Integrand size = 29, antiderivative size = 100 \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 \text {arctanh}(\cos (c+d x))}{4 d}-\frac {2 a^2 \cot ^3(c+d x)}{3 d}-\frac {a^2 \cot ^5(c+d x)}{5 d}+\frac {a^2 \cot (c+d x) \csc (c+d x)}{4 d}-\frac {a^2 \cot (c+d x) \csc ^3(c+d x)}{2 d} \] Output:
1/4*a^2*arctanh(cos(d*x+c))/d-2/3*a^2*cot(d*x+c)^3/d-1/5*a^2*cot(d*x+c)^5/ d+1/4*a^2*cot(d*x+c)*csc(d*x+c)/d-1/2*a^2*cot(d*x+c)*csc(d*x+c)^3/d
Time = 6.36 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.89 \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {a^2 \csc ^5(c+d x) \left (200 \cos (c+d x)+20 \cos (3 (c+d x))-28 \cos (5 (c+d x))-150 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)+150 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)+180 \sin (2 (c+d x))+75 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))-75 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))+30 \sin (4 (c+d x))-15 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (5 (c+d x))+15 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (5 (c+d x))\right )}{960 d} \] Input:
Integrate[Cot[c + d*x]^2*Csc[c + d*x]^4*(a + a*Sin[c + d*x])^2,x]
Output:
-1/960*(a^2*Csc[c + d*x]^5*(200*Cos[c + d*x] + 20*Cos[3*(c + d*x)] - 28*Co s[5*(c + d*x)] - 150*Log[Cos[(c + d*x)/2]]*Sin[c + d*x] + 150*Log[Sin[(c + d*x)/2]]*Sin[c + d*x] + 180*Sin[2*(c + d*x)] + 75*Log[Cos[(c + d*x)/2]]*S in[3*(c + d*x)] - 75*Log[Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] + 30*Sin[4*(c + d*x)] - 15*Log[Cos[(c + d*x)/2]]*Sin[5*(c + d*x)] + 15*Log[Sin[(c + d*x) /2]]*Sin[5*(c + d*x)]))/d
Time = 0.42 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {3042, 3352, 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.
\(\displaystyle \int \cot ^2(c+d x) \csc ^4(c+d x) (a \sin (c+d x)+a)^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^2 (a \sin (c+d x)+a)^2}{\sin (c+d x)^6}dx\) |
\(\Big \downarrow \) 3352 |
\(\displaystyle \int \left (a^2 \cot ^2(c+d x) \csc ^4(c+d x)+2 a^2 \cot ^2(c+d x) \csc ^3(c+d x)+a^2 \cot ^2(c+d x) \csc ^2(c+d x)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^2 \text {arctanh}(\cos (c+d x))}{4 d}-\frac {a^2 \cot ^5(c+d x)}{5 d}-\frac {2 a^2 \cot ^3(c+d x)}{3 d}-\frac {a^2 \cot (c+d x) \csc ^3(c+d x)}{2 d}+\frac {a^2 \cot (c+d x) \csc (c+d x)}{4 d}\) |
Input:
Int[Cot[c + d*x]^2*Csc[c + d*x]^4*(a + a*Sin[c + d*x])^2,x]
Output:
(a^2*ArcTanh[Cos[c + d*x]])/(4*d) - (2*a^2*Cot[c + d*x]^3)/(3*d) - (a^2*Co t[c + d*x]^5)/(5*d) + (a^2*Cot[c + d*x]*Csc[c + d*x])/(4*d) - (a^2*Cot[c + d*x]*Csc[c + d*x]^3)/(2*d)
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[ExpandTrig [(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x] /; F reeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0]
Time = 0.36 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.36
method | result | size |
derivativedivides | \(\frac {-\frac {a^{2} \cos \left (d x +c \right )^{3}}{3 \sin \left (d x +c \right )^{3}}+2 a^{2} \left (-\frac {\cos \left (d x +c \right )^{3}}{4 \sin \left (d x +c \right )^{4}}-\frac {\cos \left (d x +c \right )^{3}}{8 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )}{8}-\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )+a^{2} \left (-\frac {\cos \left (d x +c \right )^{3}}{5 \sin \left (d x +c \right )^{5}}-\frac {2 \cos \left (d x +c \right )^{3}}{15 \sin \left (d x +c \right )^{3}}\right )}{d}\) | \(136\) |
default | \(\frac {-\frac {a^{2} \cos \left (d x +c \right )^{3}}{3 \sin \left (d x +c \right )^{3}}+2 a^{2} \left (-\frac {\cos \left (d x +c \right )^{3}}{4 \sin \left (d x +c \right )^{4}}-\frac {\cos \left (d x +c \right )^{3}}{8 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )}{8}-\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )+a^{2} \left (-\frac {\cos \left (d x +c \right )^{3}}{5 \sin \left (d x +c \right )^{5}}-\frac {2 \cos \left (d x +c \right )^{3}}{15 \sin \left (d x +c \right )^{3}}\right )}{d}\) | \(136\) |
risch | \(-\frac {a^{2} \left (-60 i {\mathrm e}^{8 i \left (d x +c \right )}+15 \,{\mathrm e}^{9 i \left (d x +c \right )}+240 i {\mathrm e}^{6 i \left (d x +c \right )}+90 \,{\mathrm e}^{7 i \left (d x +c \right )}-40 i {\mathrm e}^{4 i \left (d x +c \right )}+80 i {\mathrm e}^{2 i \left (d x +c \right )}-90 \,{\mathrm e}^{3 i \left (d x +c \right )}-28 i-15 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{30 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{4 d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{4 d}\) | \(158\) |
Input:
int(cot(d*x+c)^2*csc(d*x+c)^4*(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(-1/3*a^2/sin(d*x+c)^3*cos(d*x+c)^3+2*a^2*(-1/4/sin(d*x+c)^4*cos(d*x+c )^3-1/8/sin(d*x+c)^2*cos(d*x+c)^3-1/8*cos(d*x+c)-1/8*ln(csc(d*x+c)-cot(d*x +c)))+a^2*(-1/5/sin(d*x+c)^5*cos(d*x+c)^3-2/15/sin(d*x+c)^3*cos(d*x+c)^3))
Leaf count of result is larger than twice the leaf count of optimal. 189 vs. \(2 (90) = 180\).
