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

Optimal. Leaf size=47 \[ \frac{\tan ^3(c+d x)}{3 a^2 d}+\frac{2 \tan (c+d x)}{a^2 d}-\frac{\cot (c+d x)}{a^2 d} \]

[Out]

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

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Rubi [A]  time = 0.0781224, antiderivative size = 47, normalized size of antiderivative = 1., 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} \[ \frac{\tan ^3(c+d x)}{3 a^2 d}+\frac{2 \tan (c+d x)}{a^2 d}-\frac{\cot (c+d x)}{a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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]

Rubi steps

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

Mathematica [A]  time = 0.038668, size = 50, normalized size = 1.06 \[ \frac{\frac{5 \tan (c+d x)}{3 d}-\frac{\cot (c+d x)}{d}+\frac{\tan (c+d x) \sec ^2(c+d x)}{3 d}}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.066, size = 37, normalized size = 0.8 \begin{align*}{\frac{1}{{a}^{2}d} \left ({\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{3}}+2\,\tan \left ( dx+c \right ) - \left ( \tan \left ( dx+c \right ) \right ) ^{-1} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.959586, size = 54, normalized size = 1.15 \begin{align*} \frac{\frac{\tan \left (d x + c\right )^{3} + 6 \, \tan \left (d x + c\right )}{a^{2}} - \frac{3}{a^{2} \tan \left (d x + c\right )}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.55082, size = 113, normalized size = 2.4 \begin{align*} -\frac{8 \, \cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{2} - 1}{3 \, a^{2} d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(8*cos(d*x + c)^4 - 4*cos(d*x + c)^2 - 1)/(a^2*d*cos(d*x + c)^3*sin(d*x + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\csc ^{2}{\left (c + d x \right )}}{\sin ^{4}{\left (c + d x \right )} - 2 \sin ^{2}{\left (c + d x \right )} + 1}\, dx}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.11487, size = 65, normalized size = 1.38 \begin{align*} -\frac{\frac{3}{a^{2} \tan \left (d x + c\right )} - \frac{a^{4} \tan \left (d x + c\right )^{3} + 6 \, a^{4} \tan \left (d x + c\right )}{a^{6}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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