3.409 \(\int \frac {\sec ^3(x)}{\sqrt {\sin (2 x)}} \, dx\)

Optimal. Leaf size=31 \[ \frac {1}{5} \sqrt {\sin (2 x)} \sec ^3(x)+\frac {4}{5} \sqrt {\sin (2 x)} \sec (x) \]

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Rubi [A]  time = 0.04, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {4299, 4291} \[ \frac {1}{5} \sqrt {\sin (2 x)} \sec ^3(x)+\frac {4}{5} \sqrt {\sin (2 x)} \sec (x) \]

Antiderivative was successfully verified.

[In]

Int[Sec[x]^3/Sqrt[Sin[2*x]],x]

[Out]

(4*Sec[x]*Sqrt[Sin[2*x]])/5 + (Sec[x]^3*Sqrt[Sin[2*x]])/5

Rule 4291

Int[(cos[(a_.) + (b_.)*(x_)]*(e_.))^(m_.)*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> -Simp[((e*Cos[a +
b*x])^m*(g*Sin[c + d*x])^(p + 1))/(b*g*m), x] /; FreeQ[{a, b, c, d, e, g, m, p}, x] && EqQ[b*c - a*d, 0] && E
qQ[d/b, 2] &&  !IntegerQ[p] && EqQ[m + 2*p + 2, 0]

Rule 4299

Int[(cos[(a_.) + (b_.)*(x_)]*(e_.))^(m_)*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> -Simp[((e*Cos[a +
b*x])^m*(g*Sin[c + d*x])^(p + 1))/(2*b*g*(m + p + 1)), x] + Dist[(m + 2*p + 2)/(e^2*(m + p + 1)), Int[(e*Cos[a
+ b*x])^(m + 2)*(g*Sin[c + d*x])^p, x], x] /; FreeQ[{a, b, c, d, e, g, p}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b
, 2] &&  !IntegerQ[p] && LtQ[m, -1] && NeQ[m + 2*p + 2, 0] && NeQ[m + p + 1, 0] && IntegersQ[2*m, 2*p]

Rubi steps

\begin {align*} \int \frac {\sec ^3(x)}{\sqrt {\sin (2 x)}} \, dx &=\frac {1}{5} \sec ^3(x) \sqrt {\sin (2 x)}+\frac {4}{5} \int \frac {\sec (x)}{\sqrt {\sin (2 x)}} \, dx\\ &=\frac {4}{5} \sec (x) \sqrt {\sin (2 x)}+\frac {1}{5} \sec ^3(x) \sqrt {\sin (2 x)}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 20, normalized size = 0.65 \[ \frac {1}{5} \sqrt {\sin (2 x)} \sec (x) \left (\sec ^2(x)+4\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[x]^3/Sqrt[Sin[2*x]],x]

[Out]

(Sec[x]*(4 + Sec[x]^2)*Sqrt[Sin[2*x]])/5

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^3(x)}{\sqrt {\sin (2 x)}} \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[Sec[x]^3/Sqrt[Sin[2*x]],x]

[Out]

Could not integrate

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fricas [A]  time = 0.87, size = 32, normalized size = 1.03 \[ \frac {4 \, \cos \relax (x)^{3} + \sqrt {2} {\left (4 \, \cos \relax (x)^{2} + 1\right )} \sqrt {\cos \relax (x) \sin \relax (x)}}{5 \, \cos \relax (x)^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(x)^3/sin(2*x)^(1/2),x, algorithm="fricas")

[Out]

1/5*(4*cos(x)^3 + sqrt(2)*(4*cos(x)^2 + 1)*sqrt(cos(x)*sin(x)))/cos(x)^3

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\cos \relax (x)^{3} \sqrt {\sin \left (2 \, x\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(x)^3/sin(2*x)^(1/2),x, algorithm="giac")

[Out]

integrate(1/(cos(x)^3*sqrt(sin(2*x))), x)

