3.20 \(\int \cot (x) \sqrt{a+b \cot ^2(x)} \, dx\)

Optimal. Leaf size=48 \[ \sqrt{a-b} \tanh ^{-1}\left (\frac{\sqrt{a+b \cot ^2(x)}}{\sqrt{a-b}}\right )-\sqrt{a+b \cot ^2(x)} \]

[Out]

Sqrt[a - b]*ArcTanh[Sqrt[a + b*Cot[x]^2]/Sqrt[a - b]] - Sqrt[a + b*Cot[x]^2]

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Rubi [A]  time = 0.0667223, antiderivative size = 48, 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 = {3670, 444, 50, 63, 208} \[ \sqrt{a-b} \tanh ^{-1}\left (\frac{\sqrt{a+b \cot ^2(x)}}{\sqrt{a-b}}\right )-\sqrt{a+b \cot ^2(x)} \]

Antiderivative was successfully verified.

[In]

Int[Cot[x]*Sqrt[a + b*Cot[x]^2],x]

[Out]

Sqrt[a - b]*ArcTanh[Sqrt[a + b*Cot[x]^2]/Sqrt[a - b]] - Sqrt[a + b*Cot[x]^2]

Rule 3670

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff)/f, Subst[Int[(((d*ff*x)/c)^m*(a + b*(ff*x)^n)^p)/(c^
2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

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 208

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

Rubi steps

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

Mathematica [A]  time = 0.0255438, size = 48, normalized size = 1. \[ \sqrt{a-b} \tanh ^{-1}\left (\frac{\sqrt{a+b \cot ^2(x)}}{\sqrt{a-b}}\right )-\sqrt{a+b \cot ^2(x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[a - b]*ArcTanh[Sqrt[a + b*Cot[x]^2]/Sqrt[a - b]] - Sqrt[a + b*Cot[x]^2]

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Maple [A]  time = 0.022, size = 71, normalized size = 1.5 \begin{align*} -\sqrt{a+b \left ( \cot \left ( x \right ) \right ) ^{2}}+{b\arctan \left ({\sqrt{a+b \left ( \cot \left ( x \right ) \right ) ^{2}}{\frac{1}{\sqrt{-a+b}}}} \right ){\frac{1}{\sqrt{-a+b}}}}-{a\arctan \left ({\sqrt{a+b \left ( \cot \left ( x \right ) \right ) ^{2}}{\frac{1}{\sqrt{-a+b}}}} \right ){\frac{1}{\sqrt{-a+b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.22684, size = 587, normalized size = 12.23 \begin{align*} \left [\frac{1}{4} \, \sqrt{a - b} \log \left (-2 \,{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (2 \, x\right )^{2} - 2 \, a^{2} + b^{2} - 2 \,{\left ({\left (a - b\right )} \cos \left (2 \, x\right )^{2} -{\left (2 \, a - b\right )} \cos \left (2 \, x\right ) + a\right )} \sqrt{a - b} \sqrt{\frac{{\left (a - b\right )} \cos \left (2 \, x\right ) - a - b}{\cos \left (2 \, x\right ) - 1}} + 4 \,{\left (a^{2} - a b\right )} \cos \left (2 \, x\right )\right ) - \sqrt{\frac{{\left (a - b\right )} \cos \left (2 \, x\right ) - a - b}{\cos \left (2 \, x\right ) - 1}}, \frac{1}{2} \, \sqrt{-a + b} \arctan \left (-\frac{\sqrt{-a + b} \sqrt{\frac{{\left (a - b\right )} \cos \left (2 \, x\right ) - a - b}{\cos \left (2 \, x\right ) - 1}}{\left (\cos \left (2 \, x\right ) - 1\right )}}{{\left (a - b\right )} \cos \left (2 \, x\right ) - a}\right ) - \sqrt{\frac{{\left (a - b\right )} \cos \left (2 \, x\right ) - a - b}{\cos \left (2 \, x\right ) - 1}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*sqrt(a - b)*log(-2*(a^2 - 2*a*b + b^2)*cos(2*x)^2 - 2*a^2 + b^2 - 2*((a - b)*cos(2*x)^2 - (2*a - b)*cos(2
*x) + a)*sqrt(a - b)*sqrt(((a - b)*cos(2*x) - a - b)/(cos(2*x) - 1)) + 4*(a^2 - a*b)*cos(2*x)) - sqrt(((a - b)
*cos(2*x) - a - b)/(cos(2*x) - 1)), 1/2*sqrt(-a + b)*arctan(-sqrt(-a + b)*sqrt(((a - b)*cos(2*x) - a - b)/(cos
(2*x) - 1))*(cos(2*x) - 1)/((a - b)*cos(2*x) - a)) - sqrt(((a - b)*cos(2*x) - a - b)/(cos(2*x) - 1))]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a + b*cot(x)**2)*cot(x), x)

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Giac [B]  time = 1.31237, size = 128, normalized size = 2.67 \begin{align*} -\frac{1}{2} \,{\left (\sqrt{a - b} \log \left ({\left (\sqrt{a - b} \sin \left (x\right ) - \sqrt{a \sin \left (x\right )^{2} - b \sin \left (x\right )^{2} + b}\right )}^{2}\right ) - \frac{4 \, \sqrt{a - b} b}{{\left (\sqrt{a - b} \sin \left (x\right ) - \sqrt{a \sin \left (x\right )^{2} - b \sin \left (x\right )^{2} + b}\right )}^{2} - b}\right )} \mathrm{sgn}\left (\sin \left (x\right )\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(sqrt(a - b)*log((sqrt(a - b)*sin(x) - sqrt(a*sin(x)^2 - b*sin(x)^2 + b))^2) - 4*sqrt(a - b)*b/((sqrt(a -
b)*sin(x) - sqrt(a*sin(x)^2 - b*sin(x)^2 + b))^2 - b))*sgn(sin(x))