\(\int \frac {1}{\sqrt {1+\sec ^2(x)}} \, dx\) [297]

Optimal result

Integrand size = 10, antiderivative size = 14 \[ \int \frac {1}{\sqrt {1+\sec ^2(x)}} \, dx=\arctan \left (\frac {\tan (x)}{\sqrt {2+\tan ^2(x)}}\right ) \]

[Out]

arctan(tan(x)/(2+tan(x)^2)^(1/2))

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {4213, 385, 209} \[ \int \frac {1}{\sqrt {1+\sec ^2(x)}} \, dx=\arctan \left (\frac {\tan (x)}{\sqrt {\tan ^2(x)+2}}\right ) \]

[In]

Int[1/Sqrt[1 + Sec[x]^2],x]

[Out]

ArcTan[Tan[x]/Sqrt[2 + Tan[x]^2]]

Rule 209

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

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 4213

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff/f, Subst[Int[(a + b + b*ff^2*x^2)^p/(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p},
x] && NeQ[a + b, 0] && NeQ[p, -1]

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(37\) vs. \(2(14)=28\).

Time = 0.12 (sec) , antiderivative size = 37, normalized size of antiderivative = 2.64 \[ \int \frac {1}{\sqrt {1+\sec ^2(x)}} \, dx=\frac {\arcsin \left (\frac {\sin (x)}{\sqrt {2}}\right ) \sqrt {3+\cos (2 x)} \sec (x)}{\sqrt {2} \sqrt {1+\sec ^2(x)}} \]

[In]

Integrate[1/Sqrt[1 + Sec[x]^2],x]

[Out]

(ArcSin[Sin[x]/Sqrt[2]]*Sqrt[3 + Cos[2*x]]*Sec[x])/(Sqrt[2]*Sqrt[1 + Sec[x]^2])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(58\) vs. \(2(12)=24\).

Time = 0.87 (sec) , antiderivative size = 59, normalized size of antiderivative = 4.21

method	result	size
default	\(\frac {\sqrt {2}\, \sqrt {\frac {\cos \left (x \right )^{2}+1}{\left (\cos \left (x \right )+1\right )^{2}}}\, \arctan \left (\frac {\sin \left (x \right )}{\left (\cos \left (x \right )+1\right ) \sqrt {\frac {\cos \left (x \right )^{2}+1}{\left (\cos \left (x \right )+1\right )^{2}}}}\right ) \left (1+\sec \left (x \right )\right )}{\sqrt {2+2 \sec \left (x \right )^{2}}}\)	\(59\)

[In]

int(1/(1+sec(x)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2^(1/2)*((cos(x)^2+1)/(cos(x)+1)^2)^(1/2)*arctan(sin(x)/(cos(x)+1)/((cos(x)^2+1)/(cos(x)+1)^2)^(1/2))/(2+2*sec
(x)^2)^(1/2)*(1+sec(x))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (12) = 24\).

Time = 0.25 (sec) , antiderivative size = 53, normalized size of antiderivative = 3.79 \[ \int \frac {1}{\sqrt {1+\sec ^2(x)}} \, dx=\frac {1}{2} \, \arctan \left (\frac {\sqrt {\frac {\cos \left (x\right )^{2} + 1}{\cos \left (x\right )^{2}}} \cos \left (x\right )^{3} \sin \left (x\right ) + \cos \left (x\right ) \sin \left (x\right )}{\cos \left (x\right )^{4} + \cos \left (x\right )^{2} - 1}\right ) - \frac {1}{2} \, \arctan \left (\frac {\sin \left (x\right )}{\cos \left (x\right )}\right ) \]

[In]

integrate(1/(1+sec(x)^2)^(1/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

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

[In]

integrate(1/(1+sec(x)**2)**(1/2),x)

[Out]

Integral(1/sqrt(sec(x)**2 + 1), x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 388 vs. \(2 (12) = 24\).

