Integrand size = 27, antiderivative size = 73 \[ \int \cot ^2(c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx=-2 a^2 x-\frac {a^2 \text {arctanh}(\cos (c+d x))}{2 d}+\frac {a^2 \cos (c+d x)}{d}-\frac {2 a^2 \cot (c+d x)}{d}-\frac {a^2 \cot (c+d x) \csc (c+d x)}{2 d} \] Output:
-2*a^2*x-1/2*a^2*arctanh(cos(d*x+c))/d+a^2*cos(d*x+c)/d-2*a^2*cot(d*x+c)/d -1/2*a^2*cot(d*x+c)*csc(d*x+c)/d
Time = 0.73 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.40 \[ \int \cot ^2(c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 \left (-16 c-16 d x+8 \cos (c+d x)-8 \cot \left (\frac {1}{2} (c+d x)\right )-\csc ^2\left (\frac {1}{2} (c+d x)\right )-4 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+4 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+\sec ^2\left (\frac {1}{2} (c+d x)\right )+8 \tan \left (\frac {1}{2} (c+d x)\right )\right )}{8 d} \] Input:
Integrate[Cot[c + d*x]^2*Csc[c + d*x]*(a + a*Sin[c + d*x])^2,x]
Output:
(a^2*(-16*c - 16*d*x + 8*Cos[c + d*x] - 8*Cot[(c + d*x)/2] - Csc[(c + d*x) /2]^2 - 4*Log[Cos[(c + d*x)/2]] + 4*Log[Sin[(c + d*x)/2]] + Sec[(c + d*x)/ 2]^2 + 8*Tan[(c + d*x)/2]))/(8*d)
Time = 0.33 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.05, 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.
\(\displaystyle \int \cot ^2(c+d x) \csc (c+d x) (a \sin (c+d x)+a)^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^2 (a \sin (c+d x)+a)^2}{\sin (c+d x)^3}dx\) |
\(\Big \downarrow \) 3351 |
\(\displaystyle \frac {\int \left (\csc ^3(c+d x) a^4+2 \csc ^2(c+d x) a^4-\sin (c+d x) a^4-2 a^4\right )dx}{a^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {a^4 \text {arctanh}(\cos (c+d x))}{2 d}+\frac {a^4 \cos (c+d x)}{d}-\frac {2 a^4 \cot (c+d x)}{d}-\frac {a^4 \cot (c+d x) \csc (c+d x)}{2 d}-2 a^4 x}{a^2}\) |
Input:
Int[Cot[c + d*x]^2*Csc[c + d*x]*(a + a*Sin[c + d*x])^2,x]
Output:
(-2*a^4*x - (a^4*ArcTanh[Cos[c + d*x]])/(2*d) + (a^4*Cos[c + d*x])/d - (2* a^4*Cot[c + d*x])/d - (a^4*Cot[c + d*x]*Csc[c + d*x])/(2*d))/a^2
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 = 0.35 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.41
method | result | size |
derivativedivides | \(\frac {a^{2} \left (\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+2 a^{2} \left (-\cot \left (d x +c \right )-d x -c \right )+a^{2} \left (-\frac {\cos \left (d x +c \right )^{3}}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )}{2}-\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )}{d}\) | \(103\) |
default | \(\frac {a^{2} \left (\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+2 a^{2} \left (-\cot \left (d x +c \right )-d x -c \right )+a^{2} \left (-\frac {\cos \left (d x +c \right )^{3}}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )}{2}-\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )}{d}\) | \(103\) |
risch | \(-2 a^{2} x +\frac {a^{2} {\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {a^{2} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {a^{2} \left ({\mathrm e}^{3 i \left (d x +c \right )}+{\mathrm e}^{i \left (d x +c \right )}-4 i {\mathrm e}^{2 i \left (d x +c \right )}+4 i\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d}\) | \(135\) |
Input:
int(cot(d*x+c)^2*csc(d*x+c)*(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(a^2*(cos(d*x+c)+ln(csc(d*x+c)-cot(d*x+c)))+2*a^2*(-cot(d*x+c)-d*x-c)+ a^2*(-1/2/sin(d*x+c)^2*cos(d*x+c)^3-1/2*cos(d*x+c)-1/2*ln(csc(d*x+c)-cot(d *x+c))))
Leaf count of result is larger than twice the leaf count of optimal. 143 vs. \(2 (69) = 138\).
