\(\int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^2} \, dx\) [552]

Optimal result

Integrand size = 27, antiderivative size = 55 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\csc ^3(c+d x)}{3 a^2 d}+\frac {\csc ^4(c+d x)}{2 a^2 d}-\frac {\csc ^5(c+d x)}{5 a^2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2915, 12, 45} \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\csc ^5(c+d x)}{5 a^2 d}+\frac {\csc ^4(c+d x)}{2 a^2 d}-\frac {\csc ^3(c+d x)}{3 a^2 d} \]

[In]

Int[(Cot[c + d*x]^5*Csc[c + d*x])/(a + a*Sin[c + d*x])^2,x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2915

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

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {a^6 (a-x)^2}{x^6} \, dx,x,a \sin (c+d x)\right )}{a^5 d} \\ & = \frac {a \text {Subst}\left (\int \frac {(a-x)^2}{x^6} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a \text {Subst}\left (\int \left (\frac {a^2}{x^6}-\frac {2 a}{x^5}+\frac {1}{x^4}\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = -\frac {\csc ^3(c+d x)}{3 a^2 d}+\frac {\csc ^4(c+d x)}{2 a^2 d}-\frac {\csc ^5(c+d x)}{5 a^2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.69 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\csc ^5(c+d x) (-11+5 \cos (2 (c+d x))+15 \sin (c+d x))}{30 a^2 d} \]

[In]

Integrate[(Cot[c + d*x]^5*Csc[c + d*x])/(a + a*Sin[c + d*x])^2,x]

[Out]

(Csc[c + d*x]^5*(-11 + 5*Cos[2*(c + d*x)] + 15*Sin[c + d*x]))/(30*a^2*d)

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.71

method	result	size
derivativedivides	\(\frac {-\frac {\left (\csc ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\csc ^{4}\left (d x +c \right )\right )}{2}-\frac {\left (\csc ^{3}\left (d x +c \right )\right )}{3}}{d \,a^{2}}\)	\(39\)
default	\(\frac {-\frac {\left (\csc ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\csc ^{4}\left (d x +c \right )\right )}{2}-\frac {\left (\csc ^{3}\left (d x +c \right )\right )}{3}}{d \,a^{2}}\)	\(39\)
parallelrisch	\(\frac {\left (-1408+640 \cos \left (2 d x +2 c \right )-45 \sin \left (5 d x +5 c \right )+1470 \sin \left (d x +c \right )+225 \sin \left (3 d x +3 c \right )\right ) \left (\sec ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\csc ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{122880 d \,a^{2}}\)	\(74\)
risch	\(\frac {8 i \left (5 \,{\mathrm e}^{7 i \left (d x +c \right )}-22 \,{\mathrm e}^{5 i \left (d x +c \right )}-15 i {\mathrm e}^{6 i \left (d x +c \right )}+5 \,{\mathrm e}^{3 i \left (d x +c \right )}+15 i {\mathrm e}^{4 i \left (d x +c \right )}\right )}{15 a^{2} d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}\)	\(81\)
norman	\(\frac {-\frac {1}{160 a d}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{80 d a}+\frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{480 d a}-\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{160 d a}-\frac {\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )}{160 d a}+\frac {\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )}{480 d a}+\frac {\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )}{80 d a}-\frac {\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )}{160 d a}+\frac {5 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}+\frac {5 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}\)	\(207\)

[In]

int(cos(d*x+c)^5*csc(d*x+c)^6/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.18 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {10 \, \cos \left (d x + c\right )^{2} + 15 \, \sin \left (d x + c\right ) - 16}{30 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} - 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d\right )} \sin \left (d x + c\right )} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^6/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/30*(10*cos(d*x + c)^2 + 15*sin(d*x + c) - 16)/((a^2*d*cos(d*x + c)^4 - 2*a^2*d*cos(d*x + c)^2 + a^2*d)*sin(d
*x + c))

Sympy [F(-1)]

Timed out. \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**5*csc(d*x+c)**6/(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.65 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {10 \, \sin \left (d x + c\right )^{2} - 15 \, \sin \left (d x + c\right ) + 6}{30 \, a^{2} d \sin \left (d x + c\right )^{5}} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^6/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/30*(10*sin(d*x + c)^2 - 15*sin(d*x + c) + 6)/(a^2*d*sin(d*x + c)^5)

Giac [A] (verification not implemented)

none

Time = 0.47 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.65 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {10 \, \sin \left (d x + c\right )^{2} - 15 \, \sin \left (d x + c\right ) + 6}{30 \, a^{2} d \sin \left (d x + c\right )^{5}} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^6/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/30*(10*sin(d*x + c)^2 - 15*sin(d*x + c) + 6)/(a^2*d*sin(d*x + c)^5)

Mupad [B] (verification not implemented)

Time = 9.75 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.65 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {{\sin \left (c+d\,x\right )}^2}{3}-\frac {\sin \left (c+d\,x\right )}{2}+\frac {1}{5}}{a^2\,d\,{\sin \left (c+d\,x\right )}^5} \]

[In]

int(cos(c + d*x)^5/(sin(c + d*x)^6*(a + a*sin(c + d*x))^2),x)

[Out]

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