3.3.89 \(\int \frac {1}{\sqrt {1+\tan ^2(x)}} \, dx\) [289]

Optimal. Leaf size=11 \[ \frac {\tan (x)}{\sqrt {\sec ^2(x)}} \]

[Out]

tan(x)/(sec(x)^2)^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 11, 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 = {3738, 4207, 197} \begin {gather*} \frac {\tan (x)}{\sqrt {\sec ^2(x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[1 + Tan[x]^2],x]

[Out]

Tan[x]/Sqrt[Sec[x]^2]

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 3738

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 4207

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 \frac {1}{\sqrt {1+\tan ^2(x)}} \, dx &=\int \frac {1}{\sqrt {\sec ^2(x)}} \, dx\\ &=\text {Subst}\left (\int \frac {1}{\left (1+x^2\right )^{3/2}} \, dx,x,\tan (x)\right )\\ &=\frac {\tan (x)}{\sqrt {\sec ^2(x)}}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 11, normalized size = 1.00 \begin {gather*} \frac {\tan (x)}{\sqrt {\sec ^2(x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[1 + Tan[x]^2],x]

[Out]

Tan[x]/Sqrt[Sec[x]^2]

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Maple [A]
time = 0.03, size = 12, normalized size = 1.09

method	result	size
derivativedivides	\(\frac {\tan \left (x \right )}{\sqrt {1+\tan ^{2}\left (x \right )}}\)	\(12\)
default	\(\frac {\tan \left (x \right )}{\sqrt {1+\tan ^{2}\left (x \right )}}\)	\(12\)
risch	\(-\frac {i {\mathrm e}^{2 i x}}{2 \sqrt {\frac {{\mathrm e}^{2 i x}}{\left ({\mathrm e}^{2 i x}+1\right )^{2}}}\, \left ({\mathrm e}^{2 i x}+1\right )}+\frac {i}{2 \sqrt {\frac {{\mathrm e}^{2 i x}}{\left ({\mathrm e}^{2 i x}+1\right )^{2}}}\, \left ({\mathrm e}^{2 i x}+1\right )}\)	\(65\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.30, size = 11, normalized size = 1.00 \begin {gather*} \frac {\tan \left (x\right )}{\sqrt {\tan \left (x\right )^{2} + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

tan(x)/sqrt(tan(x)^2 + 1)

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Fricas [A]
time = 2.72, size = 11, normalized size = 1.00 \begin {gather*} \frac {\tan \left (x\right )}{\sqrt {\tan \left (x\right )^{2} + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

tan(x)/sqrt(tan(x)^2 + 1)

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Sympy [A]
time = 0.16, size = 12, normalized size = 1.09 \begin {gather*} \frac {\tan {\left (x \right )}}{\sqrt {\tan ^{2}{\left (x \right )} + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

tan(x)/sqrt(tan(x)**2 + 1)

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Giac [A]
time = 0.41, size = 11, normalized size = 1.00 \begin {gather*} \frac {\tan \left (x\right )}{\sqrt {\tan \left (x\right )^{2} + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

tan(x)/sqrt(tan(x)^2 + 1)

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Mupad [B]
time = 0.03, size = 9, normalized size = 0.82 \begin {gather*} \mathrm {tan}\left (x\right )\,\sqrt {{\cos \left (x\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(tan(x)^2 + 1)^(1/2),x)

[Out]

tan(x)*(cos(x)^2)^(1/2)

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