3.65 \(\int \frac{1}{\sqrt{a \sec ^4(x)}} \, dx\)

Optimal. Leaf size=36 \[ \frac{x \sec ^2(x)}{2 \sqrt{a \sec ^4(x)}}+\frac{\tan (x)}{2 \sqrt{a \sec ^4(x)}} \]

[Out]

(x*Sec[x]^2)/(2*Sqrt[a*Sec[x]^4]) + Tan[x]/(2*Sqrt[a*Sec[x]^4])

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Rubi [A]  time = 0.0146596, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {4123, 2635, 8} \[ \frac{x \sec ^2(x)}{2 \sqrt{a \sec ^4(x)}}+\frac{\tan (x)}{2 \sqrt{a \sec ^4(x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[a*Sec[x]^4],x]

[Out]

(x*Sec[x]^2)/(2*Sqrt[a*Sec[x]^4]) + Tan[x]/(2*Sqrt[a*Sec[x]^4])

Rule 4123

Int[((b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> Dist[(b^IntPart[p]*(b*(c*Sec[e + f*x])^n)^
FracPart[p])/(c*Sec[e + f*x])^(n*FracPart[p]), Int[(c*Sec[e + f*x])^(n*p), x], x] /; FreeQ[{b, c, e, f, n, p},
x] &&  !IntegerQ[p]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

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

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{a \sec ^4(x)}} \, dx &=\frac{\sec ^2(x) \int \cos ^2(x) \, dx}{\sqrt{a \sec ^4(x)}}\\ &=\frac{\tan (x)}{2 \sqrt{a \sec ^4(x)}}+\frac{\sec ^2(x) \int 1 \, dx}{2 \sqrt{a \sec ^4(x)}}\\ &=\frac{x \sec ^2(x)}{2 \sqrt{a \sec ^4(x)}}+\frac{\tan (x)}{2 \sqrt{a \sec ^4(x)}}\\ \end{align*}

Mathematica [A]  time = 0.0234295, size = 23, normalized size = 0.64 \[ \frac{\tan (x)+x \sec ^2(x)}{2 \sqrt{a \sec ^4(x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[a*Sec[x]^4],x]

[Out]

(x*Sec[x]^2 + Tan[x])/(2*Sqrt[a*Sec[x]^4])

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Maple [A]  time = 0.074, size = 22, normalized size = 0.6 \begin{align*}{\frac{\cos \left ( x \right ) \sin \left ( x \right ) +x}{2\, \left ( \cos \left ( x \right ) \right ) ^{2}}{\frac{1}{\sqrt{{\frac{a}{ \left ( \cos \left ( x \right ) \right ) ^{4}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*sec(x)^4)^(1/2),x)

[Out]

1/2*(cos(x)*sin(x)+x)/cos(x)^2/(a/cos(x)^4)^(1/2)

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Maxima [A]  time = 1.66388, size = 34, normalized size = 0.94 \begin{align*} \frac{x}{2 \, \sqrt{a}} + \frac{\tan \left (x\right )}{2 \,{\left (\sqrt{a} \tan \left (x\right )^{2} + \sqrt{a}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.4198, size = 74, normalized size = 2.06 \begin{align*} \frac{{\left (\cos \left (x\right )^{3} \sin \left (x\right ) + x \cos \left (x\right )^{2}\right )} \sqrt{\frac{a}{\cos \left (x\right )^{4}}}}{2 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{a \sec ^{4}{\left (x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sec(x)**4)**(1/2),x)

[Out]

Integral(1/sqrt(a*sec(x)**4), x)

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Giac [A]  time = 1.31144, size = 53, normalized size = 1.47 \begin{align*} -\frac{1}{2} \, \sqrt{a}{\left (\frac{\pi \left \lfloor \frac{x}{\pi } + \frac{1}{2} \right \rfloor - x}{a} - \frac{\tan \left (x\right )}{{\left (\tan \left (x\right )^{2} + 1\right )} a}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(a)*((pi*floor(x/pi + 1/2) - x)/a - tan(x)/((tan(x)^2 + 1)*a))