Integrand size = 27, antiderivative size = 83 \[ \int \cos ^3(c+d x) \cot ^2(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {a \csc (c+d x)}{d}+\frac {b \log (\sin (c+d x))}{d}-\frac {2 a \sin (c+d x)}{d}-\frac {b \sin ^2(c+d x)}{d}+\frac {a \sin ^3(c+d x)}{3 d}+\frac {b \sin ^4(c+d x)}{4 d} \]
-a*csc(d*x+c)/d+b*ln(sin(d*x+c))/d-2*a*sin(d*x+c)/d-b*sin(d*x+c)^2/d+1/3*a *sin(d*x+c)^3/d+1/4*b*sin(d*x+c)^4/d
Time = 0.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00 \[ \int \cos ^3(c+d x) \cot ^2(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {a \csc (c+d x)}{d}+\frac {b \log (\sin (c+d x))}{d}-\frac {2 a \sin (c+d x)}{d}-\frac {b \sin ^2(c+d x)}{d}+\frac {a \sin ^3(c+d x)}{3 d}+\frac {b \sin ^4(c+d x)}{4 d} \]
-((a*Csc[c + d*x])/d) + (b*Log[Sin[c + d*x]])/d - (2*a*Sin[c + d*x])/d - ( b*Sin[c + d*x]^2)/d + (a*Sin[c + d*x]^3)/(3*d) + (b*Sin[c + d*x]^4)/(4*d)
Time = 0.27 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, 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.
\(\displaystyle \int \cos ^3(c+d x) \cot ^2(c+d x) (a+b \sin (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^5 (a+b \sin (c+d x))}{\sin (c+d x)^2}dx\) |
\(\Big \downarrow \) 3316 |
\(\displaystyle \frac {\int \csc ^2(c+d x) (a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^2d(b \sin (c+d x))}{b^5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\csc ^2(c+d x) (a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}{b^2}d(b \sin (c+d x))}{b^3 d}\) |
\(\Big \downarrow \) 522 |
\(\displaystyle \frac {\int \left (\sin ^3(c+d x) b^3+\csc (c+d x) b^3-2 \sin (c+d x) b^3+a \csc ^2(c+d x) b^2+a \sin ^2(c+d x) b^2-2 a b^2\right )d(b \sin (c+d x))}{b^3 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{3} a b^3 \sin ^3(c+d x)-2 a b^3 \sin (c+d x)-a b^3 \csc (c+d x)+\frac {1}{4} b^4 \sin ^4(c+d x)-b^4 \sin ^2(c+d x)+b^4 \log (b \sin (c+d x))}{b^3 d}\) |
(-(a*b^3*Csc[c + d*x]) + b^4*Log[b*Sin[c + d*x]] - 2*a*b^3*Sin[c + d*x] - b^4*Sin[c + d*x]^2 + (a*b^3*Sin[c + d*x]^3)/3 + (b^4*Sin[c + d*x]^4)/4)/(b ^3*d)
3.13.4.3.1 Defintions of rubi rules used
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 = 0.50 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.02
method | result | size |
derivativedivides | \(\frac {a \left (-\frac {\cos ^{6}\left (d x +c \right )}{\sin \left (d x +c \right )}-\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )\right )+b \left (\frac {\left (\cos ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\cos ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\) | \(85\) |
default | \(\frac {a \left (-\frac {\cos ^{6}\left (d x +c \right )}{\sin \left (d x +c \right )}-\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )\right )+b \left (\frac {\left (\cos ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\cos ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\) | \(85\) |
parallelrisch | \(\frac {-24 \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +24 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +10 \left (\cos \left (2 d x +2 c \right )+\frac {\cos \left (4 d x +4 c \right )}{20}-\frac {9}{4}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right ) a \csc \left (\frac {d x}{2}+\frac {c}{2}\right )+9 b \left (\cos \left (2 d x +2 c \right )+\frac {\cos \left (4 d x +4 c \right )}{12}-\frac {13}{12}\right )}{24 d}\) | \(103\) |
risch | \(-i x b +\frac {3 b \,{\mathrm e}^{2 i \left (d x +c \right )}}{16 d}+\frac {7 i a \,{\mathrm e}^{i \left (d x +c \right )}}{8 d}-\frac {7 i a \,{\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {3 b \,{\mathrm e}^{-2 i \left (d x +c \right )}}{16 d}-\frac {2 i b c}{d}-\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}+\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}+\frac {b \cos \left (4 d x +4 c \right )}{32 d}-\frac {a \sin \left (3 d x +3 c \right )}{12 d}\) | \(153\) |
