\(\int \frac {\csc ^6(c+d x)}{a-a \sin ^2(c+d x)} \, dx\) [48]

Optimal result

Integrand size = 24, antiderivative size = 62 \[ \int \frac {\csc ^6(c+d x)}{a-a \sin ^2(c+d x)} \, dx=-\frac {3 \cot (c+d x)}{a d}-\frac {\cot ^3(c+d x)}{a d}-\frac {\cot ^5(c+d x)}{5 a d}+\frac {\tan (c+d x)}{a d} \]

[Out]

-3*cot(d*x+c)/a/d-cot(d*x+c)^3/a/d-1/5*cot(d*x+c)^5/a/d+tan(d*x+c)/a/d

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3254, 2700, 276} \[ \int \frac {\csc ^6(c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {\tan (c+d x)}{a d}-\frac {\cot ^5(c+d x)}{5 a d}-\frac {\cot ^3(c+d x)}{a d}-\frac {3 \cot (c+d x)}{a d} \]

[In]

Int[Csc[c + d*x]^6/(a - a*Sin[c + d*x]^2),x]

[Out]

(-3*Cot[c + d*x])/(a*d) - Cot[c + d*x]^3/(a*d) - Cot[c + d*x]^5/(5*a*d) + Tan[c + d*x]/(a*d)

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 2700

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

Rule 3254

Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Dist[a^p, Int[ActivateTrig[u*cos[e + f*x
]^(2*p)], x], x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0] && IntegerQ[p]

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.13 \[ \int \frac {\csc ^6(c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {-\frac {11 \cot (c+d x)}{5 d}-\frac {3 \cot (c+d x) \csc ^2(c+d x)}{5 d}-\frac {\cot (c+d x) \csc ^4(c+d x)}{5 d}+\frac {\tan (c+d x)}{d}}{a} \]

[In]

Integrate[Csc[c + d*x]^6/(a - a*Sin[c + d*x]^2),x]

[Out]

((-11*Cot[c + d*x])/(5*d) - (3*Cot[c + d*x]*Csc[c + d*x]^2)/(5*d) - (Cot[c + d*x]*Csc[c + d*x]^4)/(5*d) + Tan[
c + d*x]/d)/a

Maple [A] (verified)

Time = 0.88 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.73

method	result	size
derivativedivides	\(\frac {\tan \left (d x +c \right )-\frac {1}{5 \tan \left (d x +c \right )^{5}}-\frac {3}{\tan \left (d x +c \right )}-\frac {1}{\tan \left (d x +c \right )^{3}}}{d a}\)	\(45\)
default	\(\frac {\tan \left (d x +c \right )-\frac {1}{5 \tan \left (d x +c \right )^{5}}-\frac {3}{\tan \left (d x +c \right )}-\frac {1}{\tan \left (d x +c \right )^{3}}}{d a}\)	\(45\)
risch	\(-\frac {32 i \left (5 \,{\mathrm e}^{4 i \left (d x +c \right )}-4 \,{\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{5 d a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}\)	\(60\)
parallelrisch	\(\frac {\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )+14 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\cot ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )+175 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+14 \left (\cot ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-700 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+175 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )}{160 d a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}\)	\(108\)
norman	\(\frac {\frac {1}{160 a d}+\frac {7 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{80 d a}+\frac {35 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}-\frac {35 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}+\frac {35 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}+\frac {7 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{80 d a}+\frac {\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )}{160 d a}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}\)	\(151\)

[In]

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

[Out]

1/d/a*(tan(d*x+c)-1/5/tan(d*x+c)^5-3/tan(d*x+c)-1/tan(d*x+c)^3)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.24 \[ \int \frac {\csc ^6(c+d x)}{a-a \sin ^2(c+d x)} \, dx=-\frac {16 \, \cos \left (d x + c\right )^{6} - 40 \, \cos \left (d x + c\right )^{4} + 30 \, \cos \left (d x + c\right )^{2} - 5}{5 \, {\left (a d \cos \left (d x + c\right )^{5} - 2 \, a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )} \]

[In]

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

[Out]

-1/5*(16*cos(d*x + c)^6 - 40*cos(d*x + c)^4 + 30*cos(d*x + c)^2 - 5)/((a*d*cos(d*x + c)^5 - 2*a*d*cos(d*x + c)
^3 + a*d*cos(d*x + c))*sin(d*x + c))

Sympy [F]

\[ \int \frac {\csc ^6(c+d x)}{a-a \sin ^2(c+d x)} \, dx=- \frac {\int \frac {\csc ^{6}{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} - 1}\, dx}{a} \]

[In]

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

[Out]

-Integral(csc(c + d*x)**6/(sin(c + d*x)**2 - 1), x)/a

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.84 \[ \int \frac {\csc ^6(c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {\frac {5 \, \tan \left (d x + c\right )}{a} - \frac {15 \, \tan \left (d x + c\right )^{4} + 5 \, \tan \left (d x + c\right )^{2} + 1}{a \tan \left (d x + c\right )^{5}}}{5 \, d} \]

[In]

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

[Out]

1/5*(5*tan(d*x + c)/a - (15*tan(d*x + c)^4 + 5*tan(d*x + c)^2 + 1)/(a*tan(d*x + c)^5))/d

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.84 \[ \int \frac {\csc ^6(c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {\frac {5 \, \tan \left (d x + c\right )}{a} - \frac {15 \, \tan \left (d x + c\right )^{4} + 5 \, \tan \left (d x + c\right )^{2} + 1}{a \tan \left (d x + c\right )^{5}}}{5 \, d} \]

[In]

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

[Out]

1/5*(5*tan(d*x + c)/a - (15*tan(d*x + c)^4 + 5*tan(d*x + c)^2 + 1)/(a*tan(d*x + c)^5))/d

Mupad [B] (verification not implemented)

Time = 13.88 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.81 \[ \int \frac {\csc ^6(c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {\mathrm {tan}\left (c+d\,x\right )}{a\,d}-\frac {3\,{\mathrm {tan}\left (c+d\,x\right )}^4+{\mathrm {tan}\left (c+d\,x\right )}^2+\frac {1}{5}}{a\,d\,{\mathrm {tan}\left (c+d\,x\right )}^5} \]

[In]

int(1/(sin(c + d*x)^6*(a - a*sin(c + d*x)^2)),x)

[Out]

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