Integrand size = 27, antiderivative size = 90 \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+b \sin (c+d x)) \, dx=\frac {b \text {arctanh}(\cos (c+d x))}{8 d}-\frac {a \cot ^3(c+d x)}{3 d}-\frac {a \cot ^5(c+d x)}{5 d}+\frac {b \cot (c+d x) \csc (c+d x)}{8 d}-\frac {b \cot (c+d x) \csc ^3(c+d x)}{4 d} \] Output:
1/8*b*arctanh(cos(d*x+c))/d-1/3*a*cot(d*x+c)^3/d-1/5*a*cot(d*x+c)^5/d+1/8* b*cot(d*x+c)*csc(d*x+c)/d-1/4*b*cot(d*x+c)*csc(d*x+c)^3/d
Time = 0.17 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.97 \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+b \sin (c+d x)) \, dx=\frac {2 a \cot (c+d x)}{15 d}+\frac {b \csc ^2\left (\frac {1}{2} (c+d x)\right )}{32 d}-\frac {b \csc ^4\left (\frac {1}{2} (c+d x)\right )}{64 d}+\frac {a \cot (c+d x) \csc ^2(c+d x)}{15 d}-\frac {a \cot (c+d x) \csc ^4(c+d x)}{5 d}+\frac {b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{8 d}-\frac {b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{8 d}-\frac {b \sec ^2\left (\frac {1}{2} (c+d x)\right )}{32 d}+\frac {b \sec ^4\left (\frac {1}{2} (c+d x)\right )}{64 d} \] Input:
Integrate[Cot[c + d*x]^2*Csc[c + d*x]^4*(a + b*Sin[c + d*x]),x]
Output:
(2*a*Cot[c + d*x])/(15*d) + (b*Csc[(c + d*x)/2]^2)/(32*d) - (b*Csc[(c + d* x)/2]^4)/(64*d) + (a*Cot[c + d*x]*Csc[c + d*x]^2)/(15*d) - (a*Cot[c + d*x] *Csc[c + d*x]^4)/(5*d) + (b*Log[Cos[(c + d*x)/2]])/(8*d) - (b*Log[Sin[(c + d*x)/2]])/(8*d) - (b*Sec[(c + d*x)/2]^2)/(32*d) + (b*Sec[(c + d*x)/2]^4)/ (64*d)
Time = 0.61 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.03, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.407, Rules used = {3042, 3317, 3042, 3087, 244, 2009, 3091, 3042, 4255, 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.
\(\displaystyle \int \cot ^2(c+d x) \csc ^4(c+d x) (a+b \sin (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^2 (a+b \sin (c+d x))}{\sin (c+d x)^6}dx\) |
\(\Big \downarrow \) 3317 |
\(\displaystyle a \int \cot ^2(c+d x) \csc ^4(c+d x)dx+b \int \cot ^2(c+d x) \csc ^3(c+d x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \int \sec \left (c+d x-\frac {\pi }{2}\right )^4 \tan \left (c+d x-\frac {\pi }{2}\right )^2dx+b \int \sec \left (c+d x-\frac {\pi }{2}\right )^3 \tan \left (c+d x-\frac {\pi }{2}\right )^2dx\) |
\(\Big \downarrow \) 3087 |
\(\displaystyle \frac {a \int \cot ^2(c+d x) \left (\cot ^2(c+d x)+1\right )d(-\cot (c+d x))}{d}+b \int \sec \left (c+d x-\frac {\pi }{2}\right )^3 \tan \left (c+d x-\frac {\pi }{2}\right )^2dx\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {a \int \left (\cot ^4(c+d x)+\cot ^2(c+d x)\right )d(-\cot (c+d x))}{d}+b \int \sec \left (c+d x-\frac {\pi }{2}\right )^3 \tan \left (c+d x-\frac {\pi }{2}\right )^2dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle b \int \sec \left (c+d x-\frac {\pi }{2}\right )^3 \tan \left (c+d x-\frac {\pi }{2}\right )^2dx+\frac {a \left (-\frac {1}{5} \cot ^5(c+d x)-\frac {1}{3} \cot ^3(c+d x)\right )}{d}\) |
\(\Big \downarrow \) 3091 |
