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

Optimal. Leaf size=14 \[ \tan ^{-1}\left (\frac{\tan (x)}{\sqrt{\tan ^2(x)+2}}\right ) \]

[Out]

ArcTan[Tan[x]/Sqrt[2 + Tan[x]^2]]

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Rubi [A]  time = 0.0197271, antiderivative size = 14, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {4128, 377, 203} \[ \tan ^{-1}\left (\frac{\tan (x)}{\sqrt{\tan ^2(x)+2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[1 + Sec[x]^2],x]

[Out]

ArcTan[Tan[x]/Sqrt[2 + Tan[x]^2]]

Rule 4128

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]

Rule 377

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 203

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

Rubi steps

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

Mathematica [B]  time = 0.025263, size = 37, normalized size = 2.64 \[ \frac{\sin ^{-1}\left (\frac{\sin (x)}{\sqrt{2}}\right ) \sqrt{\cos (2 x)+3} \sec (x)}{\sqrt{2} \sqrt{\sec ^2(x)+1}} \]

Antiderivative was successfully verified.

[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])

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Maple [C]  time = 0.227, size = 142, normalized size = 10.1 \begin{align*}{\frac{ \left ( -{\frac{1}{2}}+{\frac{i}{2}} \right ) \left ( \sin \left ( x \right ) \right ) ^{2}}{\cos \left ( x \right ) \left ( -1+\cos \left ( x \right ) \right ) }\sqrt{{\frac{i\cos \left ( x \right ) +1-i+\cos \left ( x \right ) }{\cos \left ( x \right ) +1}}}\sqrt{-{\frac{i\cos \left ( x \right ) -1-i-\cos \left ( x \right ) }{\cos \left ( x \right ) +1}}} \left ( 2\, \left ( -1 \right ) ^{3/4}{\it EllipticPi} \left ({\frac{\sqrt [4]{-1} \left ( -1+\cos \left ( x \right ) \right ) }{\sin \left ( x \right ) }},i,i \right ) -2\,\sqrt [4]{-1}{\it EllipticPi} \left ({\frac{\sqrt [4]{-1} \left ( -1+\cos \left ( x \right ) \right ) }{\sin \left ( x \right ) }},i,i \right ) +\sqrt{2}{\it EllipticF} \left ({\frac{ \left ({\frac{1}{2}}+{\frac{i}{2}} \right ) \sqrt{2} \left ( -1+\cos \left ( x \right ) \right ) }{\sin \left ( x \right ) }},i \right ) \right ){\frac{1}{\sqrt{{\frac{ \left ( \cos \left ( x \right ) \right ) ^{2}+1}{ \left ( \cos \left ( x \right ) \right ) ^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1+sec(x)^2)^(1/2),x)

[Out]

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

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Maxima [B]  time = 1.78543, size = 524, normalized size = 37.43 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [B]  time = 0.509012, size = 177, normalized size = 12.64 \begin{align*} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(sec(x)**2 + 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/sqrt(sec(x)^2 + 1), x)