Integrand size = 27, antiderivative size = 133 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {5 a^3 \csc (c+d x)}{d}+\frac {5 a^3 \csc ^2(c+d x)}{2 d}-\frac {a^3 \csc ^3(c+d x)}{3 d}-\frac {3 a^3 \csc ^4(c+d x)}{4 d}-\frac {a^3 \csc ^5(c+d x)}{5 d}+\frac {a^3 \log (\sin (c+d x))}{d}+\frac {3 a^3 \sin (c+d x)}{d}+\frac {a^3 \sin ^2(c+d x)}{2 d} \] Output:
5*a^3*csc(d*x+c)/d+5/2*a^3*csc(d*x+c)^2/d-1/3*a^3*csc(d*x+c)^3/d-3/4*a^3*c sc(d*x+c)^4/d-1/5*a^3*csc(d*x+c)^5/d+a^3*ln(sin(d*x+c))/d+3*a^3*sin(d*x+c) /d+1/2*a^3*sin(d*x+c)^2/d
Time = 0.19 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.65 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^3 \left (300 \csc (c+d x)+150 \csc ^2(c+d x)-20 \csc ^3(c+d x)-45 \csc ^4(c+d x)-12 \csc ^5(c+d x)+60 \log (\sin (c+d x))+180 \sin (c+d x)+30 \sin ^2(c+d x)\right )}{60 d} \] Input:
Integrate[Cot[c + d*x]^5*Csc[c + d*x]*(a + a*Sin[c + d*x])^3,x]
Output:
(a^3*(300*Csc[c + d*x] + 150*Csc[c + d*x]^2 - 20*Csc[c + d*x]^3 - 45*Csc[c + d*x]^4 - 12*Csc[c + d*x]^5 + 60*Log[Sin[c + d*x]] + 180*Sin[c + d*x] + 30*Sin[c + d*x]^2))/(60*d)
Time = 0.33 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.87, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3042, 3315, 27, 99, 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.
\(\displaystyle \int \cot ^5(c+d x) \csc (c+d x) (a \sin (c+d x)+a)^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^5 (a \sin (c+d x)+a)^3}{\sin (c+d x)^6}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \csc ^6(c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^5d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a \int \frac {\csc ^6(c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^5}{a^6}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {a \int \left (a \csc ^6(c+d x)+3 a \csc ^5(c+d x)+a \csc ^4(c+d x)-5 a \csc ^3(c+d x)-5 a \csc ^2(c+d x)+a \csc (c+d x)+3 a+a \sin (c+d x)\right )d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a \left (\frac {1}{2} a^2 \sin ^2(c+d x)+3 a^2 \sin (c+d x)-\frac {1}{5} a^2 \csc ^5(c+d x)-\frac {3}{4} a^2 \csc ^4(c+d x)-\frac {1}{3} a^2 \csc ^3(c+d x)+\frac {5}{2} a^2 \csc ^2(c+d x)+5 a^2 \csc (c+d x)+a^2 \log (a \sin (c+d x))\right )}{d}\) |
Input:
Int[Cot[c + d*x]^5*Csc[c + d*x]*(a + a*Sin[c + d*x])^3,x]
Output:
(a*(5*a^2*Csc[c + d*x] + (5*a^2*Csc[c + d*x]^2)/2 - (a^2*Csc[c + d*x]^3)/3 - (3*a^2*Csc[c + d*x]^4)/4 - (a^2*Csc[c + d*x]^5)/5 + a^2*Log[a*Sin[c + d *x]] + 3*a^2*Sin[c + d*x] + (a^2*Sin[c + d*x]^2)/2))/d
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
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 + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && Intege rQ[(p - 1)/2] && EqQ[a^2 - b^2, 0]
Time = 1.26 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.64
method | result | size |
derivativedivides | \(-\frac {a^{3} \left (\frac {\csc \left (d x +c \right )^{5}}{5}+\frac {3 \csc \left (d x +c \right )^{4}}{4}+\frac {\csc \left (d x +c \right )^{3}}{3}-\frac {5 \csc \left (d x +c \right )^{2}}{2}-5 \csc \left (d x +c \right )-\frac {3}{\csc \left (d x +c \right )}-\frac {1}{2 \csc \left (d x +c \right )^{2}}+\ln \left (\csc \left (d x +c \right )\right )\right )}{d}\) | \(85\) |
default | \(-\frac {a^{3} \left (\frac {\csc \left (d x +c \right )^{5}}{5}+\frac {3 \csc \left (d x +c \right )^{4}}{4}+\frac {\csc \left (d x +c \right )^{3}}{3}-\frac {5 \csc \left (d x +c \right )^{2}}{2}-5 \csc \left (d x +c \right )-\frac {3}{\csc \left (d x +c \right )}-\frac {1}{2 \csc \left (d x +c \right )^{2}}+\ln \left (\csc \left (d x +c \right )\right )\right )}{d}\) | \(85\) |
