\(\int \frac {\cot ^3(c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx\) [364]

Optimal result

Integrand size = 29, antiderivative size = 37 \[ \int \frac {\cot ^3(c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\csc ^3(c+d x)}{3 a d}-\frac {\csc ^4(c+d x)}{4 a d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2915, 12, 45} \[ \int \frac {\cot ^3(c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\csc ^3(c+d x)}{3 a d}-\frac {\csc ^4(c+d x)}{4 a d} \]

[In]

Int[(Cot[c + d*x]^3*Csc[c + d*x]^2)/(a + a*Sin[c + d*x]),x]

[Out]

Csc[c + d*x]^3/(3*a*d) - Csc[c + d*x]^4/(4*a*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^5 (a-x)}{x^5} \, dx,x,a \sin (c+d x)\right )}{a^3 d} \\ & = \frac {a^2 \text {Subst}\left (\int \frac {a-x}{x^5} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^2 \text {Subst}\left (\int \left (\frac {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 d}-\frac {\csc ^4(c+d x)}{4 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.76 \[ \int \frac {\cot ^3(c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\csc ^4(c+d x) (-3+4 \sin (c+d x))}{12 a d} \]

[In]

Integrate[(Cot[c + d*x]^3*Csc[c + d*x]^2)/(a + a*Sin[c + d*x]),x]

[Out]

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

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.81

method	result	size
derivativedivides	\(-\frac {\frac {\left (\csc ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\csc ^{3}\left (d x +c \right )\right )}{3}}{d a}\)	\(30\)
default	\(-\frac {\frac {\left (\csc ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\csc ^{3}\left (d x +c \right )\right )}{3}}{d a}\)	\(30\)
risch	\(-\frac {4 i \left (-3 i {\mathrm e}^{4 i \left (d x +c \right )}+2 \,{\mathrm e}^{5 i \left (d x +c \right )}-2 \,{\mathrm e}^{3 i \left (d x +c \right )}\right )}{3 a d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}\)	\(58\)
parallelrisch	\(-\frac {\left (36 \cos \left (2 d x +2 c \right )-256 \sin \left (d x +c \right )-9 \cos \left (4 d x +4 c \right )+165\right ) \left (\sec ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\csc ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12288 d a}\)	\(63\)
norman	\(\frac {-\frac {1}{64 a d}+\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{192 d a}-\frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{48 d a}+\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{16 d a}+\frac {\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )}{16 d a}-\frac {\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )}{48 d a}+\frac {5 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{192 d a}-\frac {\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )}{64 d a}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}\)	\(166\)

[In]

int(cos(d*x+c)^3*csc(d*x+c)^5/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.11 \[ \int \frac {\cot ^3(c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {4 \, \sin \left (d x + c\right ) - 3}{12 \, {\left (a d \cos \left (d x + c\right )^{4} - 2 \, a d \cos \left (d x + c\right )^{2} + a d\right )}} \]

[In]

integrate(cos(d*x+c)^3*csc(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/12*(4*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. \[ \int \frac {\cot ^3(c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**3*csc(d*x+c)**5/(a+a*sin(d*x+c)),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.70 \[ \int \frac {\cot ^3(c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {4 \, \sin \left (d x + c\right ) - 3}{12 \, a d \sin \left (d x + c\right )^{4}} \]

[In]

integrate(cos(d*x+c)^3*csc(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.46 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.70 \[ \int \frac {\cot ^3(c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {4 \, \sin \left (d x + c\right ) - 3}{12 \, a d \sin \left (d x + c\right )^{4}} \]

[In]

integrate(cos(d*x+c)^3*csc(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.68 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.68 \[ \int \frac {\cot ^3(c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {\sin \left (c+d\,x\right )}{3}-\frac {1}{4}}{a\,d\,{\sin \left (c+d\,x\right )}^4} \]

[In]

int(cos(c + d*x)^3/(sin(c + d*x)^5*(a + a*sin(c + d*x))),x)

[Out]

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