Integrand size = 27, antiderivative size = 74 \[ \int \cot ^4(c+d x) \csc ^2(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {3 a \text {arctanh}(\cos (c+d x))}{8 d}-\frac {a \cot ^5(c+d x)}{5 d}+\frac {3 a \cot (c+d x) \csc (c+d x)}{8 d}-\frac {a \cot ^3(c+d x) \csc (c+d x)}{4 d} \] Output:
-3/8*a*arctanh(cos(d*x+c))/d-1/5*a*cot(d*x+c)^5/d+3/8*a*cot(d*x+c)*csc(d*x +c)/d-1/4*a*cot(d*x+c)^3*csc(d*x+c)/d
Time = 0.01 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.82 \[ \int \cot ^4(c+d x) \csc ^2(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \cot ^5(c+d x)}{5 d}+\frac {5 a \csc ^2\left (\frac {1}{2} (c+d x)\right )}{32 d}-\frac {a \csc ^4\left (\frac {1}{2} (c+d x)\right )}{64 d}-\frac {3 a \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{8 d}+\frac {3 a \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{8 d}-\frac {5 a \sec ^2\left (\frac {1}{2} (c+d x)\right )}{32 d}+\frac {a \sec ^4\left (\frac {1}{2} (c+d x)\right )}{64 d} \] Input:
Integrate[Cot[c + d*x]^4*Csc[c + d*x]^2*(a + a*Sin[c + d*x]),x]
Output:
-1/5*(a*Cot[c + d*x]^5)/d + (5*a*Csc[(c + d*x)/2]^2)/(32*d) - (a*Csc[(c + d*x)/2]^4)/(64*d) - (3*a*Log[Cos[(c + d*x)/2]])/(8*d) + (3*a*Log[Sin[(c + d*x)/2]])/(8*d) - (5*a*Sec[(c + d*x)/2]^2)/(32*d) + (a*Sec[(c + d*x)/2]^4) /(64*d)
Time = 0.55 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.07, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {3042, 3317, 3042, 3087, 15, 3091, 3042, 3091, 3042, 4257}
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.
| \(\displaystyle \int \cot ^4(c+d x) \csc ^2(c+d x) (a \sin (c+d x)+a) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^4 (a \sin (c+d x)+a)}{\sin (c+d x)^6}dx\) |
| \(\Big \downarrow \) 3317 |
| \(\displaystyle a \int \cot ^4(c+d x) \csc ^2(c+d x)dx+a \int \cot ^4(c+d x) \csc (c+d x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \int \sec \left (c+d x-\frac {\pi }{2}\right ) \tan \left (c+d x-\frac {\pi }{2}\right )^4dx+a \int \sec \left (c+d x-\frac {\pi }{2}\right )^2 \tan \left (c+d x-\frac {\pi }{2}\right )^4dx\) |
| \(\Big \downarrow \) 3087 |
| \(\displaystyle \frac {a \int \cot ^4(c+d x)d(-\cot (c+d x))}{d}+a \int \sec \left (c+d x-\frac {\pi }{2}\right ) \tan \left (c+d x-\frac {\pi }{2}\right )^4dx\) |
| \(\Big \downarrow \) 15 |
| \(\displaystyle a \int \sec \left (c+d x-\frac {\pi }{2}\right ) \tan \left (c+d x-\frac {\pi }{2}\right )^4dx-\frac {a \cot ^5(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle a \left (-\frac {3}{4} \int \cot ^2(c+d x) \csc (c+d x)dx-\frac {\cot ^3(c+d x) \csc (c+d x)}{4 d}\right )-\frac {a \cot ^5(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \left (-\frac {3}{4} \int \sec \left (c+d x-\frac {\pi }{2}\right ) \tan \left (c+d x-\frac {\pi }{2}\right )^2dx-\frac {\cot ^3(c+d x) \csc (c+d x)}{4 d}\right )-\frac {a \cot ^5(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle a \left (-\frac {3}{4} \left (-\frac {1}{2} \int \csc (c+d x)dx-\frac {\cot (c+d x) \csc (c+d x)}{2 d}\right )-\frac {\cot ^3(c+d x) \csc (c+d x)}{4 d}\right )-\frac {a \cot ^5(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \left (-\frac {3}{4} \left (-\frac {1}{2} \int \csc (c+d x)dx-\frac {\cot (c+d x) \csc (c+d x)}{2 d}\right )-\frac {\cot ^3(c+d x) \csc (c+d x)}{4 d}\right )-\frac {a \cot ^5(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle a \left (-\frac {3}{4} \left (\frac {\text {arctanh}(\cos (c+d x))}{2 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 d}\right )-\frac {\cot ^3(c+d x) \csc (c+d x)}{4 d}\right )-\frac {a \cot ^5(c+d x)}{5 d}\) |
Input:
Int[Cot[c + d*x]^4*Csc[c + d*x]^2*(a + a*Sin[c + d*x]),x]
Output:
-1/5*(a*Cot[c + d*x]^5)/d + a*(-1/4*(Cot[c + d*x]^3*Csc[c + d*x])/d - (3*( ArcTanh[Cos[c + d*x]]/(2*d) - (Cot[c + d*x]*Csc[c + d*x])/(2*d)))/4)
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -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[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[b*(a*Sec[e + f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] - Simp[b^2*((n - 1)/(m + n - 1)) Int[(a*Sec[e + f*x])^m*( b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] & & NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[a Int[(g*Co s[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Simp[b/d Int[(g*Cos[e + f*x])^ p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.26 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.35
