Integrand size = 27, antiderivative size = 116 \[ \int \cot ^4(c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx=2 a^2 x+\frac {9 a^2 \text {arctanh}(\cos (c+d x))}{8 d}-\frac {a^2 \cos (c+d x)}{d}+\frac {2 a^2 \cot (c+d x)}{d}-\frac {2 a^2 \cot ^3(c+d x)}{3 d}+\frac {a^2 \cot (c+d x) \csc (c+d x)}{8 d}-\frac {a^2 \cot (c+d x) \csc ^3(c+d x)}{4 d} \] Output:
2*a^2*x+9/8*a^2*arctanh(cos(d*x+c))/d-a^2*cos(d*x+c)/d+2*a^2*cot(d*x+c)/d- 2/3*a^2*cot(d*x+c)^3/d+1/8*a^2*cot(d*x+c)*csc(d*x+c)/d-1/4*a^2*cot(d*x+c)* csc(d*x+c)^3/d
Time = 6.82 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.85 \[ \int \cot ^4(c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {a^2 \left (192 \cot (c+d x)+\csc ^4\left (\frac {1}{2} (c+d x)\right ) (8+3 \csc (c+d x))-2 \csc ^2\left (\frac {1}{2} (c+d x)\right ) (64+3 \csc (c+d x))-24 \csc (c+d x) \left (16 (c+d x)+9 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-9 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+8 (7+8 \cos (c+d x)) \sec ^4\left (\frac {1}{2} (c+d x)\right )+24 \csc ^3(c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right )-48 \csc ^5(c+d x) \sin ^4\left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x) (1+\sin (c+d x))^2}{192 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4} \] Input:
Integrate[Cot[c + d*x]^4*Csc[c + d*x]*(a + a*Sin[c + d*x])^2,x]
Output:
-1/192*(a^2*(192*Cot[c + d*x] + Csc[(c + d*x)/2]^4*(8 + 3*Csc[c + d*x]) - 2*Csc[(c + d*x)/2]^2*(64 + 3*Csc[c + d*x]) - 24*Csc[c + d*x]*(16*(c + d*x) + 9*Log[Cos[(c + d*x)/2]] - 9*Log[Sin[(c + d*x)/2]]) + 8*(7 + 8*Cos[c + d *x])*Sec[(c + d*x)/2]^4 + 24*Csc[c + d*x]^3*Sin[(c + d*x)/2]^2 - 48*Csc[c + d*x]^5*Sin[(c + d*x)/2]^4)*Sin[c + d*x]*(1 + Sin[c + d*x])^2)/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4)
Time = 0.41 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.03, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {3042, 3351, 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.
| \(\displaystyle \int \cot ^4(c+d x) \csc (c+d x) (a \sin (c+d x)+a)^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^4 (a \sin (c+d x)+a)^2}{\sin (c+d x)^5}dx\) |
| \(\Big \downarrow \) 3351 |
| \(\displaystyle \frac {\int \left (\csc ^5(c+d x) a^6+2 \csc ^4(c+d x) a^6-\csc ^3(c+d x) a^6-4 \csc ^2(c+d x) a^6-\csc (c+d x) a^6+\sin (c+d x) a^6+2 a^6\right )dx}{a^4}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {9 a^6 \text {arctanh}(\cos (c+d x))}{8 d}-\frac {a^6 \cos (c+d x)}{d}-\frac {2 a^6 \cot ^3(c+d x)}{3 d}+\frac {2 a^6 \cot (c+d x)}{d}-\frac {a^6 \cot (c+d x) \csc ^3(c+d x)}{4 d}+\frac {a^6 \cot (c+d x) \csc (c+d x)}{8 d}+2 a^6 x}{a^4}\) |
Input:
Int[Cot[c + d*x]^4*Csc[c + d*x]*(a + a*Sin[c + d*x])^2,x]
Output:
(2*a^6*x + (9*a^6*ArcTanh[Cos[c + d*x]])/(8*d) - (a^6*Cos[c + d*x])/d + (2 *a^6*Cot[c + d*x])/d - (2*a^6*Cot[c + d*x]^3)/(3*d) + (a^6*Cot[c + d*x]*Cs c[c + d*x])/(8*d) - (a^6*Cot[c + d*x]*Csc[c + d*x]^3)/(4*d))/a^4
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[1/a^p Int[Expan dTrig[(d*sin[e + f*x])^n*(a - b*sin[e + f*x])^(p/2)*(a + b*sin[e + f*x])^(m + p/2), x], x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && In tegersQ[m, n, p/2] && ((GtQ[m, 0] && GtQ[p, 0] && LtQ[-m - p, n, -1]) || (G tQ[m, 2] && LtQ[p, 0] && GtQ[m + p/2, 0]))
Time = 0.52 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.45
| method | result | size |
| derivativedivides | \(\frac {a^{2} \left (-\frac {\cos \left (d x +c \right )^{5}}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )^{3}}{2}-\frac {3 \cos \left (d x +c \right )}{2}-\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+2 a^{2} \left (-\frac {\cot \left (d x +c \right )^{3}}{3}+\cot \left (d x +c \right )+d x +c \right )+a^{2} \left (-\frac {\cos \left (d x +c \right )^{5}}{4 \sin \left (d x +c \right )^{4}}+\frac {\cos \left (d x +c \right )^{5}}{8 \sin \left (d x +c \right )^{2}}+\frac {\cos \left (d x +c \right )^{3}}{8}+\frac {3 \cos \left (d x +c \right )}{8}+\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )}{d}\) | \(168\) |
