Integrand size = 25, antiderivative size = 86 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+b \sin (c+d x)) \, dx=-\frac {a \csc (c+d x)}{d}+\frac {b \csc ^2(c+d x)}{d}+\frac {2 a \csc ^3(c+d x)}{3 d}-\frac {b \csc ^4(c+d x)}{4 d}-\frac {a \csc ^5(c+d x)}{5 d}+\frac {b \log (\sin (c+d x))}{d} \] Output:
-a*csc(d*x+c)/d+b*csc(d*x+c)^2/d+2/3*a*csc(d*x+c)^3/d-1/4*b*csc(d*x+c)^4/d -1/5*a*csc(d*x+c)^5/d+b*ln(sin(d*x+c))/d
Time = 0.04 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+b \sin (c+d x)) \, dx=-\frac {a \csc (c+d x)}{d}+\frac {b \csc ^2(c+d x)}{d}+\frac {2 a \csc ^3(c+d x)}{3 d}-\frac {b \csc ^4(c+d x)}{4 d}-\frac {a \csc ^5(c+d x)}{5 d}+\frac {b \log (\sin (c+d x))}{d} \] Input:
Integrate[Cot[c + d*x]^5*Csc[c + d*x]*(a + b*Sin[c + d*x]),x]
Output:
-((a*Csc[c + d*x])/d) + (b*Csc[c + d*x]^2)/d + (2*a*Csc[c + d*x]^3)/(3*d) - (b*Csc[c + d*x]^4)/(4*d) - (a*Csc[c + d*x]^5)/(5*d) + (b*Log[Sin[c + d*x ]])/d
Time = 0.30 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.92, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 3316, 27, 522, 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.
\(\displaystyle \int \cot ^5(c+d x) \csc (c+d x) (a+b \sin (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^5 (a+b \sin (c+d x))}{\sin (c+d x)^6}dx\) |
\(\Big \downarrow \) 3316 |
\(\displaystyle \frac {\int \csc ^6(c+d x) (a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^2d(b \sin (c+d x))}{b^5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {b \int \frac {\csc ^6(c+d x) (a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}{b^6}d(b \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 522 |
\(\displaystyle \frac {b \int \left (\frac {a \csc ^6(c+d x)}{b^2}+\frac {\csc ^5(c+d x)}{b}-\frac {2 a \csc ^4(c+d x)}{b^2}-\frac {2 \csc ^3(c+d x)}{b}+\frac {a \csc ^2(c+d x)}{b^2}+\frac {\csc (c+d x)}{b}\right )d(b \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b \left (-\frac {a \csc ^5(c+d x)}{5 b}+\frac {2 a \csc ^3(c+d x)}{3 b}-\frac {a \csc (c+d x)}{b}+\log (b \sin (c+d x))-\frac {1}{4} \csc ^4(c+d x)+\csc ^2(c+d x)\right )}{d}\) |
Input:
Int[Cot[c + d*x]^5*Csc[c + d*x]*(a + b*Sin[c + d*x]),x]
Output:
(b*(-((a*Csc[c + d*x])/b) + Csc[c + d*x]^2 + (2*a*Csc[c + d*x]^3)/(3*b) - Csc[c + d*x]^4/4 - (a*Csc[c + d*x]^5)/(5*b) + Log[b*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[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. ), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
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*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x, b* Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1) /2] && NeQ[a^2 - b^2, 0]
Time = 0.28 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.79
method | result | size |
derivativedivides | \(-\frac {\frac {a \csc \left (d x +c \right )^{5}}{5}+\frac {b \csc \left (d x +c \right )^{4}}{4}-\frac {2 a \csc \left (d x +c \right )^{3}}{3}-b \csc \left (d x +c \right )^{2}+\csc \left (d x +c \right ) a +b \ln \left (\csc \left (d x +c \right )\right )}{d}\) | \(68\) |
