Integrand size = 19, antiderivative size = 81 \[ \int \cot ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {2 a \csc (c+d x)}{d}+\frac {a \csc ^2(c+d x)}{d}-\frac {a \csc ^3(c+d x)}{3 d}-\frac {a \csc ^4(c+d x)}{4 d}+\frac {a \log (\sin (c+d x))}{d}+\frac {a \sin (c+d x)}{d} \]
2*a*csc(d*x+c)/d+a*csc(d*x+c)^2/d-1/3*a*csc(d*x+c)^3/d-1/4*a*csc(d*x+c)^4/ d+a*ln(sin(d*x+c))/d+a*sin(d*x+c)/d
Time = 0.04 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.19 \[ \int \cot ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \cot ^2(c+d x)}{2 d}-\frac {a \cot ^4(c+d x)}{4 d}+\frac {2 a \csc (c+d x)}{d}-\frac {a \csc ^3(c+d x)}{3 d}+\frac {a \log (\cos (c+d x))}{d}+\frac {a \log (\tan (c+d x))}{d}+\frac {a \sin (c+d x)}{d} \]
(a*Cot[c + d*x]^2)/(2*d) - (a*Cot[c + d*x]^4)/(4*d) + (2*a*Csc[c + d*x])/d - (a*Csc[c + d*x]^3)/(3*d) + (a*Log[Cos[c + d*x]])/d + (a*Log[Tan[c + d*x ]])/d + (a*Sin[c + d*x])/d
Time = 0.23 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.85, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3042, 3186, 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.
\(\displaystyle \int \cot ^5(c+d x) (a \sin (c+d x)+a) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {a \sin (c+d x)+a}{\tan (c+d x)^5}dx\) |
\(\Big \downarrow \) 3186 |
\(\displaystyle \frac {\int \frac {\csc ^5(c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^3}{a^5}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left (\csc ^5(c+d x)+\csc ^4(c+d x)-2 \csc ^3(c+d x)-2 \csc ^2(c+d x)+\csc (c+d x)+1\right )d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a \sin (c+d x)-\frac {1}{4} a \csc ^4(c+d x)-\frac {1}{3} a \csc ^3(c+d x)+a \csc ^2(c+d x)+2 a \csc (c+d x)+a \log (a \sin (c+d x))}{d}\) |
(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])/d
3.1.6.3.1 Defintions of rubi rules used
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[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p _.), x_Symbol] :> Simp[1/f Subst[Int[x^p*((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] && E qQ[a^2 - b^2, 0] && IntegerQ[(p + 1)/2]
Time = 1.50 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.25
method | result | size |
derivativedivides | \(\frac {a \left (-\frac {\cos ^{6}\left (d x +c \right )}{3 \sin \left (d x +c \right )^{3}}+\frac {\cos ^{6}\left (d x +c \right )}{\sin \left (d x +c \right )}+\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )\right )+a \left (-\frac {\left (\cot ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\) | \(101\) |
default | \(\frac {a \left (-\frac {\cos ^{6}\left (d x +c \right )}{3 \sin \left (d x +c \right )^{3}}+\frac {\cos ^{6}\left (d x +c \right )}{\sin \left (d x +c \right )}+\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )\right )+a \left (-\frac {\left (\cot ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\) | \(101\) |
risch | \(-i a x -\frac {i a \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {i a \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {2 i a c}{d}+\frac {4 i a \left (3 i {\mathrm e}^{6 i \left (d x +c \right )}+3 \,{\mathrm e}^{7 i \left (d x +c \right )}-3 i {\mathrm e}^{4 i \left (d x +c \right )}-7 \,{\mathrm e}^{5 i \left (d x +c \right )}+3 i {\mathrm e}^{2 i \left (d x +c \right )}+7 \,{\mathrm e}^{3 i \left (d x +c \right )}-3 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}+\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\) | \(165\) |
1/d*(a*(-1/3/sin(d*x+c)^3*cos(d*x+c)^6+1/sin(d*x+c)*cos(d*x+c)^6+(8/3+cos( d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c))+a*(-1/4*cot(d*x+c)^4+1/2*cot(d*x+c) ^2+ln(sin(d*x+c))))
