Integrand size = 27, antiderivative size = 47 \[ \int \cot (c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {2 a^2 \csc (c+d x)}{d}-\frac {a^2 \csc ^2(c+d x)}{2 d}+\frac {a^2 \log (\sin (c+d x))}{d} \] Output:
-2*a^2*csc(d*x+c)/d-1/2*a^2*csc(d*x+c)^2/d+a^2*ln(sin(d*x+c))/d
Time = 0.02 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.89 \[ \int \cot (c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=a^2 \left (-\frac {2 \csc (c+d x)}{d}-\frac {\csc ^2(c+d x)}{2 d}+\frac {\log (\sin (c+d x))}{d}\right ) \] Input:
Integrate[Cot[c + d*x]*Csc[c + d*x]^2*(a + a*Sin[c + d*x])^2,x]
Output:
a^2*((-2*Csc[c + d*x])/d - Csc[c + d*x]^2/(2*d) + Log[Sin[c + d*x]]/d)
Time = 0.28 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.79, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3042, 3312, 27, 49, 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.
\(\displaystyle \int \cot (c+d x) \csc ^2(c+d x) (a \sin (c+d x)+a)^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x) (a \sin (c+d x)+a)^2}{\sin (c+d x)^3}dx\) |
\(\Big \downarrow \) 3312 |
\(\displaystyle \frac {\int \csc ^3(c+d x) (\sin (c+d x) a+a)^2d(a \sin (c+d x))}{a d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a^2 \int \frac {\csc ^3(c+d x) (\sin (c+d x) a+a)^2}{a^3}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {a^2 \int \left (\frac {\csc ^3(c+d x)}{a}+\frac {2 \csc ^2(c+d x)}{a}+\frac {\csc (c+d x)}{a}\right )d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^2 \left (\log (a \sin (c+d x))-\frac {1}{2} \csc ^2(c+d x)-2 \csc (c+d x)\right )}{d}\) |
Input:
Int[Cot[c + d*x]*Csc[c + d*x]^2*(a + a*Sin[c + d*x])^2,x]
Output:
(a^2*(-2*Csc[c + d*x] - Csc[c + d*x]^2/2 + Log[a*Sin[c + d*x]]))/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_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*(( c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b*f) Su bst[Int[(a + x)^m*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]
Time = 0.30 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.74
method | result | size |
derivativedivides | \(-\frac {a^{2} \left (\frac {\csc \left (d x +c \right )^{2}}{2}+2 \csc \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )\right )\right )}{d}\) | \(35\) |
default | \(-\frac {a^{2} \left (\frac {\csc \left (d x +c \right )^{2}}{2}+2 \csc \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )\right )\right )}{d}\) | \(35\) |
risch | \(-i a^{2} x -\frac {2 i a^{2} c}{d}-\frac {2 i a^{2} \left (i {\mathrm e}^{2 i \left (d x +c \right )}+2 \,{\mathrm e}^{3 i \left (d x +c \right )}-2 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}+\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\) | \(95\) |
Input:
int(cot(d*x+c)*csc(d*x+c)^2*(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
-1/d*a^2*(1/2*csc(d*x+c)^2+2*csc(d*x+c)+ln(csc(d*x+c)))
Time = 0.12 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.32 \[ \int \cot (c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {4 \, a^{2} \sin \left (d x + c\right ) + a^{2} + 2 \, {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right )}{2 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )}} \] Input:
integrate(cot(d*x+c)*csc(d*x+c)^2*(a+a*sin(d*x+c))^2,x, algorithm="fricas" )
Output:
1/2*(4*a^2*sin(d*x + c) + a^2 + 2*(a^2*cos(d*x + c)^2 - a^2)*log(1/2*sin(d *x + c)))/(d*cos(d*x + c)^2 - d)
\[ \int \cot (c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=a^{2} \left (\int \cot {\left (c + d x \right )} \csc ^{2}{\left (c + d x \right )}\, dx + \int 2 \sin {\left (c + d x \right )} \cot {\left (c + d x \right )} \csc ^{2}{\left (c + d x \right )}\, dx + \int \sin ^{2}{\left (c + d x \right )} \cot {\left (c + d x \right )} \csc ^{2}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate(cot(d*x+c)*csc(d*x+c)**2*(a+a*sin(d*x+c))**2,x)
Output:
a**2*(Integral(cot(c + d*x)*csc(c + d*x)**2, x) + Integral(2*sin(c + d*x)* cot(c + d*x)*csc(c + d*x)**2, x) + Integral(sin(c + d*x)**2*cot(c + d*x)*c sc(c + d*x)**2, x))
Time = 0.03 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.91 \[ \int \cot (c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {2 \, a^{2} \log \left (\sin \left (d x + c\right )\right ) - \frac {4 \, a^{2} \sin \left (d x + c\right ) + a^{2}}{\sin \left (d x + c\right )^{2}}}{2 \, d} \] Input:
integrate(cot(d*x+c)*csc(d*x+c)^2*(a+a*sin(d*x+c))^2,x, algorithm="maxima" )
Output:
1/2*(2*a^2*log(sin(d*x + c)) - (4*a^2*sin(d*x + c) + a^2)/sin(d*x + c)^2)/ d
Time = 0.15 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.94 \[ \int \cot (c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {2 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - \frac {4 \, a^{2} \sin \left (d x + c\right ) + a^{2}}{\sin \left (d x + c\right )^{2}}}{2 \, d} \] Input:
integrate(cot(d*x+c)*csc(d*x+c)^2*(a+a*sin(d*x+c))^2,x, algorithm="giac")
Output:
1/2*(2*a^2*log(abs(sin(d*x + c))) - (4*a^2*sin(d*x + c) + a^2)/sin(d*x + c )^2)/d
Time = 18.02 (sec) , antiderivative size = 111, normalized size of antiderivative = 2.36 \[ \int \cot (c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {a^2}{8}+a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d}-\frac {a^2\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \] Input:
int((cot(c + d*x)*(a + a*sin(c + d*x))^2)/sin(c + d*x)^2,x)
Output:
(a^2*log(tan(c/2 + (d*x)/2)))/d - (a^2*tan(c/2 + (d*x)/2)^2)/(8*d) - (cot( c/2 + (d*x)/2)^2*(a^2/8 + a^2*tan(c/2 + (d*x)/2)))/d - (a^2*tan(c/2 + (d*x )/2))/d - (a^2*log(tan(c/2 + (d*x)/2)^2 + 1))/d
Time = 0.17 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.66 \[ \int \cot (c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^{2} \left (-4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{2}+4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2}+\sin \left (d x +c \right )^{2}-8 \sin \left (d x +c \right )-2\right )}{4 \sin \left (d x +c \right )^{2} d} \] Input:
int(cot(d*x+c)*csc(d*x+c)^2*(a+a*sin(d*x+c))^2,x)
Output:
(a**2*( - 4*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2 + 4*log(tan((c + d*x)/2))*sin(c + d*x)**2 + sin(c + d*x)**2 - 8*sin(c + d*x) - 2))/(4*sin(c + d*x)**2*d)