default | \(-\frac {\frac {a \csc \left (d x +c \right )^{5}}{5}+\frac {b \csc \left (d x +c \right )^{4}}{4}-\frac {2 a \csc \left (d x +c \right )^{3}}{3}-b \csc \left (d x +c \right )^{2}+\csc \left (d x +c \right ) a +b \ln \left (\csc \left (d x +c \right )\right )}{d}\) | \(68\) |
risch | \(-i x b -\frac {2 i b c}{d}-\frac {2 i \left (15 a \,{\mathrm e}^{9 i \left (d x +c \right )}-20 a \,{\mathrm e}^{7 i \left (d x +c \right )}-30 i b \,{\mathrm e}^{8 i \left (d x +c \right )}+58 a \,{\mathrm e}^{5 i \left (d x +c \right )}+60 i b \,{\mathrm e}^{6 i \left (d x +c \right )}-20 a \,{\mathrm e}^{3 i \left (d x +c \right )}-60 i b \,{\mathrm e}^{4 i \left (d x +c \right )}+15 a \,{\mathrm e}^{i \left (d x +c \right )}+30 i b \,{\mathrm e}^{2 i \left (d x +c \right )}\right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}+\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\) | \(164\) |
Input:
int(cot(d*x+c)^5*csc(d*x+c)*(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
-1/d*(1/5*a*csc(d*x+c)^5+1/4*b*csc(d*x+c)^4-2/3*a*csc(d*x+c)^3-b*csc(d*x+c )^2+csc(d*x+c)*a+b*ln(csc(d*x+c)))
Time = 0.09 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.44 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+b \sin (c+d x)) \, dx=-\frac {60 \, a \cos \left (d x + c\right )^{4} - 80 \, a \cos \left (d x + c\right )^{2} - 60 \, {\left (b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + b\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) + 15 \, {\left (4 \, b \cos \left (d x + c\right )^{2} - 3 \, b\right )} \sin \left (d x + c\right ) + 32 \, a}{60 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \] Input:
integrate(cot(d*x+c)^5*csc(d*x+c)*(a+b*sin(d*x+c)),x, algorithm="fricas")
Output:
-1/60*(60*a*cos(d*x + c)^4 - 80*a*cos(d*x + c)^2 - 60*(b*cos(d*x + c)^4 - 2*b*cos(d*x + c)^2 + b)*log(1/2*sin(d*x + c))*sin(d*x + c) + 15*(4*b*cos(d *x + c)^2 - 3*b)*sin(d*x + c) + 32*a)/((d*cos(d*x + c)^4 - 2*d*cos(d*x + c )^2 + d)*sin(d*x + c))
\[ \int \cot ^5(c+d x) \csc (c+d x) (a+b \sin (c+d x)) \, dx=\int \left (a + b \sin {\left (c + d x \right )}\right ) \cot ^{5}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx \] Input:
integrate(cot(d*x+c)**5*csc(d*x+c)*(a+b*sin(d*x+c)),x)
Output:
Integral((a + b*sin(c + d*x))*cot(c + d*x)**5*csc(c + d*x), x)
Time = 0.03 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.84 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+b \sin (c+d x)) \, dx=\frac {60 \, b \log \left (\sin \left (d x + c\right )\right ) - \frac {60 \, a \sin \left (d x + c\right )^{4} - 60 \, b \sin \left (d x + c\right )^{3} - 40 \, a \sin \left (d x + c\right )^{2} + 15 \, b \sin \left (d x + c\right ) + 12 \, a}{\sin \left (d x + c\right )^{5}}}{60 \, d} \] Input:
integrate(cot(d*x+c)^5*csc(d*x+c)*(a+b*sin(d*x+c)),x, algorithm="maxima")
Output:
1/60*(60*b*log(sin(d*x + c)) - (60*a*sin(d*x + c)^4 - 60*b*sin(d*x + c)^3 - 40*a*sin(d*x + c)^2 + 15*b*sin(d*x + c) + 12*a)/sin(d*x + c)^5)/d
