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

Optimal. Leaf size=14 \[ -\cot (x) \sqrt{a \sec ^2(x)} \]

[Out]

-(Cot[x]*Sqrt[a*Sec[x]^2])

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Rubi [A]  time = 0.0804953, antiderivative size = 14, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {3657, 4125, 2606, 8} \[ -\cot (x) \sqrt{a \sec ^2(x)} \]

Antiderivative was successfully verified.

[In]

Int[Cot[x]^2*Sqrt[a + a*Tan[x]^2],x]

[Out]

-(Cot[x]*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 4125

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

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

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

Mathematica [A]  time = 0.0141342, size = 14, normalized size = 1. \[ -\cot (x) \sqrt{a \sec ^2(x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Cot[x]*Sqrt[a*Sec[x]^2])

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Maple [A]  time = 0.083, size = 17, normalized size = 1.2 \begin{align*} -{\frac{\cos \left ( x \right ) }{\sin \left ( x \right ) }\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)^2*(a+a*tan(x)^2)^(1/2),x)

[Out]

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

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Maxima [A]  time = 1.65049, size = 23, normalized size = 1.64 \begin{align*} -\frac{\sqrt{\tan \left (x\right )^{2} + 1} \sqrt{a}}{\tan \left (x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(tan(x)^2 + 1)*sqrt(a)/tan(x)

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Fricas [A]  time = 1.28554, size = 41, normalized size = 2.93 \begin{align*} -\frac{\sqrt{a \tan \left (x\right )^{2} + a}}{\tan \left (x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(a*tan(x)^2 + a)/tan(x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.1129, size = 43, normalized size = 3.07 \begin{align*} \frac{2 \, a^{\frac{3}{2}}}{{\left (\sqrt{a} \tan \left (x\right ) - \sqrt{a \tan \left (x\right )^{2} + a}\right )}^{2} - a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a^(3/2)/((sqrt(a)*tan(x) - sqrt(a*tan(x)^2 + a))^2 - a)