3.276 \(\int \frac{\cot (x)}{\sqrt{a+a \tan ^2(x)}} \, dx\)

Optimal. Leaf size=35 \[ \frac{1}{\sqrt{a \sec ^2(x)}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a \sec ^2(x)}}{\sqrt{a}}\right )}{\sqrt{a}} \]

[Out]

-(ArcTanh[Sqrt[a*Sec[x]^2]/Sqrt[a]]/Sqrt[a]) + 1/Sqrt[a*Sec[x]^2]

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Rubi [A]  time = 0.0796105, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3657, 4124, 51, 63, 207} \[ \frac{1}{\sqrt{a \sec ^2(x)}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a \sec ^2(x)}}{\sqrt{a}}\right )}{\sqrt{a}} \]

Antiderivative was successfully verified.

[In]

Int[Cot[x]/Sqrt[a + a*Tan[x]^2],x]

[Out]

-(ArcTanh[Sqrt[a*Sec[x]^2]/Sqrt[a]]/Sqrt[a]) + 1/Sqrt[a*Sec[x]^2]

Rule 3657

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

Rule 4124

Int[((b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> Dist[b/(2*f), Subst[In
t[(-1 + x)^((m - 1)/2)*(b*x)^(p - 1), x], x, Sec[e + f*x]^2], x] /; FreeQ[{b, e, f, p}, x] &&  !IntegerQ[p] &&
IntegerQ[(m - 1)/2]

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 207

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

Rubi steps

\begin{align*} \int \frac{\cot (x)}{\sqrt{a+a \tan ^2(x)}} \, dx &=\int \frac{\cot (x)}{\sqrt{a \sec ^2(x)}} \, dx\\ &=\frac{1}{2} a \operatorname{Subst}\left (\int \frac{1}{(-1+x) (a x)^{3/2}} \, dx,x,\sec ^2(x)\right )\\ &=\frac{1}{\sqrt{a \sec ^2(x)}}+\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{a x}} \, dx,x,\sec ^2(x)\right )\\ &=\frac{1}{\sqrt{a \sec ^2(x)}}+\frac{\operatorname{Subst}\left (\int \frac{1}{-1+\frac{x^2}{a}} \, dx,x,\sqrt{a \sec ^2(x)}\right )}{a}\\ &=-\frac{\tanh ^{-1}\left (\frac{\sqrt{a \sec ^2(x)}}{\sqrt{a}}\right )}{\sqrt{a}}+\frac{1}{\sqrt{a \sec ^2(x)}}\\ \end{align*}

Mathematica [A]  time = 0.0328115, size = 32, normalized size = 0.91 \[ \frac{\sec (x) \left (\cos (x)+\log \left (\sin \left (\frac{x}{2}\right )\right )-\log \left (\cos \left (\frac{x}{2}\right )\right )\right )}{\sqrt{a \sec ^2(x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[x]/Sqrt[a + a*Tan[x]^2],x]

[Out]

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

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Maple [A]  time = 0.079, size = 29, normalized size = 0.8 \begin{align*}{\frac{1}{\cos \left ( x \right ) } \left ( \cos \left ( x \right ) +\ln \left ( -{\frac{\cos \left ( x \right ) -1}{\sin \left ( x \right ) }} \right ) +1 \right ){\frac{1}{\sqrt{{\frac{a}{ \left ( \cos \left ( x \right ) \right ) ^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(x)/(a+a*tan(x)^2)^(1/2),x)

[Out]

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

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Maxima [A]  time = 1.92286, size = 57, normalized size = 1.63 \begin{align*} \frac{2 \, \cos \left (x\right ) - \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} + 2 \, \cos \left (x\right ) + 1\right ) + \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} - 2 \, \cos \left (x\right ) + 1\right )}{2 \, \sqrt{a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)/(a+a*tan(x)^2)^(1/2),x, algorithm="maxima")

[Out]

1/2*(2*cos(x) - log(cos(x)^2 + sin(x)^2 + 2*cos(x) + 1) + log(cos(x)^2 + sin(x)^2 - 2*cos(x) + 1))/sqrt(a)

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Fricas [B]  time = 1.40237, size = 186, normalized size = 5.31 \begin{align*} \frac{{\left (\tan \left (x\right )^{2} + 1\right )} \sqrt{a} \log \left (\frac{a \tan \left (x\right )^{2} - 2 \, \sqrt{a \tan \left (x\right )^{2} + a} \sqrt{a} + 2 \, a}{\tan \left (x\right )^{2}}\right ) + 2 \, \sqrt{a \tan \left (x\right )^{2} + a}}{2 \,{\left (a \tan \left (x\right )^{2} + a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)/(a+a*tan(x)^2)^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)/(a+a*tan(x)**2)**(1/2),x)

[Out]

Integral(cot(x)/sqrt(a*(tan(x)**2 + 1)), x)

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Giac [A]  time = 1.0731, size = 58, normalized size = 1.66 \begin{align*} a{\left (\frac{\arctan \left (\frac{\sqrt{a \tan \left (x\right )^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a} + \frac{1}{\sqrt{a \tan \left (x\right )^{2} + a} a}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)/(a+a*tan(x)^2)^(1/2),x, algorithm="giac")

[Out]

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