| default | \(\frac {a^{2} \left (-\frac {\cos \left (d x +c \right )^{5}}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )^{3}}{2}-\frac {3 \cos \left (d x +c \right )}{2}-\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+2 a^{2} \left (-\frac {\cot \left (d x +c \right )^{3}}{3}+\cot \left (d x +c \right )+d x +c \right )+a^{2} \left (-\frac {\cos \left (d x +c \right )^{5}}{4 \sin \left (d x +c \right )^{4}}+\frac {\cos \left (d x +c \right )^{5}}{8 \sin \left (d x +c \right )^{2}}+\frac {\cos \left (d x +c \right )^{3}}{8}+\frac {3 \cos \left (d x +c \right )}{8}+\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )}{d}\) | \(168\) |
| risch | \(2 a^{2} x -\frac {a^{2} {\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {a^{2} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {a^{2} \left (3 \,{\mathrm e}^{7 i \left (d x +c \right )}+21 \,{\mathrm e}^{5 i \left (d x +c \right )}-96 i {\mathrm e}^{6 i \left (d x +c \right )}+21 \,{\mathrm e}^{3 i \left (d x +c \right )}+192 i {\mathrm e}^{4 i \left (d x +c \right )}+3 \,{\mathrm e}^{i \left (d x +c \right )}-160 i {\mathrm e}^{2 i \left (d x +c \right )}+64 i\right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}-\frac {9 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{8 d}+\frac {9 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{8 d}\) | \(186\) |
Input:
int(cot(d*x+c)^4*csc(d*x+c)*(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(a^2*(-1/2/sin(d*x+c)^2*cos(d*x+c)^5-1/2*cos(d*x+c)^3-3/2*cos(d*x+c)-3 /2*ln(csc(d*x+c)-cot(d*x+c)))+2*a^2*(-1/3*cot(d*x+c)^3+cot(d*x+c)+d*x+c)+a ^2*(-1/4/sin(d*x+c)^4*cos(d*x+c)^5+1/8/sin(d*x+c)^2*cos(d*x+c)^5+1/8*cos(d *x+c)^3+3/8*cos(d*x+c)+3/8*ln(csc(d*x+c)-cot(d*x+c))))
Leaf count of result is larger than twice the leaf count of optimal. 219 vs. \(2 (108) = 216\).
Time = 0.09 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.89 \[ \int \cot ^4(c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {96 \, a^{2} d x \cos \left (d x + c\right )^{4} - 48 \, a^{2} \cos \left (d x + c\right )^{5} - 192 \, a^{2} d x \cos \left (d x + c\right )^{2} + 90 \, a^{2} \cos \left (d x + c\right )^{3} + 96 \, a^{2} d x - 54 \, a^{2} \cos \left (d x + c\right ) + 27 \, {\left (a^{2} \cos \left (d x + c\right )^{4} - 2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 27 \, {\left (a^{2} \cos \left (d x + c\right )^{4} - 2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 32 \, {\left (4 \, a^{2} \cos \left (d x + c\right )^{3} - 3 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )}} \] Input:
integrate(cot(d*x+c)^4*csc(d*x+c)*(a+a*sin(d*x+c))^2,x, algorithm="fricas" )
Output:
1/48*(96*a^2*d*x*cos(d*x + c)^4 - 48*a^2*cos(d*x + c)^5 - 192*a^2*d*x*cos( d*x + c)^2 + 90*a^2*cos(d*x + c)^3 + 96*a^2*d*x - 54*a^2*cos(d*x + c) + 27 *(a^2*cos(d*x + c)^4 - 2*a^2*cos(d*x + c)^2 + a^2)*log(1/2*cos(d*x + c) + 1/2) - 27*(a^2*cos(d*x + c)^4 - 2*a^2*cos(d*x + c)^2 + a^2)*log(-1/2*cos(d *x + c) + 1/2) - 32*(4*a^2*cos(d*x + c)^3 - 3*a^2*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)
\[ \int \cot ^4(c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx=a^{2} \left (\int \cot ^{4}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx + \int 2 \sin {\left (c + d x \right )} \cot ^{4}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx + \int \sin ^{2}{\left (c + d x \right )} \cot ^{4}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx\right ) \] Input:
integrate(cot(d*x+c)**4*csc(d*x+c)*(a+a*sin(d*x+c))**2,x)
Output:
a**2*(Integral(cot(c + d*x)**4*csc(c + d*x), x) + Integral(2*sin(c + d*x)* cot(c + d*x)**4*csc(c + d*x), x) + Integral(sin(c + d*x)**2*cot(c + d*x)** 4*csc(c + d*x), x))
Time = 0.11 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.44 \[ \int \cot ^4(c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {32 \, {\left (3 \, d x + 3 \, c + \frac {3 \, \tan \left (d x + c\right )^{2} - 1}{\tan \left (d x + c\right )^{3}}\right )} a^{2} - 3 \, a^{2} {\left (\frac {2 \, {\left (5 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} + 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + 12 \, a^{2} {\left (\frac {2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} - 4 \, \cos \left (d x + c\right ) + 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{48 \, d} \] Input:
