + 1/2* (a^2*d*sin(d*x + c)^2 - 4*a^2*d*sin(d*x + c))/(a^4*d^2)
Time = 18.07 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.84 \[ \int \frac {\cos (c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {1}{a^2\,\sin \left (c+d\,x\right )+a^2}+\frac {3\,\ln \left (\sin \left (c+d\,x\right )+1\right )}{a^2}-\frac {2\,\sin \left (c+d\,x\right )}{a^2}+\frac {{\sin \left (c+d\,x\right )}^2}{2\,a^2}}{d} \] Input:
int((cos(c + d*x)*sin(c + d*x)^3)/(a + a*sin(c + d*x))^2,x)
Output:
(1/(a^2*sin(c + d*x) + a^2) + (3*log(sin(c + d*x) + 1))/a^2 - (2*sin(c + d *x))/a^2 + sin(c + d*x)^2/(2*a^2))/d
Time = 0.15 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.04 \[ \int \frac {\cos (c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {6 \,\mathrm {log}\left (\sin \left (d x +c \right )+1\right ) \sin \left (d x +c \right )+6 \,\mathrm {log}\left (\sin \left (d x +c \right )+1\right )+\sin \left (d x +c \right )^{3}-3 \sin \left (d x +c \right )^{2}-6 \sin \left (d x +c \right )}{2 a^{2} d \left (\sin \left (d x +c \right )+1\right )} \] Input:
int(cos(d*x+c)*sin(d*x+c)^3/(a+a*sin(d*x+c))^2,x)
Output:
(6*log(sin(c + d*x) + 1)*sin(c + d*x) + 6*log(sin(c + d*x) + 1) + sin(c + d*x)**3 - 3*sin(c + d*x)**2 - 6*sin(c + d*x))/(2*a**2*d*(sin(c + d*x) + 1) )