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

Optimal. Leaf size=50 \[ \frac{\sqrt{a \cos ^2(e+f x)}}{f}-\frac{\sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a \cos ^2(e+f x)}}{\sqrt{a}}\right )}{f} \]

[Out]

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

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Rubi [A]  time = 0.0790169, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {3176, 3205, 50, 63, 206} \[ \frac{\sqrt{a \cos ^2(e+f x)}}{f}-\frac{\sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a \cos ^2(e+f x)}}{\sqrt{a}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[a]*ArcTanh[Sqrt[a*Cos[e + f*x]^2]/Sqrt[a]])/f) + Sqrt[a*Cos[e + f*x]^2]/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 3205

Int[((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFact
ors[Sin[e + f*x]^2, x]}, Dist[ff^((m + 1)/2)/(2*f), Subst[Int[(x^((m - 1)/2)*(b*ff^(n/2)*x^(n/2))^p)/(1 - ff*x
)^((m + 1)/2), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{b, e, f, p}, x] && IntegerQ[(m - 1)/2] && IntegerQ[n/2
]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0692788, size = 55, normalized size = 1.1 \[ \frac{\sec (e+f x) \sqrt{a \cos ^2(e+f x)} \left (\cos (e+f x)+\log \left (\sin \left (\frac{1}{2} (e+f x)\right )\right )-\log \left (\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 1.521, size = 55, normalized size = 1.1 \begin{align*} -{\frac{1}{f}\sqrt{a}\ln \left ( 2\,{\frac{\sqrt{a}\sqrt{a \left ( \cos \left ( fx+e \right ) \right ) ^{2}}+a}{\sin \left ( fx+e \right ) }} \right ) }+{\frac{1}{f}\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)*(a-a*sin(f*x+e)^2)^(1/2),x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.65234, size = 146, normalized size = 2.92 \begin{align*} \frac{\sqrt{a \cos \left (f x + e\right )^{2}}{\left (2 \, \cos \left (f x + e\right ) - \log \left (-\frac{\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1}\right )\right )}}{2 \, f \cos \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(a*cos(f*x + e)^2)*(2*cos(f*x + e) - log(-(cos(f*x + e) + 1)/(cos(f*x + e) - 1)))/(f*cos(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{\left (e + f x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)*(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), x)

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Giac [A]  time = 1.09789, size = 69, normalized size = 1.38 \begin{align*} \frac{\frac{a \arctan \left (\frac{\sqrt{-a \sin \left (f x + e\right )^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} + \sqrt{-a \sin \left (f x + e\right )^{2} + a}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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