Integrand size = 25, antiderivative size = 78 \[ \int \cot (c+d x) \csc (c+d x) (a+a \sin (c+d x))^4 \, dx=-\frac {a^4 \csc (c+d x)}{d}+\frac {4 a^4 \log (\sin (c+d x))}{d}+\frac {6 a^4 \sin (c+d x)}{d}+\frac {2 a^4 \sin ^2(c+d x)}{d}+\frac {a^4 \sin ^3(c+d x)}{3 d} \] Output:
-a^4*csc(d*x+c)/d+4*a^4*ln(sin(d*x+c))/d+6*a^4*sin(d*x+c)/d+2*a^4*sin(d*x+ c)^2/d+1/3*a^4*sin(d*x+c)^3/d
Time = 0.03 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00 \[ \int \cot (c+d x) \csc (c+d x) (a+a \sin (c+d x))^4 \, dx=-\frac {a^4 \csc (c+d x)}{d}+\frac {4 a^4 \log (\sin (c+d x))}{d}+\frac {6 a^4 \sin (c+d x)}{d}+\frac {2 a^4 \sin ^2(c+d x)}{d}+\frac {a^4 \sin ^3(c+d x)}{3 d} \] Input:
Integrate[Cot[c + d*x]*Csc[c + d*x]*(a + a*Sin[c + d*x])^4,x]
Output:
-((a^4*Csc[c + d*x])/d) + (4*a^4*Log[Sin[c + d*x]])/d + (6*a^4*Sin[c + d*x ])/d + (2*a^4*Sin[c + d*x]^2)/d + (a^4*Sin[c + d*x]^3)/(3*d)
Time = 0.27 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.90, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 3312, 27, 49, 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.
\(\displaystyle \int \cot (c+d x) \csc (c+d x) (a \sin (c+d x)+a)^4 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x) (a \sin (c+d x)+a)^4}{\sin (c+d x)^2}dx\) |
\(\Big \downarrow \) 3312 |
\(\displaystyle \frac {\int \csc ^2(c+d x) (\sin (c+d x) a+a)^4d(a \sin (c+d x))}{a d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a \int \frac {\csc ^2(c+d x) (\sin (c+d x) a+a)^4}{a^2}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {a \int \left (\csc ^2(c+d x) a^2+\sin ^2(c+d x) a^2+4 \csc (c+d x) a^2+4 \sin (c+d x) a^2+6 a^2\right )d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a \left (\frac {1}{3} a^3 \sin ^3(c+d x)+2 a^3 \sin ^2(c+d x)+6 a^3 \sin (c+d x)-a^3 \csc (c+d x)+4 a^3 \log (a \sin (c+d x))\right )}{d}\) |
Input:
Int[Cot[c + d*x]*Csc[c + d*x]*(a + a*Sin[c + d*x])^4,x]
Output:
(a*(-(a^3*Csc[c + d*x]) + 4*a^3*Log[a*Sin[c + d*x]] + 6*a^3*Sin[c + d*x] + 2*a^3*Sin[c + d*x]^2 + (a^3*Sin[c + d*x]^3)/3))/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_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*(( c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b*f) Su bst[Int[(a + x)^m*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]
Time = 2.27 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.71
method | result | size |
derivativedivides | \(-\frac {a^{4} \left (\csc \left (d x +c \right )-\frac {1}{3 \csc \left (d x +c \right )^{3}}+4 \ln \left (\csc \left (d x +c \right )\right )-\frac {2}{\csc \left (d x +c \right )^{2}}-\frac {6}{\csc \left (d x +c \right )}\right )}{d}\) | \(55\) |
default | \(-\frac {a^{4} \left (\csc \left (d x +c \right )-\frac {1}{3 \csc \left (d x +c \right )^{3}}+4 \ln \left (\csc \left (d x +c \right )\right )-\frac {2}{\csc \left (d x +c \right )^{2}}-\frac {6}{\csc \left (d x +c \right )}\right )}{d}\) | \(55\) |
risch | \(-4 i a^{4} x -\frac {a^{4} {\mathrm e}^{2 i \left (d x +c \right )}}{2 d}-\frac {25 i a^{4} {\mathrm e}^{i \left (d x +c \right )}}{8 d}+\frac {25 i a^{4} {\mathrm e}^{-i \left (d x +c \right )}}{8 d}-\frac {a^{4} {\mathrm e}^{-2 i \left (d x +c \right )}}{2 d}-\frac {8 i a^{4} c}{d}-\frac {2 i a^{4} {\mathrm e}^{i \left (d x +c \right )}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}+\frac {4 a^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}-\frac {a^{4} \sin \left (3 d x +3 c \right )}{12 d}\) | \(157\) |
Input:
int(cot(d*x+c)*csc(d*x+c)*(a+a*sin(d*x+c))^4,x,method=_RETURNVERBOSE)
Output:
-1/d*a^4*(csc(d*x+c)-1/3/csc(d*x+c)^3+4*ln(csc(d*x+c))-2/csc(d*x+c)^2-6/cs c(d*x+c))
Time = 0.09 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.17 \[ \int \cot (c+d x) \csc (c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {a^{4} \cos \left (d x + c\right )^{4} - 20 \, a^{4} \cos \left (d x + c\right )^{2} + 12 \, a^{4} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) + 16 \, a^{4} - 3 \, {\left (2 \, a^{4} \cos \left (d x + c\right )^{2} - a^{4}\right )} \sin \left (d x + c\right )}{3 \, d \sin \left (d x + c\right )} \] Input:
integrate(cot(d*x+c)*csc(d*x+c)*(a+a*sin(d*x+c))^4,x, algorithm="fricas")
Output:
1/3*(a^4*cos(d*x + c)^4 - 20*a^4*cos(d*x + c)^2 + 12*a^4*log(1/2*sin(d*x + c))*sin(d*x + c) + 16*a^4 - 3*(2*a^4*cos(d*x + c)^2 - a^4)*sin(d*x + c))/ (d*sin(d*x + c))
