Integrand size = 25, antiderivative size = 85 \[ \int \cos (c+d x) \cot ^4(c+d x) (a+a \sin (c+d x)) \, dx=\frac {2 a \csc (c+d x)}{d}-\frac {a \csc ^2(c+d x)}{2 d}-\frac {a \csc ^3(c+d x)}{3 d}-\frac {2 a \log (\sin (c+d x))}{d}+\frac {a \sin (c+d x)}{d}+\frac {a \sin ^2(c+d x)}{2 d} \] Output:
2*a*csc(d*x+c)/d-1/2*a*csc(d*x+c)^2/d-1/3*a*csc(d*x+c)^3/d-2*a*ln(sin(d*x+ c))/d+a*sin(d*x+c)/d+1/2*a*sin(d*x+c)^2/d
Time = 0.19 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.89 \[ \int \cos (c+d x) \cot ^4(c+d x) (a+a \sin (c+d x)) \, dx=\frac {2 a \csc (c+d x)}{d}-\frac {a \csc ^3(c+d x)}{3 d}+\frac {a \sin (c+d x)}{d}-\frac {a \left (\csc ^2(c+d x)+4 \log (\sin (c+d x))-\sin ^2(c+d x)\right )}{2 d} \] Input:
Integrate[Cos[c + d*x]*Cot[c + d*x]^4*(a + a*Sin[c + d*x]),x]
Output:
(2*a*Csc[c + d*x])/d - (a*Csc[c + d*x]^3)/(3*d) + (a*Sin[c + d*x])/d - (a* (Csc[c + d*x]^2 + 4*Log[Sin[c + d*x]] - Sin[c + d*x]^2))/(2*d)
Time = 0.29 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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.
\(\displaystyle \int \cos (c+d x) \cot ^4(c+d x) (a \sin (c+d x)+a) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^5 (a \sin (c+d x)+a)}{\sin (c+d x)^4}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \csc ^4(c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^3d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\csc ^4(c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^3}{a^4}d(a \sin (c+d x))}{a d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left (a \csc ^4(c+d x)+a \csc ^3(c+d x)-2 a \csc ^2(c+d x)-2 a \csc (c+d x)+a+a \sin (c+d x)\right )d(a \sin (c+d x))}{a d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{2} a^2 \sin ^2(c+d x)+a^2 \sin (c+d x)-\frac {1}{3} a^2 \csc ^3(c+d x)-\frac {1}{2} a^2 \csc ^2(c+d x)+2 a^2 \csc (c+d x)-2 a^2 \log (a \sin (c+d x))}{a d}\) |
Input:
Int[Cos[c + d*x]*Cot[c + d*x]^4*(a + a*Sin[c + d*x]),x]
Output:
(2*a^2*Csc[c + d*x] - (a^2*Csc[c + d*x]^2)/2 - (a^2*Csc[c + d*x]^3)/3 - 2* a^2*Log[a*Sin[c + d*x]] + a^2*Sin[c + d*x] + (a^2*Sin[c + d*x]^2)/2)/(a*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 = 1.35 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.42
method | result | size |
derivativedivides | \(\frac {a \left (-\frac {\cos \left (d x +c \right )^{6}}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )^{4}}{2}-\cos \left (d x +c \right )^{2}-2 \ln \left (\sin \left (d x +c \right )\right )\right )+a \left (-\frac {\cos \left (d x +c \right )^{6}}{3 \sin \left (d x +c \right )^{3}}+\frac {\cos \left (d x +c \right )^{6}}{\sin \left (d x +c \right )}+\left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )\right )}{d}\) | \(121\) |
default | \(\frac {a \left (-\frac {\cos \left (d x +c \right )^{6}}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )^{4}}{2}-\cos \left (d x +c \right )^{2}-2 \ln \left (\sin \left (d x +c \right )\right )\right )+a \left (-\frac {\cos \left (d x +c \right )^{6}}{3 \sin \left (d x +c \right )^{3}}+\frac {\cos \left (d x +c \right )^{6}}{\sin \left (d x +c \right )}+\left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )\right )}{d}\) | \(121\) |
risch | \(2 i a x -\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {i a \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {i a \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {a \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {4 i a c}{d}+\frac {2 i a \left (6 \,{\mathrm e}^{5 i \left (d x +c \right )}-8 \,{\mathrm e}^{3 i \left (d x +c \right )}-3 i {\mathrm e}^{4 i \left (d x +c \right )}+6 \,{\mathrm e}^{i \left (d x +c \right )}+3 i {\mathrm e}^{2 i \left (d x +c \right )}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}-\frac {2 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\) | \(173\) |
Input:
int(cos(d*x+c)*cot(d*x+c)^4*(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(a*(-1/2/sin(d*x+c)^2*cos(d*x+c)^6-1/2*cos(d*x+c)^4-cos(d*x+c)^2-2*ln( sin(d*x+c)))+a*(-1/3/sin(d*x+c)^3*cos(d*x+c)^6+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)))
Time = 0.09 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.38 \[ \int \cos (c+d x) \cot ^4(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {12 \, a \cos \left (d x + c\right )^{4} - 48 \, a \cos \left (d x + c\right )^{2} + 24 \, {\left (a \cos \left (d x + c\right )^{2} - a\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) + 3 \, {\left (2 \, a \cos \left (d x + c\right )^{4} - 3 \, a \cos \left (d x + c\right )^{2} - a\right )} \sin \left (d x + c\right ) + 32 \, a}{12 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right )} \] Input:
