Integrand size = 21, antiderivative size = 102 \[ \int \cot ^3(c+d x) (a+a \sin (c+d x))^4 \, dx=-\frac {4 a^4 \csc (c+d x)}{d}-\frac {a^4 \csc ^2(c+d x)}{2 d}+\frac {5 a^4 \log (\sin (c+d x))}{d}-\frac {5 a^4 \sin ^2(c+d x)}{2 d}-\frac {4 a^4 \sin ^3(c+d x)}{3 d}-\frac {a^4 \sin ^4(c+d x)}{4 d} \]
-4*a^4*csc(d*x+c)/d-1/2*a^4*csc(d*x+c)^2/d+5*a^4*ln(sin(d*x+c))/d-5/2*a^4* sin(d*x+c)^2/d-4/3*a^4*sin(d*x+c)^3/d-1/4*a^4*sin(d*x+c)^4/d
Time = 0.11 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.67 \[ \int \cot ^3(c+d x) (a+a \sin (c+d x))^4 \, dx=-\frac {a^4 \left (48 \csc (c+d x)+6 \csc ^2(c+d x)-60 \log (\sin (c+d x))+30 \sin ^2(c+d x)+16 \sin ^3(c+d x)+3 \sin ^4(c+d x)\right )}{12 d} \]
-1/12*(a^4*(48*Csc[c + d*x] + 6*Csc[c + d*x]^2 - 60*Log[Sin[c + d*x]] + 30 *Sin[c + d*x]^2 + 16*Sin[c + d*x]^3 + 3*Sin[c + d*x]^4))/d
Time = 0.25 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.88, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3186, 84, 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.
\(\displaystyle \int \cot ^3(c+d x) (a \sin (c+d x)+a)^4 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a \sin (c+d x)+a)^4}{\tan (c+d x)^3}dx\) |
\(\Big \downarrow \) 3186 |
\(\displaystyle \frac {\int \frac {\csc ^3(c+d x) (a-a \sin (c+d x)) (\sin (c+d x) a+a)^5}{a^3}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 84 |
\(\displaystyle \frac {\int \left (\csc ^3(c+d x) a^3-\sin ^3(c+d x) a^3+4 \csc ^2(c+d x) a^3-4 \sin ^2(c+d x) a^3+5 \csc (c+d x) a^3-5 \sin (c+d x) a^3\right )d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {1}{4} a^4 \sin ^4(c+d x)-\frac {4}{3} a^4 \sin ^3(c+d x)-\frac {5}{2} a^4 \sin ^2(c+d x)-\frac {1}{2} a^4 \csc ^2(c+d x)-4 a^4 \csc (c+d x)+5 a^4 \log (a \sin (c+d x))}{d}\) |
(-4*a^4*Csc[c + d*x] - (a^4*Csc[c + d*x]^2)/2 + 5*a^4*Log[a*Sin[c + d*x]] - (5*a^4*Sin[c + d*x]^2)/2 - (4*a^4*Sin[c + d*x]^3)/3 - (a^4*Sin[c + d*x]^ 4)/4)/d
3.1.37.3.1 Defintions of rubi rules used
Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_] : > Int[ExpandIntegrand[(a + b*x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] && !(ILtQ[n + p + 2, 0 ] && GtQ[n + 2*p, 0])
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p _.), x_Symbol] :> Simp[1/f Subst[Int[x^p*((a + x)^(m - (p + 1)/2)/(a - x) ^((p + 1)/2)), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && E qQ[a^2 - b^2, 0] && IntegerQ[(p + 1)/2]
Time = 6.43 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.26
method | result | size |
derivativedivides | \(\frac {-\frac {a^{4} \left (\cos ^{4}\left (d x +c \right )\right )}{4}+\frac {4 a^{4} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+6 a^{4} \left (\frac {\left (\cos ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )+4 a^{4} \left (-\frac {\cos ^{4}\left (d x +c \right )}{\sin \left (d x +c \right )}-\left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )\right )+a^{4} \left (-\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\) | \(129\) |
