\(\int \frac {\cot ^5(c+d x)}{a+a \sin (c+d x)} \, dx\) [540]

Optimal result

Integrand size = 21, antiderivative size = 51 \[ \int \frac {\cot ^5(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\cot ^4(c+d x)}{4 a d}-\frac {\csc (c+d x)}{a d}+\frac {\csc ^3(c+d x)}{3 a d} \]

[Out]

-1/4*cot(d*x+c)^4/a/d-csc(d*x+c)/a/d+1/3*csc(d*x+c)^3/a/d

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2785, 2687, 30, 2686} \[ \int \frac {\cot ^5(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\cot ^4(c+d x)}{4 a d}+\frac {\csc ^3(c+d x)}{3 a d}-\frac {\csc (c+d x)}{a d} \]

[In]

Int[Cot[c + d*x]^5/(a + a*Sin[c + d*x]),x]

[Out]

-1/4*Cot[c + d*x]^4/(a*d) - Csc[c + d*x]/(a*d) + Csc[c + d*x]^3/(3*a*d)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2686

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 2687

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 2785

Int[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/a, Int[S
ec[e + f*x]^2*(g*Tan[e + f*x])^p, x], x] - Dist[1/(b*g), Int[Sec[e + f*x]*(g*Tan[e + f*x])^(p + 1), x], x] /;
FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {\int \cot ^3(c+d x) \csc (c+d x) \, dx}{a}+\frac {\int \cot ^3(c+d x) \csc ^2(c+d x) \, dx}{a} \\ & = -\frac {\text {Subst}\left (\int x^3 \, dx,x,-\cot (c+d x)\right )}{a d}+\frac {\text {Subst}\left (\int \left (-1+x^2\right ) \, dx,x,\csc (c+d x)\right )}{a d} \\ & = -\frac {\cot ^4(c+d x)}{4 a d}-\frac {\csc (c+d x)}{a d}+\frac {\csc ^3(c+d x)}{3 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.59 \[ \int \frac {\cot ^5(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {(-1+\csc (c+d x))^3 (5+3 \csc (c+d x))}{12 a d} \]

[In]

Integrate[Cot[c + d*x]^5/(a + a*Sin[c + d*x]),x]

[Out]

-1/12*((-1 + Csc[c + d*x])^3*(5 + 3*Csc[c + d*x]))/(a*d)

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.96

method	result	size
derivativedivides	\(\frac {\frac {1}{3 \sin \left (d x +c \right )^{3}}+\frac {1}{2 \sin \left (d x +c \right )^{2}}-\frac {1}{4 \sin \left (d x +c \right )^{4}}-\frac {1}{\sin \left (d x +c \right )}}{d a}\)	\(49\)
default	\(\frac {\frac {1}{3 \sin \left (d x +c \right )^{3}}+\frac {1}{2 \sin \left (d x +c \right )^{2}}-\frac {1}{4 \sin \left (d x +c \right )^{4}}-\frac {1}{\sin \left (d x +c \right )}}{d a}\)	\(49\)
risch	\(-\frac {2 i \left (-3 i {\mathrm e}^{6 i \left (d x +c \right )}+3 \,{\mathrm e}^{7 i \left (d x +c \right )}-5 \,{\mathrm e}^{5 i \left (d x +c \right )}-3 i {\mathrm e}^{2 i \left (d x +c \right )}+5 \,{\mathrm e}^{3 i \left (d x +c \right )}-3 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{3 a d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}\)	\(92\)
parallelrisch	\(\frac {-3 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 \left (\cot ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \left (\cot ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+12 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+12 \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-72 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-72 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )}{192 d a}\)	\(110\)
norman	\(\frac {-\frac {1}{64 a d}+\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{192 d a}+\frac {5 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{48 d a}-\frac {5 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d a}-\frac {5 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d a}+\frac {5 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{48 d a}+\frac {5 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{192 d a}-\frac {\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )}{64 d a}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}\)	\(166\)

[In]

int(cos(d*x+c)^5*csc(d*x+c)^5/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d/a*(1/3/sin(d*x+c)^3+1/2/sin(d*x+c)^2-1/4/sin(d*x+c)^4-1/sin(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.24 \[ \int \frac {\cot ^5(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {6 \, \cos \left (d x + c\right )^{2} - 4 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 2\right )} \sin \left (d x + c\right ) - 3}{12 \, {\left (a d \cos \left (d x + c\right )^{4} - 2 \, a d \cos \left (d x + c\right )^{2} + a d\right )}} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/12*(6*cos(d*x + c)^2 - 4*(3*cos(d*x + c)^2 - 2)*sin(d*x + c) - 3)/(a*d*cos(d*x + c)^4 - 2*a*d*cos(d*x + c)^
2 + a*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {\cot ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**5*csc(d*x+c)**5/(a+a*sin(d*x+c)),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.90 \[ \int \frac {\cot ^5(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {12 \, \sin \left (d x + c\right )^{3} - 6 \, \sin \left (d x + c\right )^{2} - 4 \, \sin \left (d x + c\right ) + 3}{12 \, a d \sin \left (d x + c\right )^{4}} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/12*(12*sin(d*x + c)^3 - 6*sin(d*x + c)^2 - 4*sin(d*x + c) + 3)/(a*d*sin(d*x + c)^4)

Giac [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.90 \[ \int \frac {\cot ^5(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {12 \, \sin \left (d x + c\right )^{3} - 6 \, \sin \left (d x + c\right )^{2} - 4 \, \sin \left (d x + c\right ) + 3}{12 \, a d \sin \left (d x + c\right )^{4}} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/12*(12*sin(d*x + c)^3 - 6*sin(d*x + c)^2 - 4*sin(d*x + c) + 3)/(a*d*sin(d*x + c)^4)

Mupad [B] (verification not implemented)

Time = 9.77 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.88 \[ \int \frac {\cot ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {-{\sin \left (c+d\,x\right )}^3+\frac {{\sin \left (c+d\,x\right )}^2}{2}+\frac {\sin \left (c+d\,x\right )}{3}-\frac {1}{4}}{a\,d\,{\sin \left (c+d\,x\right )}^4} \]

[In]

int(cos(c + d*x)^5/(sin(c + d*x)^5*(a + a*sin(c + d*x))),x)

[Out]

(sin(c + d*x)/3 + sin(c + d*x)^2/2 - sin(c + d*x)^3 - 1/4)/(a*d*sin(c + d*x)^4)