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

Optimal. Leaf size=28 \[ \frac{\tan (c+d x)}{a d}-\frac{\cot (c+d x)}{a d} \]

[Out]

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

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Rubi [A]  time = 0.07482, antiderivative size = 28, 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, 14} \[ \frac{\tan (c+d x)}{a d}-\frac{\cot (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Cot[c + d*x]/(a*d)) + Tan[c + d*x]/(a*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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

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

Mathematica [A]  time = 0.0270954, size = 16, normalized size = 0.57 \[ -\frac{2 \cot (2 (c+d x))}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Cot[2*(c + d*x)])/(a*d)

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Maple [A]  time = 0.06, size = 25, normalized size = 0.9 \begin{align*}{\frac{1}{da} \left ( \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),x)

[Out]

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

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Maxima [A]  time = 0.956545, size = 38, normalized size = 1.36 \begin{align*} \frac{\frac{\tan \left (d x + c\right )}{a} - \frac{1}{a \tan \left (d x + c\right )}}{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),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.58146, size = 77, normalized size = 2.75 \begin{align*} -\frac{2 \, \cos \left (d x + c\right )^{2} - 1}{a d \cos \left (d x + c\right ) \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),x, algorithm="fricas")

[Out]

-(2*cos(d*x + c)^2 - 1)/(a*d*cos(d*x + c)*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 ^{2}{\left (c + d x \right )} - 1}\, dx}{a} \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),x)

[Out]

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

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Giac [A]  time = 1.15326, size = 26, normalized size = 0.93 \begin{align*} -\frac{2}{a d \tan \left (2 \, d x + 2 \, 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),x, algorithm="giac")

[Out]

-2/(a*d*tan(2*d*x + 2*c))