norman | \(\frac {-\frac {a}{2 d}-\frac {13 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {43 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {43 a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {13 a \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {a \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {4 b \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 b \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {b \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\) | \(207\) |
1/d*(a*(-1/sin(d*x+c)*cos(d*x+c)^6-(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin (d*x+c))+b*(1/4*cos(d*x+c)^4+1/2*cos(d*x+c)^2+ln(sin(d*x+c))))
Time = 0.40 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.10 \[ \int \cos ^3(c+d x) \cot ^2(c+d x) (a+b \sin (c+d x)) \, dx=\frac {32 \, a \cos \left (d x + c\right )^{4} + 128 \, a \cos \left (d x + c\right )^{2} + 96 \, b \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) + 3 \, {\left (8 \, b \cos \left (d x + c\right )^{4} + 16 \, b \cos \left (d x + c\right )^{2} - 11 \, b\right )} \sin \left (d x + c\right ) - 256 \, a}{96 \, d \sin \left (d x + c\right )} \]
1/96*(32*a*cos(d*x + c)^4 + 128*a*cos(d*x + c)^2 + 96*b*log(1/2*sin(d*x + c))*sin(d*x + c) + 3*(8*b*cos(d*x + c)^4 + 16*b*cos(d*x + c)^2 - 11*b)*sin (d*x + c) - 256*a)/(d*sin(d*x + c))
Timed out. \[ \int \cos ^3(c+d x) \cot ^2(c+d x) (a+b \sin (c+d x)) \, dx=\text {Timed out} \]
Time = 0.20 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.83 \[ \int \cos ^3(c+d x) \cot ^2(c+d x) (a+b \sin (c+d x)) \, dx=\frac {3 \, b \sin \left (d x + c\right )^{4} + 4 \, a \sin \left (d x + c\right )^{3} - 12 \, b \sin \left (d x + c\right )^{2} + 12 \, b \log \left (\sin \left (d x + c\right )\right ) - 24 \, a \sin \left (d x + c\right ) - \frac {12 \, a}{\sin \left (d x + c\right )}}{12 \, d} \]
1/12*(3*b*sin(d*x + c)^4 + 4*a*sin(d*x + c)^3 - 12*b*sin(d*x + c)^2 + 12*b *log(sin(d*x + c)) - 24*a*sin(d*x + c) - 12*a/sin(d*x + c))/d
Time = 0.37 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.95 \[ \int \cos ^3(c+d x) \cot ^2(c+d x) (a+b \sin (c+d x)) \, dx=\frac {3 \, b \sin \left (d x + c\right )^{4} + 4 \, a \sin \left (d x + c\right )^{3} - 12 \, b \sin \left (d x + c\right )^{2} + 12 \, b \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - 24 \, a \sin \left (d x + c\right ) - \frac {12 \, {\left (b \sin \left (d x + c\right ) + a\right )}}{\sin \left (d x + c\right )}}{12 \, d} \]
1/12*(3*b*sin(d*x + c)^4 + 4*a*sin(d*x + c)^3 - 12*b*sin(d*x + c)^2 + 12*b *log(abs(sin(d*x + c))) - 24*a*sin(d*x + c) - 12*(b*sin(d*x + c) + a)/sin( d*x + c))/d
Time = 11.52 (sec) , antiderivative size = 250, normalized size of antiderivative = 3.01 \[ \int \cos ^3(c+d x) \cot ^2(c+d x) (a+b \sin (c+d x)) \, dx=\frac {b\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {b\,\ln \left (\frac {1}{{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}\right )}{d}-\frac {4\,b\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{d}+\frac {8\,b\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{d}-\frac {8\,b\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{d}+\frac {4\,b\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{d}-\frac {9\,a\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}-\frac {a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {20\,a\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}-\frac {16\,a\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{3\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {8\,a\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{3\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )} \]
(b*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d - (b*log(1/cos(c/2 + (d*x )/2)^2))/d - (4*b*cos(c/2 + (d*x)/2)^2)/d + (8*b*cos(c/2 + (d*x)/2)^4)/d - (8*b*cos(c/2 + (d*x)/2)^6)/d + (4*b*cos(c/2 + (d*x)/2)^8)/d - (9*a*cos(c/ 2 + (d*x)/2))/(2*d*sin(c/2 + (d*x)/2)) - (a*sin(c/2 + (d*x)/2))/(2*d*cos(c /2 + (d*x)/2)) + (20*a*cos(c/2 + (d*x)/2)^3)/(3*d*sin(c/2 + (d*x)/2)) - (1 6*a*cos(c/2 + (d*x)/2)^5)/(3*d*sin(c/2 + (d*x)/2)) + (8*a*cos(c/2 + (d*x)/ 2)^7)/(3*d*sin(c/2 + (d*x)/2))