3.27 \(\int \frac {1}{\sqrt {a \cot ^2(x)}} \, dx\)

Optimal. Leaf size=17 \[ -\frac {\cot (x) \log (\cos (x))}{\sqrt {a \cot ^2(x)}} \]

[Out]

-cot(x)*ln(cos(x))/(a*cot(x)^2)^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3658, 3475} \[ -\frac {\cot (x) \log (\cos (x))}{\sqrt {a \cot ^2(x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[a*Cot[x]^2],x]

[Out]

-((Cot[x]*Log[Cos[x]])/Sqrt[a*Cot[x]^2])

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3658

Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Tan[e + f*x]^n)^FracPart[p])/(Tan[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Tan[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
]])

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {a \cot ^2(x)}} \, dx &=\frac {\cot (x) \int \tan (x) \, dx}{\sqrt {a \cot ^2(x)}}\\ &=-\frac {\cot (x) \log (\cos (x))}{\sqrt {a \cot ^2(x)}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 17, normalized size = 1.00 \[ -\frac {\cot (x) \log (\cos (x))}{\sqrt {a \cot ^2(x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[a*Cot[x]^2],x]

[Out]

-((Cot[x]*Log[Cos[x]])/Sqrt[a*Cot[x]^2])

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fricas [B]  time = 0.70, size = 45, normalized size = 2.65 \[ -\frac {\sqrt {-\frac {a \cos \left (2 \, x\right ) + a}{\cos \left (2 \, x\right ) - 1}} \log \left (\frac {1}{2} \, \cos \left (2 \, x\right ) + \frac {1}{2}\right ) \sin \left (2 \, x\right )}{2 \, {\left (a \cos \left (2 \, x\right ) + a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.45, size = 19, normalized size = 1.12 \[ -\frac {\log \left ({\left | \cos \relax (x) \right |}\right )}{\sqrt {a} \mathrm {sgn}\left (\cos \relax (x)\right ) \mathrm {sgn}\left (\sin \relax (x)\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(abs(cos(x)))/(sqrt(a)*sgn(cos(x))*sgn(sin(x)))

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maple [A]  time = 0.19, size = 28, normalized size = 1.65 \[ -\frac {\cot \relax (x ) \left (2 \ln \left (\cot \relax (x )\right )-\ln \left (\cot ^{2}\relax (x )+1\right )\right )}{2 \sqrt {a \left (\cot ^{2}\relax (x )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*cot(x)*(2*ln(cot(x))-ln(cot(x)^2+1))/(a*cot(x)^2)^(1/2)

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maxima [A]  time = 0.50, size = 12, normalized size = 0.71 \[ \frac {\log \left (\tan \relax (x)^{2} + 1\right )}{2 \, \sqrt {a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*log(tan(x)^2 + 1)/sqrt(a)

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mupad [B]  time = 0.29, size = 25, normalized size = 1.47 \[ -\frac {\mathrm {atan}\left (\frac {\sqrt {-a}\,\mathrm {cot}\relax (x)}{\sqrt {a}\,\sqrt {{\mathrm {cot}\relax (x)}^2}}\right )}{\sqrt {-a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-atan(((-a)^(1/2)*cot(x))/(a^(1/2)*(cot(x)^2)^(1/2)))/(-a)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(a*cot(x)**2), x)

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