3.5.60 \(\int \cot (e+f x) \sqrt {a-a \sin ^2(e+f x)} \, dx\) [460]

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

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 50, normalized size of antiderivative = 1.00, 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 = {3255, 3284, 52, 65, 212} \begin {gather*} \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} \end {gather*}

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 52

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 65

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 212

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

Rule 3255

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 3284

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
]

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 {\text {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 \text {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 {\text {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*}

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Mathematica [A]
time = 0.05, size = 55, normalized size = 1.10 \begin {gather*} \frac {\sqrt {a \cos ^2(e+f x)} \left (\cos (e+f x)-\log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right )+\log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right )\right ) \sec (e+f x)}{f} \end {gather*}

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 = 11.22, size = 52, normalized size = 1.04

method	result	size
default	\(\frac {\sqrt {a \left (\cos ^{2}\left (f x +e \right )\right )}-\sqrt {a}\, \ln \left (\frac {2 \sqrt {a}\, \sqrt {a \left (\cos ^{2}\left (f x +e \right )\right )}+2 a}{\sin \left (f x +e \right )}\right )}{f}\)	\(52\)
risch	\(\frac {\sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, {\mathrm e}^{2 i \left (f x +e \right )}}{2 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}+\frac {\sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}}{2 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}-\frac {\ln \left ({\mathrm e}^{i f x}+{\mathrm e}^{-i e}\right ) \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, {\mathrm e}^{i \left (f x +e \right )}}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}+\frac {\ln \left ({\mathrm e}^{i f x}-{\mathrm e}^{-i e}\right ) \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, {\mathrm e}^{i \left (f x +e \right )}}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}\)	\(232\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.52, size = 74, normalized size = 1.48 \begin {gather*} -\frac {\sqrt {a} \log \left (\frac {2 \, \sqrt {-a \sin \left (f x + e\right )^{2} + a} \sqrt {a}}{{\left | \sin \left (f x + e\right ) \right |}} + \frac {2 \, a}{{\left | \sin \left (f x + e\right ) \right |}}\right ) - \sqrt {-a \sin \left (f x + e\right )^{2} + a}}{f} \end {gather*}

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]

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

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Fricas [A]
time = 0.40, size = 57, normalized size = 1.14 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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 = 0.45, size = 55, normalized size = 1.10 \begin {gather*} \frac {a {\left (\frac {\arctan \left (\frac {\sqrt {-a \sin \left (f x + e\right )^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + \frac {\sqrt {-a \sin \left (f x + e\right )^{2} + a}}{a}\right )}}{f} \end {gather*}

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)/a)/f

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \mathrm {cot}\left (e+f\,x\right )\,\sqrt {a-a\,{\sin \left (e+f\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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