\(\int \frac {\csc ^4(c+d x)}{(a-a \sin ^2(c+d x))^2} \, dx\) [60]

Optimal result

Integrand size = 24, antiderivative size = 65 \[ \int \frac {\csc ^4(c+d x)}{\left (a-a \sin ^2(c+d x)\right )^2} \, dx=-\frac {3 \cot (c+d x)}{a^2 d}-\frac {\cot ^3(c+d x)}{3 a^2 d}+\frac {3 \tan (c+d x)}{a^2 d}+\frac {\tan ^3(c+d x)}{3 a^2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 65, 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} \[ \int \frac {\csc ^4(c+d x)}{\left (a-a \sin ^2(c+d x)\right )^2} \, dx=\frac {\tan ^3(c+d x)}{3 a^2 d}+\frac {3 \tan (c+d x)}{a^2 d}-\frac {\cot ^3(c+d x)}{3 a^2 d}-\frac {3 \cot (c+d x)}{a^2 d} \]

[In]

Int[Csc[c + d*x]^4/(a - a*Sin[c + d*x]^2)^2,x]

[Out]

(-3*Cot[c + d*x])/(a^2*d) - Cot[c + d*x]^3/(3*a^2*d) + (3*Tan[c + d*x])/(a^2*d) + Tan[c + d*x]^3/(3*a^2*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 ^4(c+d x) \sec ^4(c+d x) \, dx}{a^2} \\ & = \frac {\text {Subst}\left (\int \frac {\left (1+x^2\right )^3}{x^4} \, dx,x,\tan (c+d x)\right )}{a^2 d} \\ & = \frac {\text {Subst}\left (\int \left (3+\frac {1}{x^4}+\frac {3}{x^2}+x^2\right ) \, dx,x,\tan (c+d x)\right )}{a^2 d} \\ & = -\frac {3 \cot (c+d x)}{a^2 d}-\frac {\cot ^3(c+d x)}{3 a^2 d}+\frac {3 \tan (c+d x)}{a^2 d}+\frac {\tan ^3(c+d x)}{3 a^2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.71 \[ \int \frac {\csc ^4(c+d x)}{\left (a-a \sin ^2(c+d x)\right )^2} \, dx=\frac {16 \left (-\frac {\cot (2 (c+d x))}{3 d}-\frac {\cot (2 (c+d x)) \csc ^2(2 (c+d x))}{6 d}\right )}{a^2} \]

[In]

Integrate[Csc[c + d*x]^4/(a - a*Sin[c + d*x]^2)^2,x]

[Out]

(16*(-1/3*Cot[2*(c + d*x)]/d - (Cot[2*(c + d*x)]*Csc[2*(c + d*x)]^2)/(6*d)))/a^2

Maple [A] (verified)

Time = 0.94 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.72

method	result	size
derivativedivides	\(\frac {-\frac {1}{3 \tan \left (d x +c \right )^{3}}-\frac {3}{\tan \left (d x +c \right )}+\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}+3 \tan \left (d x +c \right )}{d \,a^{2}}\)	\(47\)
default	\(\frac {-\frac {1}{3 \tan \left (d x +c \right )^{3}}-\frac {3}{\tan \left (d x +c \right )}+\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}+3 \tan \left (d x +c \right )}{d \,a^{2}}\)	\(47\)
risch	\(\frac {32 i \left (3 \,{\mathrm e}^{4 i \left (d x +c \right )}-1\right )}{3 d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}\)	\(49\)
parallelrisch	\(\frac {\left (-3 \cos \left (2 d x +2 c \right )+\cos \left (6 d x +6 c \right )\right ) \left (\sec ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\csc ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 a^{2} d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\)	\(72\)
norman	\(\frac {\frac {1}{24 a d}+\frac {5 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a}-\frac {91 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}+\frac {35 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}-\frac {91 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}+\frac {5 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a}+\frac {\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )}{24 d a}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}\)	\(154\)

[In]

int(csc(d*x+c)^4/(a-a*sin(d*x+c)^2)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.11 \[ \int \frac {\csc ^4(c+d x)}{\left (a-a \sin ^2(c+d x)\right )^2} \, dx=-\frac {16 \, \cos \left (d x + c\right )^{6} - 24 \, \cos \left (d x + c\right )^{4} + 6 \, \cos \left (d x + c\right )^{2} + 1}{3 \, {\left (a^{2} d \cos \left (d x + c\right )^{5} - a^{2} d \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )} \]

[In]

integrate(csc(d*x+c)^4/(a-a*sin(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

-1/3*(16*cos(d*x + c)^6 - 24*cos(d*x + c)^4 + 6*cos(d*x + c)^2 + 1)/((a^2*d*cos(d*x + c)^5 - a^2*d*cos(d*x + c
)^3)*sin(d*x + c))

Sympy [F]

\[ \int \frac {\csc ^4(c+d x)}{\left (a-a \sin ^2(c+d x)\right )^2} \, dx=\frac {\int \frac {\csc ^{4}{\left (c + d x \right )}}{\sin ^{4}{\left (c + d x \right )} - 2 \sin ^{2}{\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

[In]

integrate(csc(d*x+c)**4/(a-a*sin(d*x+c)**2)**2,x)

[Out]

Integral(csc(c + d*x)**4/(sin(c + d*x)**4 - 2*sin(c + d*x)**2 + 1), x)/a**2

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.80 \[ \int \frac {\csc ^4(c+d x)}{\left (a-a \sin ^2(c+d x)\right )^2} \, dx=\frac {\frac {\tan \left (d x + c\right )^{3} + 9 \, \tan \left (d x + c\right )}{a^{2}} - \frac {9 \, \tan \left (d x + c\right )^{2} + 1}{a^{2} \tan \left (d x + c\right )^{3}}}{3 \, d} \]

[In]

integrate(csc(d*x+c)^4/(a-a*sin(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

1/3*((tan(d*x + c)^3 + 9*tan(d*x + c))/a^2 - (9*tan(d*x + c)^2 + 1)/(a^2*tan(d*x + c)^3))/d

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.52 \[ \int \frac {\csc ^4(c+d x)}{\left (a-a \sin ^2(c+d x)\right )^2} \, dx=-\frac {8 \, {\left (3 \, \tan \left (2 \, d x + 2 \, c\right )^{2} + 1\right )}}{3 \, a^{2} d \tan \left (2 \, d x + 2 \, c\right )^{3}} \]

[In]

integrate(csc(d*x+c)^4/(a-a*sin(d*x+c)^2)^2,x, algorithm="giac")

[Out]

-8/3*(3*tan(2*d*x + 2*c)^2 + 1)/(a^2*d*tan(2*d*x + 2*c)^3)

Mupad [B] (verification not implemented)

Time = 13.13 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.74 \[ \int \frac {\csc ^4(c+d x)}{\left (a-a \sin ^2(c+d x)\right )^2} \, dx=-\frac {-{\mathrm {tan}\left (c+d\,x\right )}^6-9\,{\mathrm {tan}\left (c+d\,x\right )}^4+9\,{\mathrm {tan}\left (c+d\,x\right )}^2+1}{3\,a^2\,d\,{\mathrm {tan}\left (c+d\,x\right )}^3} \]

[In]

int(1/(sin(c + d*x)^4*(a - a*sin(c + d*x)^2)^2),x)

[Out]

-(9*tan(c + d*x)^2 - 9*tan(c + d*x)^4 - tan(c + d*x)^6 + 1)/(3*a^2*d*tan(c + d*x)^3)