Integrand size = 19, antiderivative size = 81 \[ \int \cot (c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {a^4 \log (\sin (c+d x))}{d}+\frac {4 a^4 \sin (c+d x)}{d}+\frac {3 a^4 \sin ^2(c+d x)}{d}+\frac {4 a^4 \sin ^3(c+d x)}{3 d}+\frac {a^4 \sin ^4(c+d x)}{4 d} \] Output:
a^4*ln(sin(d*x+c))/d+4*a^4*sin(d*x+c)/d+3*a^4*sin(d*x+c)^2/d+4/3*a^4*sin(d *x+c)^3/d+1/4*a^4*sin(d*x+c)^4/d
Time = 0.03 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00 \[ \int \cot (c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {a^4 \log (\sin (c+d x))}{d}+\frac {4 a^4 \sin (c+d x)}{d}+\frac {3 a^4 \sin ^2(c+d x)}{d}+\frac {4 a^4 \sin ^3(c+d x)}{3 d}+\frac {a^4 \sin ^4(c+d x)}{4 d} \] Input:
Integrate[Cot[c + d*x]*(a + a*Sin[c + d*x])^4,x]
Output:
(a^4*Log[Sin[c + d*x]])/d + (4*a^4*Sin[c + d*x])/d + (3*a^4*Sin[c + d*x]^2 )/d + (4*a^4*Sin[c + d*x]^3)/(3*d) + (a^4*Sin[c + d*x]^4)/(4*d)
Time = 0.24 (sec) , antiderivative size = 72, 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, 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.
\(\displaystyle \int \cot (c+d x) (a \sin (c+d x)+a)^4 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a \sin (c+d x)+a)^4}{\tan (c+d x)}dx\) |
\(\Big \downarrow \) 3186 |
\(\displaystyle \frac {\int \frac {\csc (c+d x) (\sin (c+d x) a+a)^4}{a}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\int \left (\sin ^3(c+d x) a^3+4 \sin ^2(c+d x) a^3+\csc (c+d x) a^3+6 \sin (c+d x) a^3+4 a^3\right )d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{4} a^4 \sin ^4(c+d x)+\frac {4}{3} a^4 \sin ^3(c+d x)+3 a^4 \sin ^2(c+d x)+4 a^4 \sin (c+d x)+a^4 \log (a \sin (c+d x))}{d}\) |
Input:
Int[Cot[c + d*x]*(a + a*Sin[c + d*x])^4,x]
Output:
(a^4*Log[a*Sin[c + d*x]] + 4*a^4*Sin[c + d*x] + 3*a^4*Sin[c + d*x]^2 + (4* a^4*Sin[c + d*x]^3)/3 + (a^4*Sin[c + d*x]^4)/4)/d
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[((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 = 1.90 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.83
method | result | size |
derivativedivides | \(\frac {\frac {a^{4} \sin \left (d x +c \right )^{4}}{4}+\frac {4 a^{4} \sin \left (d x +c \right )^{3}}{3}-3 a^{4} \cos \left (d x +c \right )^{2}+4 a^{4} \sin \left (d x +c \right )+a^{4} \ln \left (\sin \left (d x +c \right )\right )}{d}\) | \(67\) |
default | \(\frac {\frac {a^{4} \sin \left (d x +c \right )^{4}}{4}+\frac {4 a^{4} \sin \left (d x +c \right )^{3}}{3}-3 a^{4} \cos \left (d x +c \right )^{2}+4 a^{4} \sin \left (d x +c \right )+a^{4} \ln \left (\sin \left (d x +c \right )\right )}{d}\) | \(67\) |
risch | \(-i a^{4} x -\frac {13 a^{4} {\mathrm e}^{2 i \left (d x +c \right )}}{16 d}-\frac {13 a^{4} {\mathrm e}^{-2 i \left (d x +c \right )}}{16 d}-\frac {2 i a^{4} c}{d}+\frac {a^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}+\frac {5 a^{4} \sin \left (d x +c \right )}{d}+\frac {a^{4} \cos \left (4 d x +4 c \right )}{32 d}-\frac {a^{4} \sin \left (3 d x +3 c \right )}{3 d}\) | \(120\) |
Input:
int(cot(d*x+c)*(a+a*sin(d*x+c))^4,x,method=_RETURNVERBOSE)
Output:
1/d*(1/4*a^4*sin(d*x+c)^4+4/3*a^4*sin(d*x+c)^3-3*a^4*cos(d*x+c)^2+4*a^4*si n(d*x+c)+a^4*ln(sin(d*x+c)))
Time = 0.09 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.89 \[ \int \cot (c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {3 \, a^{4} \cos \left (d x + c\right )^{4} - 42 \, a^{4} \cos \left (d x + c\right )^{2} + 12 \, a^{4} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 16 \, {\left (a^{4} \cos \left (d x + c\right )^{2} - 4 \, a^{4}\right )} \sin \left (d x + c\right )}{12 \, d} \] Input:
integrate(cot(d*x+c)*(a+a*sin(d*x+c))^4,x, algorithm="fricas")
Output:
1/12*(3*a^4*cos(d*x + c)^4 - 42*a^4*cos(d*x + c)^2 + 12*a^4*log(1/2*sin(d* x + c)) - 16*(a^4*cos(d*x + c)^2 - 4*a^4)*sin(d*x + c))/d
\[ \int \cot (c+d x) (a+a \sin (c+d x))^4 \, dx=a^{4} \left (\int 4 \sin {\left (c + d x \right )} \cot {\left (c + d x \right )}\, dx + \int 6 \sin ^{2}{\left (c + d x \right )} \cot {\left (c + d x \right )}\, dx + \int 4 \sin ^{3}{\left (c + d x \right )} \cot {\left (c + d x \right )}\, dx + \int \sin ^{4}{\left (c + d x \right )} \cot {\left (c + d x \right )}\, dx + \int \cot {\left (c + d x \right )}\, dx\right ) \] Input:
integrate(cot(d*x+c)*(a+a*sin(d*x+c))**4,x)
Output:
a**4*(Integral(4*sin(c + d*x)*cot(c + d*x), x) + Integral(6*sin(c + d*x)** 2*cot(c + d*x), x) + Integral(4*sin(c + d*x)**3*cot(c + d*x), x) + Integra l(sin(c + d*x)**4*cot(c + d*x), x) + Integral(cot(c + d*x), x))
Time = 0.03 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.84 \[ \int \cot (c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {3 \, a^{4} \sin \left (d x + c\right )^{4} + 16 \, a^{4} \sin \left (d x + c\right )^{3} + 36 \, a^{4} \sin \left (d x + c\right )^{2} + 12 \, a^{4} \log \left (\sin \left (d x + c\right )\right ) + 48 \, a^{4} \sin \left (d x + c\right )}{12 \, d} \] Input:
integrate(cot(d*x+c)*(a+a*sin(d*x+c))^4,x, algorithm="maxima")
Output:
1/12*(3*a^4*sin(d*x + c)^4 + 16*a^4*sin(d*x + c)^3 + 36*a^4*sin(d*x + c)^2 + 12*a^4*log(sin(d*x + c)) + 48*a^4*sin(d*x + c))/d
Time = 0.15 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.85 \[ \int \cot (c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {3 \, a^{4} \sin \left (d x + c\right )^{4} + 16 \, a^{4} \sin \left (d x + c\right )^{3} + 36 \, a^{4} \sin \left (d x + c\right )^{2} + 12 \, a^{4} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 48 \, a^{4} \sin \left (d x + c\right )}{12 \, d} \] Input:
integrate(cot(d*x+c)*(a+a*sin(d*x+c))^4,x, algorithm="giac")
Output:
1/12*(3*a^4*sin(d*x + c)^4 + 16*a^4*sin(d*x + c)^3 + 36*a^4*sin(d*x + c)^2 + 12*a^4*log(abs(sin(d*x + c))) + 48*a^4*sin(d*x + c))/d
Time = 18.12 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.46 \[ \int \cot (c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {16\,a^4\,\sin \left (c+d\,x\right )}{3\,d}-\frac {a^4\,\ln \left (\frac {1}{{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}\right )}{d}+\frac {a^4\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {7\,a^4\,{\cos \left (c+d\,x\right )}^2}{2\,d}+\frac {a^4\,{\cos \left (c+d\,x\right )}^4}{4\,d}-\frac {4\,a^4\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{3\,d} \] Input:
int(cot(c + d*x)*(a + a*sin(c + d*x))^4,x)
Output:
(16*a^4*sin(c + d*x))/(3*d) - (a^4*log(1/cos(c/2 + (d*x)/2)^2))/d + (a^4*l og(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d - (7*a^4*cos(c + d*x)^2)/(2*d ) + (a^4*cos(c + d*x)^4)/(4*d) - (4*a^4*cos(c + d*x)^2*sin(c + d*x))/(3*d)
Time = 0.16 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.93 \[ \int \cot (c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {a^{4} \left (-12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )+12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 \sin \left (d x +c \right )^{4}+16 \sin \left (d x +c \right )^{3}+36 \sin \left (d x +c \right )^{2}+48 \sin \left (d x +c \right )\right )}{12 d} \] Input:
int(cot(d*x+c)*(a+a*sin(d*x+c))^4,x)
Output:
(a**4*( - 12*log(tan((c + d*x)/2)**2 + 1) + 12*log(tan((c + d*x)/2)) + 3*s in(c + d*x)**4 + 16*sin(c + d*x)**3 + 36*sin(c + d*x)**2 + 48*sin(c + d*x) ))/(12*d)