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maple [C]  time = 0.21, size = 286, normalized size = 9.23

method	result	size
default	\(\frac {\sqrt {-\frac {\tan \left (\frac {x}{2}\right )}{\tan ^{2}\left (\frac {x}{2}\right )-1}}\, \left (\tan ^{2}\left (\frac {x}{2}\right )-1\right ) \left (5 \sqrt {\tan \left (\frac {x}{2}\right )+1}\, \sqrt {-2 \tan \left (\frac {x}{2}\right )+2}\, \sqrt {-\tan \left (\frac {x}{2}\right )}\, \EllipticF \left (\sqrt {\tan \left (\frac {x}{2}\right )+1}, \frac {\sqrt {2}}{2}\right ) \left (\tan ^{6}\left (\frac {x}{2}\right )\right )+15 \sqrt {\tan \left (\frac {x}{2}\right )+1}\, \sqrt {-2 \tan \left (\frac {x}{2}\right )+2}\, \sqrt {-\tan \left (\frac {x}{2}\right )}\, \EllipticF \left (\sqrt {\tan \left (\frac {x}{2}\right )+1}, \frac {\sqrt {2}}{2}\right ) \left (\tan ^{4}\left (\frac {x}{2}\right )\right )-14 \left (\tan ^{7}\left (\frac {x}{2}\right )\right )+15 \sqrt {\tan \left (\frac {x}{2}\right )+1}\, \sqrt {-2 \tan \left (\frac {x}{2}\right )+2}\, \sqrt {-\tan \left (\frac {x}{2}\right )}\, \EllipticF \left (\sqrt {\tan \left (\frac {x}{2}\right )+1}, \frac {\sqrt {2}}{2}\right ) \left (\tan ^{2}\left (\frac {x}{2}\right )\right )+2 \left (\tan ^{5}\left (\frac {x}{2}\right )\right )+5 \sqrt {\tan \left (\frac {x}{2}\right )+1}\, \sqrt {-2 \tan \left (\frac {x}{2}\right )+2}\, \sqrt {-\tan \left (\frac {x}{2}\right )}\, \EllipticF \left (\sqrt {\tan \left (\frac {x}{2}\right )+1}, \frac {\sqrt {2}}{2}\right )-2 \left (\tan ^{3}\left (\frac {x}{2}\right )\right )+14 \tan \left (\frac {x}{2}\right )\right )}{12 \sqrt {\tan \left (\frac {x}{2}\right ) \left (\tan ^{2}\left (\frac {x}{2}\right )-1\right )}\, \left (1+\tan ^{2}\left (\frac {x}{2}\right )\right )^{3} \sqrt {\tan ^{3}\left (\frac {x}{2}\right )-\tan \left (\frac {x}{2}\right )}}\)	\(286\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cos(x)^3/sin(2*x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/12*(-tan(1/2*x)/(tan(1/2*x)^2-1))^(1/2)*(tan(1/2*x)^2-1)*(5*(tan(1/2*x)+1)^(1/2)*(-2*tan(1/2*x)+2)^(1/2)*(-t
an(1/2*x))^(1/2)*EllipticF((tan(1/2*x)+1)^(1/2),1/2*2^(1/2))*tan(1/2*x)^6+15*(tan(1/2*x)+1)^(1/2)*(-2*tan(1/2*
x)+2)^(1/2)*(-tan(1/2*x))^(1/2)*EllipticF((tan(1/2*x)+1)^(1/2),1/2*2^(1/2))*tan(1/2*x)^4-14*tan(1/2*x)^7+15*(t
an(1/2*x)+1)^(1/2)*(-2*tan(1/2*x)+2)^(1/2)*(-tan(1/2*x))^(1/2)*EllipticF((tan(1/2*x)+1)^(1/2),1/2*2^(1/2))*tan
(1/2*x)^2+2*tan(1/2*x)^5+5*(tan(1/2*x)+1)^(1/2)*(-2*tan(1/2*x)+2)^(1/2)*(-tan(1/2*x))^(1/2)*EllipticF((tan(1/2
*x)+1)^(1/2),1/2*2^(1/2))-2*tan(1/2*x)^3+14*tan(1/2*x))/(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)/(1+tan(1/2*x)^2)^3
/(tan(1/2*x)^3-tan(1/2*x))^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\cos \relax (x)^{3} \sqrt {\sin \left (2 \, x\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(x)^3/sin(2*x)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(cos(x)^3*sqrt(sin(2*x))), x)

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mupad [B]  time = 0.39, size = 20, normalized size = 0.65 \[ \frac {\sqrt {\sin \left (2\,x\right )}\,\left (2\,\cos \left (2\,x\right )+3\right )}{5\,{\cos \relax (x)}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(2*x)^(1/2)*cos(x)^3),x)

[Out]

(sin(2*x)^(1/2)*(2*cos(2*x) + 3))/(5*cos(x)^3)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(x)**3/sin(2*x)**(1/2),x)

[Out]

Timed out

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