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

Optimal. Leaf size=65 \[ \frac {\tan ^3(c+d x)}{3 a^2 d}+\frac {3 \tan (c+d x)}{a^2 d}-\frac {\cot ^3(c+d x)}{3 a^2 d}-\frac {3 \cot (c+d x)}{a^2 d} \]

[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

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Rubi [A]  time = 0.08, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3175, 2620, 270} \[ \frac {\tan ^3(c+d x)}{3 a^2 d}+\frac {3 \tan (c+d x)}{a^2 d}-\frac {\cot ^3(c+d x)}{3 a^2 d}-\frac {3 \cot (c+d x)}{a^2 d} \]

Antiderivative was successfully verified.

[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 270

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 2620

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 3175

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*} \int \frac {\csc ^4(c+d x)}{\left (a-a \sin ^2(c+d x)\right )^2} \, dx &=\frac {\int \csc ^4(c+d x) \sec ^4(c+d x) \, dx}{a^2}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^3}{x^4} \, dx,x,\tan (c+d x)\right )}{a^2 d}\\ &=\frac {\operatorname {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*}

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Mathematica [A]  time = 0.03, size = 46, normalized size = 0.71 \[ \frac {16 \left (-\frac {\cot (2 (c+d x))}{3 d}-\frac {\cot (2 (c+d x)) \csc ^2(2 (c+d x))}{6 d}\right )}{a^2} \]

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.41, size = 72, normalized size = 1.11 \[ -\frac {16 \, \cos \left (d x + c\right )^{6} - 24 \, \cos \left (d x + c\right )^{4} + 6 \, \cos \left (d x + c\right )^{2} + 1}{3 \, {\left (a^{2} d \cos \left (d x + c\right )^{5} - a^{2} d \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [A]  time = 0.15, size = 34, normalized size = 0.52 \[ -\frac {8 \, {\left (3 \, \tan \left (2 \, d x + 2 \, c\right )^{2} + 1\right )}}{3 \, a^{2} d \tan \left (2 \, d x + 2 \, c\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.51, size = 47, normalized size = 0.72 \[ \frac {\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}+3 \tan \left (d x +c \right )-\frac {3}{\tan \left (d x +c \right )}-\frac {1}{3 \tan \left (d x +c \right )^{3}}}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.34, size = 52, normalized size = 0.80 \[ \frac {\frac {\tan \left (d x + c\right )^{3} + 9 \, \tan \left (d x + c\right )}{a^{2}} - \frac {9 \, \tan \left (d x + c\right )^{2} + 1}{a^{2} \tan \left (d x + c\right )^{3}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 13.68, size = 48, normalized size = 0.74 \[ -\frac {-{\mathrm {tan}\left (c+d\,x\right )}^6-9\,{\mathrm {tan}\left (c+d\,x\right )}^4+9\,{\mathrm {tan}\left (c+d\,x\right )}^2+1}{3\,a^2\,d\,{\mathrm {tan}\left (c+d\,x\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\csc ^{4}{\left (c + d x \right )}}{\sin ^{4}{\left (c + d x \right )} - 2 \sin ^{2}{\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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