integrate(cos(d*x+c)*cot(d*x+c)^4*(a+a*sin(d*x+c)),x, algorithm="fricas")
Output:
-1/12*(12*a*cos(d*x + c)^4 - 48*a*cos(d*x + c)^2 + 24*(a*cos(d*x + c)^2 - a)*log(1/2*sin(d*x + c))*sin(d*x + c) + 3*(2*a*cos(d*x + c)^4 - 3*a*cos(d* x + c)^2 - a)*sin(d*x + c) + 32*a)/((d*cos(d*x + c)^2 - d)*sin(d*x + c))
\[ \int \cos (c+d x) \cot ^4(c+d x) (a+a \sin (c+d x)) \, dx=a \left (\int \cos {\left (c + d x \right )} \cot ^{4}{\left (c + d x \right )}\, dx + \int \sin {\left (c + d x \right )} \cos {\left (c + d x \right )} \cot ^{4}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate(cos(d*x+c)*cot(d*x+c)**4*(a+a*sin(d*x+c)),x)
Output:
a*(Integral(cos(c + d*x)*cot(c + d*x)**4, x) + Integral(sin(c + d*x)*cos(c + d*x)*cot(c + d*x)**4, x))
Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.81 \[ \int \cos (c+d x) \cot ^4(c+d x) (a+a \sin (c+d x)) \, dx=\frac {3 \, a \sin \left (d x + c\right )^{2} - 12 \, a \log \left (\sin \left (d x + c\right )\right ) + 6 \, a \sin \left (d x + c\right ) + \frac {12 \, a \sin \left (d x + c\right )^{2} - 3 \, a \sin \left (d x + c\right ) - 2 \, a}{\sin \left (d x + c\right )^{3}}}{6 \, d} \] Input:
integrate(cos(d*x+c)*cot(d*x+c)^4*(a+a*sin(d*x+c)),x, algorithm="maxima")
Output:
1/6*(3*a*sin(d*x + c)^2 - 12*a*log(sin(d*x + c)) + 6*a*sin(d*x + c) + (12* a*sin(d*x + c)^2 - 3*a*sin(d*x + c) - 2*a)/sin(d*x + c)^3)/d
Time = 0.15 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.82 \[ \int \cos (c+d x) \cot ^4(c+d x) (a+a \sin (c+d x)) \, dx=\frac {3 \, a \sin \left (d x + c\right )^{2} - 12 \, a \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 6 \, a \sin \left (d x + c\right ) + \frac {12 \, a \sin \left (d x + c\right )^{2} - 3 \, a \sin \left (d x + c\right ) - 2 \, a}{\sin \left (d x + c\right )^{3}}}{6 \, d} \] Input:
integrate(cos(d*x+c)*cot(d*x+c)^4*(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
1/6*(3*a*sin(d*x + c)^2 - 12*a*log(abs(sin(d*x + c))) + 6*a*sin(d*x + c) + (12*a*sin(d*x + c)^2 - 3*a*sin(d*x + c) - 2*a)/sin(d*x + c)^3)/d
Time = 18.24 (sec) , antiderivative size = 218, normalized size of antiderivative = 2.56 \[ \int \cos (c+d x) \cot ^4(c+d x) (a+a \sin (c+d x)) \, dx=\frac {7\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,d}+\frac {2\,a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}-\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,d}-\frac {2\,a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {23\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {89\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}-2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {19\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}-a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {a}{3}}{d\,\left (8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\right )} \] Input:
int(cos(c + d*x)*cot(c + d*x)^4*(a + a*sin(c + d*x)),x)
Output:
(7*a*tan(c/2 + (d*x)/2))/(8*d) + (2*a*log(tan(c/2 + (d*x)/2)^2 + 1))/d - ( a*tan(c/2 + (d*x)/2)^2)/(8*d) - (a*tan(c/2 + (d*x)/2)^3)/(24*d) - (2*a*log (tan(c/2 + (d*x)/2)))/d + ((19*a*tan(c/2 + (d*x)/2)^2)/3 - a*tan(c/2 + (d* x)/2) - a/3 - 2*a*tan(c/2 + (d*x)/2)^3 + (89*a*tan(c/2 + (d*x)/2)^4)/3 + 1 5*a*tan(c/2 + (d*x)/2)^5 + 23*a*tan(c/2 + (d*x)/2)^6)/(d*(8*tan(c/2 + (d*x )/2)^3 + 16*tan(c/2 + (d*x)/2)^5 + 8*tan(c/2 + (d*x)/2)^7))
Time = 0.15 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.27 \[ \int \cos (c+d x) \cot ^4(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \left (12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{3}-12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{3}+3 \sin \left (d x +c \right )^{5}+6 \sin \left (d x +c \right )^{4}+3 \sin \left (d x +c \right )^{3}+12 \sin \left (d x +c \right )^{2}-3 \sin \left (d x +c \right )-2\right )}{6 \sin \left (d x +c \right )^{3} d} \] Input:
int(cos(d*x+c)*cot(d*x+c)^4*(a+a*sin(d*x+c)),x)
Output:
(a*(12*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**3 - 12*log(tan((c + d*x) /2))*sin(c + d*x)**3 + 3*sin(c + d*x)**5 + 6*sin(c + d*x)**4 + 3*sin(c + d *x)**3 + 12*sin(c + d*x)**2 - 3*sin(c + d*x) - 2))/(6*sin(c + d*x)**3*d)