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

Optimal. Leaf size=24 \[ \tan ^{-1}\left (\frac {\tan (x)}{\sqrt {\tan ^2(x)+2}}\right )+\sinh ^{-1}\left (\frac {\tan (x)}{\sqrt {2}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4128, 402, 215, 377, 203} \[ \tan ^{-1}\left (\frac {\tan (x)}{\sqrt {\tan ^2(x)+2}}\right )+\sinh ^{-1}\left (\frac {\tan (x)}{\sqrt {2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 + Sec[x]^2],x]

[Out]

ArcSinh[Tan[x]/Sqrt[2]] + ArcTan[Tan[x]/Sqrt[2 + Tan[x]^2]]

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])

Rule 215

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

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 402

Int[((a_) + (b_.)*(x_)^2)^(p_.)/((c_) + (d_.)*(x_)^2), x_Symbol] :> Dist[b/d, Int[(a + b*x^2)^(p - 1), x], x]
- Dist[(b*c - a*d)/d, Int[(a + b*x^2)^(p - 1)/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d,
0] && GtQ[p, 0] && (EqQ[p, 1/2] || EqQ[Denominator[p], 4])

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]

Rubi steps

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

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Mathematica [B]  time = 0.05, size = 57, normalized size = 2.38 \[ \frac {\sqrt {2} \cos (x) \sqrt {\sec ^2(x)+1} \left (\sin ^{-1}\left (\frac {\sin (x)}{\sqrt {2}}\right )+\tanh ^{-1}\left (\frac {\sqrt {2} \sin (x)}{\sqrt {\cos (2 x)+3}}\right )\right )}{\sqrt {\cos (2 x)+3}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 + Sec[x]^2],x]

[Out]

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

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fricas [B]  time = 0.57, size = 131, normalized size = 5.46 \[ \frac {1}{2} \, \arctan \left (\frac {\sqrt {\frac {\cos \relax (x)^{2} + 1}{\cos \relax (x)^{2}}} \cos \relax (x)^{3} \sin \relax (x) + \cos \relax (x) \sin \relax (x)}{\cos \relax (x)^{4} + \cos \relax (x)^{2} - 1}\right ) - \frac {1}{2} \, \arctan \left (\frac {\sin \relax (x)}{\cos \relax (x)}\right ) + \frac {1}{2} \, \log \left (\cos \relax (x)^{2} + \cos \relax (x) \sin \relax (x) + {\left (\cos \relax (x)^{2} + \cos \relax (x) \sin \relax (x)\right )} \sqrt {\frac {\cos \relax (x)^{2} + 1}{\cos \relax (x)^{2}}} + 1\right ) - \frac {1}{2} \, \log \left (\cos \relax (x)^{2} - \cos \relax (x) \sin \relax (x) + {\left (\cos \relax (x)^{2} - \cos \relax (x) \sin \relax (x)\right )} \sqrt {\frac {\cos \relax (x)^{2} + 1}{\cos \relax (x)^{2}}} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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)) + 1/2*log(cos(x)^2 + cos(x)*sin(x) + (cos(x)^2 + cos(x)*sin(x))*sqrt((cos(x)^2 + 1)/cos(x)
^2) + 1) - 1/2*log(cos(x)^2 - cos(x)*sin(x) + (cos(x)^2 - cos(x)*sin(x))*sqrt((cos(x)^2 + 1)/cos(x)^2) + 1)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {\sec \relax (x)^{2} + 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(sec(x)^2 + 1), x)

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maple [C]  time = 2.70, size = 190, normalized size = 7.92 \[ \frac {\left (-1+i\right ) \left (\left (-1\right )^{\frac {3}{4}} \EllipticPi \left (\frac {\left (-1\right )^{\frac {1}{4}} \left (-1+\cos \relax (x )\right )}{\sin \relax (x )}, -i, i\right )+\left (-1\right )^{\frac {3}{4}} \EllipticPi \left (\frac {\left (-1\right )^{\frac {1}{4}} \left (-1+\cos \relax (x )\right )}{\sin \relax (x )}, i, i\right )+\sqrt {2}\, \EllipticF \left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {2}\, \left (-1+\cos \relax (x )\right )}{\sin \relax (x )}, i\right )-\left (-1\right )^{\frac {1}{4}} \EllipticPi \left (\frac {\left (-1\right )^{\frac {1}{4}} \left (-1+\cos \relax (x )\right )}{\sin \relax (x )}, -i, i\right )-\left (-1\right )^{\frac {1}{4}} \EllipticPi \left (\frac {\left (-1\right )^{\frac {1}{4}} \left (-1+\cos \relax (x )\right )}{\sin \relax (x )}, i, i\right )\right ) \cos \relax (x ) \left (\sin ^{2}\relax (x )\right ) \sqrt {\frac {1+\cos ^{2}\relax (x )}{\cos \relax (x )^{2}}}\, \sqrt {\frac {i \cos \relax (x )+1-i+\cos \relax (x )}{\cos \relax (x )+1}}\, \sqrt {-\frac {i \cos \relax (x )-\cos \relax (x )-1-i}{\cos \relax (x )+1}}}{\left (-1+\cos \relax (x )\right ) \left (1+\cos ^{2}\relax (x )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: sign: argument cannot be imaginary
; found %i

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/cos(x)^2 + 1)^(1/2),x)

[Out]

int((1/cos(x)^2 + 1)^(1/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(sec(x)**2 + 1), x)

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