\(\int \cot ^5(c+d x) \csc ^4(c+d x) (a+a \sin (c+d x)) \, dx\) [507]

Optimal result

Integrand size = 27, antiderivative size = 81 \[ \int \cot ^5(c+d x) \csc ^4(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \cot ^6(c+d x)}{6 d}-\frac {a \cot ^8(c+d x)}{8 d}-\frac {a \csc ^3(c+d x)}{3 d}+\frac {2 a \csc ^5(c+d x)}{5 d}-\frac {a \csc ^7(c+d x)}{7 d} \]

[Out]

-1/6*a*cot(d*x+c)^6/d-1/8*a*cot(d*x+c)^8/d-1/3*a*csc(d*x+c)^3/d+2/5*a*csc(d*x+c)^5/d-1/7*a*csc(d*x+c)^7/d

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2913, 2687, 14, 2686, 276} \[ \int \cot ^5(c+d x) \csc ^4(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \cot ^8(c+d x)}{8 d}-\frac {a \cot ^6(c+d x)}{6 d}-\frac {a \csc ^7(c+d x)}{7 d}+\frac {2 a \csc ^5(c+d x)}{5 d}-\frac {a \csc ^3(c+d x)}{3 d} \]

[In]

Int[Cot[c + d*x]^5*Csc[c + d*x]^4*(a + a*Sin[c + d*x]),x]

[Out]

-1/6*(a*Cot[c + d*x]^6)/d - (a*Cot[c + d*x]^8)/(8*d) - (a*Csc[c + d*x]^3)/(3*d) + (2*a*Csc[c + d*x]^5)/(5*d) -
(a*Csc[c + d*x]^7)/(7*d)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 2686

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 2687

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 2913

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

Rubi steps \begin{align*} \text {integral}& = a \int \cot ^5(c+d x) \csc ^3(c+d x) \, dx+a \int \cot ^5(c+d x) \csc ^4(c+d x) \, dx \\ & = -\frac {a \text {Subst}\left (\int x^2 \left (-1+x^2\right )^2 \, dx,x,\csc (c+d x)\right )}{d}-\frac {a \text {Subst}\left (\int x^5 \left (1+x^2\right ) \, dx,x,-\cot (c+d x)\right )}{d} \\ & = -\frac {a \text {Subst}\left (\int \left (x^2-2 x^4+x^6\right ) \, dx,x,\csc (c+d x)\right )}{d}-\frac {a \text {Subst}\left (\int \left (x^5+x^7\right ) \, dx,x,-\cot (c+d x)\right )}{d} \\ & = -\frac {a \cot ^6(c+d x)}{6 d}-\frac {a \cot ^8(c+d x)}{8 d}-\frac {a \csc ^3(c+d x)}{3 d}+\frac {2 a \csc ^5(c+d x)}{5 d}-\frac {a \csc ^7(c+d x)}{7 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.20 \[ \int \cot ^5(c+d x) \csc ^4(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \csc ^3(c+d x)}{3 d}-\frac {a \csc ^4(c+d x)}{4 d}+\frac {2 a \csc ^5(c+d x)}{5 d}+\frac {a \csc ^6(c+d x)}{3 d}-\frac {a \csc ^7(c+d x)}{7 d}-\frac {a \csc ^8(c+d x)}{8 d} \]

[In]

Integrate[Cot[c + d*x]^5*Csc[c + d*x]^4*(a + a*Sin[c + d*x]),x]

[Out]

-1/3*(a*Csc[c + d*x]^3)/d - (a*Csc[c + d*x]^4)/(4*d) + (2*a*Csc[c + d*x]^5)/(5*d) + (a*Csc[c + d*x]^6)/(3*d) -
(a*Csc[c + d*x]^7)/(7*d) - (a*Csc[c + d*x]^8)/(8*d)

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.84

method	result	size
derivativedivides	\(-\frac {a \left (\frac {\left (\csc ^{8}\left (d x +c \right )\right )}{8}+\frac {\left (\csc ^{7}\left (d x +c \right )\right )}{7}-\frac {\left (\csc ^{6}\left (d x +c \right )\right )}{3}-\frac {2 \left (\csc ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\csc ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\csc ^{3}\left (d x +c \right )\right )}{3}\right )}{d}\)	\(68\)
default	\(-\frac {a \left (\frac {\left (\csc ^{8}\left (d x +c \right )\right )}{8}+\frac {\left (\csc ^{7}\left (d x +c \right )\right )}{7}-\frac {\left (\csc ^{6}\left (d x +c \right )\right )}{3}-\frac {2 \left (\csc ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\csc ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\csc ^{3}\left (d x +c \right )\right )}{3}\right )}{d}\)	\(68\)
parallelrisch	\(-\frac {a \left (716520 \cos \left (2 d x +2 c \right )-2555 \cos \left (8 d x +8 c \right )+286720 \sin \left (5 d x +5 c \right )+20440 \cos \left (6 d x +6 c \right )+704512 \sin \left (d x +c \right )[phone] \sin \left (3 d x +3 c \right )+358540 \cos \left (4 d x +4 c \right )\right ) \left (\sec ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\csc ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3523215360 d}\)	\(105\)
risch	\(\frac {4 i a \left (105 i {\mathrm e}^{12 i \left (d x +c \right )}+70 \,{\mathrm e}^{13 i \left (d x +c \right )}+140 i {\mathrm e}^{10 i \left (d x +c \right )}-14 \,{\mathrm e}^{11 i \left (d x +c \right )}+350 i {\mathrm e}^{8 i \left (d x +c \right )}+172 \,{\mathrm e}^{9 i \left (d x +c \right )}+140 i {\mathrm e}^{6 i \left (d x +c \right )}-172 \,{\mathrm e}^{7 i \left (d x +c \right )}+105 i {\mathrm e}^{4 i \left (d x +c \right )}+14 \,{\mathrm e}^{5 i \left (d x +c \right )}-70 \,{\mathrm e}^{3 i \left (d x +c \right )}\right )}{105 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{8}}\)	\(148\)

