Integrand size = 29, antiderivative size = 119 \[ \int \cot ^5(c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {2 a^2 \csc (c+d x)}{d}+\frac {a^2 \csc ^2(c+d x)}{2 d}+\frac {4 a^2 \csc ^3(c+d x)}{3 d}+\frac {a^2 \csc ^4(c+d x)}{4 d}-\frac {2 a^2 \csc ^5(c+d x)}{5 d}-\frac {a^2 \csc ^6(c+d x)}{6 d}+\frac {a^2 \log (\sin (c+d x))}{d} \] Output:
-2*a^2*csc(d*x+c)/d+1/2*a^2*csc(d*x+c)^2/d+4/3*a^2*csc(d*x+c)^3/d+1/4*a^2* csc(d*x+c)^4/d-2/5*a^2*csc(d*x+c)^5/d-1/6*a^2*csc(d*x+c)^6/d+a^2*ln(sin(d* x+c))/d
Time = 0.05 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.86 \[ \int \cot ^5(c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=a^2 \left (-\frac {2 \csc (c+d x)}{d}+\frac {\csc ^2(c+d x)}{2 d}+\frac {4 \csc ^3(c+d x)}{3 d}+\frac {\csc ^4(c+d x)}{4 d}-\frac {2 \csc ^5(c+d x)}{5 d}-\frac {\csc ^6(c+d x)}{6 d}+\frac {\log (\sin (c+d x))}{d}\right ) \] Input:
Integrate[Cot[c + d*x]^5*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) + (4*Csc[c + d*x]^3)/(3*d) + Csc[c + d*x]^4/(4*d) - (2*Csc[c + d*x]^5)/(5*d) - Csc[c + d*x]^6/(6*d) + Log[Sin[c + d*x]]/d)
Time = 0.34 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.71, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 3315, 27, 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.
\(\displaystyle \int \cot ^5(c+d x) \csc ^2(c+d x) (a \sin (c+d x)+a)^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^5 (a \sin (c+d x)+a)^2}{\sin (c+d x)^7}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \csc ^7(c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^4d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a^2 \int \frac {\csc ^7(c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^4}{a^7}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {a^2 \int \left (\frac {\csc ^7(c+d x)}{a}+\frac {2 \csc ^6(c+d x)}{a}-\frac {\csc ^5(c+d x)}{a}-\frac {4 \csc ^4(c+d x)}{a}-\frac {\csc ^3(c+d x)}{a}+\frac {2 \csc ^2(c+d x)}{a}+\frac {\csc (c+d x)}{a}\right )d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^2 \left (\log (a \sin (c+d x))-\frac {1}{6} \csc ^6(c+d x)-\frac {2}{5} \csc ^5(c+d x)+\frac {1}{4} \csc ^4(c+d x)+\frac {4}{3} \csc ^3(c+d x)+\frac {1}{2} \csc ^2(c+d x)-2 \csc (c+d x)\right )}{d}\) |
Input:
Int[Cot[c + d*x]^5*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 + (4*Csc[c + d*x]^3)/3 + Csc[c + d*x]^4/4 - (2*Csc[c + d*x]^5)/5 - Csc[c + d*x]^6/6 + 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_)*((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[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && Intege rQ[(p - 1)/2] && EqQ[a^2 - b^2, 0]
Time = 0.26 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.63
method | result | size |
derivativedivides | \(-\frac {a^{2} \left (\frac {\csc \left (d x +c \right )^{6}}{6}+\frac {2 \csc \left (d x +c \right )^{5}}{5}-\frac {\csc \left (d x +c \right )^{4}}{4}-\frac {4 \csc \left (d x +c \right )^{3}}{3}-\frac {\csc \left (d x +c \right )^{2}}{2}+2 \csc \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )\right )\right )}{d}\) | \(75\) |