Time = 0.32 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.36 \[ \int \cot ^5(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {12 \, a \cos \left (d x + c\right )^{2} - 12 \, {\left (a \cos \left (d x + c\right )^{4} - 2 \, a \cos \left (d x + c\right )^{2} + a\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 4 \, {\left (3 \, a \cos \left (d x + c\right )^{4} - 12 \, a \cos \left (d x + c\right )^{2} + 8 \, a\right )} \sin \left (d x + c\right ) - 9 \, a}{12 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )}} \]
-1/12*(12*a*cos(d*x + c)^2 - 12*(a*cos(d*x + c)^4 - 2*a*cos(d*x + c)^2 + a )*log(1/2*sin(d*x + c)) - 4*(3*a*cos(d*x + c)^4 - 12*a*cos(d*x + c)^2 + 8* a)*sin(d*x + c) - 9*a)/(d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)
\[ \int \cot ^5(c+d x) (a+a \sin (c+d x)) \, dx=a \left (\int \sin {\left (c + d x \right )} \cot ^{5}{\left (c + d x \right )}\, dx + \int \cot ^{5}{\left (c + d x \right )}\, dx\right ) \]
Time = 0.20 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.85 \[ \int \cot ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {12 \, a \log \left (\sin \left (d x + c\right )\right ) + 12 \, a \sin \left (d x + c\right ) + \frac {24 \, a \sin \left (d x + c\right )^{3} + 12 \, a \sin \left (d x + c\right )^{2} - 4 \, a \sin \left (d x + c\right ) - 3 \, a}{\sin \left (d x + c\right )^{4}}}{12 \, d} \]
1/12*(12*a*log(sin(d*x + c)) + 12*a*sin(d*x + c) + (24*a*sin(d*x + c)^3 + 12*a*sin(d*x + c)^2 - 4*a*sin(d*x + c) - 3*a)/sin(d*x + c)^4)/d
Time = 0.42 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.01 \[ \int \cot ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {12 \, a \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 12 \, a \sin \left (d x + c\right ) - \frac {25 \, a \sin \left (d x + c\right )^{4} - 24 \, a \sin \left (d x + c\right )^{3} - 12 \, a \sin \left (d x + c\right )^{2} + 4 \, a \sin \left (d x + c\right ) + 3 \, a}{\sin \left (d x + c\right )^{4}}}{12 \, d} \]
1/12*(12*a*log(abs(sin(d*x + c))) + 12*a*sin(d*x + c) - (25*a*sin(d*x + c) ^4 - 24*a*sin(d*x + c)^3 - 12*a*sin(d*x + c)^2 + 4*a*sin(d*x + c) + 3*a)/s in(d*x + c)^4)/d
Time = 5.97 (sec) , antiderivative size = 207, normalized size of antiderivative = 2.56 \[ \int \cot ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {7\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,d}+\frac {46\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {40\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {11\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4}-\frac {2\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}-\frac {a}{4}}{d\,\left (16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )}-\frac {a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}+\frac {3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{16\,d}-\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,d}-\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,d}+\frac {a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d} \]
(7*a*tan(c/2 + (d*x)/2))/(8*d) + ((11*a*tan(c/2 + (d*x)/2)^2)/4 - (2*a*tan (c/2 + (d*x)/2))/3 - a/4 + (40*a*tan(c/2 + (d*x)/2)^3)/3 + 3*a*tan(c/2 + ( d*x)/2)^4 + 46*a*tan(c/2 + (d*x)/2)^5)/(d*(16*tan(c/2 + (d*x)/2)^4 + 16*ta n(c/2 + (d*x)/2)^6)) - (a*log(tan(c/2 + (d*x)/2)^2 + 1))/d + (3*a*tan(c/2 + (d*x)/2)^2)/(16*d) - (a*tan(c/2 + (d*x)/2)^3)/(24*d) - (a*tan(c/2 + (d*x )/2)^4)/(64*d) + (a*log(tan(c/2 + (d*x)/2)))/d