Time = 0.24 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.85 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+b \sin (c+d x)) \, dx=\frac {60 \, b \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - \frac {60 \, a \sin \left (d x + c\right )^{4} - 60 \, b \sin \left (d x + c\right )^{3} - 40 \, a \sin \left (d x + c\right )^{2} + 15 \, b \sin \left (d x + c\right ) + 12 \, a}{\sin \left (d x + c\right )^{5}}}{60 \, d} \] Input:
integrate(cot(d*x+c)^5*csc(d*x+c)*(a+b*sin(d*x+c)),x, algorithm="giac")
Output:
1/60*(60*b*log(abs(sin(d*x + c))) - (60*a*sin(d*x + c)^4 - 60*b*sin(d*x + c)^3 - 40*a*sin(d*x + c)^2 + 15*b*sin(d*x + c) + 12*a)/sin(d*x + c)^5)/d
Time = 21.75 (sec) , antiderivative size = 193, normalized size of antiderivative = 2.24 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+b \sin (c+d x)) \, dx=\frac {5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96\,d}-\frac {b\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {5\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,d}-\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,d}+\frac {3\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{16\,d}-\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,d}+\frac {b\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (10\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-6\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-\frac {5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}+\frac {a}{5}\right )}{32\,d} \] Input:
int((cot(c + d*x)^5*(a + b*sin(c + d*x)))/sin(c + d*x),x)
Output:
(5*a*tan(c/2 + (d*x)/2)^3)/(96*d) - (b*log(tan(c/2 + (d*x)/2)^2 + 1))/d - (5*a*tan(c/2 + (d*x)/2))/(16*d) - (a*tan(c/2 + (d*x)/2)^5)/(160*d) + (3*b* tan(c/2 + (d*x)/2)^2)/(16*d) - (b*tan(c/2 + (d*x)/2)^4)/(64*d) + (b*log(ta n(c/2 + (d*x)/2)))/d - (cot(c/2 + (d*x)/2)^5*(a/5 + (b*tan(c/2 + (d*x)/2)) /2 - (5*a*tan(c/2 + (d*x)/2)^2)/3 + 10*a*tan(c/2 + (d*x)/2)^4 - 6*b*tan(c/ 2 + (d*x)/2)^3))/(32*d)
Time = 0.17 (sec) , antiderivative size = 184, normalized size of antiderivative = 2.14 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+b \sin (c+d x)) \, dx=\frac {-3 \cos \left (d x +c \right ) \cot \left (d x +c \right )^{3} \csc \left (d x +c \right ) b +13 \cos \left (d x +c \right ) \cot \left (d x +c \right ) \csc \left (d x +c \right ) b -12 \cot \left (d x +c \right )^{4} \csc \left (d x +c \right ) \sin \left (d x +c \right ) b -12 \cot \left (d x +c \right )^{4} \csc \left (d x +c \right ) a +17 \cot \left (d x +c \right )^{2} \csc \left (d x +c \right ) \sin \left (d x +c \right ) b +16 \cot \left (d x +c \right )^{2} \csc \left (d x +c \right ) a +13 \csc \left (d x +c \right ) \sin \left (d x +c \right ) b -32 \csc \left (d x +c \right ) a -60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) b +60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b}{60 d} \] Input:
int(cot(d*x+c)^5*csc(d*x+c)*(a+b*sin(d*x+c)),x)
Output:
( - 3*cos(c + d*x)*cot(c + d*x)**3*csc(c + d*x)*b + 13*cos(c + d*x)*cot(c + d*x)*csc(c + d*x)*b - 12*cot(c + d*x)**4*csc(c + d*x)*sin(c + d*x)*b - 1 2*cot(c + d*x)**4*csc(c + d*x)*a + 17*cot(c + d*x)**2*csc(c + d*x)*sin(c + d*x)*b + 16*cot(c + d*x)**2*csc(c + d*x)*a + 13*csc(c + d*x)*sin(c + d*x) *b - 32*csc(c + d*x)*a - 60*log(tan((c + d*x)/2)**2 + 1)*b + 60*log(tan((c + d*x)/2))*b)/(60*d)