\(\displaystyle b \left (-\frac {1}{4} \int \csc ^3(c+d x)dx-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 d}\right )+\frac {a \left (-\frac {1}{5} \cot ^5(c+d x)-\frac {1}{3} \cot ^3(c+d x)\right )}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b \left (-\frac {1}{4} \int \csc (c+d x)^3dx-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 d}\right )+\frac {a \left (-\frac {1}{5} \cot ^5(c+d x)-\frac {1}{3} \cot ^3(c+d x)\right )}{d}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle b \left (\frac {1}{4} \left (\frac {\cot (c+d x) \csc (c+d x)}{2 d}-\frac {1}{2} \int \csc (c+d x)dx\right )-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 d}\right )+\frac {a \left (-\frac {1}{5} \cot ^5(c+d x)-\frac {1}{3} \cot ^3(c+d x)\right )}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b \left (\frac {1}{4} \left (\frac {\cot (c+d x) \csc (c+d x)}{2 d}-\frac {1}{2} \int \csc (c+d x)dx\right )-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 d}\right )+\frac {a \left (-\frac {1}{5} \cot ^5(c+d x)-\frac {1}{3} \cot ^3(c+d x)\right )}{d}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {a \left (-\frac {1}{5} \cot ^5(c+d x)-\frac {1}{3} \cot ^3(c+d x)\right )}{d}+b \left (\frac {1}{4} \left (\frac {\text {arctanh}(\cos (c+d x))}{2 d}+\frac {\cot (c+d x) \csc (c+d x)}{2 d}\right )-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 d}\right )\) |
Input:
Int[Cot[c + d*x]^2*Csc[c + d*x]^4*(a + b*Sin[c + d*x]),x]
Output:
(a*(-1/3*Cot[c + d*x]^3 - Cot[c + d*x]^5/5))/d + b*(-1/4*(Cot[c + d*x]*Csc [c + d*x]^3)/d + (ArcTanh[Cos[c + d*x]]/(2*d) + (Cot[c + d*x]*Csc[c + d*x] )/(2*d))/4)
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[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_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.37 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.22
method | result | size |
derivativedivides | \(\frac {a \left (-\frac {\cos \left (d x +c \right )^{3}}{5 \sin \left (d x +c \right )^{5}}-\frac {2 \cos \left (d x +c \right )^{3}}{15 \sin \left (d x +c \right )^{3}}\right )+b \left (-\frac {\cos \left (d x +c \right )^{3}}{4 \sin \left (d x +c \right )^{4}}-\frac {\cos \left (d x +c \right )^{3}}{8 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )}{8}-\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )}{d}\) | \(110\) |
default | \(\frac {a \left (-\frac {\cos \left (d x +c \right )^{3}}{5 \sin \left (d x +c \right )^{5}}-\frac {2 \cos \left (d x +c \right )^{3}}{15 \sin \left (d x +c \right )^{3}}\right )+b \left (-\frac {\cos \left (d x +c \right )^{3}}{4 \sin \left (d x +c \right )^{4}}-\frac {\cos \left (d x +c \right )^{3}}{8 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )}{8}-\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )}{d}\) | \(110\) |
risch | \(-\frac {15 b \,{\mathrm e}^{9 i \left (d x +c \right )}+240 i a \,{\mathrm e}^{6 i \left (d x +c \right )}+90 b \,{\mathrm e}^{7 i \left (d x +c \right )}+80 i a \,{\mathrm e}^{4 i \left (d x +c \right )}+80 i a \,{\mathrm e}^{2 i \left (d x +c \right )}-90 b \,{\mathrm e}^{3 i \left (d x +c \right )}-16 i a -15 b \,{\mathrm e}^{i \left (d x +c \right )}}{60 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{8 d}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{8 d}\) | \(148\) |
Input:
int(cot(d*x+c)^2*csc(d*x+c)^4*(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(a*(-1/5/sin(d*x+c)^5*cos(d*x+c)^3-2/15/sin(d*x+c)^3*cos(d*x+c)^3)+b*( -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))))
Leaf count of result is larger than twice the leaf count of optimal. 169 vs. \(2 (80) = 160\).