default | \(-\frac {a^{2} \left (\frac {\csc \left (d x +c \right )^{6}}{6}+\frac {2 \csc \left (d x +c \right )^{5}}{5}-\frac {\csc \left (d x +c \right )^{4}}{4}-\frac {4 \csc \left (d x +c \right )^{3}}{3}-\frac {\csc \left (d x +c \right )^{2}}{2}+2 \csc \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )\right )\right )}{d}\) | \(75\) |
risch | \(-i a^{2} x -\frac {2 i a^{2} c}{d}-\frac {2 i a^{2} \left (-15 i {\mathrm e}^{10 i \left (d x +c \right )}+30 \,{\mathrm e}^{11 i \left (d x +c \right )}+90 i {\mathrm e}^{8 i \left (d x +c \right )}-70 \,{\mathrm e}^{9 i \left (d x +c \right )}-70 i {\mathrm e}^{6 i \left (d x +c \right )}+156 \,{\mathrm e}^{7 i \left (d x +c \right )}+90 i {\mathrm e}^{4 i \left (d x +c \right )}-156 \,{\mathrm e}^{5 i \left (d x +c \right )}-15 i {\mathrm e}^{2 i \left (d x +c \right )}+70 \,{\mathrm e}^{3 i \left (d x +c \right )}-30 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{6}}+\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\) | \(187\) |
Input:
int(cot(d*x+c)^5*csc(d*x+c)^2*(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
-a^2/d*(1/6*csc(d*x+c)^6+2/5*csc(d*x+c)^5-1/4*csc(d*x+c)^4-4/3*csc(d*x+c)^ 3-1/2*csc(d*x+c)^2+2*csc(d*x+c)+ln(csc(d*x+c)))
Time = 0.09 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.40 \[ \int \cot ^5(c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {30 \, a^{2} \cos \left (d x + c\right )^{4} - 75 \, a^{2} \cos \left (d x + c\right )^{2} + 35 \, a^{2} - 60 \, {\left (a^{2} \cos \left (d x + c\right )^{6} - 3 \, a^{2} \cos \left (d x + c\right )^{4} + 3 \, a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 8 \, {\left (15 \, a^{2} \cos \left (d x + c\right )^{4} - 20 \, a^{2} \cos \left (d x + c\right )^{2} + 8 \, a^{2}\right )} \sin \left (d x + c\right )}{60 \, {\left (d \cos \left (d x + c\right )^{6} - 3 \, d \cos \left (d x + c\right )^{4} + 3 \, d \cos \left (d x + c\right )^{2} - d\right )}} \] Input:
integrate(cot(d*x+c)^5*csc(d*x+c)^2*(a+a*sin(d*x+c))^2,x, algorithm="frica s")
Output:
-1/60*(30*a^2*cos(d*x + c)^4 - 75*a^2*cos(d*x + c)^2 + 35*a^2 - 60*(a^2*co s(d*x + c)^6 - 3*a^2*cos(d*x + c)^4 + 3*a^2*cos(d*x + c)^2 - a^2)*log(1/2* sin(d*x + c)) - 8*(15*a^2*cos(d*x + c)^4 - 20*a^2*cos(d*x + c)^2 + 8*a^2)* sin(d*x + c))/(d*cos(d*x + c)^6 - 3*d*cos(d*x + c)^4 + 3*d*cos(d*x + c)^2 - d)
Timed out. \[ \int \cot ^5(c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\text {Timed out} \] Input:
integrate(cot(d*x+c)**5*csc(d*x+c)**2*(a+a*sin(d*x+c))**2,x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.82 \[ \int \cot ^5(c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {60 \, a^{2} \log \left (\sin \left (d x + c\right )\right ) - \frac {120 \, a^{2} \sin \left (d x + c\right )^{5} - 30 \, a^{2} \sin \left (d x + c\right )^{4} - 80 \, a^{2} \sin \left (d x + c\right )^{3} - 15 \, a^{2} \sin \left (d x + c\right )^{2} + 24 \, a^{2} \sin \left (d x + c\right ) + 10 \, a^{2}}{\sin \left (d x + c\right )^{6}}}{60 \, d} \] Input:
integrate(cot(d*x+c)^5*csc(d*x+c)^2*(a+a*sin(d*x+c))^2,x, algorithm="maxim a")
Output:
1/60*(60*a^2*log(sin(d*x + c)) - (120*a^2*sin(d*x + c)^5 - 30*a^2*sin(d*x + c)^4 - 80*a^2*sin(d*x + c)^3 - 15*a^2*sin(d*x + c)^2 + 24*a^2*sin(d*x + c) + 10*a^2)/sin(d*x + c)^6)/d
