Integrand size = 19, antiderivative size = 54 \[ \int \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \csc (c+d x)}{d}-\frac {a \csc ^2(c+d x)}{2 d}-\frac {a \log (\sin (c+d x))}{d}-\frac {a \sin (c+d x)}{d} \] Output:
-a*csc(d*x+c)/d-1/2*a*csc(d*x+c)^2/d-a*ln(sin(d*x+c))/d-a*sin(d*x+c)/d
Time = 0.01 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00 \[ \int \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \csc (c+d x)}{d}-\frac {a \csc ^2(c+d x)}{2 d}-\frac {a \log (\sin (c+d x))}{d}-\frac {a \sin (c+d x)}{d} \] Input:
Integrate[Cot[c + d*x]^3*(a + a*Sin[c + d*x]),x]
Output:
-((a*Csc[c + d*x])/d) - (a*Csc[c + d*x]^2)/(2*d) - (a*Log[Sin[c + d*x]])/d - (a*Sin[c + d*x])/d
Time = 0.23 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.89, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, 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.
\(\displaystyle \int \cot ^3(c+d x) (a \sin (c+d x)+a) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {a \sin (c+d x)+a}{\tan (c+d x)^3}dx\) |
\(\Big \downarrow \) 3186 |
\(\displaystyle \frac {\int \frac {\csc ^3(c+d x) (a-a \sin (c+d x)) (\sin (c+d x) a+a)^2}{a^3}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 84 |
\(\displaystyle \frac {\int \left (\csc ^3(c+d x)+\csc ^2(c+d x)-\csc (c+d x)-1\right )d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-a \sin (c+d x)-\frac {1}{2} a \csc ^2(c+d x)-a \csc (c+d x)-a \log (a \sin (c+d x))}{d}\) |
Input:
Int[Cot[c + d*x]^3*(a + a*Sin[c + d*x]),x]
Output:
(-(a*Csc[c + d*x]) - (a*Csc[c + d*x]^2)/2 - a*Log[a*Sin[c + d*x]] - a*Sin[ c + d*x])/d
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 = 0.55 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.24
method | result | size |
derivativedivides | \(\frac {a \left (-\frac {\cos \left (d x +c \right )^{4}}{\sin \left (d x +c \right )}-\left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )\right )+a \left (-\frac {\cot \left (d x +c \right )^{2}}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\) | \(67\) |
default | \(\frac {a \left (-\frac {\cos \left (d x +c \right )^{4}}{\sin \left (d x +c \right )}-\left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )\right )+a \left (-\frac {\cot \left (d x +c \right )^{2}}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\) | \(67\) |
risch | \(i a x +\frac {i a \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {i a \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {2 i a c}{d}-\frac {2 i a \left (i {\mathrm e}^{2 i \left (d x +c \right )}+{\mathrm e}^{3 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\) | \(118\) |
Input:
int(cot(d*x+c)^3*(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(a*(-1/sin(d*x+c)*cos(d*x+c)^4-(2+cos(d*x+c)^2)*sin(d*x+c))+a*(-1/2*co t(d*x+c)^2-ln(sin(d*x+c))))
Time = 0.08 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.28 \[ \int \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {2 \, {\left (a \cos \left (d x + c\right )^{2} - a\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 2 \, {\left (a \cos \left (d x + c\right )^{2} - 2 \, a\right )} \sin \left (d x + c\right ) - a}{2 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )}} \] Input:
integrate(cot(d*x+c)^3*(a+a*sin(d*x+c)),x, algorithm="fricas")
Output:
-1/2*(2*(a*cos(d*x + c)^2 - a)*log(1/2*sin(d*x + c)) + 2*(a*cos(d*x + c)^2 - 2*a)*sin(d*x + c) - a)/(d*cos(d*x + c)^2 - d)
\[ \int \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx=a \left (\int \sin {\left (c + d x \right )} \cot ^{3}{\left (c + d x \right )}\, dx + \int \cot ^{3}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate(cot(d*x+c)**3*(a+a*sin(d*x+c)),x)
Output:
a*(Integral(sin(c + d*x)*cot(c + d*x)**3, x) + Integral(cot(c + d*x)**3, x ))
Time = 0.03 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.83 \[ \int \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {2 \, a \log \left (\sin \left (d x + c\right )\right ) + 2 \, a \sin \left (d x + c\right ) + \frac {2 \, a \sin \left (d x + c\right ) + a}{\sin \left (d x + c\right )^{2}}}{2 \, d} \] Input:
integrate(cot(d*x+c)^3*(a+a*sin(d*x+c)),x, algorithm="maxima")
Output:
-1/2*(2*a*log(sin(d*x + c)) + 2*a*sin(d*x + c) + (2*a*sin(d*x + c) + a)/si n(d*x + c)^2)/d
Time = 0.17 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.85 \[ \int \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {2 \, a \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 2 \, a \sin \left (d x + c\right ) + \frac {2 \, a \sin \left (d x + c\right ) + a}{\sin \left (d x + c\right )^{2}}}{2 \, d} \] Input:
integrate(cot(d*x+c)^3*(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
-1/2*(2*a*log(abs(sin(d*x + c))) + 2*a*sin(d*x + c) + (2*a*sin(d*x + c) + a)/sin(d*x + c)^2)/d
Time = 17.74 (sec) , antiderivative size = 146, normalized size of antiderivative = 2.70 \[ \int \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d}-\frac {10\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}+2\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a}{2}}{d\,\left (4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}-\frac {a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d} \] Input:
int(cot(c + d*x)^3*(a + a*sin(c + d*x)),x)
Output:
(a*log(tan(c/2 + (d*x)/2)^2 + 1))/d - (a*tan(c/2 + (d*x)/2))/(2*d) - (a/2 + 2*a*tan(c/2 + (d*x)/2) + (a*tan(c/2 + (d*x)/2)^2)/2 + 10*a*tan(c/2 + (d* x)/2)^3)/(d*(4*tan(c/2 + (d*x)/2)^2 + 4*tan(c/2 + (d*x)/2)^4)) - (a*tan(c/ 2 + (d*x)/2)^2)/(8*d) - (a*log(tan(c/2 + (d*x)/2)))/d
Time = 0.17 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.63 \[ \int \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \left (8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{2}-8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2}-8 \sin \left (d x +c \right )^{3}+3 \sin \left (d x +c \right )^{2}-8 \sin \left (d x +c \right )-4\right )}{8 \sin \left (d x +c \right )^{2} d} \] Input:
int(cot(d*x+c)^3*(a+a*sin(d*x+c)),x)
Output:
(a*(8*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2 - 8*log(tan((c + d*x)/2 ))*sin(c + d*x)**2 - 8*sin(c + d*x)**3 + 3*sin(c + d*x)**2 - 8*sin(c + d*x ) - 4))/(8*sin(c + d*x)**2*d)