| method | result | size |
| derivativedivides | \(\frac {a \left (-\frac {\cos \left (d x +c \right )^{5}}{4 \sin \left (d x +c \right )^{4}}+\frac {\cos \left (d x +c \right )^{5}}{8 \sin \left (d x +c \right )^{2}}+\frac {\cos \left (d x +c \right )^{3}}{8}+\frac {3 \cos \left (d x +c \right )}{8}+\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )-\frac {a \cos \left (d x +c \right )^{5}}{5 \sin \left (d x +c \right )^{5}}}{d}\) | \(100\) |
| default | \(\frac {a \left (-\frac {\cos \left (d x +c \right )^{5}}{4 \sin \left (d x +c \right )^{4}}+\frac {\cos \left (d x +c \right )^{5}}{8 \sin \left (d x +c \right )^{2}}+\frac {\cos \left (d x +c \right )^{3}}{8}+\frac {3 \cos \left (d x +c \right )}{8}+\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )-\frac {a \cos \left (d x +c \right )^{5}}{5 \sin \left (d x +c \right )^{5}}}{d}\) | \(100\) |
| risch | \(-\frac {a \left (40 i {\mathrm e}^{8 i \left (d x +c \right )}+25 \,{\mathrm e}^{9 i \left (d x +c \right )}-10 \,{\mathrm e}^{7 i \left (d x +c \right )}+80 i {\mathrm e}^{4 i \left (d x +c \right )}+10 \,{\mathrm e}^{3 i \left (d x +c \right )}+8 i-25 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{20 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{8 d}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{8 d}\) | \(128\) |
Input:
int(cot(d*x+c)^4*csc(d*x+c)^2*(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(a*(-1/4/sin(d*x+c)^4*cos(d*x+c)^5+1/8/sin(d*x+c)^2*cos(d*x+c)^5+1/8*c os(d*x+c)^3+3/8*cos(d*x+c)+3/8*ln(csc(d*x+c)-cot(d*x+c)))-1/5*a/sin(d*x+c) ^5*cos(d*x+c)^5)
Leaf count of result is larger than twice the leaf count of optimal. 160 vs. \(2 (66) = 132\).
Time = 0.09 (sec) , antiderivative size = 160, normalized size of antiderivative = 2.16 \[ \int \cot ^4(c+d x) \csc ^2(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {16 \, a \cos \left (d x + c\right )^{5} + 15 \, {\left (a \cos \left (d x + c\right )^{4} - 2 \, a \cos \left (d x + c\right )^{2} + a\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 15 \, {\left (a \cos \left (d x + c\right )^{4} - 2 \, a \cos \left (d x + c\right )^{2} + a\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 10 \, {\left (5 \, a \cos \left (d x + c\right )^{3} - 3 \, a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{80 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \] Input:
integrate(cot(d*x+c)^4*csc(d*x+c)^2*(a+a*sin(d*x+c)),x, algorithm="fricas" )
Output:
-1/80*(16*a*cos(d*x + c)^5 + 15*(a*cos(d*x + c)^4 - 2*a*cos(d*x + c)^2 + a )*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 15*(a*cos(d*x + c)^4 - 2*a*co s(d*x + c)^2 + a)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 10*(5*a*cos( d*x + c)^3 - 3*a*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))
\[ \int \cot ^4(c+d x) \csc ^2(c+d x) (a+a \sin (c+d x)) \, dx=a \left (\int \cot ^{4}{\left (c + d x \right )} \csc ^{2}{\left (c + d x \right )}\, dx + \int \sin {\left (c + d x \right )} \cot ^{4}{\left (c + d x \right )} \csc ^{2}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate(cot(d*x+c)**4*csc(d*x+c)**2*(a+a*sin(d*x+c)),x)
Output:
a*(Integral(cot(c + d*x)**4*csc(c + d*x)**2, x) + Integral(sin(c + d*x)*co t(c + d*x)**4*csc(c + d*x)**2, x))
Time = 0.03 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.16 \[ \int \cot ^4(c+d x) \csc ^2(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {16 \, a \cot \left (d x + c\right )^{5} + 5 \, a {\left (\frac {2 \, {\left (5 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} + 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{80 \, d} \] Input:
integrate(cot(d*x+c)^4*csc(d*x+c)^2*(a+a*sin(d*x+c)),x, algorithm="maxima" )
Output:
-1/80*(16*a*cot(d*x + c)^5 + 5*a*(2*(5*cos(d*x + c)^3 - 3*cos(d*x + c))/(c os(d*x + c)^4 - 2*cos(d*x + c)^2 + 1) + 3*log(cos(d*x + c) + 1) - 3*log(co s(d*x + c) - 1)))/d
Leaf count of result is larger than twice the leaf count of optimal. 173 vs. \(2 (66) = 132\).