default | \(\frac {-\frac {a^{4} \left (\cos ^{4}\left (d x +c \right )\right )}{4}+\frac {4 a^{4} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+6 a^{4} \left (\frac {\left (\cos ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )+4 a^{4} \left (-\frac {\cos ^{4}\left (d x +c \right )}{\sin \left (d x +c \right )}-\left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )\right )+a^{4} \left (-\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\) | \(129\) |
risch | \(-5 i a^{4} x -\frac {i a^{4} {\mathrm e}^{3 i \left (d x +c \right )}}{6 d}+\frac {11 a^{4} {\mathrm e}^{2 i \left (d x +c \right )}}{16 d}+\frac {i a^{4} {\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {i a^{4} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {11 a^{4} {\mathrm e}^{-2 i \left (d x +c \right )}}{16 d}+\frac {i a^{4} {\mathrm e}^{-3 i \left (d x +c \right )}}{6 d}-\frac {10 i a^{4} c}{d}-\frac {2 i a^{4} \left (i {\mathrm e}^{2 i \left (d x +c \right )}+4 \,{\mathrm e}^{3 i \left (d x +c \right )}-4 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}+\frac {5 a^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}-\frac {a^{4} \cos \left (4 d x +4 c \right )}{32 d}\) | \(219\) |
1/d*(-1/4*a^4*cos(d*x+c)^4+4/3*a^4*(2+cos(d*x+c)^2)*sin(d*x+c)+6*a^4*(1/2* cos(d*x+c)^2+ln(sin(d*x+c)))+4*a^4*(-1/sin(d*x+c)*cos(d*x+c)^4-(2+cos(d*x+ c)^2)*sin(d*x+c))+a^4*(-1/2*cot(d*x+c)^2-ln(sin(d*x+c))))
Time = 0.31 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.28 \[ \int \cot ^3(c+d x) (a+a \sin (c+d x))^4 \, dx=-\frac {24 \, a^{4} \cos \left (d x + c\right )^{6} - 312 \, a^{4} \cos \left (d x + c\right )^{4} + 423 \, a^{4} \cos \left (d x + c\right )^{2} - 183 \, a^{4} - 480 \, {\left (a^{4} \cos \left (d x + c\right )^{2} - a^{4}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 128 \, {\left (a^{4} \cos \left (d x + c\right )^{4} - 2 \, a^{4} \cos \left (d x + c\right )^{2} + 4 \, a^{4}\right )} \sin \left (d x + c\right )}{96 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )}} \]
-1/96*(24*a^4*cos(d*x + c)^6 - 312*a^4*cos(d*x + c)^4 + 423*a^4*cos(d*x + c)^2 - 183*a^4 - 480*(a^4*cos(d*x + c)^2 - a^4)*log(1/2*sin(d*x + c)) - 12 8*(a^4*cos(d*x + c)^4 - 2*a^4*cos(d*x + c)^2 + 4*a^4)*sin(d*x + c))/(d*cos (d*x + c)^2 - d)
\[ \int \cot ^3(c+d x) (a+a \sin (c+d x))^4 \, dx=a^{4} \left (\int 4 \sin {\left (c + d x \right )} \cot ^{3}{\left (c + d x \right )}\, dx + \int 6 \sin ^{2}{\left (c + d x \right )} \cot ^{3}{\left (c + d x \right )}\, dx + \int 4 \sin ^{3}{\left (c + d x \right )} \cot ^{3}{\left (c + d x \right )}\, dx + \int \sin ^{4}{\left (c + d x \right )} \cot ^{3}{\left (c + d x \right )}\, dx + \int \cot ^{3}{\left (c + d x \right )}\, dx\right ) \]
a**4*(Integral(4*sin(c + d*x)*cot(c + d*x)**3, x) + Integral(6*sin(c + d*x )**2*cot(c + d*x)**3, x) + Integral(4*sin(c + d*x)**3*cot(c + d*x)**3, x) + Integral(sin(c + d*x)**4*cot(c + d*x)**3, x) + Integral(cot(c + d*x)**3, x))