risch | \(-i a^{3} x -\frac {a^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {3 i a^{3} {\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {3 i a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {a^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {2 i a^{3} c}{d}+\frac {2 i a^{3} \left (75 \,{\mathrm e}^{9 i \left (d x +c \right )}-280 \,{\mathrm e}^{7 i \left (d x +c \right )}+75 i {\mathrm e}^{8 i \left (d x +c \right )}+362 \,{\mathrm e}^{5 i \left (d x +c \right )}-135 i {\mathrm e}^{6 i \left (d x +c \right )}-280 \,{\mathrm e}^{3 i \left (d x +c \right )}+135 i {\mathrm e}^{4 i \left (d x +c \right )}+75 \,{\mathrm e}^{i \left (d x +c \right )}-75 i {\mathrm e}^{2 i \left (d x +c \right )}\right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}+\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\) | \(234\) |
Input:
int(cot(d*x+c)^5*csc(d*x+c)*(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
-a^3/d*(1/5*csc(d*x+c)^5+3/4*csc(d*x+c)^4+1/3*csc(d*x+c)^3-5/2*csc(d*x+c)^ 2-5*csc(d*x+c)-3/csc(d*x+c)-1/2/csc(d*x+c)^2+ln(csc(d*x+c)))
Time = 0.10 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.35 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {180 \, a^{3} \cos \left (d x + c\right )^{6} - 840 \, a^{3} \cos \left (d x + c\right )^{4} + 1120 \, a^{3} \cos \left (d x + c\right )^{2} - 448 \, a^{3} - 60 \, {\left (a^{3} \cos \left (d x + c\right )^{4} - 2 \, a^{3} \cos \left (d x + c\right )^{2} + a^{3}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) + 15 \, {\left (2 \, a^{3} \cos \left (d x + c\right )^{6} - 5 \, a^{3} \cos \left (d x + c\right )^{4} + 14 \, a^{3} \cos \left (d x + c\right )^{2} - 8 \, a^{3}\right )} \sin \left (d x + c\right )}{60 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \] Input:
integrate(cot(d*x+c)^5*csc(d*x+c)*(a+a*sin(d*x+c))^3,x, algorithm="fricas" )
Output:
-1/60*(180*a^3*cos(d*x + c)^6 - 840*a^3*cos(d*x + c)^4 + 1120*a^3*cos(d*x + c)^2 - 448*a^3 - 60*(a^3*cos(d*x + c)^4 - 2*a^3*cos(d*x + c)^2 + a^3)*lo g(1/2*sin(d*x + c))*sin(d*x + c) + 15*(2*a^3*cos(d*x + c)^6 - 5*a^3*cos(d* x + c)^4 + 14*a^3*cos(d*x + c)^2 - 8*a^3)*sin(d*x + c))/((d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)*sin(d*x + c))
Timed out. \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx=\text {Timed out} \] Input:
integrate(cot(d*x+c)**5*csc(d*x+c)*(a+a*sin(d*x+c))**3,x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.80 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {30 \, a^{3} \sin \left (d x + c\right )^{2} + 60 \, a^{3} \log \left (\sin \left (d x + c\right )\right ) + 180 \, a^{3} \sin \left (d x + c\right ) + \frac {300 \, a^{3} \sin \left (d x + c\right )^{4} + 150 \, a^{3} \sin \left (d x + c\right )^{3} - 20 \, a^{3} \sin \left (d x + c\right )^{2} - 45 \, a^{3} \sin \left (d x + c\right ) - 12 \, a^{3}}{\sin \left (d x + c\right )^{5}}}{60 \, d} \] Input:
integrate(cot(d*x+c)^5*csc(d*x+c)*(a+a*sin(d*x+c))^3,x, algorithm="maxima" )
Output:
1/60*(30*a^3*sin(d*x + c)^2 + 60*a^3*log(sin(d*x + c)) + 180*a^3*sin(d*x + c) + (300*a^3*sin(d*x + c)^4 + 150*a^3*sin(d*x + c)^3 - 20*a^3*sin(d*x + c)^2 - 45*a^3*sin(d*x + c) - 12*a^3)/sin(d*x + c)^5)/d