Time = 0.09 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.96 \[ \int \cot ^2(c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {8 \, a^{2} d x \cos \left (d x + c\right )^{2} - 4 \, a^{2} \cos \left (d x + c\right )^{3} - 8 \, a^{2} d x - 8 \, a^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 2 \, a^{2} \cos \left (d x + c\right ) + {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{4 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )}} \] Input:
integrate(cot(d*x+c)^2*csc(d*x+c)*(a+a*sin(d*x+c))^2,x, algorithm="fricas" )
Output:
-1/4*(8*a^2*d*x*cos(d*x + c)^2 - 4*a^2*cos(d*x + c)^3 - 8*a^2*d*x - 8*a^2* cos(d*x + c)*sin(d*x + c) + 2*a^2*cos(d*x + c) + (a^2*cos(d*x + c)^2 - a^2 )*log(1/2*cos(d*x + c) + 1/2) - (a^2*cos(d*x + c)^2 - a^2)*log(-1/2*cos(d* x + c) + 1/2))/(d*cos(d*x + c)^2 - d)
\[ \int \cot ^2(c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx=a^{2} \left (\int \cot ^{2}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx + \int 2 \sin {\left (c + d x \right )} \cot ^{2}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx + \int \sin ^{2}{\left (c + d x \right )} \cot ^{2}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx\right ) \] Input:
integrate(cot(d*x+c)**2*csc(d*x+c)*(a+a*sin(d*x+c))**2,x)
Output:
a**2*(Integral(cot(c + d*x)**2*csc(c + d*x), x) + Integral(2*sin(c + d*x)* cot(c + d*x)**2*csc(c + d*x), x) + Integral(sin(c + d*x)**2*cot(c + d*x)** 2*csc(c + d*x), x))
Time = 0.11 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.42 \[ \int \cot ^2(c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {8 \, {\left (d x + c + \frac {1}{\tan \left (d x + c\right )}\right )} a^{2} - a^{2} {\left (\frac {2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} + \log \left (\cos \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 2 \, a^{2} {\left (2 \, \cos \left (d x + c\right ) - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{4 \, d} \] Input:
integrate(cot(d*x+c)^2*csc(d*x+c)*(a+a*sin(d*x+c))^2,x, algorithm="maxima" )
Output:
-1/4*(8*(d*x + c + 1/tan(d*x + c))*a^2 - a^2*(2*cos(d*x + c)/(cos(d*x + c) ^2 - 1) + log(cos(d*x + c) + 1) - log(cos(d*x + c) - 1)) - 2*a^2*(2*cos(d* x + c) - log(cos(d*x + c) + 1) + log(cos(d*x + c) - 1)))/d
Time = 0.16 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.75 \[ \int \cot ^2(c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 16 \, {\left (d x + c\right )} a^{2} + 4 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 8 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {16 \, a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} - \frac {6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 8 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{8 \, d} \] Input:
integrate(cot(d*x+c)^2*csc(d*x+c)*(a+a*sin(d*x+c))^2,x, algorithm="giac")
Output:
1/8*(a^2*tan(1/2*d*x + 1/2*c)^2 - 16*(d*x + c)*a^2 + 4*a^2*log(abs(tan(1/2 *d*x + 1/2*c))) + 8*a^2*tan(1/2*d*x + 1/2*c) + 16*a^2/(tan(1/2*d*x + 1/2*c )^2 + 1) - (6*a^2*tan(1/2*d*x + 1/2*c)^2 + 8*a^2*tan(1/2*d*x + 1/2*c) + a^ 2)/tan(1/2*d*x + 1/2*c)^2)/d
Time = 17.80 (sec) , antiderivative size = 213, normalized size of antiderivative = 2.92 \[ \int \cot ^2(c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}-\frac {4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-\frac {15\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}+4\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a^2}{2}}{d\,\left (4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}+\frac {a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,d}+\frac {4\,a^2\,\mathrm {atan}\left (\frac {16\,a^4}{4\,a^4+16\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}-\frac {4\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,a^4+16\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d} \] Input:
int((cot(c + d*x)^2*(a + a*sin(c + d*x))^2)/sin(c + d*x),x)
Output:
(a^2*tan(c/2 + (d*x)/2)^2)/(8*d) - (4*a^2*tan(c/2 + (d*x)/2)^3 - (15*a^2*t an(c/2 + (d*x)/2)^2)/2 + a^2/2 + 4*a^2*tan(c/2 + (d*x)/2))/(d*(4*tan(c/2 + (d*x)/2)^2 + 4*tan(c/2 + (d*x)/2)^4)) + (a^2*log(tan(c/2 + (d*x)/2)))/(2* d) + (4*a^2*atan((16*a^4)/(4*a^4 + 16*a^4*tan(c/2 + (d*x)/2)) - (4*a^4*tan (c/2 + (d*x)/2))/(4*a^4 + 16*a^4*tan(c/2 + (d*x)/2))))/d + (a^2*tan(c/2 + (d*x)/2))/d
Time = 0.16 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.33 \[ \int \cot ^2(c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^{2} \left (8 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}-16 \cos \left (d x +c \right ) \sin \left (d x +c \right )-4 \cos \left (d x +c \right )+4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2}-16 \sin \left (d x +c \right )^{2} d x -7 \sin \left (d x +c \right )^{2}\right )}{8 \sin \left (d x +c \right )^{2} d} \] Input:
int(cot(d*x+c)^2*csc(d*x+c)*(a+a*sin(d*x+c))^2,x)
Output:
(a**2*(8*cos(c + d*x)*sin(c + d*x)**2 - 16*cos(c + d*x)*sin(c + d*x) - 4*c os(c + d*x) + 4*log(tan((c + d*x)/2))*sin(c + d*x)**2 - 16*sin(c + d*x)**2 *d*x - 7*sin(c + d*x)**2))/(8*sin(c + d*x)**2*d)