Integrand size = 29, antiderivative size = 97 \[ \int \frac {\cos ^4(c+d x) \cot ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\csc (c+d x)}{a d}-\frac {\csc ^2(c+d x)}{2 a d}-\frac {2 \log (\sin (c+d x))}{a d}+\frac {2 \sin (c+d x)}{a d}+\frac {\sin ^2(c+d x)}{2 a d}-\frac {\sin ^3(c+d x)}{3 a d} \] Output:
csc(d*x+c)/a/d-1/2*csc(d*x+c)^2/a/d-2*ln(sin(d*x+c))/a/d+2*sin(d*x+c)/a/d+ 1/2*sin(d*x+c)^2/a/d-1/3*sin(d*x+c)^3/a/d
Time = 0.13 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.68 \[ \int \frac {\cos ^4(c+d x) \cot ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {6 \csc (c+d x)-3 \csc ^2(c+d x)-12 \log (\sin (c+d x))+12 \sin (c+d x)+3 \sin ^2(c+d x)-2 \sin ^3(c+d x)}{6 a d} \] Input:
Integrate[(Cos[c + d*x]^4*Cot[c + d*x]^3)/(a + a*Sin[c + d*x]),x]
Output:
(6*Csc[c + d*x] - 3*Csc[c + d*x]^2 - 12*Log[Sin[c + d*x]] + 12*Sin[c + d*x ] + 3*Sin[c + d*x]^2 - 2*Sin[c + d*x]^3)/(6*a*d)
Time = 0.33 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.91, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 3315, 27, 99, 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.
\(\displaystyle \int \frac {\cos ^4(c+d x) \cot ^3(c+d x)}{a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^7}{\sin (c+d x)^3 (a \sin (c+d x)+a)}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \csc ^3(c+d x) (a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^2d(a \sin (c+d x))}{a^7 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\csc ^3(c+d x) (a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^2}{a^3}d(a \sin (c+d x))}{a^4 d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left (a^2 \csc ^3(c+d x)-a^2 \csc ^2(c+d x)-2 a^2 \csc (c+d x)+2 a^2-a^2 \sin ^2(c+d x)+a^2 \sin (c+d x)\right )d(a \sin (c+d x))}{a^4 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {1}{3} a^3 \sin ^3(c+d x)+\frac {1}{2} a^3 \sin ^2(c+d x)+2 a^3 \sin (c+d x)-\frac {1}{2} a^3 \csc ^2(c+d x)+a^3 \csc (c+d x)-2 a^3 \log (a \sin (c+d x))}{a^4 d}\) |
Input:
Int[(Cos[c + d*x]^4*Cot[c + d*x]^3)/(a + a*Sin[c + d*x]),x]
Output:
(a^3*Csc[c + d*x] - (a^3*Csc[c + d*x]^2)/2 - 2*a^3*Log[a*Sin[c + d*x]] + 2 *a^3*Sin[c + d*x] + (a^3*Sin[c + d*x]^2)/2 - (a^3*Sin[c + d*x]^3)/3)/(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_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && Intege rQ[(p - 1)/2] && EqQ[a^2 - b^2, 0]
Time = 2.50 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.66
method | result | size |
derivativedivides | \(\frac {-\frac {\sin \left (d x +c \right )^{3}}{3}+\frac {\sin \left (d x +c \right )^{2}}{2}+2 \sin \left (d x +c \right )+\frac {1}{\sin \left (d x +c \right )}-\frac {1}{2 \sin \left (d x +c \right )^{2}}-2 \ln \left (\sin \left (d x +c \right )\right )}{d a}\) | \(64\) |
default | \(\frac {-\frac {\sin \left (d x +c \right )^{3}}{3}+\frac {\sin \left (d x +c \right )^{2}}{2}+2 \sin \left (d x +c \right )+\frac {1}{\sin \left (d x +c \right )}-\frac {1}{2 \sin \left (d x +c \right )^{2}}-2 \ln \left (\sin \left (d x +c \right )\right )}{d a}\) | \(64\) |
risch | \(\frac {2 i x}{a}-\frac {i {\mathrm e}^{3 i \left (d x +c \right )}}{24 d a}-\frac {{\mathrm e}^{2 i \left (d x +c \right )}}{8 a d}-\frac {7 i {\mathrm e}^{i \left (d x +c \right )}}{8 a d}+\frac {7 i {\mathrm e}^{-i \left (d x +c \right )}}{8 d a}-\frac {{\mathrm e}^{-2 i \left (d x +c \right )}}{8 a d}+\frac {i {\mathrm e}^{-3 i \left (d x +c \right )}}{24 d a}+\frac {4 i c}{a d}+\frac {2 i \left (-i {\mathrm e}^{2 i \left (d x +c \right )}+{\mathrm e}^{3 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}\right )}{a d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {2 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d a}\) | \(200\) |
Input:
int(cos(d*x+c)^4*cot(d*x+c)^3/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d/a*(-1/3*sin(d*x+c)^3+1/2*sin(d*x+c)^2+2*sin(d*x+c)+1/sin(d*x+c)-1/2/si n(d*x+c)^2-2*ln(sin(d*x+c)))
