Integrand size = 29, antiderivative size = 91 \[ \int \frac {\cot ^7(c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\cot ^6(c+d x)}{6 a d}-\frac {\cot ^8(c+d x)}{8 a d}+\frac {\csc ^3(c+d x)}{3 a d}-\frac {2 \csc ^5(c+d x)}{5 a d}+\frac {\csc ^7(c+d x)}{7 a d} \] Output:
-1/6*cot(d*x+c)^6/a/d-1/8*cot(d*x+c)^8/a/d+1/3*csc(d*x+c)^3/a/d-2/5*csc(d* x+c)^5/a/d+1/7*csc(d*x+c)^7/a/d
Time = 0.16 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.75 \[ \int \frac {\cot ^7(c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\csc ^3(c+d x) \left (280-210 \csc (c+d x)-336 \csc ^2(c+d x)+280 \csc ^3(c+d x)+120 \csc ^4(c+d x)-105 \csc ^5(c+d x)\right )}{840 a d} \] Input:
Integrate[(Cot[c + d*x]^7*Csc[c + d*x]^2)/(a + a*Sin[c + d*x]),x]
Output:
(Csc[c + d*x]^3*(280 - 210*Csc[c + d*x] - 336*Csc[c + d*x]^2 + 280*Csc[c + d*x]^3 + 120*Csc[c + d*x]^4 - 105*Csc[c + d*x]^5))/(840*a*d)
Time = 0.51 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.86, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.345, Rules used = {3042, 3314, 3042, 25, 3086, 244, 2009, 3087, 244, 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.
\(\displaystyle \int \frac {\cot ^7(c+d x) \csc ^2(c+d x)}{a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^7}{\sin (c+d x)^9 (a \sin (c+d x)+a)}dx\) |
\(\Big \downarrow \) 3314 |
\(\displaystyle \frac {\int \cot ^5(c+d x) \csc ^4(c+d x)dx}{a}-\frac {\int \cot ^5(c+d x) \csc ^3(c+d x)dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int -\sec \left (c+d x-\frac {\pi }{2}\right )^4 \tan \left (c+d x-\frac {\pi }{2}\right )^5dx}{a}-\frac {\int -\sec \left (c+d x-\frac {\pi }{2}\right )^3 \tan \left (c+d x-\frac {\pi }{2}\right )^5dx}{a}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right )^3 \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )^5dx}{a}-\frac {\int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right )^4 \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )^5dx}{a}\) |
\(\Big \downarrow \) 3086 |
\(\displaystyle \frac {\int \csc ^2(c+d x) \left (1-\csc ^2(c+d x)\right )^2d\csc (c+d x)}{a d}-\frac {\int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right )^4 \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )^5dx}{a}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {\int \left (\csc ^6(c+d x)-2 \csc ^4(c+d x)+\csc ^2(c+d x)\right )d\csc (c+d x)}{a d}-\frac {\int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right )^4 \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )^5dx}{a}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{7} \csc ^7(c+d x)-\frac {2}{5} \csc ^5(c+d x)+\frac {1}{3} \csc ^3(c+d x)}{a d}-\frac {\int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right )^4 \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )^5dx}{a}\) |
\(\Big \downarrow \) 3087 |
\(\displaystyle \frac {\frac {1}{7} \csc ^7(c+d x)-\frac {2}{5} \csc ^5(c+d x)+\frac {1}{3} \csc ^3(c+d x)}{a d}-\frac {\int -\cot ^5(c+d x) \left (\cot ^2(c+d x)+1\right )d(-\cot (c+d x))}{a d}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {\frac {1}{7} \csc ^7(c+d x)-\frac {2}{5} \csc ^5(c+d x)+\frac {1}{3} \csc ^3(c+d x)}{a d}-\frac {\int \left (-\cot ^7(c+d x)-\cot ^5(c+d x)\right )d(-\cot (c+d x))}{a d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{7} \csc ^7(c+d x)-\frac {2}{5} \csc ^5(c+d x)+\frac {1}{3} \csc ^3(c+d x)}{a d}-\frac {\frac {1}{8} \cot ^8(c+d x)+\frac {1}{6} \cot ^6(c+d x)}{a d}\) |