[In]

int(cos(d*x+c)^5*csc(d*x+c)^9*(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

-a/d*(1/8*csc(d*x+c)^8+1/7*csc(d*x+c)^7-1/3*csc(d*x+c)^6-2/5*csc(d*x+c)^5+1/4*csc(d*x+c)^4+1/3*csc(d*x+c)^3)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.35 \[ \int \cot ^5(c+d x) \csc ^4(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {210 \, a \cos \left (d x + c\right )^{4} - 140 \, a \cos \left (d x + c\right )^{2} + 8 \, {\left (35 \, a \cos \left (d x + c\right )^{4} - 28 \, a \cos \left (d x + c\right )^{2} + 8 \, a\right )} \sin \left (d x + c\right ) + 35 \, a}{840 \, {\left (d \cos \left (d x + c\right )^{8} - 4 \, d \cos \left (d x + c\right )^{6} + 6 \, d \cos \left (d x + c\right )^{4} - 4 \, d \cos \left (d x + c\right )^{2} + d\right )}} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^9*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/840*(210*a*cos(d*x + c)^4 - 140*a*cos(d*x + c)^2 + 8*(35*a*cos(d*x + c)^4 - 28*a*cos(d*x + c)^2 + 8*a)*sin(
d*x + c) + 35*a)/(d*cos(d*x + c)^8 - 4*d*cos(d*x + c)^6 + 6*d*cos(d*x + c)^4 - 4*d*cos(d*x + c)^2 + d)

Sympy [F(-1)]

Timed out. \[ \int \cot ^5(c+d x) \csc ^4(c+d x) (a+a \sin (c+d x)) \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**5*csc(d*x+c)**9*(a+a*sin(d*x+c)),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.86 \[ \int \cot ^5(c+d x) \csc ^4(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {280 \, a \sin \left (d x + c\right )^{5} + 210 \, a \sin \left (d x + c\right )^{4} - 336 \, a \sin \left (d x + c\right )^{3} - 280 \, a \sin \left (d x + c\right )^{2} + 120 \, a \sin \left (d x + c\right ) + 105 \, a}{840 \, d \sin \left (d x + c\right )^{8}} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^9*(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/840*(280*a*sin(d*x + c)^5 + 210*a*sin(d*x + c)^4 - 336*a*sin(d*x + c)^3 - 280*a*sin(d*x + c)^2 + 120*a*sin(
d*x + c) + 105*a)/(d*sin(d*x + c)^8)

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.86 \[ \int \cot ^5(c+d x) \csc ^4(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {280 \, a \sin \left (d x + c\right )^{5} + 210 \, a \sin \left (d x + c\right )^{4} - 336 \, a \sin \left (d x + c\right )^{3} - 280 \, a \sin \left (d x + c\right )^{2} + 120 \, a \sin \left (d x + c\right ) + 105 \, a}{840 \, d \sin \left (d x + c\right )^{8}} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^9*(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/840*(280*a*sin(d*x + c)^5 + 210*a*sin(d*x + c)^4 - 336*a*sin(d*x + c)^3 - 280*a*sin(d*x + c)^2 + 120*a*sin(
d*x + c) + 105*a)/(d*sin(d*x + c)^8)

Mupad [B] (verification not implemented)

Time = 9.68 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.86 \[ \int \cot ^5(c+d x) \csc ^4(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {280\,a\,{\sin \left (c+d\,x\right )}^5+210\,a\,{\sin \left (c+d\,x\right )}^4-336\,a\,{\sin \left (c+d\,x\right )}^3-280\,a\,{\sin \left (c+d\,x\right )}^2+120\,a\,\sin \left (c+d\,x\right )+105\,a}{840\,d\,{\sin \left (c+d\,x\right )}^8} \]

[In]

int((cos(c + d*x)^5*(a + a*sin(c + d*x)))/sin(c + d*x)^9,x)

[Out]

-(105*a + 120*a*sin(c + d*x) - 280*a*sin(c + d*x)^2 - 336*a*sin(c + d*x)^3 + 210*a*sin(c + d*x)^4 + 280*a*sin(
c + d*x)^5)/(840*d*sin(c + d*x)^8)