Integrand size = 29, antiderivative size = 94 \[ \int \frac {\cos ^2(c+d x) \cot ^5(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {2 \csc (c+d x)}{a d}+\frac {\csc ^2(c+d x)}{a d}+\frac {\csc ^3(c+d x)}{3 a d}-\frac {\csc ^4(c+d x)}{4 a d}+\frac {\log (\sin (c+d x))}{a d}-\frac {\sin (c+d x)}{a d} \] Output:
-2*csc(d*x+c)/a/d+csc(d*x+c)^2/a/d+1/3*csc(d*x+c)^3/a/d-1/4*csc(d*x+c)^4/a /d+ln(sin(d*x+c))/a/d-sin(d*x+c)/a/d
Time = 0.32 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.70 \[ \int \frac {\cos ^2(c+d x) \cot ^5(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {24 \csc (c+d x)-12 \csc ^2(c+d x)-4 \csc ^3(c+d x)+3 \csc ^4(c+d x)-12 \log (\sin (c+d x))+12 \sin (c+d x)}{12 a d} \] Input:
Integrate[(Cos[c + d*x]^2*Cot[c + d*x]^5)/(a + a*Sin[c + d*x]),x]
Output:
-1/12*(24*Csc[c + d*x] - 12*Csc[c + d*x]^2 - 4*Csc[c + d*x]^3 + 3*Csc[c + d*x]^4 - 12*Log[Sin[c + d*x]] + 12*Sin[c + d*x])/(a*d)
Time = 0.33 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.78, 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.
\(\displaystyle \int \frac {\cos ^2(c+d x) \cot ^5(c+d x)}{a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^7}{\sin (c+d x)^5 (a \sin (c+d x)+a)}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \csc ^5(c+d x) (a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^2d(a \sin (c+d x))}{a^7 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\csc ^5(c+d x) (a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^2}{a^5}d(a \sin (c+d x))}{a^2 d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left (\csc ^5(c+d x)-\csc ^4(c+d x)-2 \csc ^3(c+d x)+2 \csc ^2(c+d x)+\csc (c+d x)-1\right )d(a \sin (c+d x))}{a^2 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-a \sin (c+d x)-\frac {1}{4} a \csc ^4(c+d x)+\frac {1}{3} a \csc ^3(c+d x)+a \csc ^2(c+d x)-2 a \csc (c+d x)+a \log (a \sin (c+d x))}{a^2 d}\) |
Input:
Int[(Cos[c + d*x]^2*Cot[c + d*x]^5)/(a + a*Sin[c + d*x]),x]
Output:
(-2*a*Csc[c + d*x] + a*Csc[c + d*x]^2 + (a*Csc[c + d*x]^3)/3 - (a*Csc[c + d*x]^4)/4 + a*Log[a*Sin[c + d*x]] - a*Sin[c + d*x])/(a^2*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 = 1.24 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.66
method | result | size |
derivativedivides | \(\frac {-\sin \left (d x +c \right )-\frac {2}{\sin \left (d x +c \right )}+\ln \left (\sin \left (d x +c \right )\right )-\frac {1}{4 \sin \left (d x +c \right )^{4}}+\frac {1}{\sin \left (d x +c \right )^{2}}+\frac {1}{3 \sin \left (d x +c \right )^{3}}}{d a}\) | \(62\) |
default | \(\frac {-\sin \left (d x +c \right )-\frac {2}{\sin \left (d x +c \right )}+\ln \left (\sin \left (d x +c \right )\right )-\frac {1}{4 \sin \left (d x +c \right )^{4}}+\frac {1}{\sin \left (d x +c \right )^{2}}+\frac {1}{3 \sin \left (d x +c \right )^{3}}}{d a}\) | \(62\) |
risch | \(-\frac {i x}{a}+\frac {i {\mathrm e}^{i \left (d x +c \right )}}{2 a d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{2 d a}-\frac {2 i c}{a d}-\frac {4 i \left (-3 i {\mathrm e}^{6 i \left (d x +c \right )}+3 \,{\mathrm e}^{7 i \left (d x +c \right )}+3 i {\mathrm e}^{4 i \left (d x +c \right )}-7 \,{\mathrm e}^{5 i \left (d x +c \right )}-3 i {\mathrm e}^{2 i \left (d x +c \right )}+7 \,{\mathrm e}^{3 i \left (d x +c \right )}-3 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{3 a d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d a}\) | \(177\) |
Input:
int(cos(d*x+c)^2*cot(d*x+c)^5/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d/a*(-sin(d*x+c)-2/sin(d*x+c)+ln(sin(d*x+c))-1/4/sin(d*x+c)^4+1/sin(d*x+ c)^2+1/3/sin(d*x+c)^3)
Time = 0.09 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.11 \[ \int \frac {\cos ^2(c+d x) \cot ^5(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {12 \, \cos \left (d x + c\right )^{2} - 12 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 4 \, {\left (3 \, \cos \left (d x + c\right )^{4} - 12 \, \cos \left (d x + c\right )^{2} + 8\right )} \sin \left (d x + c\right ) - 9}{12 \, {\left (a d \cos \left (d x + c\right )^{4} - 2 \, a d \cos \left (d x + c\right )^{2} + a d\right )}} \] Input:
