\(\int \frac {\cot ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx\) [551]

Optimal result

Integrand size = 21, antiderivative size = 55 \[ \int \frac {\cot ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\csc ^2(c+d x)}{2 a^2 d}+\frac {2 \csc ^3(c+d x)}{3 a^2 d}-\frac {\csc ^4(c+d x)}{4 a^2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2786, 45} \[ \int \frac {\cot ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\csc ^4(c+d x)}{4 a^2 d}+\frac {2 \csc ^3(c+d x)}{3 a^2 d}-\frac {\csc ^2(c+d x)}{2 a^2 d} \]

[In]

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

[Out]

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

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 2786

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[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]
&& EqQ[a^2 - b^2, 0] && IntegerQ[(p + 1)/2]

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.69 \[ \int \frac {\cot ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\csc ^4(c+d x) (-6+3 \cos (2 (c+d x))+8 \sin (c+d x))}{12 a^2 d} \]

[In]

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

[Out]

(Csc[c + d*x]^4*(-6 + 3*Cos[2*(c + d*x)] + 8*Sin[c + d*x]))/(12*a^2*d)

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.71

method	result	size
derivativedivides	\(\frac {\frac {2}{3 \sin \left (d x +c \right )^{3}}-\frac {1}{4 \sin \left (d x +c \right )^{4}}-\frac {1}{2 \sin \left (d x +c \right )^{2}}}{d \,a^{2}}\)	\(39\)
default	\(\frac {\frac {2}{3 \sin \left (d x +c \right )^{3}}-\frac {1}{4 \sin \left (d x +c \right )^{4}}-\frac {1}{2 \sin \left (d x +c \right )^{2}}}{d \,a^{2}}\)	\(39\)
parallelrisch	\(\frac {\left (60 \cos \left (2 d x +2 c \right )+512 \sin \left (d x +c \right )+33 \cos \left (4 d x +4 c \right )-285\right ) \left (\sec ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\csc ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12288 d \,a^{2}}\)	\(63\)
risch	\(\frac {2 \,{\mathrm e}^{6 i \left (d x +c \right )}-8 \,{\mathrm e}^{4 i \left (d x +c \right )}-\frac {16 i {\mathrm e}^{5 i \left (d x +c \right )}}{3}+2 \,{\mathrm e}^{2 i \left (d x +c \right )}+\frac {16 i {\mathrm e}^{3 i \left (d x +c \right )}}{3}}{a^{2} d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}\)	\(80\)
norman	\(\frac {-\frac {1}{64 a d}+\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{192 d a}+\frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{64 d a}-\frac {5 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d a}-\frac {5 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d a}+\frac {\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )}{64 d a}+\frac {7 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{192 d a}-\frac {\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )}{64 d a}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}\)	\(169\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.04 \[ \int \frac {\cot ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {6 \, \cos \left (d x + c\right )^{2} + 8 \, \sin \left (d x + c\right ) - 9}{12 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} - 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d\right )}} \]

[In]

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

[Out]

1/12*(6*cos(d*x + c)^2 + 8*sin(d*x + c) - 9)/(a^2*d*cos(d*x + c)^4 - 2*a^2*d*cos(d*x + c)^2 + a^2*d)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.65 \[ \int \frac {\cot ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {6 \, \sin \left (d x + c\right )^{2} - 8 \, \sin \left (d x + c\right ) + 3}{12 \, a^{2} d \sin \left (d x + c\right )^{4}} \]

[In]

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

[Out]

-1/12*(6*sin(d*x + c)^2 - 8*sin(d*x + c) + 3)/(a^2*d*sin(d*x + c)^4)

Giac [A] (verification not implemented)

none

Time = 0.45 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.65 \[ \int \frac {\cot ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {6 \, \sin \left (d x + c\right )^{2} - 8 \, \sin \left (d x + c\right ) + 3}{12 \, a^{2} d \sin \left (d x + c\right )^{4}} \]

[In]

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

[Out]

-1/12*(6*sin(d*x + c)^2 - 8*sin(d*x + c) + 3)/(a^2*d*sin(d*x + c)^4)

Mupad [B] (verification not implemented)

Time = 10.65 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.65 \[ \int \frac {\cot ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {{\sin \left (c+d\,x\right )}^2}{2}-\frac {2\,\sin \left (c+d\,x\right )}{3}+\frac {1}{4}}{a^2\,d\,{\sin \left (c+d\,x\right )}^4} \]

[In]

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

[Out]

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