\(\int \cot (c+d x) \csc ^5(c+d x) (a+a \sin (c+d x))^3 \, dx\) [214]

Optimal result

Integrand size = 27, antiderivative size = 61 \[ \int \cot (c+d x) \csc ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {\csc ^4(c+d x) (a+a \sin (c+d x))^4}{20 a d}-\frac {\csc ^5(c+d x) (a+a \sin (c+d x))^4}{5 a d} \]

[Out]

1/20*csc(d*x+c)^4*(a+a*sin(d*x+c))^4/a/d-1/5*csc(d*x+c)^5*(a+a*sin(d*x+c))^4/a/d

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2912, 12, 47, 37} \[ \int \cot (c+d x) \csc ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {\csc ^4(c+d x) (a \sin (c+d x)+a)^4}{20 a d}-\frac {\csc ^5(c+d x) (a \sin (c+d x)+a)^4}{5 a d} \]

[In]

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

[Out]

(Csc[c + d*x]^4*(a + a*Sin[c + d*x])^4)/(20*a*d) - (Csc[c + d*x]^5*(a + a*Sin[c + d*x])^4)/(5*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 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*(Simplify[m + n + 2]/((b*c - a*d)*(m + 1))), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rule 2912

Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)
])^(n_.), x_Symbol] :> Dist[1/(b*f), Subst[Int[(a + x)^m*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[
{a, b, c, d, e, f, m, n}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {a^6 (a+x)^3}{x^6} \, dx,x,a \sin (c+d x)\right )}{a d} \\ & = \frac {a^5 \text {Subst}\left (\int \frac {(a+x)^3}{x^6} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = -\frac {\csc ^5(c+d x) (a+a \sin (c+d x))^4}{5 a d}-\frac {a^4 \text {Subst}\left (\int \frac {(a+x)^3}{x^5} \, dx,x,a \sin (c+d x)\right )}{5 d} \\ & = \frac {\csc ^4(c+d x) (a+a \sin (c+d x))^4}{20 a d}-\frac {\csc ^5(c+d x) (a+a \sin (c+d x))^4}{5 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.16 \[ \int \cot (c+d x) \csc ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {a^3 \csc ^2(c+d x)}{2 d}-\frac {a^3 \csc ^3(c+d x)}{d}-\frac {3 a^3 \csc ^4(c+d x)}{4 d}-\frac {a^3 \csc ^5(c+d x)}{5 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.79

method	result	size
derivativedivides	\(-\frac {a^{3} \left (\frac {\left (\csc ^{5}\left (d x +c \right )\right )}{5}+\frac {3 \left (\csc ^{4}\left (d x +c \right )\right )}{4}+\csc ^{3}\left (d x +c \right )+\frac {\left (\csc ^{2}\left (d x +c \right )\right )}{2}\right )}{d}\)	\(48\)
default	\(-\frac {a^{3} \left (\frac {\left (\csc ^{5}\left (d x +c \right )\right )}{5}+\frac {3 \left (\csc ^{4}\left (d x +c \right )\right )}{4}+\csc ^{3}\left (d x +c \right )+\frac {\left (\csc ^{2}\left (d x +c \right )\right )}{2}\right )}{d}\)	\(48\)
parallelrisch	\(-\frac {a^{3} \left (1792-1280 \cos \left (2 d x +2 c \right )-85 \sin \left (5 d x +5 c \right )+2030 \sin \left (d x +c \right )+105 \sin \left (3 d x +3 c \right )\right ) \left (\sec ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\csc ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{81920 d}\)	\(74\)
risch	\(\frac {2 a^{3} \left (20 i {\mathrm e}^{7 i \left (d x +c \right )}+5 \,{\mathrm e}^{8 i \left (d x +c \right )}-56 i {\mathrm e}^{5 i \left (d x +c \right )}-45 \,{\mathrm e}^{6 i \left (d x +c \right )}+20 i {\mathrm e}^{3 i \left (d x +c \right )}+45 \,{\mathrm e}^{4 i \left (d x +c \right )}-5 \,{\mathrm e}^{2 i \left (d x +c \right )}\right )}{5 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}\)	\(103\)
norman	\(\frac {-\frac {a^{3}}{160 d}-\frac {3 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{64 d}-\frac {7 a^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d}-\frac {29 a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d}-\frac {37 a^{3} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d}-\frac {89 a^{3} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d}-\frac {47 a^{3} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d}-\frac {89 a^{3} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d}-\frac {37 a^{3} \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d}-\frac {29 a^{3} \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d}-\frac {7 a^{3} \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d}-\frac {3 a^{3} \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d}-\frac {a^{3} \left (\tan ^{16}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{160 d}+\frac {31 a^{3} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d}+\frac {31 a^{3} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}\)	\(301\)

[In]

int(cos(d*x+c)*csc(d*x+c)^6*(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.33 \[ \int \cot (c+d x) \csc ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {20 \, a^{3} \cos \left (d x + c\right )^{2} - 24 \, a^{3} + 5 \, {\left (2 \, a^{3} \cos \left (d x + c\right )^{2} - 5 \, a^{3}\right )} \sin \left (d x + c\right )}{20 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \]

[In]

integrate(cos(d*x+c)*csc(d*x+c)^6*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/20*(20*a^3*cos(d*x + c)^2 - 24*a^3 + 5*(2*a^3*cos(d*x + c)^2 - 5*a^3)*sin(d*x + c))/((d*cos(d*x + c)^4 - 2*d
*cos(d*x + c)^2 + d)*sin(d*x + c))

Sympy [F(-1)]

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

[In]

integrate(cos(d*x+c)*csc(d*x+c)**6*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.92 \[ \int \cot (c+d x) \csc ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {10 \, a^{3} \sin \left (d x + c\right )^{3} + 20 \, a^{3} \sin \left (d x + c\right )^{2} + 15 \, a^{3} \sin \left (d x + c\right ) + 4 \, a^{3}}{20 \, d \sin \left (d x + c\right )^{5}} \]

[In]

integrate(cos(d*x+c)*csc(d*x+c)^6*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/20*(10*a^3*sin(d*x + c)^3 + 20*a^3*sin(d*x + c)^2 + 15*a^3*sin(d*x + c) + 4*a^3)/(d*sin(d*x + c)^5)

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.92 \[ \int \cot (c+d x) \csc ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {10 \, a^{3} \sin \left (d x + c\right )^{3} + 20 \, a^{3} \sin \left (d x + c\right )^{2} + 15 \, a^{3} \sin \left (d x + c\right ) + 4 \, a^{3}}{20 \, d \sin \left (d x + c\right )^{5}} \]

[In]

integrate(cos(d*x+c)*csc(d*x+c)^6*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/20*(10*a^3*sin(d*x + c)^3 + 20*a^3*sin(d*x + c)^2 + 15*a^3*sin(d*x + c) + 4*a^3)/(d*sin(d*x + c)^5)

Mupad [B] (verification not implemented)

Time = 9.86 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.92 \[ \int \cot (c+d x) \csc ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {10\,a^3\,{\sin \left (c+d\,x\right )}^3+20\,a^3\,{\sin \left (c+d\,x\right )}^2+15\,a^3\,\sin \left (c+d\,x\right )+4\,a^3}{20\,d\,{\sin \left (c+d\,x\right )}^5} \]

[In]

int((cos(c + d*x)*(a + a*sin(c + d*x))^3)/sin(c + d*x)^6,x)

[Out]

-(15*a^3*sin(c + d*x) + 4*a^3 + 20*a^3*sin(c + d*x)^2 + 10*a^3*sin(c + d*x)^3)/(20*d*sin(c + d*x)^5)