Time = 0.09 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.94 \[ \int \frac {\cos ^4(c+d x) \cot ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {6 \, \cos \left (d x + c\right )^{4} - 9 \, \cos \left (d x + c\right )^{2} + 24 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 4 \, {\left (\cos \left (d x + c\right )^{4} + 4 \, \cos \left (d x + c\right )^{2} - 8\right )} \sin \left (d x + c\right ) - 3}{12 \, {\left (a d \cos \left (d x + c\right )^{2} - a d\right )}} \] Input:
integrate(cos(d*x+c)^4*cot(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="fricas" )
Output:
-1/12*(6*cos(d*x + c)^4 - 9*cos(d*x + c)^2 + 24*(cos(d*x + c)^2 - 1)*log(1 /2*sin(d*x + c)) - 4*(cos(d*x + c)^4 + 4*cos(d*x + c)^2 - 8)*sin(d*x + c) - 3)/(a*d*cos(d*x + c)^2 - a*d)
\[ \int \frac {\cos ^4(c+d x) \cot ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {\cos ^{4}{\left (c + d x \right )} \cot ^{3}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \] Input:
integrate(cos(d*x+c)**4*cot(d*x+c)**3/(a+a*sin(d*x+c)),x)
Output:
Integral(cos(c + d*x)**4*cot(c + d*x)**3/(sin(c + d*x) + 1), x)/a
Time = 0.04 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.76 \[ \int \frac {\cos ^4(c+d x) \cot ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {2 \, \sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )^{2} - 12 \, \sin \left (d x + c\right )}{a} + \frac {12 \, \log \left (\sin \left (d x + c\right )\right )}{a} - \frac {3 \, {\left (2 \, \sin \left (d x + c\right ) - 1\right )}}{a \sin \left (d x + c\right )^{2}}}{6 \, d} \] Input:
integrate(cos(d*x+c)^4*cot(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="maxima" )
Output:
-1/6*((2*sin(d*x + c)^3 - 3*sin(d*x + c)^2 - 12*sin(d*x + c))/a + 12*log(s in(d*x + c))/a - 3*(2*sin(d*x + c) - 1)/(a*sin(d*x + c)^2))/d
Time = 0.18 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.69 \[ \int \frac {\cos ^4(c+d x) \cot ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {2 \, \sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )^{2} - \frac {3 \, {\left (2 \, \sin \left (d x + c\right ) - 1\right )}}{\sin \left (d x + c\right )^{2}} + 12 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - 12 \, \sin \left (d x + c\right )}{6 \, a d} \] Input:
integrate(cos(d*x+c)^4*cot(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
-1/6*(2*sin(d*x + c)^3 - 3*sin(d*x + c)^2 - 3*(2*sin(d*x + c) - 1)/sin(d*x + c)^2 + 12*log(abs(sin(d*x + c))) - 12*sin(d*x + c))/(a*d)
Time = 31.56 (sec) , antiderivative size = 231, normalized size of antiderivative = 2.38 \[ \int \frac {\cos ^4(c+d x) \cot ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {18\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\frac {15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{2}+\frac {82\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{3}+\frac {13\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{2}+22\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {1}{2}}{d\,\left (4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+12\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+12\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a\,d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a\,d}+\frac {2\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a\,d} \] Input:
int((cos(c + d*x)^4*cot(c + d*x)^3)/(a + a*sin(c + d*x)),x)
Output:
(2*tan(c/2 + (d*x)/2) - (3*tan(c/2 + (d*x)/2)^2)/2 + 22*tan(c/2 + (d*x)/2) ^3 + (13*tan(c/2 + (d*x)/2)^4)/2 + (82*tan(c/2 + (d*x)/2)^5)/3 + (15*tan(c /2 + (d*x)/2)^6)/2 + 18*tan(c/2 + (d*x)/2)^7 - 1/2)/(d*(4*a*tan(c/2 + (d*x )/2)^2 + 12*a*tan(c/2 + (d*x)/2)^4 + 12*a*tan(c/2 + (d*x)/2)^6 + 4*a*tan(c /2 + (d*x)/2)^8)) - (2*log(tan(c/2 + (d*x)/2)))/(a*d) - tan(c/2 + (d*x)/2) ^2/(8*a*d) + tan(c/2 + (d*x)/2)/(2*a*d) + (2*log(tan(c/2 + (d*x)/2)^2 + 1) )/(a*d)
\[ \int \frac {\cos ^4(c+d x) \cot ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\int \frac {\cos \left (d x +c \right )^{4} \cot \left (d x +c \right )^{3}}{\sin \left (d x +c \right ) a +a}d x \] Input:
int(cos(d*x+c)^4*cot(d*x+c)^3/(a+a*sin(d*x+c)),x)
Output:
int(cos(d*x+c)^4*cot(d*x+c)^3/(a+a*sin(d*x+c)),x)