\[ \int \cot (c+d x) \csc (c+d x) (a+a \sin (c+d x))^4 \, dx=a^{4} \left (\int \cot {\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx + \int 4 \sin {\left (c + d x \right )} \cot {\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx + \int 6 \sin ^{2}{\left (c + d x \right )} \cot {\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx + \int 4 \sin ^{3}{\left (c + d x \right )} \cot {\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx + \int \sin ^{4}{\left (c + d x \right )} \cot {\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx\right ) \] Input:
integrate(cot(d*x+c)*csc(d*x+c)*(a+a*sin(d*x+c))**4,x)
Output:
a**4*(Integral(cot(c + d*x)*csc(c + d*x), x) + Integral(4*sin(c + d*x)*cot (c + d*x)*csc(c + d*x), x) + Integral(6*sin(c + d*x)**2*cot(c + d*x)*csc(c + d*x), x) + Integral(4*sin(c + d*x)**3*cot(c + d*x)*csc(c + d*x), x) + I ntegral(sin(c + d*x)**4*cot(c + d*x)*csc(c + d*x), x))
Time = 0.03 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.86 \[ \int \cot (c+d x) \csc (c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {a^{4} \sin \left (d x + c\right )^{3} + 6 \, a^{4} \sin \left (d x + c\right )^{2} + 12 \, a^{4} \log \left (\sin \left (d x + c\right )\right ) + 18 \, a^{4} \sin \left (d x + c\right ) - \frac {3 \, a^{4}}{\sin \left (d x + c\right )}}{3 \, d} \] Input:
integrate(cot(d*x+c)*csc(d*x+c)*(a+a*sin(d*x+c))^4,x, algorithm="maxima")
Output:
1/3*(a^4*sin(d*x + c)^3 + 6*a^4*sin(d*x + c)^2 + 12*a^4*log(sin(d*x + c)) + 18*a^4*sin(d*x + c) - 3*a^4/sin(d*x + c))/d
Time = 0.15 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.87 \[ \int \cot (c+d x) \csc (c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {a^{4} \sin \left (d x + c\right )^{3} + 6 \, a^{4} \sin \left (d x + c\right )^{2} + 12 \, a^{4} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 18 \, a^{4} \sin \left (d x + c\right ) - \frac {3 \, a^{4}}{\sin \left (d x + c\right )}}{3 \, d} \] Input:
integrate(cot(d*x+c)*csc(d*x+c)*(a+a*sin(d*x+c))^4,x, algorithm="giac")
Output:
1/3*(a^4*sin(d*x + c)^3 + 6*a^4*sin(d*x + c)^2 + 12*a^4*log(abs(sin(d*x + c))) + 18*a^4*sin(d*x + c) - 3*a^4/sin(d*x + c))/d
Time = 18.09 (sec) , antiderivative size = 235, normalized size of antiderivative = 3.01 \[ \int \cot (c+d x) \csc (c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {8\,a^4\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{d}-\frac {8\,a^4\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{d}-\frac {4\,a^4\,\ln \left (\frac {1}{{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}\right )}{d}+\frac {4\,a^4\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {28\,a^4\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}-\frac {16\,a^4\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{3\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {8\,a^4\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{3\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {23\,a^4\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}-\frac {a^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )} \] Input:
int((cot(c + d*x)*(a + a*sin(c + d*x))^4)/sin(c + d*x),x)
Output:
(8*a^4*cos(c/2 + (d*x)/2)^2)/d - (8*a^4*cos(c/2 + (d*x)/2)^4)/d - (4*a^4*l og(1/cos(c/2 + (d*x)/2)^2))/d + (4*a^4*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d *x)/2)))/d - (28*a^4*cos(c/2 + (d*x)/2)^3)/(3*d*sin(c/2 + (d*x)/2)) - (16* a^4*cos(c/2 + (d*x)/2)^5)/(3*d*sin(c/2 + (d*x)/2)) + (8*a^4*cos(c/2 + (d*x )/2)^7)/(3*d*sin(c/2 + (d*x)/2)) + (23*a^4*cos(c/2 + (d*x)/2))/(2*d*sin(c/ 2 + (d*x)/2)) - (a^4*sin(c/2 + (d*x)/2))/(2*d*cos(c/2 + (d*x)/2))
Time = 0.16 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.10 \[ \int \cot (c+d x) \csc (c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {a^{4} \left (-12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )+12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )+\sin \left (d x +c \right )^{4}+6 \sin \left (d x +c \right )^{3}+18 \sin \left (d x +c \right )^{2}-3\right )}{3 \sin \left (d x +c \right ) d} \] Input:
int(cot(d*x+c)*csc(d*x+c)*(a+a*sin(d*x+c))^4,x)
Output:
(a**4*( - 12*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x) + 12*log(tan((c + d *x)/2))*sin(c + d*x) + sin(c + d*x)**4 + 6*sin(c + d*x)**3 + 18*sin(c + d* x)**2 - 3))/(3*sin(c + d*x)*d)