Integrand size = 27, antiderivative size = 73 \[ \int \frac {\cot ^7(c+d x) \csc (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\cot ^6(c+d x)}{6 a d}-\frac {\csc ^3(c+d x)}{3 a d}+\frac {2 \csc ^5(c+d x)}{5 a d}-\frac {\csc ^7(c+d x)}{7 a d} \] Output:
1/6*cot(d*x+c)^6/a/d-1/3*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 = 0.17 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.93 \[ \int \frac {\cot ^7(c+d x) \csc (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\csc ^2(c+d x) \left (105-70 \csc (c+d x)-105 \csc ^2(c+d x)+84 \csc ^3(c+d x)+35 \csc ^4(c+d x)-30 \csc ^5(c+d x)\right )}{210 a d} \] Input:
Integrate[(Cot[c + d*x]^7*Csc[c + d*x])/(a + a*Sin[c + d*x]),x]
Output:
(Csc[c + d*x]^2*(105 - 70*Csc[c + d*x] - 105*Csc[c + d*x]^2 + 84*Csc[c + d *x]^3 + 35*Csc[c + d*x]^4 - 30*Csc[c + d*x]^5))/(210*a*d)
Time = 0.46 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.88, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3314, 3042, 25, 3086, 244, 2009, 3087, 15}
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.
\(\displaystyle \int \frac {\cot ^7(c+d x) \csc (c+d x)}{a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^7}{\sin (c+d x)^8 (a \sin (c+d x)+a)}dx\) |
\(\Big \downarrow \) 3314 |
\(\displaystyle \frac {\int \cot ^5(c+d x) \csc ^3(c+d x)dx}{a}-\frac {\int \cot ^5(c+d x) \csc ^2(c+d x)dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int -\sec \left (c+d x-\frac {\pi }{2}\right )^3 \tan \left (c+d x-\frac {\pi }{2}\right )^5dx}{a}-\frac {\int -\sec \left (c+d x-\frac {\pi }{2}\right )^2 \tan \left (c+d x-\frac {\pi }{2}\right )^5dx}{a}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right )^2 \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )^5dx}{a}-\frac {\int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right )^3 \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )^5dx}{a}\) |
\(\Big \downarrow \) 3086 |
\(\displaystyle \frac {\int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right )^2 \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )^5dx}{a}-\frac {\int \csc ^2(c+d x) \left (1-\csc ^2(c+d x)\right )^2d\csc (c+d x)}{a d}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {\int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right )^2 \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )^5dx}{a}-\frac {\int \left (\csc ^6(c+d x)-2 \csc ^4(c+d x)+\csc ^2(c+d x)\right )d\csc (c+d x)}{a d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right )^2 \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )^5dx}{a}-\frac {\frac {1}{7} \csc ^7(c+d x)-\frac {2}{5} \csc ^5(c+d x)+\frac {1}{3} \csc ^3(c+d x)}{a d}\) |
\(\Big \downarrow \) 3087 |
\(\displaystyle \frac {\int -\cot ^5(c+d x)d(-\cot (c+d x))}{a d}-\frac {\frac {1}{7} \csc ^7(c+d x)-\frac {2}{5} \csc ^5(c+d x)+\frac {1}{3} \csc ^3(c+d x)}{a d}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {\cot ^6(c+d x)}{6 a d}-\frac {\frac {1}{7} \csc ^7(c+d x)-\frac {2}{5} \csc ^5(c+d x)+\frac {1}{3} \csc ^3(c+d x)}{a d}\) |
Input:
Int[(Cot[c + d*x]^7*Csc[c + d*x])/(a + a*Sin[c + d*x]),x]
Output:
Cot[c + d*x]^6/(6*a*d) - (Csc[c + d*x]^3/3 - (2*Csc[c + d*x]^5)/5 + Csc[c + d*x]^7/7)/(a*d)
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
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 = 1.09 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.96
method | result | size |
derivativedivides | \(-\frac {\frac {\csc \left (d x +c \right )^{7}}{7}-\frac {\csc \left (d x +c \right )^{6}}{6}-\frac {2 \csc \left (d x +c \right )^{5}}{5}+\frac {\csc \left (d x +c \right )^{4}}{2}+\frac {\csc \left (d x +c \right )^{3}}{3}-\frac {\csc \left (d x +c \right )^{2}}{2}}{d a}\) | \(70\) |
default | \(-\frac {\frac {\csc \left (d x +c \right )^{7}}{7}-\frac {\csc \left (d x +c \right )^{6}}{6}-\frac {2 \csc \left (d x +c \right )^{5}}{5}+\frac {\csc \left (d x +c \right )^{4}}{2}+\frac {\csc \left (d x +c \right )^{3}}{3}-\frac {\csc \left (d x +c \right )^{2}}{2}}{d a}\) | \(70\) |
