\(\int \frac {\cot ^5(c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx\) [553]

Optimal result

Integrand size = 29, antiderivative size = 55 \[ \int \frac {\cot ^5(c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\csc ^4(c+d x)}{4 a^2 d}+\frac {2 \csc ^5(c+d x)}{5 a^2 d}-\frac {\csc ^6(c+d x)}{6 a^2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2915, 12, 45} \[ \int \frac {\cot ^5(c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\csc ^6(c+d x)}{6 a^2 d}+\frac {2 \csc ^5(c+d x)}{5 a^2 d}-\frac {\csc ^4(c+d x)}{4 a^2 d} \]

[In]

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

[Out]

-1/4*Csc[c + d*x]^4/(a^2*d) + (2*Csc[c + d*x]^5)/(5*a^2*d) - Csc[c + d*x]^6/(6*a^2*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 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^7 (a-x)^2}{x^7} \, dx,x,a \sin (c+d x)\right )}{a^5 d} \\ & = \frac {a^2 \text {Subst}\left (\int \frac {(a-x)^2}{x^7} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^2 \text {Subst}\left (\int \left (\frac {a^2}{x^7}-\frac {2 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^2 d}+\frac {2 \csc ^5(c+d x)}{5 a^2 d}-\frac {\csc ^6(c+d x)}{6 a^2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.69 \[ \int \frac {\cot ^5(c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\csc ^4(c+d x) \left (15-24 \csc (c+d x)+10 \csc ^2(c+d x)\right )}{60 a^2 d} \]

[In]

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

[Out]

-1/60*(Csc[c + d*x]^4*(15 - 24*Csc[c + d*x] + 10*Csc[c + d*x]^2))/(a^2*d)

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.73

method	result	size
derivativedivides	\(-\frac {\frac {\left (\csc ^{6}\left (d x +c \right )\right )}{6}-\frac {2 \left (\csc ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\csc ^{4}\left (d x +c \right )\right )}{4}}{d \,a^{2}}\)	\(40\)
default	\(-\frac {\frac {\left (\csc ^{6}\left (d x +c \right )\right )}{6}-\frac {2 \left (\csc ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\csc ^{4}\left (d x +c \right )\right )}{4}}{d \,a^{2}}\)	\(40\)
parallelrisch	\(\frac {\left (435 \cos \left (2 d x +2 c \right )-35 \cos \left (6 d x +6 c \right )+3072 \sin \left (d x +c \right )+210 \cos \left (4 d x +4 c \right )-1890\right ) \left (\sec ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\csc ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{491520 d \,a^{2}}\)	\(74\)
risch	\(-\frac {4 \left (15 \,{\mathrm e}^{8 i \left (d x +c \right )}-70 \,{\mathrm e}^{6 i \left (d x +c \right )}-48 i {\mathrm e}^{7 i \left (d x +c \right )}+15 \,{\mathrm e}^{4 i \left (d x +c \right )}+48 i {\mathrm e}^{5 i \left (d x +c \right )}\right )}{15 a^{2} d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{6}}\)	\(80\)
norman	\(\frac {-\frac {1}{384 a d}+\frac {3 \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{640 d a}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{640 d a}-\frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{640 d a}+\frac {7 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{1920 d a}+\frac {3 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{640 d a}-\frac {3 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{128 d a}-\frac {3 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{128 d a}+\frac {3 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{640 d a}+\frac {7 \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{1920 d a}-\frac {\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )}{640 d a}-\frac {\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )}{384 d a}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}\)	\(245\)

[In]

int(cos(d*x+c)^5*csc(d*x+c)^7/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.31 \[ \int \frac {\cot ^5(c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {15 \, \cos \left (d x + c\right )^{2} + 24 \, \sin \left (d x + c\right ) - 25}{60 \, {\left (a^{2} d \cos \left (d x + c\right )^{6} - 3 \, a^{2} d \cos \left (d x + c\right )^{4} + 3 \, a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d\right )}} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^7/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/60*(15*cos(d*x + c)^2 + 24*sin(d*x + c) - 25)/(a^2*d*cos(d*x + c)^6 - 3*a^2*d*cos(d*x + c)^4 + 3*a^2*d*cos(
d*x + c)^2 - a^2*d)

Sympy [F(-1)]

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

[In]

integrate(cos(d*x+c)**5*csc(d*x+c)**7/(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.65 \[ \int \frac {\cot ^5(c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {15 \, \sin \left (d x + c\right )^{2} - 24 \, \sin \left (d x + c\right ) + 10}{60 \, a^{2} d \sin \left (d x + c\right )^{6}} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^7/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/60*(15*sin(d*x + c)^2 - 24*sin(d*x + c) + 10)/(a^2*d*sin(d*x + c)^6)

Giac [A] (verification not implemented)

none

Time = 0.50 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.65 \[ \int \frac {\cot ^5(c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {15 \, \sin \left (d x + c\right )^{2} - 24 \, \sin \left (d x + c\right ) + 10}{60 \, a^{2} d \sin \left (d x + c\right )^{6}} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^7/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/60*(15*sin(d*x + c)^2 - 24*sin(d*x + c) + 10)/(a^2*d*sin(d*x + c)^6)

Mupad [B] (verification not implemented)

Time = 10.46 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.65 \[ \int \frac {\cot ^5(c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {{\sin \left (c+d\,x\right )}^2}{4}-\frac {2\,\sin \left (c+d\,x\right )}{5}+\frac {1}{6}}{a^2\,d\,{\sin \left (c+d\,x\right )}^6} \]

[In]

int(cos(c + d*x)^5/(sin(c + d*x)^7*(a + a*sin(c + d*x))^2),x)

[Out]

-(sin(c + d*x)^2/4 - (2*sin(c + d*x))/5 + 1/6)/(a^2*d*sin(c + d*x)^6)