\(\int \frac {\cot ^5(c+d x) \csc ^3(c+d x)}{a+a \sin (c+d x)} \, dx\) [543]

Optimal result

Integrand size = 29, antiderivative size = 73 \[ \int \frac {\cot ^5(c+d x) \csc ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\csc ^4(c+d x)}{4 a d}+\frac {\csc ^5(c+d x)}{5 a d}+\frac {\csc ^6(c+d x)}{6 a d}-\frac {\csc ^7(c+d x)}{7 a d} \]

[Out]

-1/4*csc(d*x+c)^4/a/d+1/5*csc(d*x+c)^5/a/d+1/6*csc(d*x+c)^6/a/d-1/7*csc(d*x+c)^7/a/d

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 73, 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, 76} \[ \int \frac {\cot ^5(c+d x) \csc ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\csc ^7(c+d x)}{7 a d}+\frac {\csc ^6(c+d x)}{6 a d}+\frac {\csc ^5(c+d x)}{5 a d}-\frac {\csc ^4(c+d x)}{4 a d} \]

[In]

Int[(Cot[c + d*x]^5*Csc[c + d*x]^3)/(a + a*Sin[c + d*x]),x]

[Out]

-1/4*Csc[c + d*x]^4/(a*d) + Csc[c + d*x]^5/(5*a*d) + Csc[c + d*x]^6/(6*a*d) - Csc[c + d*x]^7/(7*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 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n
+ p + 2, 0] && GtQ[n + 2*p, 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^8 (a-x)^2 (a+x)}{x^8} \, dx,x,a \sin (c+d x)\right )}{a^5 d} \\ & = \frac {a^3 \text {Subst}\left (\int \frac {(a-x)^2 (a+x)}{x^8} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^3 \text {Subst}\left (\int \left (\frac {a^3}{x^8}-\frac {a^2}{x^7}-\frac {a}{x^6}+\frac {1}{x^5}\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = -\frac {\csc ^4(c+d x)}{4 a d}+\frac {\csc ^5(c+d x)}{5 a d}+\frac {\csc ^6(c+d x)}{6 a d}-\frac {\csc ^7(c+d x)}{7 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.66 \[ \int \frac {\cot ^5(c+d x) \csc ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\csc ^4(c+d x) \left (-105+84 \csc (c+d x)+70 \csc ^2(c+d x)-60 \csc ^3(c+d x)\right )}{420 a d} \]

[In]

Integrate[(Cot[c + d*x]^5*Csc[c + d*x]^3)/(a + a*Sin[c + d*x]),x]

[Out]

(Csc[c + d*x]^4*(-105 + 84*Csc[c + d*x] + 70*Csc[c + d*x]^2 - 60*Csc[c + d*x]^3))/(420*a*d)

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.68

method	result	size
derivativedivides	\(-\frac {\frac {\left (\csc ^{7}\left (d x +c \right )\right )}{7}-\frac {\left (\csc ^{6}\left (d x +c \right )\right )}{6}-\frac {\left (\csc ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\csc ^{4}\left (d x +c \right )\right )}{4}}{d a}\)	\(50\)
default	\(-\frac {\frac {\left (\csc ^{7}\left (d x +c \right )\right )}{7}-\frac {\left (\csc ^{6}\left (d x +c \right )\right )}{6}-\frac {\left (\csc ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\csc ^{4}\left (d x +c \right )\right )}{4}}{d a}\)	\(50\)
parallelrisch	\(-\frac {\left (2304+5376 \cos \left (2 d x +2 c \right )+35 \sin \left (7 d x +7 c \right )-245 \sin \left (5 d x +5 c \right )-105 \sin \left (d x +c \right )-2625 \sin \left (3 d x +3 c \right )\right ) \left (\sec ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\csc ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6881280 d a}\)	\(85\)
risch	\(-\frac {4 \left (-168 i {\mathrm e}^{9 i \left (d x +c \right )}+105 \,{\mathrm e}^{10 i \left (d x +c \right )}-144 i {\mathrm e}^{7 i \left (d x +c \right )}-35 \,{\mathrm e}^{8 i \left (d x +c \right )}-168 i {\mathrm e}^{5 i \left (d x +c \right )}+35 \,{\mathrm e}^{6 i \left (d x +c \right )}-105 \,{\mathrm e}^{4 i \left (d x +c \right )}\right )}{105 a d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{7}}\)	\(103\)
norman	\(\frac {-\frac {1}{896 a d}+\frac {\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )}{672 d a}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{672 d a}+\frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{960 d a}-\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{640 d a}+\frac {\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )}{128 d a}-\frac {\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )}{64 d a}-\frac {\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )}{64 d a}+\frac {\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )}{128 d a}-\frac {\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )}{640 d a}+\frac {\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )}{960 d a}-\frac {\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )}{896 d a}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}\)	\(242\)

[In]

int(cos(d*x+c)^5*csc(d*x+c)^8/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

-1/d/a*(1/7*csc(d*x+c)^7-1/6*csc(d*x+c)^6-1/5*csc(d*x+c)^5+1/4*csc(d*x+c)^4)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.15 \[ \int \frac {\cot ^5(c+d x) \csc ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {84 \, \cos \left (d x + c\right )^{2} - 35 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - 24}{420 \, {\left (a d \cos \left (d x + c\right )^{6} - 3 \, a d \cos \left (d x + c\right )^{4} + 3 \, a d \cos \left (d x + c\right )^{2} - a d\right )} \sin \left (d x + c\right )} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^8/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/420*(84*cos(d*x + c)^2 - 35*(3*cos(d*x + c)^2 - 1)*sin(d*x + c) - 24)/((a*d*cos(d*x + c)^6 - 3*a*d*cos(d*x +
c)^4 + 3*a*d*cos(d*x + c)^2 - a*d)*sin(d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \frac {\cot ^5(c+d x) \csc ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**5*csc(d*x+c)**8/(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 = 0.63 \[ \int \frac {\cot ^5(c+d x) \csc ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {105 \, \sin \left (d x + c\right )^{3} - 84 \, \sin \left (d x + c\right )^{2} - 70 \, \sin \left (d x + c\right ) + 60}{420 \, a d \sin \left (d x + c\right )^{7}} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^8/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/420*(105*sin(d*x + c)^3 - 84*sin(d*x + c)^2 - 70*sin(d*x + c) + 60)/(a*d*sin(d*x + c)^7)

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.63 \[ \int \frac {\cot ^5(c+d x) \csc ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {105 \, \sin \left (d x + c\right )^{3} - 84 \, \sin \left (d x + c\right )^{2} - 70 \, \sin \left (d x + c\right ) + 60}{420 \, a d \sin \left (d x + c\right )^{7}} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^8/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/420*(105*sin(d*x + c)^3 - 84*sin(d*x + c)^2 - 70*sin(d*x + c) + 60)/(a*d*sin(d*x + c)^7)

Mupad [B] (verification not implemented)

Time = 9.71 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.63 \[ \int \frac {\cot ^5(c+d x) \csc ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {-105\,{\sin \left (c+d\,x\right )}^3+84\,{\sin \left (c+d\,x\right )}^2+70\,\sin \left (c+d\,x\right )-60}{420\,a\,d\,{\sin \left (c+d\,x\right )}^7} \]

[In]

int(cos(c + d*x)^5/(sin(c + d*x)^8*(a + a*sin(c + d*x))),x)

[Out]

(70*sin(c + d*x) + 84*sin(c + d*x)^2 - 105*sin(c + d*x)^3 - 60)/(420*a*d*sin(c + d*x)^7)