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]

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

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Rubi [A]  time = 0.05, antiderivative size = 26, normalized size of antiderivative = 1.00, 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 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]

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])

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*}

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Mathematica [B]  time = 0.12, 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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fricas [B]  time = 1.05, size = 72, normalized size = 2.77 \[ -\frac {\arctan \left (\frac {{\left (3 \, \cos \relax (x)^{3} - 2 \, \cos \relax (x)\right )} \sqrt {\frac {2 \, \cos \relax (x)^{2} - 1}{\cos \relax (x)^{2}}}}{2 \, {\left (2 \, \cos \relax (x)^{2} - 1\right )} \sin \relax (x)}\right ) \cos \relax (x) - 2 \, \sqrt {\frac {2 \, \cos \relax (x)^{2} - 1}{\cos \relax (x)^{2}}} \sin \relax (x)}{4 \, \cos \relax (x)} \]

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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giac [A]  time = 0.15, size = 20, normalized size = 0.77 \[ \frac {1}{2} \, \sqrt {-\tan \relax (x)^{2} + 1} \tan \relax (x) + \frac {1}{2} \, \arcsin \left (\tan \relax (x)\right ) \]

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))

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maple [C]  time = 0.43, size = 492, normalized size = 18.92 \[ \frac {\sin \relax (x ) \left (2 \left (\cos ^{2}\relax (x )\right ) \sin \relax (x ) \sqrt {2}\, \sqrt {\frac {\cos \relax (x ) \sqrt {2}-\sqrt {2}+2 \cos \relax (x )-1}{1+\cos \relax (x )}}\, \sqrt {-\frac {2 \left (\cos \relax (x ) \sqrt {2}-\sqrt {2}-2 \cos \relax (x )+1\right )}{1+\cos \relax (x )}}\, \EllipticPi \left (\frac {\sqrt {3+2 \sqrt {2}}\, \left (-1+\cos \relax (x )\right )}{\sin \relax (x )}, \frac {1}{3+2 \sqrt {2}}, \frac {\sqrt {3-2 \sqrt {2}}}{\sqrt {3+2 \sqrt {2}}}\right )-\left (\cos ^{2}\relax (x )\right ) \sin \relax (x ) \sqrt {2}\, \sqrt {\frac {\cos \relax (x ) \sqrt {2}-\sqrt {2}+2 \cos \relax (x )-1}{1+\cos \relax (x )}}\, \sqrt {-\frac {2 \left (\cos \relax (x ) \sqrt {2}-\sqrt {2}-2 \cos \relax (x )+1\right )}{1+\cos \relax (x )}}\, \EllipticF \left (\frac {\left (1+\sqrt {2}\right ) \left (-1+\cos \relax (x )\right )}{\sin \relax (x )}, 3-2 \sqrt {2}\right )+4 \left (\cos ^{2}\relax (x )\right ) \sin \relax (x ) \sqrt {\frac {\cos \relax (x ) \sqrt {2}-\sqrt {2}+2 \cos \relax (x )-1}{1+\cos \relax (x )}}\, \sqrt {-\frac {2 \left (\cos \relax (x ) \sqrt {2}-\sqrt {2}-2 \cos \relax (x )+1\right )}{1+\cos \relax (x )}}\, \EllipticPi \left (\frac {\sqrt {3+2 \sqrt {2}}\, \left (-1+\cos \relax (x )\right )}{\sin \relax (x )}, \frac {1}{3+2 \sqrt {2}}, \frac {\sqrt {3-2 \sqrt {2}}}{\sqrt {3+2 \sqrt {2}}}\right )-2 \left (\cos ^{2}\relax (x )\right ) \sin \relax (x ) \sqrt {\frac {\cos \relax (x ) \sqrt {2}-\sqrt {2}+2 \cos \relax (x )-1}{1+\cos \relax (x )}}\, \sqrt {-\frac {2 \left (\cos \relax (x ) \sqrt {2}-\sqrt {2}-2 \cos \relax (x )+1\right )}{1+\cos \relax (x )}}\, \EllipticF \left (\frac {\left (1+\sqrt {2}\right ) \left (-1+\cos \relax (x )\right )}{\sin \relax (x )}, 3-2 \sqrt {2}\right )+4 \left (\cos ^{3}\relax (x )\right ) \sqrt {2}-4 \left (\cos ^{2}\relax (x )\right ) \sqrt {2}+6 \left (\cos ^{3}\relax (x )\right )-2 \cos \relax (x ) \sqrt {2}-6 \left (\cos ^{2}\relax (x )\right )+2 \sqrt {2}-3 \cos \relax (x )+3\right ) \sqrt {\frac {2 \left (\cos ^{2}\relax (x )\right )-1}{\cos \relax (x )^{2}}}}{2 \left (-1+\cos \relax (x )\right ) \left (2 \left (\cos ^{2}\relax (x )\right )-1\right ) \cos \relax (x ) \left (1+\sqrt {2}\right ) \sqrt {3+2 \sqrt {2}}} \]

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

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maxima [A]  time = 0.41, size = 20, normalized size = 0.77 \[ \frac {1}{2} \, \sqrt {-\tan \relax (x)^{2} + 1} \tan \relax (x) + \frac {1}{2} \, \arcsin \left (\tan \relax (x)\right ) \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \[ \int \frac {\sqrt {1-{\mathrm {tan}\relax (x)}^2}}{{\cos \relax (x)}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1 - tan(x)^2)^(1/2)/cos(x)^2,x)

[Out]

int((1 - tan(x)^2)^(1/2)/cos(x)^2, x)

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

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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