risch | \(-\frac {2 \left (-140 i {\mathrm e}^{11 i \left (d x +c \right )}+105 \,{\mathrm e}^{12 i \left (d x +c \right )}-112 i {\mathrm e}^{9 i \left (d x +c \right )}-105 \,{\mathrm e}^{10 i \left (d x +c \right )}-456 i {\mathrm e}^{7 i \left (d x +c \right )}+350 \,{\mathrm e}^{8 i \left (d x +c \right )}-112 i {\mathrm e}^{5 i \left (d x +c \right )}-350 \,{\mathrm e}^{6 i \left (d x +c \right )}-140 i {\mathrm e}^{3 i \left (d x +c \right )}+105 \,{\mathrm e}^{4 i \left (d x +c \right )}-105 \,{\mathrm e}^{2 i \left (d x +c \right )}\right )}{105 a d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{7}}\) | \(149\) |
Input:
int(cot(d*x+c)^7*csc(d*x+c)/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
-1/d/a*(1/7*csc(d*x+c)^7-1/6*csc(d*x+c)^6-2/5*csc(d*x+c)^5+1/2*csc(d*x+c)^ 4+1/3*csc(d*x+c)^3-1/2*csc(d*x+c)^2)
Time = 0.08 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.42 \[ \int \frac {\cot ^7(c+d x) \csc (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {70 \, \cos \left (d x + c\right )^{4} - 56 \, \cos \left (d x + c\right )^{2} - 35 \, {\left (3 \, \cos \left (d x + c\right )^{4} - 3 \, \cos \left (d x + c\right )^{2} + 1\right )} \sin \left (d x + c\right ) + 16}{210 \, {\left (a d \cos \left (d x + c\right )^{6} - 3 \, a d \cos \left (d x + c\right )^{4} + 3 \, a d \cos \left (d x + c\right )^{2} - a d\right )} \sin \left (d x + c\right )} \] Input:
integrate(cot(d*x+c)^7*csc(d*x+c)/(a+a*sin(d*x+c)),x, algorithm="fricas")
Output:
1/210*(70*cos(d*x + c)^4 - 56*cos(d*x + c)^2 - 35*(3*cos(d*x + c)^4 - 3*co s(d*x + c)^2 + 1)*sin(d*x + c) + 16)/((a*d*cos(d*x + c)^6 - 3*a*d*cos(d*x + c)^4 + 3*a*d*cos(d*x + c)^2 - a*d)*sin(d*x + c))
\[ \int \frac {\cot ^7(c+d x) \csc (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {\cot ^{7}{\left (c + d x \right )} \csc {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \] Input:
integrate(cot(d*x+c)**7*csc(d*x+c)/(a+a*sin(d*x+c)),x)
Output:
Integral(cot(c + d*x)**7*csc(c + d*x)/(sin(c + d*x) + 1), x)/a
Time = 0.04 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.90 \[ \int \frac {\cot ^7(c+d x) \csc (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {105 \, \sin \left (d x + c\right )^{5} - 70 \, \sin \left (d x + c\right )^{4} - 105 \, \sin \left (d x + c\right )^{3} + 84 \, \sin \left (d x + c\right )^{2} + 35 \, \sin \left (d x + c\right ) - 30}{210 \, a d \sin \left (d x + c\right )^{7}} \] Input:
integrate(cot(d*x+c)^7*csc(d*x+c)/(a+a*sin(d*x+c)),x, algorithm="maxima")
Output:
1/210*(105*sin(d*x + c)^5 - 70*sin(d*x + c)^4 - 105*sin(d*x + c)^3 + 84*si n(d*x + c)^2 + 35*sin(d*x + c) - 30)/(a*d*sin(d*x + c)^7)
Time = 0.14 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.90 \[ \int \frac {\cot ^7(c+d x) \csc (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {105 \, \sin \left (d x + c\right )^{5} - 70 \, \sin \left (d x + c\right )^{4} - 105 \, \sin \left (d x + c\right )^{3} + 84 \, \sin \left (d x + c\right )^{2} + 35 \, \sin \left (d x + c\right ) - 30}{210 \, a d \sin \left (d x + c\right )^{7}} \] Input:
integrate(cot(d*x+c)^7*csc(d*x+c)/(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
1/210*(105*sin(d*x + c)^5 - 70*sin(d*x + c)^4 - 105*sin(d*x + c)^3 + 84*si n(d*x + c)^2 + 35*sin(d*x + c) - 30)/(a*d*sin(d*x + c)^7)
Time = 31.26 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.90 \[ \int \frac {\cot ^7(c+d x) \csc (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {105\,{\sin \left (c+d\,x\right )}^5-70\,{\sin \left (c+d\,x\right )}^4-105\,{\sin \left (c+d\,x\right )}^3+84\,{\sin \left (c+d\,x\right )}^2+35\,\sin \left (c+d\,x\right )-30}{210\,a\,d\,{\sin \left (c+d\,x\right )}^7} \] Input:
int(cot(c + d*x)^7/(sin(c + d*x)*(a + a*sin(c + d*x))),x)
Output:
(35*sin(c + d*x) + 84*sin(c + d*x)^2 - 105*sin(c + d*x)^3 - 70*sin(c + d*x )^4 + 105*sin(c + d*x)^5 - 30)/(210*a*d*sin(c + d*x)^7)
Time = 0.17 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.04 \[ \int \frac {\cot ^7(c+d x) \csc (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {-385 \sin \left (d x +c \right )^{7}+1680 \sin \left (d x +c \right )^{5}-1120 \sin \left (d x +c \right )^{4}-1680 \sin \left (d x +c \right )^{3}+1344 \sin \left (d x +c \right )^{2}+560 \sin \left (d x +c \right )-480}{3360 \sin \left (d x +c \right )^{7} a d} \] Input:
int(cot(d*x+c)^7*csc(d*x+c)/(a+a*sin(d*x+c)),x)
Output:
( - 385*sin(c + d*x)**7 + 1680*sin(c + d*x)**5 - 1120*sin(c + d*x)**4 - 16 80*sin(c + d*x)**3 + 1344*sin(c + d*x)**2 + 560*sin(c + d*x) - 480)/(3360* sin(c + d*x)**7*a*d)