Time = 0.09 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.89 \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {56 \, a^{2} \cos \left (d x + c\right )^{5} - 80 \, a^{2} \cos \left (d x + c\right )^{3} + 15 \, {\left (a^{2} \cos \left (d x + c\right )^{4} - 2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 15 \, {\left (a^{2} \cos \left (d x + c\right )^{4} - 2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 30 \, {\left (a^{2} \cos \left (d x + c\right )^{3} + a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \] Input:
integrate(cot(d*x+c)^2*csc(d*x+c)^4*(a+a*sin(d*x+c))^2,x, algorithm="frica s")
Output:
1/120*(56*a^2*cos(d*x + c)^5 - 80*a^2*cos(d*x + c)^3 + 15*(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)*sin(d*x + c) - 15*(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)*sin(d*x + c) - 30*(a^2*cos(d*x + c)^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)*sin(d*x + c))
Timed out. \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\text {Timed out} \] Input:
integrate(cot(d*x+c)**2*csc(d*x+c)**4*(a+a*sin(d*x+c))**2,x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.09 \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {15 \, a^{2} {\left (\frac {2 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + \frac {40 \, a^{2}}{\tan \left (d x + c\right )^{3}} + \frac {8 \, {\left (5 \, \tan \left (d x + c\right )^{2} + 3\right )} a^{2}}{\tan \left (d x + c\right )^{5}}}{120 \, d} \] Input:
integrate(cot(d*x+c)^2*csc(d*x+c)^4*(a+a*sin(d*x+c))^2,x, algorithm="maxim a")
Output:
-1/120*(15*a^2*(2*(cos(d*x + c)^3 + cos(d*x + c))/(cos(d*x + c)^4 - 2*cos( d*x + c)^2 + 1) - log(cos(d*x + c) + 1) + log(cos(d*x + c) - 1)) + 40*a^2/ tan(d*x + c)^3 + 8*(5*tan(d*x + c)^2 + 3)*a^2/tan(d*x + c)^5)/d
Time = 0.18 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.64 \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 25 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 120 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - 90 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {274 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 90 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 25 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{480 \, d} \] Input:
integrate(cot(d*x+c)^2*csc(d*x+c)^4*(a+a*sin(d*x+c))^2,x, algorithm="giac" )
Output:
1/480*(3*a^2*tan(1/2*d*x + 1/2*c)^5 + 15*a^2*tan(1/2*d*x + 1/2*c)^4 + 25*a ^2*tan(1/2*d*x + 1/2*c)^3 - 120*a^2*log(abs(tan(1/2*d*x + 1/2*c))) - 90*a^ 2*tan(1/2*d*x + 1/2*c) + (274*a^2*tan(1/2*d*x + 1/2*c)^5 + 90*a^2*tan(1/2* d*x + 1/2*c)^4 - 25*a^2*tan(1/2*d*x + 1/2*c)^2 - 15*a^2*tan(1/2*d*x + 1/2* c) - 3*a^2)/tan(1/2*d*x + 1/2*c)^5)/d
Time = 17.70 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.60 \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {5\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96\,d}+\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{32\,d}+\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,d}-\frac {a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (-6\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {5\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a^2}{5}\right )}{32\,d}-\frac {3\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,d} \] Input:
int((cot(c + d*x)^2*(a + a*sin(c + d*x))^2)/sin(c + d*x)^4,x)
Output:
(5*a^2*tan(c/2 + (d*x)/2)^3)/(96*d) + (a^2*tan(c/2 + (d*x)/2)^4)/(32*d) + (a^2*tan(c/2 + (d*x)/2)^5)/(160*d) - (a^2*log(tan(c/2 + (d*x)/2)))/(4*d) - (cot(c/2 + (d*x)/2)^5*((5*a^2*tan(c/2 + (d*x)/2)^2)/3 - 6*a^2*tan(c/2 + ( d*x)/2)^4 + a^2/5 + a^2*tan(c/2 + (d*x)/2)))/(32*d) - (3*a^2*tan(c/2 + (d* x)/2))/(16*d)
Time = 0.16 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.07 \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^{2} \left (28 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}+15 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}-16 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}-30 \cos \left (d x +c \right ) \sin \left (d x +c \right )-12 \cos \left (d x +c \right )-15 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{5}\right )}{60 \sin \left (d x +c \right )^{5} d} \] Input:
int(cot(d*x+c)^2*csc(d*x+c)^4*(a+a*sin(d*x+c))^2,x)
Output:
(a**2*(28*cos(c + d*x)*sin(c + d*x)**4 + 15*cos(c + d*x)*sin(c + d*x)**3 - 16*cos(c + d*x)*sin(c + d*x)**2 - 30*cos(c + d*x)*sin(c + d*x) - 12*cos(c + d*x) - 15*log(tan((c + d*x)/2))*sin(c + d*x)**5))/(60*sin(c + d*x)**5*d )