Integrand size = 21, antiderivative size = 36 \[ \int \frac {\cos ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\log (1+\sin (c+d x))}{a^3 d}-\frac {2}{a^3 d (1+\sin (c+d x))} \] Output:
-ln(1+sin(d*x+c))/a^3/d-2/a^3/d/(1+sin(d*x+c))
Time = 0.07 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.61 \[ \int \frac {\cos ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {2+\log (1+\sin (c+d x))+\log (1+\sin (c+d x)) \sin (c+d x)}{a^3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2} \] Input:
Integrate[Cos[c + d*x]^3/(a + a*Sin[c + d*x])^3,x]
Output:
-((2 + Log[1 + Sin[c + d*x]] + Log[1 + Sin[c + d*x]]*Sin[c + d*x])/(a^3*d* (Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2))
Time = 0.24 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3146, 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.
\(\displaystyle \int \frac {\cos ^3(c+d x)}{(a \sin (c+d x)+a)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^3}{(a \sin (c+d x)+a)^3}dx\) |
\(\Big \downarrow \) 3146 |
\(\displaystyle \frac {\int \frac {a-a \sin (c+d x)}{(\sin (c+d x) a+a)^2}d(a \sin (c+d x))}{a^3 d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\int \left (\frac {2 a}{(\sin (c+d x) a+a)^2}+\frac {1}{-\sin (c+d x) a-a}\right )d(a \sin (c+d x))}{a^3 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {2 a}{a \sin (c+d x)+a}-\log (a \sin (c+d x)+a)}{a^3 d}\) |
Input:
Int[Cos[c + d*x]^3/(a + a*Sin[c + d*x])^3,x]
Output:
(-Log[a + a*Sin[c + d*x]] - (2*a)/(a + a*Sin[c + d*x]))/(a^3*d)
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_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x )^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && I ntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] || !IntegerQ[m + 1/ 2])
Time = 0.28 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.89
method | result | size |
derivativedivides | \(\frac {-\frac {2}{1+\sin \left (d x +c \right )}-\ln \left (1+\sin \left (d x +c \right )\right )}{d \,a^{3}}\) | \(32\) |
default | \(\frac {-\frac {2}{1+\sin \left (d x +c \right )}-\ln \left (1+\sin \left (d x +c \right )\right )}{d \,a^{3}}\) | \(32\) |
parallelrisch | \(\frac {\left (1+\sin \left (d x +c \right )\right ) \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )+\left (-2-2 \sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+2 \sin \left (d x +c \right )}{\left (1+\sin \left (d x +c \right )\right ) d \,a^{3}}\) | \(71\) |
risch | \(\frac {i x}{a^{3}}+\frac {2 i c}{d \,a^{3}}-\frac {4 i {\mathrm e}^{i \left (d x +c \right )}}{d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{2}}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{3}}\) | \(72\) |
norman | \(\frac {\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}+\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{d a}+\frac {12 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a d}+\frac {12 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d a}+\frac {24 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d a}+\frac {24 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d a}+\frac {40 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d a}+\frac {40 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d a}+\frac {48 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d a}+\frac {48 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {\ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{d \,a^{3}}-\frac {2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{3}}\) | \(264\) |
Input:
int(cos(d*x+c)^3/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/d/a^3*(-2/(1+sin(d*x+c))-ln(1+sin(d*x+c)))
Time = 0.13 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.14 \[ \int \frac {\cos ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {{\left (\sin \left (d x + c\right ) + 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 2}{a^{3} d \sin \left (d x + c\right ) + a^{3} d} \] Input:
integrate(cos(d*x+c)^3/(a+a*sin(d*x+c))^3,x, algorithm="fricas")
Output:
-((sin(d*x + c) + 1)*log(sin(d*x + c) + 1) + 2)/(a^3*d*sin(d*x + c) + a^3* d)
Leaf count of result is larger than twice the leaf count of optimal. 299 vs. \(2 (31) = 62\).
Time = 0.68 (sec) , antiderivative size = 299, normalized size of antiderivative = 8.31 \[ \int \frac {\cos ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\begin {cases} - \frac {2 \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin ^{2}{\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} - \frac {4 \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin {\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} - \frac {2 \log {\left (\sin {\left (c + d x \right )} + 1 \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} - \frac {2 \sin {\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} - \frac {\cos ^{2}{\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} - \frac {2}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \cos ^{3}{\left (c \right )}}{\left (a \sin {\left (c \right )} + a\right )^{3}} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**3/(a+a*sin(d*x+c))**3,x)
Output:
Piecewise((-2*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) - 4*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) - 2*log (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)/(2*a**3*d*sin(c + d*x)**2 + 4*a**3*d*sin(c + d*x) + 2*a**3*d) - cos(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) - 2/(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*cos(c)**3/(a*sin(c) + a)**3, True))
Time = 0.03 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.03 \[ \int \frac {\cos ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {2}{a^{3} \sin \left (d x + c\right ) + a^{3}} + \frac {\log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}}}{d} \] Input:
integrate(cos(d*x+c)^3/(a+a*sin(d*x+c))^3,x, algorithm="maxima")
Output:
-(2/(a^3*sin(d*x + c) + a^3) + log(sin(d*x + c) + 1)/a^3)/d
Time = 0.12 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.03 \[ \int \frac {\cos ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3} d} - \frac {2}{a^{3} d {\left (\sin \left (d x + c\right ) + 1\right )}} \] Input:
integrate(cos(d*x+c)^3/(a+a*sin(d*x+c))^3,x, algorithm="giac")
Output:
-log(abs(sin(d*x + c) + 1))/(a^3*d) - 2/(a^3*d*(sin(d*x + c) + 1))
Time = 25.37 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00 \[ \int \frac {\cos ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {2}{a^3\,d\,\left (\sin \left (c+d\,x\right )+1\right )}-\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )}{a^3\,d} \] Input:
int(cos(c + d*x)^3/(a + a*sin(c + d*x))^3,x)
Output:
- 2/(a^3*d*(sin(c + d*x) + 1)) - log(sin(c + d*x) + 1)/(a^3*d)
Time = 0.17 (sec) , antiderivative size = 95, normalized size of antiderivative = 2.64 \[ \int \frac {\cos ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {-\cos \left (d x +c \right )^{2}-2 \,\mathrm {log}\left (\sin \left (d x +c \right )+1\right ) \sin \left (d x +c \right )^{2}-4 \,\mathrm {log}\left (\sin \left (d x +c \right )+1\right ) \sin \left (d x +c \right )-2 \,\mathrm {log}\left (\sin \left (d x +c \right )+1\right )+\sin \left (d x +c \right )^{2}-1}{2 a^{3} d \left (\sin \left (d x +c \right )^{2}+2 \sin \left (d x +c \right )+1\right )} \] Input:
int(cos(d*x+c)^3/(a+a*sin(d*x+c))^3,x)
Output:
( - cos(c + d*x)**2 - 2*log(sin(c + d*x) + 1)*sin(c + d*x)**2 - 4*log(sin( c + d*x) + 1)*sin(c + d*x) - 2*log(sin(c + d*x) + 1) + sin(c + d*x)**2 - 1 )/(2*a**3*d*(sin(c + d*x)**2 + 2*sin(c + d*x) + 1))