Input:
Int[(Cot[c + d*x]^7*Csc[c + d*x]^2)/(a + a*Sin[c + d*x]),x]
Output:
-((Cot[c + d*x]^6/6 + Cot[c + d*x]^8/8)/(a*d)) + (Csc[c + d*x]^3/3 - (2*Cs c[c + d*x]^5)/5 + Csc[c + d*x]^7/7)/(a*d)
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_.), x_Symbol] :> Simp[a/f Subst[Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2 ), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2 ] && !(IntegerQ[m/2] && LtQ[0, m, n + 1])
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_S ymbol] :> Simp[1/f Subst[Int[(b*x)^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] && !(IntegerQ[(n - 1) /2] && LtQ[0, n, m - 1])
Int[(cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/(( a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[1/a Int[Cos[e + f *x]^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Simp[1/(b*d) Int[Cos[e + f*x]^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] & & IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && IntegerQ[n] && (LtQ[0, n, (p + 1)/2] || (LeQ[p, -n] && LtQ[-n, 2*p - 3]) || (GtQ[n, 0] && LeQ[n, -p]))
Time = 1.54 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.76
method | result | size |
derivativedivides | \(\frac {-\frac {\csc \left (d x +c \right )^{8}}{8}+\frac {\csc \left (d x +c \right )^{7}}{7}+\frac {\csc \left (d x +c \right )^{6}}{3}-\frac {2 \csc \left (d x +c \right )^{5}}{5}-\frac {\csc \left (d x +c \right )^{4}}{4}+\frac {\csc \left (d x +c \right )^{3}}{3}}{d a}\) | \(69\) |
default | \(\frac {-\frac {\csc \left (d x +c \right )^{8}}{8}+\frac {\csc \left (d x +c \right )^{7}}{7}+\frac {\csc \left (d x +c \right )^{6}}{3}-\frac {2 \csc \left (d x +c \right )^{5}}{5}-\frac {\csc \left (d x +c \right )^{4}}{4}+\frac {\csc \left (d x +c \right )^{3}}{3}}{d a}\) | \(69\) |
risch | \(-\frac {4 i \left (-105 i {\mathrm e}^{12 i \left (d x +c \right )}+70 \,{\mathrm e}^{13 i \left (d x +c \right )}-140 i {\mathrm e}^{10 i \left (d x +c \right )}-14 \,{\mathrm e}^{11 i \left (d x +c \right )}-350 i {\mathrm e}^{8 i \left (d x +c \right )}+172 \,{\mathrm e}^{9 i \left (d x +c \right )}-140 i {\mathrm e}^{6 i \left (d x +c \right )}-172 \,{\mathrm e}^{7 i \left (d x +c \right )}-105 i {\mathrm e}^{4 i \left (d x +c \right )}+14 \,{\mathrm e}^{5 i \left (d x +c \right )}-70 \,{\mathrm e}^{3 i \left (d x +c \right )}\right )}{105 a d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{8}}\) | \(150\) |
Input:
int(cot(d*x+c)^7*csc(d*x+c)^2/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d/a*(-1/8*csc(d*x+c)^8+1/7*csc(d*x+c)^7+1/3*csc(d*x+c)^6-2/5*csc(d*x+c)^ 5-1/4*csc(d*x+c)^4+1/3*csc(d*x+c)^3)
Time = 0.08 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.18 \[ \int \frac {\cot ^7(c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {210 \, \cos \left (d x + c\right )^{4} - 140 \, \cos \left (d x + c\right )^{2} - 8 \, {\left (35 \, \cos \left (d x + c\right )^{4} - 28 \, \cos \left (d x + c\right )^{2} + 8\right )} \sin \left (d x + c\right ) + 35}{840 \, {\left (a d \cos \left (d x + c\right )^{8} - 4 \, a d \cos \left (d x + c\right )^{6} + 6 \, a d \cos \left (d x + c\right )^{4} - 4 \, a d \cos \left (d x + c\right )^{2} + a d\right )}} \] Input:
