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

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

[Out]

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

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Rubi [A]  time = 0.0413526, antiderivative size = 31, normalized size of antiderivative = 1., 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 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]

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]

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*}

Mathematica [A]  time = 0.028768, 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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Maple [C]  time = 0.05, size = 286, normalized size = 9.2 \begin{align*}{\frac{1}{12}\sqrt{-{\tan \left ({\frac{x}{2}} \right ) \left ( \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}-1 \right ) ^{-1}}} \left ( \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}-1 \right ) \left ( 5\,\sqrt{1+\tan \left ( x/2 \right ) }\sqrt{-2\,\tan \left ( x/2 \right ) +2}\sqrt{-\tan \left ( x/2 \right ) }{\it EllipticF} \left ( \sqrt{1+\tan \left ( x/2 \right ) },1/2\,\sqrt{2} \right ) \left ( \tan \left ( x/2 \right ) \right ) ^{6}+15\,\sqrt{1+\tan \left ( x/2 \right ) }\sqrt{-2\,\tan \left ( x/2 \right ) +2}\sqrt{-\tan \left ( x/2 \right ) }{\it EllipticF} \left ( \sqrt{1+\tan \left ( x/2 \right ) },1/2\,\sqrt{2} \right ) \left ( \tan \left ( x/2 \right ) \right ) ^{4}-14\, \left ( \tan \left ( x/2 \right ) \right ) ^{7}+15\,\sqrt{1+\tan \left ( x/2 \right ) }\sqrt{-2\,\tan \left ( x/2 \right ) +2}\sqrt{-\tan \left ( x/2 \right ) }{\it EllipticF} \left ( \sqrt{1+\tan \left ( x/2 \right ) },1/2\,\sqrt{2} \right ) \left ( \tan \left ( x/2 \right ) \right ) ^{2}+2\, \left ( \tan \left ( x/2 \right ) \right ) ^{5}+5\,\sqrt{1+\tan \left ( x/2 \right ) }\sqrt{-2\,\tan \left ( x/2 \right ) +2}\sqrt{-\tan \left ( x/2 \right ) }{\it EllipticF} \left ( \sqrt{1+\tan \left ( x/2 \right ) },1/2\,\sqrt{2} \right ) -2\, \left ( \tan \left ( x/2 \right ) \right ) ^{3}+14\,\tan \left ( x/2 \right ) \right ){\frac{1}{\sqrt{\tan \left ({\frac{x}{2}} \right ) \left ( \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}-1 \right ) }}}{\frac{1}{\sqrt{ \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{3}-\tan \left ({\frac{x}{2}} \right ) }}} \left ( \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\cos \left (x\right )^{3} \sqrt{\sin \left (2 \, x\right )}}\,{d x} \end{align*}

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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Fricas [A]  time = 1.95203, size = 100, normalized size = 3.23 \begin{align*} \frac{4 \, \cos \left (x\right )^{3} + \sqrt{2}{\left (4 \, \cos \left (x\right )^{2} + 1\right )} \sqrt{\cos \left (x\right ) \sin \left (x\right )}}{5 \, \cos \left (x\right )^{3}} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\cos \left (x\right )^{3} \sqrt{\sin \left (2 \, x\right )}}\,{d x} \end{align*}

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)