\(\int \frac {\cot ^7(c+d x) \csc ^4(c+d x)}{a+a \sin (c+d x)} \, dx\) [694]

Optimal result

Integrand size = 29, antiderivative size = 109 \[ \int \frac {\cot ^7(c+d x) \csc ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\csc ^5(c+d x)}{5 a d}-\frac {\csc ^6(c+d x)}{6 a d}-\frac {2 \csc ^7(c+d x)}{7 a d}+\frac {\csc ^8(c+d x)}{4 a d}+\frac {\csc ^9(c+d x)}{9 a d}-\frac {\csc ^{10}(c+d x)}{10 a d} \]

[Out]

1/5*csc(d*x+c)^5/a/d-1/6*csc(d*x+c)^6/a/d-2/7*csc(d*x+c)^7/a/d+1/4*csc(d*x+c)^8/a/d+1/9*csc(d*x+c)^9/a/d-1/10*
csc(d*x+c)^10/a/d

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 109, 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, 90} \[ \int \frac {\cot ^7(c+d x) \csc ^4(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\csc ^{10}(c+d x)}{10 a d}+\frac {\csc ^9(c+d x)}{9 a d}+\frac {\csc ^8(c+d x)}{4 a d}-\frac {2 \csc ^7(c+d x)}{7 a d}-\frac {\csc ^6(c+d x)}{6 a d}+\frac {\csc ^5(c+d x)}{5 a d} \]

[In]

Int[(Cot[c + d*x]^7*Csc[c + d*x]^4)/(a + a*Sin[c + d*x]),x]

[Out]

Csc[c + d*x]^5/(5*a*d) - Csc[c + d*x]^6/(6*a*d) - (2*Csc[c + d*x]^7)/(7*a*d) + Csc[c + d*x]^8/(4*a*d) + Csc[c
+ d*x]^9/(9*a*d) - Csc[c + d*x]^10/(10*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 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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^{11} (a-x)^3 (a+x)^2}{x^{11}} \, dx,x,a \sin (c+d x)\right )}{a^7 d} \\ & = \frac {a^4 \text {Subst}\left (\int \frac {(a-x)^3 (a+x)^2}{x^{11}} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^4 \text {Subst}\left (\int \left (\frac {a^5}{x^{11}}-\frac {a^4}{x^{10}}-\frac {2 a^3}{x^9}+\frac {2 a^2}{x^8}+\frac {a}{x^7}-\frac {1}{x^6}\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {\csc ^5(c+d x)}{5 a d}-\frac {\csc ^6(c+d x)}{6 a d}-\frac {2 \csc ^7(c+d x)}{7 a d}+\frac {\csc ^8(c+d x)}{4 a d}+\frac {\csc ^9(c+d x)}{9 a d}-\frac {\csc ^{10}(c+d x)}{10 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.62 \[ \int \frac {\cot ^7(c+d x) \csc ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\csc ^5(c+d x) \left (252-210 \csc (c+d x)-360 \csc ^2(c+d x)+315 \csc ^3(c+d x)+140 \csc ^4(c+d x)-126 \csc ^5(c+d x)\right )}{1260 a d} \]

[In]

Integrate[(Cot[c + d*x]^7*Csc[c + d*x]^4)/(a + a*Sin[c + d*x]),x]

[Out]

(Csc[c + d*x]^5*(252 - 210*Csc[c + d*x] - 360*Csc[c + d*x]^2 + 315*Csc[c + d*x]^3 + 140*Csc[c + d*x]^4 - 126*C
sc[c + d*x]^5))/(1260*a*d)

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.64

method	result	size
derivativedivides	\(-\frac {\frac {\left (\csc ^{10}\left (d x +c \right )\right )}{10}-\frac {\left (\csc ^{9}\left (d x +c \right )\right )}{9}-\frac {\left (\csc ^{8}\left (d x +c \right )\right )}{4}+\frac {2 \left (\csc ^{7}\left (d x +c \right )\right )}{7}+\frac {\left (\csc ^{6}\left (d x +c \right )\right )}{6}-\frac {\left (\csc ^{5}\left (d x +c \right )\right )}{5}}{d a}\)	\(70\)
default	\(-\frac {\frac {\left (\csc ^{10}\left (d x +c \right )\right )}{10}-\frac {\left (\csc ^{9}\left (d x +c \right )\right )}{9}-\frac {\left (\csc ^{8}\left (d x +c \right )\right )}{4}+\frac {2 \left (\csc ^{7}\left (d x +c \right )\right )}{7}+\frac {\left (\csc ^{6}\left (d x +c \right )\right )}{6}-\frac {\left (\csc ^{5}\left (d x +c \right )\right )}{5}}{d a}\)	\(70\)
risch	\(\frac {32 i \left (-105 i {\mathrm e}^{14 i \left (d x +c \right )}+63 \,{\mathrm e}^{15 i \left (d x +c \right )}-210 i {\mathrm e}^{12 i \left (d x +c \right )}+45 \,{\mathrm e}^{13 i \left (d x +c \right )}-378 i {\mathrm e}^{10 i \left (d x +c \right )}+110 \,{\mathrm e}^{11 i \left (d x +c \right )}-210 i {\mathrm e}^{8 i \left (d x +c \right )}-110 \,{\mathrm e}^{9 i \left (d x +c \right )}-105 i {\mathrm e}^{6 i \left (d x +c \right )}-45 \,{\mathrm e}^{7 i \left (d x +c \right )}-63 \,{\mathrm e}^{5 i \left (d x +c \right )}\right )}{315 d a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{10}}\)	\(150\)
parallelrisch	\(\frac {-63 \left (\tan ^{20}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+140 \left (\tan ^{19}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-180 \left (\tan ^{17}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+525 \left (\tan ^{16}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1008 \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-63+1680 \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3150 \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+7560 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+7560 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3150 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1680 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1008 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+525 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-180 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+140 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{645120 d a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}\)	\(215\)

