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 = 0.0112337, antiderivative size = 3, 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 = {3657, 4122, 215} \[ \sinh ^{-1}(\tan (x)) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 + Tan[x]^2],x]

[Out]

ArcSinh[Tan[x]]

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
]

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]

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*}

Mathematica [B]  time = 0.0087386, 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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Maple [A]  time = 0.021, size = 4, normalized size = 1.3 \begin{align*}{\it Arcsinh} \left ( \tan \left ( x \right ) \right ) \end{align*}

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 = 1.61566, size = 4, normalized size = 1.33 \begin{align*} \operatorname{arsinh}\left (\tan \left (x\right )\right ) \end{align*}

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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Fricas [B]  time = 1.62911, size = 185, normalized size = 61.67 \begin{align*} \frac{1}{2} \, \log \left (\frac{\tan \left (x\right )^{2} + \sqrt{\tan \left (x\right )^{2} + 1} \tan \left (x\right ) + 1}{\tan \left (x\right )^{2} + 1}\right ) - \frac{1}{2} \, \log \left (\frac{\tan \left (x\right )^{2} - \sqrt{\tan \left (x\right )^{2} + 1} \tan \left (x\right ) + 1}{\tan \left (x\right )^{2} + 1}\right ) \end{align*}

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

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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Giac [B]  time = 1.13204, size = 22, normalized size = 7.33 \begin{align*} -\log \left (\sqrt{\tan \left (x\right )^{2} + 1} - \tan \left (x\right )\right ) \end{align*}

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))