Time = 0.19 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.80 \[ \int \cot ^3(c+d x) (a+a \sin (c+d x))^4 \, dx=-\frac {3 \, a^{4} \sin \left (d x + c\right )^{4} + 16 \, a^{4} \sin \left (d x + c\right )^{3} + 30 \, a^{4} \sin \left (d x + c\right )^{2} - 60 \, a^{4} \log \left (\sin \left (d x + c\right )\right ) + \frac {6 \, {\left (8 \, a^{4} \sin \left (d x + c\right ) + a^{4}\right )}}{\sin \left (d x + c\right )^{2}}}{12 \, d} \]
-1/12*(3*a^4*sin(d*x + c)^4 + 16*a^4*sin(d*x + c)^3 + 30*a^4*sin(d*x + c)^ 2 - 60*a^4*log(sin(d*x + c)) + 6*(8*a^4*sin(d*x + c) + a^4)/sin(d*x + c)^2 )/d
Time = 0.57 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.94 \[ \int \cot ^3(c+d x) (a+a \sin (c+d x))^4 \, dx=-\frac {3 \, a^{4} \sin \left (d x + c\right )^{4} + 16 \, a^{4} \sin \left (d x + c\right )^{3} + 30 \, a^{4} \sin \left (d x + c\right )^{2} - 60 \, a^{4} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + \frac {6 \, {\left (15 \, a^{4} \sin \left (d x + c\right )^{2} + 8 \, a^{4} \sin \left (d x + c\right ) + a^{4}\right )}}{\sin \left (d x + c\right )^{2}}}{12 \, d} \]
-1/12*(3*a^4*sin(d*x + c)^4 + 16*a^4*sin(d*x + c)^3 + 30*a^4*sin(d*x + c)^ 2 - 60*a^4*log(abs(sin(d*x + c))) + 6*(15*a^4*sin(d*x + c)^2 + 8*a^4*sin(d *x + c) + a^4)/sin(d*x + c)^2)/d
Time = 6.48 (sec) , antiderivative size = 298, normalized size of antiderivative = 2.92 \[ \int \cot ^3(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {5\,a^4\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}-\frac {8\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\frac {81\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{2}+\frac {224\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{3}+98\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {272\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{3}+43\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+32\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+8\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a^4}{2}}{d\,\left (4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+24\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {2\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d}-\frac {5\,a^4\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \]
(5*a^4*log(tan(c/2 + (d*x)/2)))/d - (a^4*tan(c/2 + (d*x)/2)^2)/(8*d) - (2* a^4*tan(c/2 + (d*x)/2)^2 + 32*a^4*tan(c/2 + (d*x)/2)^3 + 43*a^4*tan(c/2 + (d*x)/2)^4 + (272*a^4*tan(c/2 + (d*x)/2)^5)/3 + 98*a^4*tan(c/2 + (d*x)/2)^ 6 + (224*a^4*tan(c/2 + (d*x)/2)^7)/3 + (81*a^4*tan(c/2 + (d*x)/2)^8)/2 + 8 *a^4*tan(c/2 + (d*x)/2)^9 + a^4/2 + 8*a^4*tan(c/2 + (d*x)/2))/(d*(4*tan(c/ 2 + (d*x)/2)^2 + 16*tan(c/2 + (d*x)/2)^4 + 24*tan(c/2 + (d*x)/2)^6 + 16*ta n(c/2 + (d*x)/2)^8 + 4*tan(c/2 + (d*x)/2)^10)) - (2*a^4*tan(c/2 + (d*x)/2) )/d - (5*a^4*log(tan(c/2 + (d*x)/2)^2 + 1))/d