3.319 \(\int \frac{1}{\sqrt{d \sec (e+f x)} \sqrt{b \tan (e+f x)}} \, dx\)

Optimal. Leaf size=32 \[ \frac{2 \sqrt{b \tan (e+f x)}}{b f \sqrt{d \sec (e+f x)}} \]

[Out]

(2*Sqrt[b*Tan[e + f*x]])/(b*f*Sqrt[d*Sec[e + f*x]])

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Rubi [A]  time = 0.045485, antiderivative size = 32, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.04, Rules used = {2605} \[ \frac{2 \sqrt{b \tan (e+f x)}}{b f \sqrt{d \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[d*Sec[e + f*x]]*Sqrt[b*Tan[e + f*x]]),x]

[Out]

(2*Sqrt[b*Tan[e + f*x]])/(b*f*Sqrt[d*Sec[e + f*x]])

Rule 2605

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> -Simp[((a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n + 1))/(b*f*m), x] /; FreeQ[{a, b, e, f, m, n}, x] && EqQ[m + n + 1, 0]

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{d \sec (e+f x)} \sqrt{b \tan (e+f x)}} \, dx &=\frac{2 \sqrt{b \tan (e+f x)}}{b f \sqrt{d \sec (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.38786, size = 32, normalized size = 1. \[ \frac{2 \sqrt{b \tan (e+f x)}}{b f \sqrt{d \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[d*Sec[e + f*x]]*Sqrt[b*Tan[e + f*x]]),x]

[Out]

(2*Sqrt[b*Tan[e + f*x]])/(b*f*Sqrt[d*Sec[e + f*x]])

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Maple [A]  time = 0.165, size = 50, normalized size = 1.6 \begin{align*} 2\,{\frac{\sin \left ( fx+e \right ) }{f\cos \left ( fx+e \right ) }{\frac{1}{\sqrt{{\frac{d}{\cos \left ( fx+e \right ) }}}}}{\frac{1}{\sqrt{{\frac{b\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(d*sec(f*x+e))^(1/2)/(b*tan(f*x+e))^(1/2),x)

[Out]

2/f*sin(f*x+e)/(d/cos(f*x+e))^(1/2)/(b*sin(f*x+e)/cos(f*x+e))^(1/2)/cos(f*x+e)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{d \sec \left (f x + e\right )} \sqrt{b \tan \left (f x + e\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*sec(f*x+e))^(1/2)/(b*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(d*sec(f*x + e))*sqrt(b*tan(f*x + e))), x)

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Fricas [A]  time = 1.67059, size = 107, normalized size = 3.34 \begin{align*} \frac{2 \, \sqrt{\frac{b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt{\frac{d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{b d f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*sec(f*x+e))^(1/2)/(b*tan(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

2*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))*cos(f*x + e)/(b*d*f)

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Sympy [A]  time = 22.735, size = 51, normalized size = 1.59 \begin{align*} \begin{cases} \frac{2 \sqrt{\tan{\left (e + f x \right )}}}{\sqrt{b} \sqrt{d} f \sqrt{\sec{\left (e + f x \right )}}} & \text{for}\: f \neq 0 \\\frac{x}{\sqrt{b \tan{\left (e \right )}} \sqrt{d \sec{\left (e \right )}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*sec(f*x+e))**(1/2)/(b*tan(f*x+e))**(1/2),x)

[Out]

Piecewise((2*sqrt(tan(e + f*x))/(sqrt(b)*sqrt(d)*f*sqrt(sec(e + f*x))), Ne(f, 0)), (x/(sqrt(b*tan(e))*sqrt(d*s
ec(e))), True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{d \sec \left (f x + e\right )} \sqrt{b \tan \left (f x + e\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*sec(f*x+e))^(1/2)/(b*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(d*sec(f*x + e))*sqrt(b*tan(f*x + e))), x)