\(\int \frac {\cot ^5(c+d x) \csc (c+d x)}{a+a \sin (c+d x)} \, dx\) [541]

Optimal result

Integrand size = 27, antiderivative size = 55 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\cot ^4(c+d x)}{4 a d}+\frac {\csc ^3(c+d x)}{3 a d}-\frac {\csc ^5(c+d x)}{5 a d} \]

[Out]

1/4*cot(d*x+c)^4/a/d+1/3*csc(d*x+c)^3/a/d-1/5*csc(d*x+c)^5/a/d

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2914, 2686, 14, 2687, 30} \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\cot ^4(c+d x)}{4 a d}-\frac {\csc ^5(c+d x)}{5 a d}+\frac {\csc ^3(c+d x)}{3 a d} \]

[In]

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

[Out]

Cot[c + d*x]^4/(4*a*d) + Csc[c + d*x]^3/(3*a*d) - Csc[c + d*x]^5/(5*a*d)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 2914

Int[(cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]
), x_Symbol] :> Dist[1/a, Int[Cos[e + f*x]^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[1/(b*d), Int[Cos[e + f*x]
^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2
- b^2, 0] && IntegerQ[n] && (LtQ[0, n, (p + 1)/2] || (LeQ[p, -n] && LtQ[-n, 2*p - 3]) || (GtQ[n, 0] && LeQ[n,
-p]))

Rubi steps \begin{align*} \text {integral}& = -\frac {\int \cot ^3(c+d x) \csc ^2(c+d x) \, dx}{a}+\frac {\int \cot ^3(c+d x) \csc ^3(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 x^2 \left (-1+x^2\right ) \, dx,x,\csc (c+d x)\right )}{a d} \\ & = \frac {\cot ^4(c+d x)}{4 a d}-\frac {\text {Subst}\left (\int \left (-x^2+x^4\right ) \, dx,x,\csc (c+d x)\right )}{a d} \\ & = \frac {\cot ^4(c+d x)}{4 a d}+\frac {\csc ^3(c+d x)}{3 a d}-\frac {\csc ^5(c+d x)}{5 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.87 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\csc ^2(c+d x) \left (-30+20 \csc (c+d x)+15 \csc ^2(c+d x)-12 \csc ^3(c+d x)\right )}{60 a d} \]

[In]

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

[Out]

(Csc[c + d*x]^2*(-30 + 20*Csc[c + d*x] + 15*Csc[c + d*x]^2 - 12*Csc[c + d*x]^3))/(60*a*d)

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.89

method	result	size
derivativedivides	\(\frac {-\frac {\left (\csc ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\csc ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\csc ^{3}\left (d x +c \right )\right )}{3}-\frac {\left (\csc ^{2}\left (d x +c \right )\right )}{2}}{d a}\)	\(49\)
default	\(\frac {-\frac {\left (\csc ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\csc ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\csc ^{3}\left (d x +c \right )\right )}{3}-\frac {\left (\csc ^{2}\left (d x +c \right )\right )}{2}}{d a}\)	\(49\)
parallelrisch	\(-\frac {\left (256+1280 \cos \left (2 d x +2 c \right )-75 \sin \left (5 d x +5 c \right )+210 \sin \left (d x +c \right )-585 \sin \left (3 d x +3 c \right )\right ) \left (\sec ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\csc ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{245760 d a}\)	\(74\)
risch	\(\frac {-\frac {8 i {\mathrm e}^{7 i \left (d x +c \right )}}{3}+2 \,{\mathrm e}^{8 i \left (d x +c \right )}-\frac {16 i {\mathrm e}^{5 i \left (d x +c \right )}}{15}-2 \,{\mathrm e}^{6 i \left (d x +c \right )}-\frac {8 i {\mathrm e}^{3 i \left (d x +c \right )}}{3}+2 \,{\mathrm e}^{4 i \left (d x +c \right )}-2 \,{\mathrm e}^{2 i \left (d x +c \right )}}{a d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}\)	\(103\)
norman	\(\frac {-\frac {1}{160 a d}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{320 d a}+\frac {5 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{192 d a}-\frac {5 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{96 d a}-\frac {5 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{96 d a}+\frac {5 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{192 d a}+\frac {3 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{320 d a}-\frac {\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )}{160 d a}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}\)	\(166\)

[In]

int(cos(d*x+c)^5*csc(d*x+c)^6/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d/a*(-1/5*csc(d*x+c)^5+1/4*csc(d*x+c)^4+1/3*csc(d*x+c)^3-1/2*csc(d*x+c)^2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.29 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {20 \, \cos \left (d x + c\right )^{2} - 15 \, {\left (2 \, \cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - 8}{60 \, {\left (a d \cos \left (d x + c\right )^{4} - 2 \, a d \cos \left (d x + c\right )^{2} + a d\right )} \sin \left (d x + c\right )} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^6/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/60*(20*cos(d*x + c)^2 - 15*(2*cos(d*x + c)^2 - 1)*sin(d*x + c) - 8)/((a*d*cos(d*x + c)^4 - 2*a*d*cos(d*x +
c)^2 + a*d)*sin(d*x + c))

Sympy [F(-1)]

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

[In]

integrate(cos(d*x+c)**5*csc(d*x+c)**6/(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 = 0.84 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {30 \, \sin \left (d x + c\right )^{3} - 20 \, \sin \left (d x + c\right )^{2} - 15 \, \sin \left (d x + c\right ) + 12}{60 \, a d \sin \left (d x + c\right )^{5}} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^6/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/60*(30*sin(d*x + c)^3 - 20*sin(d*x + c)^2 - 15*sin(d*x + c) + 12)/(a*d*sin(d*x + c)^5)

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.84 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {30 \, \sin \left (d x + c\right )^{3} - 20 \, \sin \left (d x + c\right )^{2} - 15 \, \sin \left (d x + c\right ) + 12}{60 \, a d \sin \left (d x + c\right )^{5}} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^6/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/60*(30*sin(d*x + c)^3 - 20*sin(d*x + c)^2 - 15*sin(d*x + c) + 12)/(a*d*sin(d*x + c)^5)

Mupad [B] (verification not implemented)

Time = 9.93 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.84 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {-30\,{\sin \left (c+d\,x\right )}^3+20\,{\sin \left (c+d\,x\right )}^2+15\,\sin \left (c+d\,x\right )-12}{60\,a\,d\,{\sin \left (c+d\,x\right )}^5} \]

[In]

int(cos(c + d*x)^5/(sin(c + d*x)^6*(a + a*sin(c + d*x))),x)

[Out]

(15*sin(c + d*x) + 20*sin(c + d*x)^2 - 30*sin(c + d*x)^3 - 12)/(60*a*d*sin(c + d*x)^5)