\(\int \sec ^2(x) \sqrt {1-\tan ^2(x)} \, dx\) [713]

Optimal result

Integrand size = 17, antiderivative size = 26 \[ \int \sec ^2(x) \sqrt {1-\tan ^2(x)} \, dx=\frac {1}{2} \arcsin (\tan (x))+\frac {1}{2} \tan (x) \sqrt {1-\tan ^2(x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {3756, 201, 222} \[ \int \sec ^2(x) \sqrt {1-\tan ^2(x)} \, dx=\frac {1}{2} \arcsin (\tan (x))+\frac {1}{2} \tan (x) \sqrt {1-\tan ^2(x)} \]

[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 201

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 222

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 3756

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*} \text {integral}& = \text {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} \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2}} \, dx,x,\tan (x)\right ) \\ & = \frac {1}{2} \arcsin (\tan (x))+\frac {1}{2} \tan (x) \sqrt {1-\tan ^2(x)} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(63\) vs. \(2(26)=52\).

Time = 0.11 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.42 \[ \int \sec ^2(x) \sqrt {1-\tan ^2(x)} \, dx=\frac {\cos (2 x) \tan (x)+\arcsin \left (\frac {\sin (x)}{\sqrt {\cos ^2(x)}}\right ) \cos (x) \sqrt {\cos ^2(x)} \sqrt {1-\tan ^2(x)}}{2 \sqrt {\cos ^2(x)} \sqrt {\cos (2 x)}} \]

[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]])

Maple [A] (verified)

Time = 0.96 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81

method	result	size
derivativedivides	\(\frac {\arcsin \left (\tan \left (x \right )\right )}{2}+\frac {\sqrt {1-\tan \left (x \right )^{2}}\, \tan \left (x \right )}{2}\)	\(21\)
default	\(\frac {\arcsin \left (\tan \left (x \right )\right )}{2}+\frac {\sqrt {1-\tan \left (x \right )^{2}}\, \tan \left (x \right )}{2}\)	\(21\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 72 vs. \(2 (20) = 40\).

Time = 0.26 (sec) , antiderivative size = 72, normalized size of antiderivative = 2.77 \[ \int \sec ^2(x) \sqrt {1-\tan ^2(x)} \, dx=-\frac {\arctan \left (\frac {{\left (3 \, \cos \left (x\right )^{3} - 2 \, \cos \left (x\right )\right )} \sqrt {\frac {2 \, \cos \left (x\right )^{2} - 1}{\cos \left (x\right )^{2}}}}{2 \, {\left (2 \, \cos \left (x\right )^{2} - 1\right )} \sin \left (x\right )}\right ) \cos \left (x\right ) - 2 \, \sqrt {\frac {2 \, \cos \left (x\right )^{2} - 1}{\cos \left (x\right )^{2}}} \sin \left (x\right )}{4 \, \cos \left (x\right )} \]

[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)

Sympy [F]

\[ \int \sec ^2(x) \sqrt {1-\tan ^2(x)} \, dx=\int \sqrt {- \left (\tan {\left (x \right )} - 1\right ) \left (\tan {\left (x \right )} + 1\right )} \sec ^{2}{\left (x \right )}\, dx \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \sec ^2(x) \sqrt {1-\tan ^2(x)} \, dx=\frac {1}{2} \, \sqrt {-\tan \left (x\right )^{2} + 1} \tan \left (x\right ) + \frac {1}{2} \, \arcsin \left (\tan \left (x\right )\right ) \]

[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))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \sec ^2(x) \sqrt {1-\tan ^2(x)} \, dx=\frac {1}{2} \, \sqrt {-\tan \left (x\right )^{2} + 1} \tan \left (x\right ) + \frac {1}{2} \, \arcsin \left (\tan \left (x\right )\right ) \]

[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))

Mupad [F(-1)]

Timed out. \[ \int \sec ^2(x) \sqrt {1-\tan ^2(x)} \, dx=\int \frac {\sqrt {1-{\mathrm {tan}\left (x\right )}^2}}{{\cos \left (x\right )}^2} \,d x \]

[In]

int((1 - tan(x)^2)^(1/2)/cos(x)^2,x)

[Out]

int((1 - tan(x)^2)^(1/2)/cos(x)^2, x)