Integrand size = 25, antiderivative size = 86 \[ \int \cos ^4(c+d x) \cot (c+d x) (a+b \sin (c+d x)) \, dx=\frac {a \log (\sin (c+d x))}{d}+\frac {b \sin (c+d x)}{d}-\frac {a \sin ^2(c+d x)}{d}-\frac {2 b \sin ^3(c+d x)}{3 d}+\frac {a \sin ^4(c+d x)}{4 d}+\frac {b \sin ^5(c+d x)}{5 d} \] Output:
a*ln(sin(d*x+c))/d+b*sin(d*x+c)/d-a*sin(d*x+c)^2/d-2/3*b*sin(d*x+c)^3/d+1/ 4*a*sin(d*x+c)^4/d+1/5*b*sin(d*x+c)^5/d
Time = 0.03 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00 \[ \int \cos ^4(c+d x) \cot (c+d x) (a+b \sin (c+d x)) \, dx=\frac {a \log (\sin (c+d x))}{d}+\frac {b \sin (c+d x)}{d}-\frac {a \sin ^2(c+d x)}{d}-\frac {2 b \sin ^3(c+d x)}{3 d}+\frac {a \sin ^4(c+d x)}{4 d}+\frac {b \sin ^5(c+d x)}{5 d} \] Input:
Integrate[Cos[c + d*x]^4*Cot[c + d*x]*(a + b*Sin[c + d*x]),x]
Output:
(a*Log[Sin[c + d*x]])/d + (b*Sin[c + d*x])/d - (a*Sin[c + d*x]^2)/d - (2*b *Sin[c + d*x]^3)/(3*d) + (a*Sin[c + d*x]^4)/(4*d) + (b*Sin[c + d*x]^5)/(5* d)
Time = 0.28 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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.
\(\displaystyle \int \cos ^4(c+d x) \cot (c+d x) (a+b \sin (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^5 (a+b \sin (c+d x))}{\sin (c+d x)}dx\) |
\(\Big \downarrow \) 3316 |
\(\displaystyle \frac {\int \csc (c+d x) (a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^2d(b \sin (c+d x))}{b^5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\csc (c+d x) (a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}{b}d(b \sin (c+d x))}{b^4 d}\) |
\(\Big \downarrow \) 522 |
\(\displaystyle \frac {\int \left (\sin ^4(c+d x) b^4-2 \sin ^2(c+d x) b^4+b^4+a \sin ^3(c+d x) b^3+a \csc (c+d x) b^3-2 a \sin (c+d x) b^3\right )d(b \sin (c+d x))}{b^4 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{4} a b^4 \sin ^4(c+d x)-a b^4 \sin ^2(c+d x)+a b^4 \log (b \sin (c+d x))+\frac {1}{5} b^5 \sin ^5(c+d x)-\frac {2}{3} b^5 \sin ^3(c+d x)+b^5 \sin (c+d x)}{b^4 d}\) |
Input:
Int[Cos[c + d*x]^4*Cot[c + d*x]*(a + b*Sin[c + d*x]),x]
Output:
(a*b^4*Log[b*Sin[c + d*x]] + b^5*Sin[c + d*x] - a*b^4*Sin[c + d*x]^2 - (2* b^5*Sin[c + d*x]^3)/3 + (a*b^4*Sin[c + d*x]^4)/4 + (b^5*Sin[c + d*x]^5)/5) /(b^4*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 = 1.99 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.76
method | result | size |
derivativedivides | \(\frac {a \left (\frac {\cos \left (d x +c \right )^{4}}{4}+\frac {\cos \left (d x +c \right )^{2}}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )+\frac {b \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}}{d}\) | \(65\) |
default | \(\frac {a \left (\frac {\cos \left (d x +c \right )^{4}}{4}+\frac {\cos \left (d x +c \right )^{2}}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )+\frac {b \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}}{d}\) | \(65\) |
risch | \(-i a x +\frac {3 a \,{\mathrm e}^{2 i \left (d x +c \right )}}{16 d}+\frac {3 a \,{\mathrm e}^{-2 i \left (d x +c \right )}}{16 d}-\frac {2 i a c}{d}+\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}+\frac {5 b \sin \left (d x +c \right )}{8 d}+\frac {b \sin \left (5 d x +5 c \right )}{80 d}+\frac {a \cos \left (4 d x +4 c \right )}{32 d}+\frac {5 b \sin \left (3 d x +3 c \right )}{48 d}\) | \(119\) |
Input:
int(cos(d*x+c)^4*cot(d*x+c)*(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(a*(1/4*cos(d*x+c)^4+1/2*cos(d*x+c)^2+ln(sin(d*x+c)))+1/5*b*(8/3+cos(d *x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c))