Time = 0.38 (sec) , antiderivative size = 388, normalized size of antiderivative = 27.71 \[ \int \frac {1}{\sqrt {1+\sec ^2(x)}} \, dx=-\frac {1}{2} \, \arctan \left (2 \, {\left (2 \, {\left (6 \, \cos \left (2 \, x\right ) + 1\right )} \cos \left (4 \, x\right ) + \cos \left (4 \, x\right )^{2} + 36 \, \cos \left (2 \, x\right )^{2} + \sin \left (4 \, x\right )^{2} + 12 \, \sin \left (4 \, x\right ) \sin \left (2 \, x\right ) + 36 \, \sin \left (2 \, x\right )^{2} + 12 \, \cos \left (2 \, x\right ) + 1\right )}^{\frac {1}{4}} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (4 \, x\right ) + 6 \, \sin \left (2 \, x\right ), \cos \left (4 \, x\right ) + 6 \, \cos \left (2 \, x\right ) + 1\right )\right ), 2 \, {\left (2 \, {\left (6 \, \cos \left (2 \, x\right ) + 1\right )} \cos \left (4 \, x\right ) + \cos \left (4 \, x\right )^{2} + 36 \, \cos \left (2 \, x\right )^{2} + \sin \left (4 \, x\right )^{2} + 12 \, \sin \left (4 \, x\right ) \sin \left (2 \, x\right ) + 36 \, \sin \left (2 \, x\right )^{2} + 12 \, \cos \left (2 \, x\right ) + 1\right )}^{\frac {1}{4}} \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (4 \, x\right ) + 6 \, \sin \left (2 \, x\right ), \cos \left (4 \, x\right ) + 6 \, \cos \left (2 \, x\right ) + 1\right )\right ) + 8\right ) + \frac {1}{2} \, \arctan \left (2 \, {\left (2 \, {\left (6 \, \cos \left (2 \, x\right ) + 1\right )} \cos \left (4 \, x\right ) + \cos \left (4 \, x\right )^{2} + 36 \, \cos \left (2 \, x\right )^{2} + \sin \left (4 \, x\right )^{2} + 12 \, \sin \left (4 \, x\right ) \sin \left (2 \, x\right ) + 36 \, \sin \left (2 \, x\right )^{2} + 12 \, \cos \left (2 \, x\right ) + 1\right )}^{\frac {1}{4}} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (4 \, x\right ) + 6 \, \sin \left (2 \, x\right ), \cos \left (4 \, x\right ) + 6 \, \cos \left (2 \, x\right ) + 1\right )\right ) + 2 \, \sin \left (2 \, x\right ), 2 \, {\left (2 \, {\left (6 \, \cos \left (2 \, x\right ) + 1\right )} \cos \left (4 \, x\right ) + \cos \left (4 \, x\right )^{2} + 36 \, \cos \left (2 \, x\right )^{2} + \sin \left (4 \, x\right )^{2} + 12 \, \sin \left (4 \, x\right ) \sin \left (2 \, x\right ) + 36 \, \sin \left (2 \, x\right )^{2} + 12 \, \cos \left (2 \, x\right ) + 1\right )}^{\frac {1}{4}} \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (4 \, x\right ) + 6 \, \sin \left (2 \, x\right ), \cos \left (4 \, x\right ) + 6 \, \cos \left (2 \, x\right ) + 1\right )\right ) + 2 \, \cos \left (2 \, x\right ) + 6\right ) \]

[In]

integrate(1/(1+sec(x)^2)^(1/2),x, algorithm="maxima")

[Out]

-1/2*arctan2(2*(2*(6*cos(2*x) + 1)*cos(4*x) + cos(4*x)^2 + 36*cos(2*x)^2 + sin(4*x)^2 + 12*sin(4*x)*sin(2*x) +
36*sin(2*x)^2 + 12*cos(2*x) + 1)^(1/4)*sin(1/2*arctan2(sin(4*x) + 6*sin(2*x), cos(4*x) + 6*cos(2*x) + 1)), 2*
(2*(6*cos(2*x) + 1)*cos(4*x) + cos(4*x)^2 + 36*cos(2*x)^2 + sin(4*x)^2 + 12*sin(4*x)*sin(2*x) + 36*sin(2*x)^2
+ 12*cos(2*x) + 1)^(1/4)*cos(1/2*arctan2(sin(4*x) + 6*sin(2*x), cos(4*x) + 6*cos(2*x) + 1)) + 8) + 1/2*arctan2
(2*(2*(6*cos(2*x) + 1)*cos(4*x) + cos(4*x)^2 + 36*cos(2*x)^2 + sin(4*x)^2 + 12*sin(4*x)*sin(2*x) + 36*sin(2*x)
^2 + 12*cos(2*x) + 1)^(1/4)*sin(1/2*arctan2(sin(4*x) + 6*sin(2*x), cos(4*x) + 6*cos(2*x) + 1)) + 2*sin(2*x), 2
*(2*(6*cos(2*x) + 1)*cos(4*x) + cos(4*x)^2 + 36*cos(2*x)^2 + sin(4*x)^2 + 12*sin(4*x)*sin(2*x) + 36*sin(2*x)^2
+ 12*cos(2*x) + 1)^(1/4)*cos(1/2*arctan2(sin(4*x) + 6*sin(2*x), cos(4*x) + 6*cos(2*x) + 1)) + 2*cos(2*x) + 6)

Giac [F]

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

[In]

integrate(1/(1+sec(x)^2)^(1/2),x, algorithm="giac")

[Out]

integrate(1/sqrt(sec(x)^2 + 1), x)

Mupad [F(-1)]

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

[In]

int(1/(1/cos(x)^2 + 1)^(1/2),x)

[Out]

int(1/(1/cos(x)^2 + 1)^(1/2), x)