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((a*sec(x)^2)^(1/2)/a^(1/2))/a^(1/2)+1/(a*sec(x)^2)^(1/2)

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Rubi [A]  time = 0.08, antiderivative size = 35, normalized size of antiderivative = 1.00, 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 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])

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]

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*}

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Mathematica [A]  time = 0.03, 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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fricas [B]  time = 0.44, size = 66, normalized size = 1.89 \[ \frac {{\left (\tan \relax (x)^{2} + 1\right )} \sqrt {a} \log \left (\frac {a \tan \relax (x)^{2} - 2 \, \sqrt {a \tan \relax (x)^{2} + a} \sqrt {a} + 2 \, a}{\tan \relax (x)^{2}}\right ) + 2 \, \sqrt {a \tan \relax (x)^{2} + a}}{2 \, {\left (a \tan \relax (x)^{2} + a\right )}} \]

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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giac [A]  time = 0.35, size = 34, normalized size = 0.97 \[ \frac {\arctan \left (\frac {\sqrt {a \tan \relax (x)^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + \frac {1}{\sqrt {a \tan \relax (x)^{2} + a}} \]

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]

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

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maple [A]  time = 0.62, size = 29, normalized size = 0.83 \[ \frac {\cos \relax (x )+\ln \left (-\frac {-1+\cos \relax (x )}{\sin \relax (x )}\right )+1}{\sqrt {\frac {a}{\cos \relax (x )^{2}}}\, \cos \relax (x )} \]

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

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maxima [A]  time = 0.90, size = 42, normalized size = 1.20 \[ \frac {2 \, \cos \relax (x) - \log \left (\cos \relax (x)^{2} + \sin \relax (x)^{2} + 2 \, \cos \relax (x) + 1\right ) + \log \left (\cos \relax (x)^{2} + \sin \relax (x)^{2} - 2 \, \cos \relax (x) + 1\right )}{2 \, \sqrt {a}} \]

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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mupad [B]  time = 0.14, size = 31, normalized size = 0.89 \[ \frac {1}{\sqrt {a\,{\mathrm {tan}\relax (x)}^2+a}}-\frac {\mathrm {atanh}\left (\frac {\sqrt {a\,{\mathrm {tan}\relax (x)}^2+a}}{\sqrt {a}}\right )}{\sqrt {a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(x)/(a + a*tan(x)^2)^(1/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot {\relax (x )}}{\sqrt {a \left (\tan ^{2}{\relax (x )} + 1\right )}}\, dx \]

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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