Time = 0.09 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.86 \[ \int \cos ^4(c+d x) \cot (c+d x) (a+b \sin (c+d x)) \, dx=\frac {15 \, a \cos \left (d x + c\right )^{4} + 30 \, a \cos \left (d x + c\right )^{2} + 60 \, a \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 4 \, {\left (3 \, b \cos \left (d x + c\right )^{4} + 4 \, b \cos \left (d x + c\right )^{2} + 8 \, b\right )} \sin \left (d x + c\right )}{60 \, d} \] Input:
integrate(cos(d*x+c)^4*cot(d*x+c)*(a+b*sin(d*x+c)),x, algorithm="fricas")
Output:
1/60*(15*a*cos(d*x + c)^4 + 30*a*cos(d*x + c)^2 + 60*a*log(1/2*sin(d*x + c )) + 4*(3*b*cos(d*x + c)^4 + 4*b*cos(d*x + c)^2 + 8*b)*sin(d*x + c))/d
\[ \int \cos ^4(c+d x) \cot (c+d x) (a+b \sin (c+d x)) \, dx=\int \left (a + b \sin {\left (c + d x \right )}\right ) \cos ^{4}{\left (c + d x \right )} \cot {\left (c + d x \right )}\, dx \] Input:
integrate(cos(d*x+c)**4*cot(d*x+c)*(a+b*sin(d*x+c)),x)
Output:
Integral((a + b*sin(c + d*x))*cos(c + d*x)**4*cot(c + d*x), x)
Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.80 \[ \int \cos ^4(c+d x) \cot (c+d x) (a+b \sin (c+d x)) \, dx=\frac {12 \, b \sin \left (d x + c\right )^{5} + 15 \, a \sin \left (d x + c\right )^{4} - 40 \, b \sin \left (d x + c\right )^{3} - 60 \, a \sin \left (d x + c\right )^{2} + 60 \, a \log \left (\sin \left (d x + c\right )\right ) + 60 \, b \sin \left (d x + c\right )}{60 \, d} \] Input:
integrate(cos(d*x+c)^4*cot(d*x+c)*(a+b*sin(d*x+c)),x, algorithm="maxima")
Output:
1/60*(12*b*sin(d*x + c)^5 + 15*a*sin(d*x + c)^4 - 40*b*sin(d*x + c)^3 - 60 *a*sin(d*x + c)^2 + 60*a*log(sin(d*x + c)) + 60*b*sin(d*x + c))/d
Time = 0.23 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.81 \[ \int \cos ^4(c+d x) \cot (c+d x) (a+b \sin (c+d x)) \, dx=\frac {12 \, b \sin \left (d x + c\right )^{5} + 15 \, a \sin \left (d x + c\right )^{4} - 40 \, b \sin \left (d x + c\right )^{3} - 60 \, a \sin \left (d x + c\right )^{2} + 60 \, a \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 60 \, b \sin \left (d x + c\right )}{60 \, d} \] Input:
integrate(cos(d*x+c)^4*cot(d*x+c)*(a+b*sin(d*x+c)),x, algorithm="giac")
Output:
1/60*(12*b*sin(d*x + c)^5 + 15*a*sin(d*x + c)^4 - 40*b*sin(d*x + c)^3 - 60 *a*sin(d*x + c)^2 + 60*a*log(abs(sin(d*x + c))) + 60*b*sin(d*x + c))/d
Time = 22.39 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.47 \[ \int \cos ^4(c+d x) \cot (c+d x) (a+b \sin (c+d x)) \, dx=\frac {a\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {a\,\ln \left (\frac {1}{{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}\right )}{d}+\frac {a\,{\cos \left (c+d\,x\right )}^2}{2\,d}+\frac {a\,{\cos \left (c+d\,x\right )}^4}{4\,d}+\frac {8\,b\,\sin \left (c+d\,x\right )}{15\,d}+\frac {4\,b\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{15\,d}+\frac {b\,{\cos \left (c+d\,x\right )}^4\,\sin \left (c+d\,x\right )}{5\,d} \] Input:
int(cos(c + d*x)^4*cot(c + d*x)*(a + b*sin(c + d*x)),x)
Output:
(a*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d - (a*log(1/cos(c/2 + (d*x )/2)^2))/d + (a*cos(c + d*x)^2)/(2*d) + (a*cos(c + d*x)^4)/(4*d) + (8*b*si n(c + d*x))/(15*d) + (4*b*cos(c + d*x)^2*sin(c + d*x))/(15*d) + (b*cos(c + d*x)^4*sin(c + d*x))/(5*d)
Time = 0.18 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.03 \[ \int \cos ^4(c+d x) \cot (c+d x) (a+b \sin (c+d x)) \, dx=\frac {-60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) a +60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +12 \sin \left (d x +c \right )^{5} b +15 \sin \left (d x +c \right )^{4} a -40 \sin \left (d x +c \right )^{3} b -60 \sin \left (d x +c \right )^{2} a +60 \sin \left (d x +c \right ) b}{60 d} \] Input:
int(cos(d*x+c)^4*cot(d*x+c)*(a+b*sin(d*x+c)),x)
Output:
( - 60*log(tan((c + d*x)/2)**2 + 1)*a + 60*log(tan((c + d*x)/2))*a + 12*si n(c + d*x)**5*b + 15*sin(c + d*x)**4*a - 40*sin(c + d*x)**3*b - 60*sin(c + d*x)**2*a + 60*sin(c + d*x)*b)/(60*d)