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

Optimal. Leaf size=31 \[ -\frac {\csc (x) \sec (x)}{\sqrt {a \sec ^2(x)}}-\frac {\tan (x)}{\sqrt {a \sec ^2(x)}} \]

[Out]

-csc(x)*sec(x)/(a*sec(x)^2)^(1/2)-tan(x)/(a*sec(x)^2)^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {3657, 4125, 2590, 14} \[ -\frac {\csc (x) \sec (x)}{\sqrt {a \sec ^2(x)}}-\frac {\tan (x)}{\sqrt {a \sec ^2(x)}} \]

Antiderivative was successfully verified.

[In]

Int[Cot[x]^2/Sqrt[a + a*Tan[x]^2],x]

[Out]

-((Csc[x]*Sec[x])/Sqrt[a*Sec[x]^2]) - Tan[x]/Sqrt[a*Sec[x]^2]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2590

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[f^(-1), Subst[Int[(1 - x^2
)^((m + n - 1)/2)/x^n, x], x, Cos[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/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
]])

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 22, normalized size = 0.71 \[ \frac {-\tan (x)-\csc (x) \sec (x)}{\sqrt {a \sec ^2(x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(Csc[x]*Sec[x]) - Tan[x])/Sqrt[a*Sec[x]^2]

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fricas [A]  time = 0.39, size = 33, normalized size = 1.06 \[ -\frac {\sqrt {a \tan \relax (x)^{2} + a} {\left (2 \, \tan \relax (x)^{2} + 1\right )}}{a \tan \relax (x)^{3} + a \tan \relax (x)} \]

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)*(2*tan(x)^2 + 1)/(a*tan(x)^3 + a*tan(x))

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giac [A]  time = 0.32, size = 47, normalized size = 1.52 \[ -\frac {\tan \relax (x)}{\sqrt {a \tan \relax (x)^{2} + a}} + \frac {2 \, \sqrt {a}}{{\left (\sqrt {a} \tan \relax (x) - \sqrt {a \tan \relax (x)^{2} + a}\right )}^{2} - a} \]

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]

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

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maple [A]  time = 0.58, size = 24, normalized size = 0.77 \[ \frac {\cos ^{2}\relax (x )-2}{\sin \relax (x ) \cos \relax (x ) \sqrt {\frac {a}{\cos \relax (x )^{2}}}} \]

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

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maxima [B]  time = 1.48, size = 128, normalized size = 4.13 \[ \frac {{\left ({\left (\sin \left (3 \, x\right ) - \sin \relax (x)\right )} \cos \left (4 \, x\right ) - {\left (\cos \left (3 \, x\right ) - \cos \relax (x)\right )} \sin \left (4 \, x\right ) - {\left (6 \, \cos \left (2 \, x\right ) - 1\right )} \sin \left (3 \, x\right ) + 6 \, \cos \left (3 \, x\right ) \sin \left (2 \, x\right ) - 6 \, \cos \relax (x) \sin \left (2 \, x\right ) + 6 \, \cos \left (2 \, x\right ) \sin \relax (x) - \sin \relax (x)\right )} \sqrt {a}}{2 \, {\left (a \cos \left (3 \, x\right )^{2} - 2 \, a \cos \left (3 \, x\right ) \cos \relax (x) + a \cos \relax (x)^{2} + a \sin \left (3 \, x\right )^{2} - 2 \, a \sin \left (3 \, x\right ) \sin \relax (x) + a \sin \relax (x)^{2}\right )}} \]

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]

1/2*((sin(3*x) - sin(x))*cos(4*x) - (cos(3*x) - cos(x))*sin(4*x) - (6*cos(2*x) - 1)*sin(3*x) + 6*cos(3*x)*sin(
2*x) - 6*cos(x)*sin(2*x) + 6*cos(2*x)*sin(x) - sin(x))*sqrt(a)/(a*cos(3*x)^2 - 2*a*cos(3*x)*cos(x) + a*cos(x)^
2 + a*sin(3*x)^2 - 2*a*sin(3*x)*sin(x) + a*sin(x)^2)

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mupad [B]  time = 11.82, size = 40, normalized size = 1.29 \[ \frac {\sqrt {2}\,\left (6\,\sin \left (2\,x\right )-2\,\sin \left (2\,x\right )\,\left (2\,{\cos \relax (x)}^2-1\right )\right )}{8\,\sqrt {a}\,\sqrt {2\,{\cos \relax (x)}^2}\,\left ({\cos \relax (x)}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2^(1/2)*(6*sin(2*x) - 2*sin(2*x)*(2*cos(x)^2 - 1)))/(8*a^(1/2)*(2*cos(x)^2)^(1/2)*(cos(x)^2 - 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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