Time = 0.17 (sec) , antiderivative size = 173, normalized size of antiderivative = 2.34 \[ \int \cot ^4(c+d x) \csc ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 5 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 10 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 120 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 20 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {274 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 20 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 40 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 10 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 5 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{320 \, d} \] Input:
integrate(cot(d*x+c)^4*csc(d*x+c)^2*(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
1/320*(2*a*tan(1/2*d*x + 1/2*c)^5 + 5*a*tan(1/2*d*x + 1/2*c)^4 - 10*a*tan( 1/2*d*x + 1/2*c)^3 - 40*a*tan(1/2*d*x + 1/2*c)^2 + 120*a*log(abs(tan(1/2*d *x + 1/2*c))) + 20*a*tan(1/2*d*x + 1/2*c) - (274*a*tan(1/2*d*x + 1/2*c)^5 + 20*a*tan(1/2*d*x + 1/2*c)^4 - 40*a*tan(1/2*d*x + 1/2*c)^3 - 10*a*tan(1/2 *d*x + 1/2*c)^2 + 5*a*tan(1/2*d*x + 1/2*c) + 2*a)/tan(1/2*d*x + 1/2*c)^5)/ d
Time = 18.69 (sec) , antiderivative size = 289, normalized size of antiderivative = 3.91 \[ \int \cot ^4(c+d x) \csc ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a\,\left (2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-2\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-5\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-10\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-40\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+20\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-20\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+40\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+10\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+120\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\right )}{320\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5} \] Input:
int((cot(c + d*x)^4*(a + a*sin(c + d*x)))/sin(c + d*x)^2,x)
Output:
(a*(2*sin(c/2 + (d*x)/2)^10 - 2*cos(c/2 + (d*x)/2)^10 + 5*cos(c/2 + (d*x)/ 2)*sin(c/2 + (d*x)/2)^9 - 5*cos(c/2 + (d*x)/2)^9*sin(c/2 + (d*x)/2) - 10*c os(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^8 - 40*cos(c/2 + (d*x)/2)^3*sin(c/2 + (d*x)/2)^7 + 20*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^6 - 20*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^4 + 40*cos(c/2 + (d*x)/2)^7*sin(c/2 + (d*x )/2)^3 + 10*cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2)^2 + 120*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(c/2 + (d*x)/2)^5*sin(c/2 + (d*x)/2)^5))/( 320*d*cos(c/2 + (d*x)/2)^5*sin(c/2 + (d*x)/2)^5)
Time = 0.16 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.42 \[ \int \cot ^4(c+d x) \csc ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \left (-8 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}+25 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}+16 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}-10 \cos \left (d x +c \right ) \sin \left (d x +c \right )-8 \cos \left (d x +c \right )+15 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{5}\right )}{40 \sin \left (d x +c \right )^{5} d} \] Input:
int(cot(d*x+c)^4*csc(d*x+c)^2*(a+a*sin(d*x+c)),x)
Output:
(a*( - 8*cos(c + d*x)*sin(c + d*x)**4 + 25*cos(c + d*x)*sin(c + d*x)**3 + 16*cos(c + d*x)*sin(c + d*x)**2 - 10*cos(c + d*x)*sin(c + d*x) - 8*cos(c + d*x) + 15*log(tan((c + d*x)/2))*sin(c + d*x)**5))/(40*sin(c + d*x)**5*d)