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

Optimal. Leaf size=57 \[ -\frac{\tan (e+f x) \sqrt{a \cos ^2(e+f x)}}{f}-\frac{\csc (e+f x) \sec (e+f x) \sqrt{a \cos ^2(e+f x)}}{f} \]

[Out]

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

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Rubi [A]  time = 0.111669, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3176, 3207, 2590, 14} \[ -\frac{\tan (e+f x) \sqrt{a \cos ^2(e+f x)}}{f}-\frac{\csc (e+f x) \sec (e+f x) \sqrt{a \cos ^2(e+f x)}}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2590

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[f^(-1), Subst[Int[(1 - x^2
)^((m + n - 1)/2)/x^n, x], x, Cos[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/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 \cot ^2(e+f x) \sqrt{a-a \sin ^2(e+f x)} \, dx &=\int \sqrt{a \cos ^2(e+f x)} \cot ^2(e+f x) \, dx\\ &=\left (\sqrt{a \cos ^2(e+f x)} \sec (e+f x)\right ) \int \cos (e+f x) \cot ^2(e+f x) \, dx\\ &=-\frac{\left (\sqrt{a \cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1-x^2}{x^2} \, dx,x,-\sin (e+f x)\right )}{f}\\ &=-\frac{\left (\sqrt{a \cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname{Subst}\left (\int \left (-1+\frac{1}{x^2}\right ) \, dx,x,-\sin (e+f x)\right )}{f}\\ &=-\frac{\sqrt{a \cos ^2(e+f x)} \csc (e+f x) \sec (e+f x)}{f}-\frac{\sqrt{a \cos ^2(e+f x)} \tan (e+f x)}{f}\\ \end{align*}

Mathematica [A]  time = 0.0753841, size = 35, normalized size = 0.61 \[ -\frac{\tan (e+f x) \left (\csc ^2(e+f x)+1\right ) \sqrt{a \cos ^2(e+f x)}}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[a*Cos[e + f*x]^2]*(1 + Csc[e + f*x]^2)*Tan[e + f*x])/f)

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Maple [A]  time = 0.707, size = 43, normalized size = 0.8 \begin{align*} -{\frac{\cos \left ( fx+e \right ) a \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{2}+1 \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)*a*(sin(f*x+e)^2+1)/sin(f*x+e)/(a*cos(f*x+e)^2)^(1/2)/f

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Maxima [A]  time = 1.48685, size = 57, normalized size = 1. \begin{align*} -\frac{2 \, \sqrt{a} \tan \left (f x + e\right )^{2} + \sqrt{a}}{\sqrt{\tan \left (f x + e\right )^{2} + 1} f \tan \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="maxima")

[Out]

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

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Fricas [A]  time = 1.57825, size = 101, normalized size = 1.77 \begin{align*} \frac{\sqrt{a \cos \left (f x + e\right )^{2}}{\left (\cos \left (f x + e\right )^{2} - 2\right )}}{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)*(cos(f*x + e)^2 - 2)/(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 \sqrt{- a \left (\sin{\left (e + f x \right )} - 1\right ) \left (\sin{\left (e + f x \right )} + 1\right )} \cot ^{2}{\left (e + f x \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(sqrt(-a*(sin(e + f*x) - 1)*(sin(e + f*x) + 1))*cot(e + f*x)**2, x)

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Giac [A]  time = 1.19357, size = 122, normalized size = 2.14 \begin{align*} \frac{{\left ({\left (\frac{1}{\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )} + \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 1\right ) + \frac{4 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 1\right )}{\frac{1}{\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )} + \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}\right )} \sqrt{a}}{2 \, 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*((1/tan(1/2*f*x + 1/2*e) + tan(1/2*f*x + 1/2*e))*sgn(tan(1/2*f*x + 1/2*e)^4 - 1) + 4*sgn(tan(1/2*f*x + 1/2
*e)^4 - 1)/(1/tan(1/2*f*x + 1/2*e) + tan(1/2*f*x + 1/2*e)))*sqrt(a)/f