[In]

int(cos(d*x+c)^7*csc(d*x+c)^11/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

-1/d/a*(1/10*csc(d*x+c)^10-1/9*csc(d*x+c)^9-1/4*csc(d*x+c)^8+2/7*csc(d*x+c)^7+1/6*csc(d*x+c)^6-1/5*csc(d*x+c)^
5)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.10 \[ \int \frac {\cot ^7(c+d x) \csc ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {210 \, \cos \left (d x + c\right )^{4} - 105 \, \cos \left (d x + c\right )^{2} - 4 \, {\left (63 \, \cos \left (d x + c\right )^{4} - 36 \, \cos \left (d x + c\right )^{2} + 8\right )} \sin \left (d x + c\right ) + 21}{1260 \, {\left (a d \cos \left (d x + c\right )^{10} - 5 \, a d \cos \left (d x + c\right )^{8} + 10 \, a d \cos \left (d x + c\right )^{6} - 10 \, a d \cos \left (d x + c\right )^{4} + 5 \, a d \cos \left (d x + c\right )^{2} - a d\right )}} \]

[In]

integrate(cos(d*x+c)^7*csc(d*x+c)^11/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/1260*(210*cos(d*x + c)^4 - 105*cos(d*x + c)^2 - 4*(63*cos(d*x + c)^4 - 36*cos(d*x + c)^2 + 8)*sin(d*x + c) +
21)/(a*d*cos(d*x + c)^10 - 5*a*d*cos(d*x + c)^8 + 10*a*d*cos(d*x + c)^6 - 10*a*d*cos(d*x + c)^4 + 5*a*d*cos(d
*x + c)^2 - a*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {\cot ^7(c+d x) \csc ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**7*csc(d*x+c)**11/(a+a*sin(d*x+c)),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.61 \[ \int \frac {\cot ^7(c+d x) \csc ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {252 \, \sin \left (d x + c\right )^{5} - 210 \, \sin \left (d x + c\right )^{4} - 360 \, \sin \left (d x + c\right )^{3} + 315 \, \sin \left (d x + c\right )^{2} + 140 \, \sin \left (d x + c\right ) - 126}{1260 \, a d \sin \left (d x + c\right )^{10}} \]

[In]

integrate(cos(d*x+c)^7*csc(d*x+c)^11/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/1260*(252*sin(d*x + c)^5 - 210*sin(d*x + c)^4 - 360*sin(d*x + c)^3 + 315*sin(d*x + c)^2 + 140*sin(d*x + c) -
126)/(a*d*sin(d*x + c)^10)

Giac [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.61 \[ \int \frac {\cot ^7(c+d x) \csc ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {252 \, \sin \left (d x + c\right )^{5} - 210 \, \sin \left (d x + c\right )^{4} - 360 \, \sin \left (d x + c\right )^{3} + 315 \, \sin \left (d x + c\right )^{2} + 140 \, \sin \left (d x + c\right ) - 126}{1260 \, a d \sin \left (d x + c\right )^{10}} \]

[In]

integrate(cos(d*x+c)^7*csc(d*x+c)^11/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/1260*(252*sin(d*x + c)^5 - 210*sin(d*x + c)^4 - 360*sin(d*x + c)^3 + 315*sin(d*x + c)^2 + 140*sin(d*x + c) -
126)/(a*d*sin(d*x + c)^10)

Mupad [B] (verification not implemented)

Time = 10.36 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.61 \[ \int \frac {\cot ^7(c+d x) \csc ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {252\,{\sin \left (c+d\,x\right )}^5-210\,{\sin \left (c+d\,x\right )}^4-360\,{\sin \left (c+d\,x\right )}^3+315\,{\sin \left (c+d\,x\right )}^2+140\,\sin \left (c+d\,x\right )-126}{1260\,a\,d\,{\sin \left (c+d\,x\right )}^{10}} \]

[In]

int(cos(c + d*x)^7/(sin(c + d*x)^11*(a + a*sin(c + d*x))),x)

[Out]

(140*sin(c + d*x) + 315*sin(c + d*x)^2 - 360*sin(c + d*x)^3 - 210*sin(c + d*x)^4 + 252*sin(c + d*x)^5 - 126)/(
1260*a*d*sin(c + d*x)^10)