3.713 \(\int \sec ^2(x) \sqrt{1-\tan ^2(x)} \, dx\)

Optimal. Leaf size=26 \[ \frac{1}{2} \tan (x) \sqrt{1-\tan ^2(x)}+\frac{1}{2} \sin ^{-1}(\tan (x)) \]

[Out]

ArcSin[Tan[x]]/2 + (Tan[x]*Sqrt[1 - Tan[x]^2])/2

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Rubi [A]  time = 0.0458562, antiderivative size = 26, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {3675, 195, 216} \[ \frac{1}{2} \tan (x) \sqrt{1-\tan ^2(x)}+\frac{1}{2} \sin ^{-1}(\tan (x)) \]

Antiderivative was successfully verified.

[In]

Int[Sec[x]^2*Sqrt[1 - Tan[x]^2],x]

[Out]

ArcSin[Tan[x]]/2 + (Tan[x]*Sqrt[1 - Tan[x]^2])/2

Rule 3675

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/(c^(m - 1)*f), Subst[Int[(c^2 + ff^2*x^2)^(m/2 - 1)*(a + b*(ff*x)
^n)^p, x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2] && (IntegersQ[n, p
] || IGtQ[m, 0] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

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

Mathematica [B]  time = 0.112636, size = 63, normalized size = 2.42 \[ \frac{\cos (2 x) \tan (x)+\sqrt{\cos ^2(x)} \cos (x) \sqrt{1-\tan ^2(x)} \sin ^{-1}\left (\frac{\sin (x)}{\sqrt{\cos ^2(x)}}\right )}{2 \sqrt{\cos ^2(x)} \sqrt{\cos (2 x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[x]^2*Sqrt[1 - Tan[x]^2],x]

[Out]

(Cos[2*x]*Tan[x] + ArcSin[Sin[x]/Sqrt[Cos[x]^2]]*Cos[x]*Sqrt[Cos[x]^2]*Sqrt[1 - Tan[x]^2])/(2*Sqrt[Cos[x]^2]*S
qrt[Cos[2*x]])

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Maple [C]  time = 0.21, size = 492, normalized size = 18.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(x)^2*(1-tan(x)^2)^(1/2),x)

[Out]

1/2/(1+2^(1/2))/(3+2*2^(1/2))^(1/2)*sin(x)*(-2^(1/2)*((cos(x)*2^(1/2)-2^(1/2)+2*cos(x)-1)/(1+cos(x)))^(1/2)*(-
2*(cos(x)*2^(1/2)-2^(1/2)-2*cos(x)+1)/(1+cos(x)))^(1/2)*EllipticF((1+2^(1/2))*(-1+cos(x))/sin(x),3-2*2^(1/2))*
cos(x)^2*sin(x)+2*2^(1/2)*((cos(x)*2^(1/2)-2^(1/2)+2*cos(x)-1)/(1+cos(x)))^(1/2)*(-2*(cos(x)*2^(1/2)-2^(1/2)-2
*cos(x)+1)/(1+cos(x)))^(1/2)*EllipticPi((3+2*2^(1/2))^(1/2)*(-1+cos(x))/sin(x),1/(3+2*2^(1/2)),(3-2*2^(1/2))^(
1/2)/(3+2*2^(1/2))^(1/2))*cos(x)^2*sin(x)-2*((cos(x)*2^(1/2)-2^(1/2)+2*cos(x)-1)/(1+cos(x)))^(1/2)*(-2*(cos(x)
*2^(1/2)-2^(1/2)-2*cos(x)+1)/(1+cos(x)))^(1/2)*EllipticF((1+2^(1/2))*(-1+cos(x))/sin(x),3-2*2^(1/2))*cos(x)^2*
sin(x)+4*((cos(x)*2^(1/2)-2^(1/2)+2*cos(x)-1)/(1+cos(x)))^(1/2)*(-2*(cos(x)*2^(1/2)-2^(1/2)-2*cos(x)+1)/(1+cos
(x)))^(1/2)*EllipticPi((3+2*2^(1/2))^(1/2)*(-1+cos(x))/sin(x),1/(3+2*2^(1/2)),(3-2*2^(1/2))^(1/2)/(3+2*2^(1/2)
)^(1/2))*cos(x)^2*sin(x)+4*cos(x)^3*2^(1/2)-4*cos(x)^2*2^(1/2)+6*cos(x)^3-2*cos(x)*2^(1/2)-6*cos(x)^2+2*2^(1/2
)-3*cos(x)+3)*((2*cos(x)^2-1)/cos(x)^2)^(1/2)/(-1+cos(x))/(2*cos(x)^2-1)/cos(x)

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Maxima [A]  time = 1.46187, size = 27, normalized size = 1.04 \begin{align*} \frac{1}{2} \, \sqrt{-\tan \left (x\right )^{2} + 1} \tan \left (x\right ) + \frac{1}{2} \, \arcsin \left (\tan \left (x\right )\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^2*(1-tan(x)^2)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.34886, size = 215, normalized size = 8.27 \begin{align*} -\frac{\arctan \left (\frac{{\left (3 \, \cos \left (x\right )^{3} - 2 \, \cos \left (x\right )\right )} \sqrt{\frac{2 \, \cos \left (x\right )^{2} - 1}{\cos \left (x\right )^{2}}}}{2 \,{\left (2 \, \cos \left (x\right )^{2} - 1\right )} \sin \left (x\right )}\right ) \cos \left (x\right ) - 2 \, \sqrt{\frac{2 \, \cos \left (x\right )^{2} - 1}{\cos \left (x\right )^{2}}} \sin \left (x\right )}{4 \, \cos \left (x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^2*(1-tan(x)^2)^(1/2),x, algorithm="fricas")

[Out]

-1/4*(arctan(1/2*(3*cos(x)^3 - 2*cos(x))*sqrt((2*cos(x)^2 - 1)/cos(x)^2)/((2*cos(x)^2 - 1)*sin(x)))*cos(x) - 2
*sqrt((2*cos(x)^2 - 1)/cos(x)^2)*sin(x))/cos(x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)**2*(1-tan(x)**2)**(1/2),x)

[Out]

Integral(sqrt(-(tan(x) - 1)*(tan(x) + 1))*sec(x)**2, x)

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Giac [A]  time = 1.10653, size = 27, normalized size = 1.04 \begin{align*} \frac{1}{2} \, \sqrt{-\tan \left (x\right )^{2} + 1} \tan \left (x\right ) + \frac{1}{2} \, \arcsin \left (\tan \left (x\right )\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^2*(1-tan(x)^2)^(1/2),x, algorithm="giac")

[Out]

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