Time = 0.19 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.81 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {30 \, a^{3} \sin \left (d x + c\right )^{2} + 60 \, a^{3} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 180 \, a^{3} \sin \left (d x + c\right ) + \frac {300 \, a^{3} \sin \left (d x + c\right )^{4} + 150 \, a^{3} \sin \left (d x + c\right )^{3} - 20 \, a^{3} \sin \left (d x + c\right )^{2} - 45 \, a^{3} \sin \left (d x + c\right ) - 12 \, a^{3}}{\sin \left (d x + c\right )^{5}}}{60 \, d} \] Input:
integrate(cot(d*x+c)^5*csc(d*x+c)*(a+a*sin(d*x+c))^3,x, algorithm="giac")
Output:
1/60*(30*a^3*sin(d*x + c)^2 + 60*a^3*log(abs(sin(d*x + c))) + 180*a^3*sin( d*x + c) + (300*a^3*sin(d*x + c)^4 + 150*a^3*sin(d*x + c)^3 - 20*a^3*sin(d *x + c)^2 - 45*a^3*sin(d*x + c) - 12*a^3)/sin(d*x + c)^5)/d
Time = 32.49 (sec) , antiderivative size = 311, normalized size of antiderivative = 2.34 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {7\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{16\,d}-\frac {7\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96\,d}-\frac {3\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,d}-\frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,d}+\frac {a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {266\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+78\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\frac {1013\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3}+\frac {53\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{2}+\frac {1037\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{15}+11\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-\frac {41\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{15}-\frac {3\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}-\frac {a^3}{5}}{d\,\left (32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+64\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\right )}+\frac {37\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,d}-\frac {a^3\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \] Input:
int((cot(c + d*x)^5*(a + a*sin(c + d*x))^3)/sin(c + d*x),x)
Output:
(7*a^3*tan(c/2 + (d*x)/2)^2)/(16*d) - (7*a^3*tan(c/2 + (d*x)/2)^3)/(96*d) - (3*a^3*tan(c/2 + (d*x)/2)^4)/(64*d) - (a^3*tan(c/2 + (d*x)/2)^5)/(160*d) + (a^3*log(tan(c/2 + (d*x)/2)))/d + (11*a^3*tan(c/2 + (d*x)/2)^3 - (41*a^ 3*tan(c/2 + (d*x)/2)^2)/15 + (1037*a^3*tan(c/2 + (d*x)/2)^4)/15 + (53*a^3* tan(c/2 + (d*x)/2)^5)/2 + (1013*a^3*tan(c/2 + (d*x)/2)^6)/3 + 78*a^3*tan(c /2 + (d*x)/2)^7 + 266*a^3*tan(c/2 + (d*x)/2)^8 - a^3/5 - (3*a^3*tan(c/2 + (d*x)/2))/2)/(d*(32*tan(c/2 + (d*x)/2)^5 + 64*tan(c/2 + (d*x)/2)^7 + 32*ta n(c/2 + (d*x)/2)^9)) + (37*a^3*tan(c/2 + (d*x)/2))/(16*d) - (a^3*log(tan(c /2 + (d*x)/2)^2 + 1))/d
Time = 0.16 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.98 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^{3} \left (-960 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{5}+960 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{5}+480 \sin \left (d x +c \right )^{7}+2880 \sin \left (d x +c \right )^{6}-1725 \sin \left (d x +c \right )^{5}+4800 \sin \left (d x +c \right )^{4}+2400 \sin \left (d x +c \right )^{3}-320 \sin \left (d x +c \right )^{2}-720 \sin \left (d x +c \right )-192\right )}{960 \sin \left (d x +c \right )^{5} d} \] Input:
int(cot(d*x+c)^5*csc(d*x+c)*(a+a*sin(d*x+c))^3,x)
Output:
(a**3*( - 960*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**5 + 960*log(tan(( c + d*x)/2))*sin(c + d*x)**5 + 480*sin(c + d*x)**7 + 2880*sin(c + d*x)**6 - 1725*sin(c + d*x)**5 + 4800*sin(c + d*x)**4 + 2400*sin(c + d*x)**3 - 320 *sin(c + d*x)**2 - 720*sin(c + d*x) - 192))/(960*sin(c + d*x)**5*d)