3.288 \(\int \sqrt {1+\tan ^2(x)} \, dx\)

Optimal. Leaf size=3 \[ \sinh ^{-1}(\tan (x)) \]

[Out]

arcsinh(tan(x))

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Rubi [A]  time = 0.01, antiderivative size = 3, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3657, 4122, 215} \[ \sinh ^{-1}(\tan (x)) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 + Tan[x]^2],x]

[Out]

ArcSinh[Tan[x]]

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 3657

Int[(u_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[ActivateTrig[u*(a*sec[e + f*x]^2)^p]
, x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a, b]

Rule 4122

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

Rubi steps

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

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Mathematica [B]  time = 0.01, size = 44, normalized size = 14.67 \[ \cos (x) \sqrt {\sec ^2(x)} \left (\log \left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right )-\log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 + Tan[x]^2],x]

[Out]

Cos[x]*(-Log[Cos[x/2] - Sin[x/2]] + Log[Cos[x/2] + Sin[x/2]])*Sqrt[Sec[x]^2]

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fricas [B]  time = 0.41, size = 60, normalized size = 20.00 \[ \frac {1}{2} \, \log \left (\frac {\tan \relax (x)^{2} + \sqrt {\tan \relax (x)^{2} + 1} \tan \relax (x) + 1}{\tan \relax (x)^{2} + 1}\right ) - \frac {1}{2} \, \log \left (\frac {\tan \relax (x)^{2} - \sqrt {\tan \relax (x)^{2} + 1} \tan \relax (x) + 1}{\tan \relax (x)^{2} + 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*log((tan(x)^2 + sqrt(tan(x)^2 + 1)*tan(x) + 1)/(tan(x)^2 + 1)) - 1/2*log((tan(x)^2 - sqrt(tan(x)^2 + 1)*ta
n(x) + 1)/(tan(x)^2 + 1))

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giac [B]  time = 0.39, size = 16, normalized size = 5.33 \[ -\log \left (\sqrt {\tan \relax (x)^{2} + 1} - \tan \relax (x)\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(sqrt(tan(x)^2 + 1) - tan(x))

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maple [A]  time = 0.14, size = 4, normalized size = 1.33 \[ \arcsinh \left (\tan \relax (x )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arcsinh(tan(x))

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maxima [A]  time = 0.75, size = 3, normalized size = 1.00 \[ \operatorname {arsinh}\left (\tan \relax (x)\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arcsinh(tan(x))

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mupad [B]  time = 0.06, size = 3, normalized size = 1.00 \[ \mathrm {asinh}\left (\mathrm {tan}\relax (x)\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((tan(x)^2 + 1)^(1/2),x)

[Out]

asinh(tan(x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(tan(x)**2 + 1), x)

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