integrate(cot(d*x+c)^7*csc(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="fricas" )
Output:
-1/840*(210*cos(d*x + c)^4 - 140*cos(d*x + c)^2 - 8*(35*cos(d*x + c)^4 - 2 8*cos(d*x + c)^2 + 8)*sin(d*x + c) + 35)/(a*d*cos(d*x + c)^8 - 4*a*d*cos(d *x + c)^6 + 6*a*d*cos(d*x + c)^4 - 4*a*d*cos(d*x + c)^2 + a*d)
Timed out. \[ \int \frac {\cot ^7(c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Timed out} \] Input:
integrate(cot(d*x+c)**7*csc(d*x+c)**2/(a+a*sin(d*x+c)),x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.73 \[ \int \frac {\cot ^7(c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {280 \, \sin \left (d x + c\right )^{5} - 210 \, \sin \left (d x + c\right )^{4} - 336 \, \sin \left (d x + c\right )^{3} + 280 \, \sin \left (d x + c\right )^{2} + 120 \, \sin \left (d x + c\right ) - 105}{840 \, a d \sin \left (d x + c\right )^{8}} \] Input:
integrate(cot(d*x+c)^7*csc(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="maxima" )
Output:
1/840*(280*sin(d*x + c)^5 - 210*sin(d*x + c)^4 - 336*sin(d*x + c)^3 + 280* sin(d*x + c)^2 + 120*sin(d*x + c) - 105)/(a*d*sin(d*x + c)^8)
Time = 0.15 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.73 \[ \int \frac {\cot ^7(c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {280 \, \sin \left (d x + c\right )^{5} - 210 \, \sin \left (d x + c\right )^{4} - 336 \, \sin \left (d x + c\right )^{3} + 280 \, \sin \left (d x + c\right )^{2} + 120 \, \sin \left (d x + c\right ) - 105}{840 \, a d \sin \left (d x + c\right )^{8}} \] Input:
integrate(cot(d*x+c)^7*csc(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
1/840*(280*sin(d*x + c)^5 - 210*sin(d*x + c)^4 - 336*sin(d*x + c)^3 + 280* sin(d*x + c)^2 + 120*sin(d*x + c) - 105)/(a*d*sin(d*x + c)^8)
Time = 31.27 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.73 \[ \int \frac {\cot ^7(c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {280\,{\sin \left (c+d\,x\right )}^5-210\,{\sin \left (c+d\,x\right )}^4-336\,{\sin \left (c+d\,x\right )}^3+280\,{\sin \left (c+d\,x\right )}^2+120\,\sin \left (c+d\,x\right )-105}{840\,a\,d\,{\sin \left (c+d\,x\right )}^8} \] Input:
int(cot(c + d*x)^7/(sin(c + d*x)^2*(a + a*sin(c + d*x))),x)
Output:
(120*sin(c + d*x) + 280*sin(c + d*x)^2 - 336*sin(c + d*x)^3 - 210*sin(c + d*x)^4 + 280*sin(c + d*x)^5 - 105)/(840*a*d*sin(c + d*x)^8)
Time = 0.17 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.84 \[ \int \frac {\cot ^7(c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {2555 \sin \left (d x +c \right )^{8}+35840 \sin \left (d x +c \right )^{5}-26880 \sin \left (d x +c \right )^{4}-43008 \sin \left (d x +c \right )^{3}+35840 \sin \left (d x +c \right )^{2}+15360 \sin \left (d x +c \right )-13440}{107520 \sin \left (d x +c \right )^{8} a d} \] Input:
int(cot(d*x+c)^7*csc(d*x+c)^2/(a+a*sin(d*x+c)),x)
Output:
(2555*sin(c + d*x)**8 + 35840*sin(c + d*x)**5 - 26880*sin(c + d*x)**4 - 43 008*sin(c + d*x)**3 + 35840*sin(c + d*x)**2 + 15360*sin(c + d*x) - 13440)/ (107520*sin(c + d*x)**8*a*d)