Time = 0.09 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.88 \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+b \sin (c+d x)) \, dx=\frac {32 \, a \cos \left (d x + c\right )^{5} - 80 \, a \cos \left (d x + c\right )^{3} + 15 \, {\left (b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + b\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 15 \, {\left (b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + b\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 30 \, {\left (b \cos \left (d x + c\right )^{3} + b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \] Input:
integrate(cot(d*x+c)^2*csc(d*x+c)^4*(a+b*sin(d*x+c)),x, algorithm="fricas" )
Output:
1/240*(32*a*cos(d*x + c)^5 - 80*a*cos(d*x + c)^3 + 15*(b*cos(d*x + c)^4 - 2*b*cos(d*x + c)^2 + b)*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 15*(b*c os(d*x + c)^4 - 2*b*cos(d*x + c)^2 + b)*log(-1/2*cos(d*x + c) + 1/2)*sin(d *x + c) - 30*(b*cos(d*x + c)^3 + b*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))
\[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+b \sin (c+d x)) \, dx=\int \left (a + b \sin {\left (c + d x \right )}\right ) \cot ^{2}{\left (c + d x \right )} \csc ^{4}{\left (c + d x \right )}\, dx \] Input:
integrate(cot(d*x+c)**2*csc(d*x+c)**4*(a+b*sin(d*x+c)),x)
Output:
Integral((a + b*sin(c + d*x))*cot(c + d*x)**2*csc(c + d*x)**4, x)
Time = 0.04 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.02 \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {15 \, b {\left (\frac {2 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + \frac {16 \, {\left (5 \, \tan \left (d x + c\right )^{2} + 3\right )} a}{\tan \left (d x + c\right )^{5}}}{240 \, d} \] Input:
integrate(cot(d*x+c)^2*csc(d*x+c)^4*(a+b*sin(d*x+c)),x, algorithm="maxima" )
Output:
-1/240*(15*b*(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)) + 16*(5*tan (d*x + c)^2 + 3)*a/tan(d*x + c)^5)/d
Time = 0.16 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.60 \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+b \sin (c+d x)) \, dx=\frac {6 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 10 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 120 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - 60 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {274 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 60 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 10 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{960 \, d} \] Input:
integrate(cot(d*x+c)^2*csc(d*x+c)^4*(a+b*sin(d*x+c)),x, algorithm="giac")
Output:
1/960*(6*a*tan(1/2*d*x + 1/2*c)^5 + 15*b*tan(1/2*d*x + 1/2*c)^4 + 10*a*tan (1/2*d*x + 1/2*c)^3 - 120*b*log(abs(tan(1/2*d*x + 1/2*c))) - 60*a*tan(1/2* d*x + 1/2*c) + (274*b*tan(1/2*d*x + 1/2*c)^5 + 60*a*tan(1/2*d*x + 1/2*c)^4 - 10*a*tan(1/2*d*x + 1/2*c)^2 - 15*b*tan(1/2*d*x + 1/2*c) - 6*a)/tan(1/2* d*x + 1/2*c)^5)/d
Time = 33.73 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.59 \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+b \sin (c+d x)) \, dx=\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96\,d}-\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,d}+\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,d}+\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,d}-\frac {b\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{8\,d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (-2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}+\frac {a}{5}\right )}{32\,d} \] Input:
int((cot(c + d*x)^2*(a + b*sin(c + d*x)))/sin(c + d*x)^4,x)
Output:
(a*tan(c/2 + (d*x)/2)^3)/(96*d) - (a*tan(c/2 + (d*x)/2))/(16*d) + (a*tan(c /2 + (d*x)/2)^5)/(160*d) + (b*tan(c/2 + (d*x)/2)^4)/(64*d) - (b*log(tan(c/ 2 + (d*x)/2)))/(8*d) - (cot(c/2 + (d*x)/2)^5*(a/5 + (b*tan(c/2 + (d*x)/2)) /2 + (a*tan(c/2 + (d*x)/2)^2)/3 - 2*a*tan(c/2 + (d*x)/2)^4))/(32*d)
Time = 0.17 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.22 \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+b \sin (c+d x)) \, dx=\frac {16 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a +15 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} b +8 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a -30 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b -24 \cos \left (d x +c \right ) a -15 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{5} b}{120 \sin \left (d x +c \right )^{5} d} \] Input:
int(cot(d*x+c)^2*csc(d*x+c)^4*(a+b*sin(d*x+c)),x)
Output:
(16*cos(c + d*x)*sin(c + d*x)**4*a + 15*cos(c + d*x)*sin(c + d*x)**3*b + 8 *cos(c + d*x)*sin(c + d*x)**2*a - 30*cos(c + d*x)*sin(c + d*x)*b - 24*cos( c + d*x)*a - 15*log(tan((c + d*x)/2))*sin(c + d*x)**5*b)/(120*sin(c + d*x) **5*d)