integrate(cos(d*x+c)^2*cot(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="fricas" )
Output:
-1/12*(12*cos(d*x + c)^2 - 12*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*log( 1/2*sin(d*x + c)) + 4*(3*cos(d*x + c)^4 - 12*cos(d*x + c)^2 + 8)*sin(d*x + c) - 9)/(a*d*cos(d*x + c)^4 - 2*a*d*cos(d*x + c)^2 + a*d)
\[ \int \frac {\cos ^2(c+d x) \cot ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {\cos ^{2}{\left (c + d x \right )} \cot ^{5}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \] Input:
integrate(cos(d*x+c)**2*cot(d*x+c)**5/(a+a*sin(d*x+c)),x)
Output:
Integral(cos(c + d*x)**2*cot(c + d*x)**5/(sin(c + d*x) + 1), x)/a
Time = 0.03 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.77 \[ \int \frac {\cos ^2(c+d x) \cot ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {12 \, \log \left (\sin \left (d x + c\right )\right )}{a} - \frac {12 \, \sin \left (d x + c\right )}{a} - \frac {24 \, \sin \left (d x + c\right )^{3} - 12 \, \sin \left (d x + c\right )^{2} - 4 \, \sin \left (d x + c\right ) + 3}{a \sin \left (d x + c\right )^{4}}}{12 \, d} \] Input:
integrate(cos(d*x+c)^2*cot(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="maxima" )
Output:
1/12*(12*log(sin(d*x + c))/a - 12*sin(d*x + c)/a - (24*sin(d*x + c)^3 - 12 *sin(d*x + c)^2 - 4*sin(d*x + c) + 3)/(a*sin(d*x + c)^4))/d
Time = 0.14 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.70 \[ \int \frac {\cos ^2(c+d x) \cot ^5(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {24 \, \sin \left (d x + c\right )^{3} - 12 \, \sin \left (d x + c\right )^{2} - 4 \, \sin \left (d x + c\right ) + 3}{\sin \left (d x + c\right )^{4}} - 12 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 12 \, \sin \left (d x + c\right )}{12 \, a d} \] Input:
integrate(cos(d*x+c)^2*cot(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
-1/12*((24*sin(d*x + c)^3 - 12*sin(d*x + c)^2 - 4*sin(d*x + c) + 3)/sin(d* x + c)^4 - 12*log(abs(sin(d*x + c))) + 12*sin(d*x + c))/(a*d)
Time = 31.70 (sec) , antiderivative size = 214, normalized size of antiderivative = 2.28 \[ \int \frac {\cos ^2(c+d x) \cot ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{16\,a\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,a\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,a\,d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a\,d}+\frac {-46\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-\frac {40\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {11\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4}+\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}-\frac {1}{4}}{d\,\left (16\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+16\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )}-\frac {7\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,a\,d}-\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a\,d} \] Input:
int((cos(c + d*x)^2*cot(c + d*x)^5)/(a + a*sin(c + d*x)),x)
Output:
(3*tan(c/2 + (d*x)/2)^2)/(16*a*d) + tan(c/2 + (d*x)/2)^3/(24*a*d) - tan(c/ 2 + (d*x)/2)^4/(64*a*d) + log(tan(c/2 + (d*x)/2))/(a*d) + ((2*tan(c/2 + (d *x)/2))/3 + (11*tan(c/2 + (d*x)/2)^2)/4 - (40*tan(c/2 + (d*x)/2)^3)/3 + 3* tan(c/2 + (d*x)/2)^4 - 46*tan(c/2 + (d*x)/2)^5 - 1/4)/(d*(16*a*tan(c/2 + ( d*x)/2)^4 + 16*a*tan(c/2 + (d*x)/2)^6)) - (7*tan(c/2 + (d*x)/2))/(8*a*d) - log(tan(c/2 + (d*x)/2)^2 + 1)/(a*d)
\[ \int \frac {\cos ^2(c+d x) \cot ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\int \frac {\cos \left (d x +c \right )^{2} \cot \left (d x +c \right )^{5}}{\sin \left (d x +c \right ) a +a}d x \] Input:
int(cos(d*x+c)^2*cot(d*x+c)^5/(a+a*sin(d*x+c)),x)
Output:
int(cos(d*x+c)^2*cot(d*x+c)^5/(a+a*sin(d*x+c)),x)