Time = 0.17 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.82 \[ \int \cot ^5(c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {60 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - \frac {120 \, a^{2} \sin \left (d x + c\right )^{5} - 30 \, a^{2} \sin \left (d x + c\right )^{4} - 80 \, a^{2} \sin \left (d x + c\right )^{3} - 15 \, a^{2} \sin \left (d x + c\right )^{2} + 24 \, a^{2} \sin \left (d x + c\right ) + 10 \, a^{2}}{\sin \left (d x + c\right )^{6}}}{60 \, d} \] Input:
integrate(cot(d*x+c)^5*csc(d*x+c)^2*(a+a*sin(d*x+c))^2,x, algorithm="giac" )
Output:
1/60*(60*a^2*log(abs(sin(d*x + c))) - (120*a^2*sin(d*x + c)^5 - 30*a^2*sin (d*x + c)^4 - 80*a^2*sin(d*x + c)^3 - 15*a^2*sin(d*x + c)^2 + 24*a^2*sin(d *x + c) + 10*a^2)/sin(d*x + c)^6)/d
Time = 32.49 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.82 \[ \int \cot ^5(c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {19\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{128\,d}+\frac {5\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{48\,d}-\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{80\,d}-\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{384\,d}+\frac {a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (40\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-\frac {19\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{2}-\frac {20\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {4\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{5}+\frac {a^2}{6}\right )}{64\,d}-\frac {5\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,d}-\frac {a^2\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \] Input:
int((cot(c + d*x)^5*(a + a*sin(c + d*x))^2)/sin(c + d*x)^2,x)
Output:
(19*a^2*tan(c/2 + (d*x)/2)^2)/(128*d) + (5*a^2*tan(c/2 + (d*x)/2)^3)/(48*d ) - (a^2*tan(c/2 + (d*x)/2)^5)/(80*d) - (a^2*tan(c/2 + (d*x)/2)^6)/(384*d) + (a^2*log(tan(c/2 + (d*x)/2)))/d - (cot(c/2 + (d*x)/2)^6*(40*a^2*tan(c/2 + (d*x)/2)^5 - (19*a^2*tan(c/2 + (d*x)/2)^4)/2 - (20*a^2*tan(c/2 + (d*x)/ 2)^3)/3 + a^2/6 + (4*a^2*tan(c/2 + (d*x)/2))/5))/(64*d) - (5*a^2*tan(c/2 + (d*x)/2))/(8*d) - (a^2*log(tan(c/2 + (d*x)/2)^2 + 1))/d
Time = 0.15 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.01 \[ \int \cot ^5(c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^{2} \left (-120 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{6}+120 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{6}-35 \sin \left (d x +c \right )^{6}-240 \sin \left (d x +c \right )^{5}+60 \sin \left (d x +c \right )^{4}+160 \sin \left (d x +c \right )^{3}+30 \sin \left (d x +c \right )^{2}-48 \sin \left (d x +c \right )-20\right )}{120 \sin \left (d x +c \right )^{6} d} \] Input:
int(cot(d*x+c)^5*csc(d*x+c)^2*(a+a*sin(d*x+c))^2,x)
Output:
(a**2*( - 120*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**6 + 120*log(tan(( c + d*x)/2))*sin(c + d*x)**6 - 35*sin(c + d*x)**6 - 240*sin(c + d*x)**5 + 60*sin(c + d*x)**4 + 160*sin(c + d*x)**3 + 30*sin(c + d*x)**2 - 48*sin(c + d*x) - 20))/(120*sin(c + d*x)**6*d)