\(\int \cot (c+d x) \csc ^5(c+d x) (a+a \sin (c+d x))^2 \, dx\) [204]

Optimal result

Integrand size = 27, antiderivative size = 55 \[ \int \cot (c+d x) \csc ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {a^2 \csc ^3(c+d x)}{3 d}-\frac {a^2 \csc ^4(c+d x)}{2 d}-\frac {a^2 \csc ^5(c+d x)}{5 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2912, 12, 45} \[ \int \cot (c+d x) \csc ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {a^2 \csc ^5(c+d x)}{5 d}-\frac {a^2 \csc ^4(c+d x)}{2 d}-\frac {a^2 \csc ^3(c+d x)}{3 d} \]

[In]

Int[Cot[c + d*x]*Csc[c + d*x]^5*(a + a*Sin[c + d*x])^2,x]

[Out]

-1/3*(a^2*Csc[c + d*x]^3)/d - (a^2*Csc[c + d*x]^4)/(2*d) - (a^2*Csc[c + d*x]^5)/(5*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 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)^2}{x^6} \, dx,x,a \sin (c+d x)\right )}{a d} \\ & = \frac {a^5 \text {Subst}\left (\int \frac {(a+x)^2}{x^6} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^5 \text {Subst}\left (\int \left (\frac {a^2}{x^6}+\frac {2 a}{x^5}+\frac {1}{x^4}\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = -\frac {a^2 \csc ^3(c+d x)}{3 d}-\frac {a^2 \csc ^4(c+d x)}{2 d}-\frac {a^2 \csc ^5(c+d x)}{5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00 \[ \int \cot (c+d x) \csc ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {a^2 \csc ^3(c+d x)}{3 d}-\frac {a^2 \csc ^4(c+d x)}{2 d}-\frac {a^2 \csc ^5(c+d x)}{5 d} \]

[In]

Integrate[Cot[c + d*x]*Csc[c + d*x]^5*(a + a*Sin[c + d*x])^2,x]

[Out]

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

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.73

method	result	size
derivativedivides	\(-\frac {a^{2} \left (\frac {\left (\csc ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\csc ^{4}\left (d x +c \right )\right )}{2}+\frac {\left (\csc ^{3}\left (d x +c \right )\right )}{3}\right )}{d}\)	\(40\)
default	\(-\frac {a^{2} \left (\frac {\left (\csc ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\csc ^{4}\left (d x +c \right )\right )}{2}+\frac {\left (\csc ^{3}\left (d x +c \right )\right )}{3}\right )}{d}\)	\(40\)
parallelrisch	\(\frac {a^{2} \left (\sec ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\csc ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (-1408+640 \cos \left (2 d x +2 c \right )+45 \sin \left (5 d x +5 c \right )-1470 \sin \left (d x +c \right )-225 \sin \left (3 d x +3 c \right )\right )}{122880 d}\)	\(74\)
risch	\(\frac {8 i a^{2} \left (5 \,{\mathrm e}^{7 i \left (d x +c \right )}-22 \,{\mathrm e}^{5 i \left (d x +c \right )}+15 i {\mathrm e}^{6 i \left (d x +c \right )}+5 \,{\mathrm e}^{3 i \left (d x +c \right )}-15 i {\mathrm e}^{4 i \left (d x +c \right )}\right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}\)	\(81\)
norman	\(\frac {-\frac {a^{2}}{160 d}-\frac {a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{32 d}-\frac {41 a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{480 d}-\frac {3 a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d}-\frac {163 a^{2} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{480 d}-\frac {61 a^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{96 d}-\frac {61 a^{2} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{96 d}-\frac {163 a^{2} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{480 d}-\frac {3 a^{2} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d}-\frac {41 a^{2} \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{480 d}-\frac {a^{2} \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d}-\frac {a^{2} \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{160 d}+\frac {5 a^{2} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}\)	\(263\)

[In]

int(cos(d*x+c)*csc(d*x+c)^6*(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.18 \[ \int \cot (c+d x) \csc ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {10 \, a^{2} \cos \left (d x + c\right )^{2} - 15 \, a^{2} \sin \left (d x + c\right ) - 16 \, a^{2}}{30 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \]

[In]

integrate(cos(d*x+c)*csc(d*x+c)^6*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/30*(10*a^2*cos(d*x + c)^2 - 15*a^2*sin(d*x + c) - 16*a^2)/((d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)*sin(d
*x + c))

Sympy [F(-1)]

Timed out. \[ \int \cot (c+d x) \csc ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)*csc(d*x+c)**6*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.78 \[ \int \cot (c+d x) \csc ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {10 \, a^{2} \sin \left (d x + c\right )^{2} + 15 \, a^{2} \sin \left (d x + c\right ) + 6 \, a^{2}}{30 \, d \sin \left (d x + c\right )^{5}} \]

[In]

integrate(cos(d*x+c)*csc(d*x+c)^6*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/30*(10*a^2*sin(d*x + c)^2 + 15*a^2*sin(d*x + c) + 6*a^2)/(d*sin(d*x + c)^5)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.78 \[ \int \cot (c+d x) \csc ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {10 \, a^{2} \sin \left (d x + c\right )^{2} + 15 \, a^{2} \sin \left (d x + c\right ) + 6 \, a^{2}}{30 \, d \sin \left (d x + c\right )^{5}} \]

[In]

integrate(cos(d*x+c)*csc(d*x+c)^6*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/30*(10*a^2*sin(d*x + c)^2 + 15*a^2*sin(d*x + c) + 6*a^2)/(d*sin(d*x + c)^5)

Mupad [B] (verification not implemented)

Time = 9.22 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.78 \[ \int \cot (c+d x) \csc ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {\frac {a^2\,{\sin \left (c+d\,x\right )}^2}{3}+\frac {a^2\,\sin \left (c+d\,x\right )}{2}+\frac {a^2}{5}}{d\,{\sin \left (c+d\,x\right )}^5} \]

[In]

int((cos(c + d*x)*(a + a*sin(c + d*x))^2)/sin(c + d*x)^6,x)

[Out]

-((a^2*sin(c + d*x))/2 + a^2/5 + (a^2*sin(c + d*x)^2)/3)/(d*sin(c + d*x)^5)