3.862 \(\int \sec (x) \sqrt {4+3 \sec (x)} \tan (x) \, dx\)

Optimal. Leaf size=14 \[ \frac {2}{9} (3 \sec (x)+4)^{3/2} \]

[Out]

2/9*(4+3*sec(x))^(3/2)

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Rubi [A]  time = 0.04, antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {4339, 261} \[ \frac {2}{9} (3 \sec (x)+4)^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[Sec[x]*Sqrt[4 + 3*Sec[x]]*Tan[x],x]

[Out]

(2*(4 + 3*Sec[x])^(3/2))/9

Rule 261

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

Rule 4339

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> With[{d = FreeFactors[Cos[c*(a + b*x)], x]}, -Dist[(b*
c)^(-1), Subst[Int[SubstFor[1/x, Cos[c*(a + b*x)]/d, u, x], x], x, Cos[c*(a + b*x)]/d], x] /; FunctionOfQ[Cos[
c*(a + b*x)]/d, u, x, True]] /; FreeQ[{a, b, c}, x] && (EqQ[F, Tan] || EqQ[F, tan])

Rubi steps

\begin {align*} \int \sec (x) \sqrt {4+3 \sec (x)} \tan (x) \, dx &=-\operatorname {Subst}\left (\int \frac {\sqrt {4+\frac {3}{x}}}{x^2} \, dx,x,\cos (x)\right )\\ &=\frac {2}{9} (4+3 \sec (x))^{3/2}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 14, normalized size = 1.00 \[ \frac {2}{9} (3 \sec (x)+4)^{3/2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[x]*Sqrt[4 + 3*Sec[x]]*Tan[x],x]

[Out]

(2*(4 + 3*Sec[x])^(3/2))/9

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fricas [B]  time = 1.26, size = 25, normalized size = 1.79 \[ \frac {2 \, \sqrt {\frac {4 \, \cos \relax (x) + 3}{\cos \relax (x)}} {\left (4 \, \cos \relax (x) + 3\right )}}{9 \, \cos \relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)*(4+3*sec(x))^(1/2)*tan(x),x, algorithm="fricas")

[Out]

2/9*sqrt((4*cos(x) + 3)/cos(x))*(4*cos(x) + 3)/cos(x)

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giac [B]  time = 0.16, size = 68, normalized size = 4.86 \[ \frac {2 \, {\left (4 \, {\left (\sqrt {4 \, \cos \relax (x)^{2} + 3 \, \cos \relax (x)} - 2 \, \cos \relax (x)\right )}^{2} - 6 \, \sqrt {4 \, \cos \relax (x)^{2} + 3 \, \cos \relax (x)} + 12 \, \cos \relax (x) + 3\right )} \mathrm {sgn}\left (\cos \relax (x)\right )}{{\left (\sqrt {4 \, \cos \relax (x)^{2} + 3 \, \cos \relax (x)} - 2 \, \cos \relax (x)\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)*(4+3*sec(x))^(1/2)*tan(x),x, algorithm="giac")

[Out]

2*(4*(sqrt(4*cos(x)^2 + 3*cos(x)) - 2*cos(x))^2 - 6*sqrt(4*cos(x)^2 + 3*cos(x)) + 12*cos(x) + 3)*sgn(cos(x))/(
sqrt(4*cos(x)^2 + 3*cos(x)) - 2*cos(x))^3

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maple [A]  time = 0.06, size = 11, normalized size = 0.79 \[ \frac {2 \left (4+3 \sec \relax (x )\right )^{\frac {3}{2}}}{9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(x)*(4+3*sec(x))^(1/2)*tan(x),x)

[Out]

2/9*(4+3*sec(x))^(3/2)

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maxima [A]  time = 0.32, size = 10, normalized size = 0.71 \[ \frac {2}{9} \, {\left (3 \, \sec \relax (x) + 4\right )}^{\frac {3}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)*(4+3*sec(x))^(1/2)*tan(x),x, algorithm="maxima")

[Out]

2/9*(3*sec(x) + 4)^(3/2)

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mupad [B]  time = 3.23, size = 29, normalized size = 2.07 \[ \frac {8\,\sqrt {\frac {3}{\cos \relax (x)}+4}}{9}+\frac {2\,\sqrt {\frac {3}{\cos \relax (x)}+4}}{3\,\cos \relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((tan(x)*(3/cos(x) + 4)^(1/2))/cos(x),x)

[Out]

(8*(3/cos(x) + 4)^(1/2))/9 + (2*(3/cos(x) + 4)^(1/2))/(3*cos(x))

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sympy [B]  time = 0.68, size = 29, normalized size = 2.07 \[ \frac {2 \sqrt {3 \sec {\relax (x )} + 4} \sec {\relax (x )}}{3} + \frac {8 \sqrt {3 \sec {\relax (x )} + 4}}{9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)*(4+3*sec(x))**(1/2)*tan(x),x)

[Out]

2*sqrt(3*sec(x) + 4)*sec(x)/3 + 8*sqrt(3*sec(x) + 4)/9

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