3.475 \(\int \frac{\cot ^2(e+f x)}{\sqrt{a-a \sin ^2(e+f x)}} \, dx\)

Optimal. Leaf size=25 \[ -\frac{\cot (e+f x)}{f \sqrt{a \cos ^2(e+f x)}} \]

[Out]

-(Cot[e + f*x]/(f*Sqrt[a*Cos[e + f*x]^2]))

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Rubi [A]  time = 0.101829, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3176, 3207, 2606, 8} \[ -\frac{\cot (e+f x)}{f \sqrt{a \cos ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^2/Sqrt[a - a*Sin[e + f*x]^2],x]

[Out]

-(Cot[e + f*x]/(f*Sqrt[a*Cos[e + f*x]^2]))

Rule 3176

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

Rule 3207

Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Sin[e + f*x]^n)^FracPart[p])/(Sin[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Sin[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{\cot ^2(e+f x)}{\sqrt{a-a \sin ^2(e+f x)}} \, dx &=\int \frac{\cot ^2(e+f x)}{\sqrt{a \cos ^2(e+f x)}} \, dx\\ &=\frac{\cos (e+f x) \int \cot (e+f x) \csc (e+f x) \, dx}{\sqrt{a \cos ^2(e+f x)}}\\ &=-\frac{\cos (e+f x) \operatorname{Subst}(\int 1 \, dx,x,\csc (e+f x))}{f \sqrt{a \cos ^2(e+f x)}}\\ &=-\frac{\cot (e+f x)}{f \sqrt{a \cos ^2(e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.0291194, size = 25, normalized size = 1. \[ -\frac{\cot (e+f x)}{f \sqrt{a \cos ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]^2/Sqrt[a - a*Sin[e + f*x]^2],x]

[Out]

-(Cot[e + f*x]/(f*Sqrt[a*Cos[e + f*x]^2]))

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Maple [A]  time = 0.246, size = 32, normalized size = 1.3 \begin{align*} -{\frac{\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) f}{\frac{1}{\sqrt{a \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)^2/(a-a*sin(f*x+e)^2)^(1/2),x)

[Out]

-cos(f*x+e)/sin(f*x+e)/(a*cos(f*x+e)^2)^(1/2)/f

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Maxima [B]  time = 1.72, size = 122, normalized size = 4.88 \begin{align*} -\frac{2 \,{\left (\cos \left (f x + e\right ) \sin \left (2 \, f x + 2 \, e\right ) - \cos \left (2 \, f x + 2 \, e\right ) \sin \left (f x + e\right ) + \sin \left (f x + e\right )\right )} \sqrt{a}}{{\left (a \cos \left (2 \, f x + 2 \, e\right )^{2} + a \sin \left (2 \, f x + 2 \, e\right )^{2} - 2 \, a \cos \left (2 \, f x + 2 \, e\right ) + a\right )} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^2/(a-a*sin(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

-2*(cos(f*x + e)*sin(2*f*x + 2*e) - cos(2*f*x + 2*e)*sin(f*x + e) + sin(f*x + e))*sqrt(a)/((a*cos(2*f*x + 2*e)
^2 + a*sin(2*f*x + 2*e)^2 - 2*a*cos(2*f*x + 2*e) + a)*f)

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Fricas [A]  time = 1.54424, size = 77, normalized size = 3.08 \begin{align*} -\frac{\sqrt{a \cos \left (f x + e\right )^{2}}}{a f \cos \left (f x + e\right ) \sin \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^2/(a-a*sin(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

-sqrt(a*cos(f*x + e)^2)/(a*f*cos(f*x + e)*sin(f*x + e))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)**2/(a-a*sin(f*x+e)**2)**(1/2),x)

[Out]

Integral(cot(e + f*x)**2/sqrt(-a*(sin(e + f*x) - 1)*(sin(e + f*x) + 1)), x)

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Giac [B]  time = 1.2792, size = 90, normalized size = 3.6 \begin{align*} \frac{\frac{\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{\mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 1\right )} + \frac{1}{\mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 1\right ) \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}}{2 \, \sqrt{a} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^2/(a-a*sin(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

1/2*(tan(1/2*f*x + 1/2*e)/sgn(tan(1/2*f*x + 1/2*e)^4 - 1) + 1/(sgn(tan(1/2*f*x + 1/2*e)^4 - 1)*tan(1/2*f*x + 1
/2*e)))/(sqrt(a)*f)