Integrand size = 21, antiderivative size = 116 \[ \int \cot ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {4 a^2 \csc (c+d x)}{d}+\frac {a^2 \csc ^2(c+d x)}{2 d}-\frac {2 a^2 \csc ^3(c+d x)}{3 d}-\frac {a^2 \csc ^4(c+d x)}{4 d}-\frac {a^2 \log (\sin (c+d x))}{d}+\frac {2 a^2 \sin (c+d x)}{d}+\frac {a^2 \sin ^2(c+d x)}{2 d} \]
4*a^2*csc(d*x+c)/d+1/2*a^2*csc(d*x+c)^2/d-2/3*a^2*csc(d*x+c)^3/d-1/4*a^2*c sc(d*x+c)^4/d-a^2*ln(sin(d*x+c))/d+2*a^2*sin(d*x+c)/d+1/2*a^2*sin(d*x+c)^2 /d
Time = 0.29 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.66 \[ \int \cot ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 \left (48 \csc (c+d x)+6 \csc ^2(c+d x)-8 \csc ^3(c+d x)-3 \csc ^4(c+d x)-12 \log (\sin (c+d x))+24 \sin (c+d x)+6 \sin ^2(c+d x)\right )}{12 d} \]
(a^2*(48*Csc[c + d*x] + 6*Csc[c + d*x]^2 - 8*Csc[c + d*x]^3 - 3*Csc[c + d* x]^4 - 12*Log[Sin[c + d*x]] + 24*Sin[c + d*x] + 6*Sin[c + d*x]^2))/(12*d)
Time = 0.25 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.87, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, 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.
| \(\displaystyle \int \cot ^5(c+d x) (a \sin (c+d x)+a)^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (c+d x)+a)^2}{\tan (c+d x)^5}dx\) |
| \(\Big \downarrow \) 3186 |
| \(\displaystyle \frac {\int \frac {\csc ^5(c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^4}{a^5}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {\int \left (a \csc ^5(c+d x)+2 a \csc ^4(c+d x)-a \csc ^3(c+d x)-4 a \csc ^2(c+d x)-a \csc (c+d x)+2 a+a \sin (c+d x)\right )d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {1}{2} a^2 \sin ^2(c+d x)+2 a^2 \sin (c+d x)-\frac {1}{4} a^2 \csc ^4(c+d x)-\frac {2}{3} a^2 \csc ^3(c+d x)+\frac {1}{2} a^2 \csc ^2(c+d x)+4 a^2 \csc (c+d x)-a^2 \log (a \sin (c+d x))}{d}\) |
(4*a^2*Csc[c + d*x] + (a^2*Csc[c + d*x]^2)/2 - (2*a^2*Csc[c + d*x]^3)/3 - (a^2*Csc[c + d*x]^4)/4 - a^2*Log[a*Sin[c + d*x]] + 2*a^2*Sin[c + d*x] + (a ^2*Sin[c + d*x]^2)/2)/d
3.6.18.3.1 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 = 0.33 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.36
| method | result | size |
| derivativedivides | \(\frac {a^{2} \left (-\frac {\cos ^{6}\left (d x +c \right )}{2 \sin \left (d x +c \right )^{2}}-\frac {\left (\cos ^{4}\left (d x +c \right )\right )}{2}-\left (\cos ^{2}\left (d x +c \right )\right )-2 \ln \left (\sin \left (d x +c \right )\right )\right )+2 a^{2} \left (-\frac {\cos ^{6}\left (d x +c \right )}{3 \sin \left (d x +c \right )^{3}}+\frac {\cos ^{6}\left (d x +c \right )}{\sin \left (d x +c \right )}+\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )\right )+a^{2} \left (-\frac {\left (\cot ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\) | \(158\) |
| default | \(\frac {a^{2} \left (-\frac {\cos ^{6}\left (d x +c \right )}{2 \sin \left (d x +c \right )^{2}}-\frac {\left (\cos ^{4}\left (d x +c \right )\right )}{2}-\left (\cos ^{2}\left (d x +c \right )\right )-2 \ln \left (\sin \left (d x +c \right )\right )\right )+2 a^{2} \left (-\frac {\cos ^{6}\left (d x +c \right )}{3 \sin \left (d x +c \right )^{3}}+\frac {\cos ^{6}\left (d x +c \right )}{\sin \left (d x +c \right )}+\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )\right )+a^{2} \left (-\frac {\left (\cot ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\) | \(158\) |
| parallelrisch | \(\frac {\left (\csc ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\left (3+\cos \left (4 d x +4 c \right )-4 \cos \left (2 d x +2 c \right )\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-\cos \left (4 d x +4 c \right )-3+4 \cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin \left (5 d x +5 c \right )-\frac {45 \cos \left (2 d x +2 c \right )}{16}+\frac {31 \cos \left (4 d x +4 c \right )}{64}-\frac {\cos \left (6 d x +6 c \right )}{8}+\frac {86 \sin \left (d x +c \right )}{3}-13 \sin \left (3 d x +3 c \right )+\frac {29}{64}\right ) \left (\sec ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{128 d}\) | \(164\) |
| risch | \(i a^{2} x -\frac {a^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {i a^{2} {\mathrm e}^{i \left (d x +c \right )}}{d}+\frac {i a^{2} {\mathrm e}^{-i \left (d x +c \right )}}{d}-\frac {a^{2} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {2 i a^{2} c}{d}+\frac {2 i a^{2} \left (3 i {\mathrm e}^{6 i \left (d x +c \right )}+12 \,{\mathrm e}^{7 i \left (d x +c \right )}-28 \,{\mathrm e}^{5 i \left (d x +c \right )}+3 i {\mathrm e}^{2 i \left (d x +c \right )}+28 \,{\mathrm e}^{3 i \left (d x +c \right )}-12 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}-\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\) | \(200\) |
