3.845 \(\int \frac {\sec (\sqrt {x}) \tan (\sqrt {x})}{\sqrt {x}} \, dx\)

Optimal. Leaf size=8 \[ 2 \sec \left (\sqrt {x}\right ) \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 8, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6715, 2606, 8} \[ 2 \sec \left (\sqrt {x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(Sec[Sqrt[x]]*Tan[Sqrt[x]])/Sqrt[x],x]

[Out]

2*Sec[Sqrt[x]]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 8, normalized size = 1.00 \[ 2 \sec \left (\sqrt {x}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[Sqrt[x]]*Tan[Sqrt[x]])/Sqrt[x],x]

[Out]

2*Sec[Sqrt[x]]

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fricas [A]  time = 0.57, size = 8, normalized size = 1.00 \[ \frac {2}{\cos \left (\sqrt {x}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/cos(sqrt(x))

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giac [A]  time = 0.14, size = 8, normalized size = 1.00 \[ \frac {2}{\cos \left (\sqrt {x}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/cos(sqrt(x))

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maple [A]  time = 0.06, size = 7, normalized size = 0.88 \[ 2 \sec \left (\sqrt {x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(x^(1/2))*tan(x^(1/2))/x^(1/2),x)

[Out]

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

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maxima [A]  time = 0.31, size = 8, normalized size = 1.00 \[ \frac {2}{\cos \left (\sqrt {x}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/cos(sqrt(x))

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mupad [B]  time = 3.00, size = 8, normalized size = 1.00 \[ \frac {2}{\cos \left (\sqrt {x}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.29, size = 7, normalized size = 0.88 \[ 2 \sec {\left (\sqrt {x} \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sec(sqrt(x))

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