integrate(cot(d*x+c)^4*csc(d*x+c)*(a+a*sin(d*x+c))^2,x, algorithm="maxima" )
Output:
1/48*(32*(3*d*x + 3*c + (3*tan(d*x + c)^2 - 1)/tan(d*x + c)^3)*a^2 - 3*a^2 *(2*(5*cos(d*x + c)^3 - 3*cos(d*x + c))/(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1) + 3*log(cos(d*x + c) + 1) - 3*log(cos(d*x + c) - 1)) + 12*a^2*(2*cos (d*x + c)/(cos(d*x + c)^2 - 1) - 4*cos(d*x + c) + 3*log(cos(d*x + c) + 1) - 3*log(cos(d*x + c) - 1)))/d
Time = 0.22 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.40 \[ \int \cot ^4(c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 16 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 384 \, {\left (d x + c\right )} a^{2} - 216 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - 240 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {384 \, a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + \frac {450 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 240 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 16 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{192 \, d} \] Input:
integrate(cot(d*x+c)^4*csc(d*x+c)*(a+a*sin(d*x+c))^2,x, algorithm="giac")
Output:
1/192*(3*a^2*tan(1/2*d*x + 1/2*c)^4 + 16*a^2*tan(1/2*d*x + 1/2*c)^3 + 384* (d*x + c)*a^2 - 216*a^2*log(abs(tan(1/2*d*x + 1/2*c))) - 240*a^2*tan(1/2*d *x + 1/2*c) - 384*a^2/(tan(1/2*d*x + 1/2*c)^2 + 1) + (450*a^2*tan(1/2*d*x + 1/2*c)^4 + 240*a^2*tan(1/2*d*x + 1/2*c)^3 - 16*a^2*tan(1/2*d*x + 1/2*c) - 3*a^2)/tan(1/2*d*x + 1/2*c)^4)/d
Time = 17.64 (sec) , antiderivative size = 265, normalized size of antiderivative = 2.28 \[ \int \cot ^4(c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{12\,d}+\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,d}-\frac {9\,a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{8\,d}-\frac {4\,a^2\,\mathrm {atan}\left (\frac {16\,a^4}{9\,a^4+16\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}-\frac {9\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{9\,a^4+16\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {5\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,d}-\frac {-20\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+32\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-\frac {56\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4}+\frac {4\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}+\frac {a^2}{4}}{d\,\left (16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )} \] Input:
int((cot(c + d*x)^4*(a + a*sin(c + d*x))^2)/sin(c + d*x),x)
Output:
(a^2*tan(c/2 + (d*x)/2)^3)/(12*d) + (a^2*tan(c/2 + (d*x)/2)^4)/(64*d) - (9 *a^2*log(tan(c/2 + (d*x)/2)))/(8*d) - (4*a^2*atan((16*a^4)/(9*a^4 + 16*a^4 *tan(c/2 + (d*x)/2)) - (9*a^4*tan(c/2 + (d*x)/2))/(9*a^4 + 16*a^4*tan(c/2 + (d*x)/2))))/d - (5*a^2*tan(c/2 + (d*x)/2))/(4*d) - ((a^2*tan(c/2 + (d*x) /2)^2)/4 - (56*a^2*tan(c/2 + (d*x)/2)^3)/3 + 32*a^2*tan(c/2 + (d*x)/2)^4 - 20*a^2*tan(c/2 + (d*x)/2)^5 + a^2/4 + (4*a^2*tan(c/2 + (d*x)/2))/3)/(d*(1 6*tan(c/2 + (d*x)/2)^4 + 16*tan(c/2 + (d*x)/2)^6))
Time = 0.15 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.11 \[ \int \cot ^4(c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^{2} \left (-24 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}+64 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}+3 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}-16 \cos \left (d x +c \right ) \sin \left (d x +c \right )-6 \cos \left (d x +c \right )-27 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{4}+48 \sin \left (d x +c \right )^{4} d x +24 \sin \left (d x +c \right )^{4}\right )}{24 \sin \left (d x +c \right )^{4} d} \] Input:
int(cot(d*x+c)^4*csc(d*x+c)*(a+a*sin(d*x+c))^2,x)
Output:
(a**2*( - 24*cos(c + d*x)*sin(c + d*x)**4 + 64*cos(c + d*x)*sin(c + d*x)** 3 + 3*cos(c + d*x)*sin(c + d*x)**2 - 16*cos(c + d*x)*sin(c + d*x) - 6*cos( c + d*x) - 27*log(tan((c + d*x)/2))*sin(c + d*x)**4 + 48*sin(c + d*x)**4*d *x + 24*sin(c + d*x)**4))/(24*sin(c + d*x)**4*d)