| norman | \(\frac {-\frac {a^{2}}{64 d}-\frac {a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{12 d}+\frac {a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d}+\frac {19 a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {55 a^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}+\frac {55 a^{2} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}+\frac {19 a^{2} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {a^{2} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d}-\frac {a^{2} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}-\frac {a^{2} \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d}+\frac {61 a^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {a^{2} \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\) | \(265\) |
1/d*(a^2*(-1/2/sin(d*x+c)^2*cos(d*x+c)^6-1/2*cos(d*x+c)^4-cos(d*x+c)^2-2*l n(sin(d*x+c)))+2*a^2*(-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^2*(-1/4*cot(d*x+c)^ 4+1/2*cot(d*x+c)^2+ln(sin(d*x+c))))
Time = 0.27 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.31 \[ \int \cot ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {6 \, a^{2} \cos \left (d x + c\right )^{6} - 15 \, a^{2} \cos \left (d x + c\right )^{4} + 18 \, a^{2} \cos \left (d x + c\right )^{2} - 6 \, a^{2} + 12 \, {\left (a^{2} \cos \left (d x + c\right )^{4} - 2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 8 \, {\left (3 \, a^{2} \cos \left (d x + c\right )^{4} - 12 \, a^{2} \cos \left (d x + c\right )^{2} + 8 \, a^{2}\right )} \sin \left (d x + c\right )}{12 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )}} \]
-1/12*(6*a^2*cos(d*x + c)^6 - 15*a^2*cos(d*x + c)^4 + 18*a^2*cos(d*x + c)^ 2 - 6*a^2 + 12*(a^2*cos(d*x + c)^4 - 2*a^2*cos(d*x + c)^2 + a^2)*log(1/2*s in(d*x + c)) - 8*(3*a^2*cos(d*x + c)^4 - 12*a^2*cos(d*x + c)^2 + 8*a^2)*si n(d*x + c))/(d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)
Timed out. \[ \int \cot ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=\text {Timed out} \]
Time = 0.22 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.81 \[ \int \cot ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {6 \, a^{2} \sin \left (d x + c\right )^{2} - 12 \, a^{2} \log \left (\sin \left (d x + c\right )\right ) + 24 \, a^{2} \sin \left (d x + c\right ) + \frac {48 \, a^{2} \sin \left (d x + c\right )^{3} + 6 \, a^{2} \sin \left (d x + c\right )^{2} - 8 \, a^{2} \sin \left (d x + c\right ) - 3 \, a^{2}}{\sin \left (d x + c\right )^{4}}}{12 \, d} \]
1/12*(6*a^2*sin(d*x + c)^2 - 12*a^2*log(sin(d*x + c)) + 24*a^2*sin(d*x + c ) + (48*a^2*sin(d*x + c)^3 + 6*a^2*sin(d*x + c)^2 - 8*a^2*sin(d*x + c) - 3 *a^2)/sin(d*x + c)^4)/d
Time = 0.37 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.93 \[ \int \cot ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {6 \, a^{2} \sin \left (d x + c\right )^{2} - 12 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 24 \, a^{2} \sin \left (d x + c\right ) + \frac {25 \, a^{2} \sin \left (d x + c\right )^{4} + 48 \, a^{2} \sin \left (d x + c\right )^{3} + 6 \, a^{2} \sin \left (d x + c\right )^{2} - 8 \, a^{2} \sin \left (d x + c\right ) - 3 \, a^{2}}{\sin \left (d x + c\right )^{4}}}{12 \, d} \]
1/12*(6*a^2*sin(d*x + c)^2 - 12*a^2*log(abs(sin(d*x + c))) + 24*a^2*sin(d* x + c) + (25*a^2*sin(d*x + c)^4 + 48*a^2*sin(d*x + c)^3 + 6*a^2*sin(d*x + c)^2 - 8*a^2*sin(d*x + c) - 3*a^2)/sin(d*x + c)^4)/d
Time = 10.00 (sec) , antiderivative size = 276, normalized size of antiderivative = 2.38 \[ \int \cot ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{16\,d}-\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{12\,d}-\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,d}-\frac {a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {7\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,d}+\frac {a^2\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}+\frac {92\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+33\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {356\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{3}+\frac {7\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{4}+\frac {76\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}-\frac {4\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}-\frac {a^2}{4}}{d\,\left (16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )} \]
(a^2*tan(c/2 + (d*x)/2)^2)/(16*d) - (a^2*tan(c/2 + (d*x)/2)^3)/(12*d) - (a ^2*tan(c/2 + (d*x)/2)^4)/(64*d) - (a^2*log(tan(c/2 + (d*x)/2)))/d + (7*a^2 *tan(c/2 + (d*x)/2))/(4*d) + (a^2*log(tan(c/2 + (d*x)/2)^2 + 1))/d + ((a^2 *tan(c/2 + (d*x)/2)^2)/2 + (76*a^2*tan(c/2 + (d*x)/2)^3)/3 + (7*a^2*tan(c/ 2 + (d*x)/2)^4)/4 + (356*a^2*tan(c/2 + (d*x)/2)^5)/3 + 33*a^2*tan(c/2 + (d *x)/2)^6 + 92*a^2*tan(c/2 + (d*x)/2)^7 - a^2/4 - (4*a^2*tan(c/2 + (d*x)/2) )/3)/(d*(16*tan(c/2 + (d*x)/2)^4 + 32*tan(c/2 + (d*x)/2)^6 + 16*tan(c/2 + (d*x)/2)^8))