3.295 \(\int \frac{\sqrt{b \tan (e+f x)}}{(d \sec (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=34 \[ \frac{2 (b \tan (e+f x))^{3/2}}{3 b f (d \sec (e+f x))^{3/2}} \]

[Out]

(2*(b*Tan[e + f*x])^(3/2))/(3*b*f*(d*Sec[e + f*x])^(3/2))

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Rubi [A]  time = 0.0510625, antiderivative size = 34, 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} \[ \frac{2 (b \tan (e+f x))^{3/2}}{3 b f (d \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*Tan[e + f*x]]/(d*Sec[e + f*x])^(3/2),x]

[Out]

(2*(b*Tan[e + f*x])^(3/2))/(3*b*f*(d*Sec[e + f*x])^(3/2))

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{\sqrt{b \tan (e+f x)}}{(d \sec (e+f x))^{3/2}} \, dx &=\frac{2 (b \tan (e+f x))^{3/2}}{3 b f (d \sec (e+f x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.119699, size = 34, normalized size = 1. \[ \frac{2 (b \tan (e+f x))^{3/2}}{3 b f (d \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b*Tan[e + f*x]]/(d*Sec[e + f*x])^(3/2),x]

[Out]

(2*(b*Tan[e + f*x])^(3/2))/(3*b*f*(d*Sec[e + f*x])^(3/2))

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Maple [A]  time = 0.18, size = 50, normalized size = 1.5 \begin{align*}{\frac{2\,\sin \left ( fx+e \right ) }{3\,f\cos \left ( fx+e \right ) }\sqrt{{\frac{b\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }}} \left ({\frac{d}{\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(f*x+e))^(1/2)/(d*sec(f*x+e))^(3/2),x)

[Out]

2/3/f*sin(f*x+e)*(b*sin(f*x+e)/cos(f*x+e))^(1/2)/cos(f*x+e)/(d/cos(f*x+e))^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{b \tan \left (f x + e\right )}}{\left (d \sec \left (f x + e\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e))^(1/2)/(d*sec(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*tan(f*x + e))/(d*sec(f*x + e))^(3/2), x)

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Fricas [A]  time = 1.7156, size = 127, normalized size = 3.74 \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 ) \sin \left (f x + e\right )}{3 \, d^{2} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e))^(1/2)/(d*sec(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

2/3*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))*cos(f*x + e)*sin(f*x + e)/(d^2*f)

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Sympy [A]  time = 42.9155, size = 53, normalized size = 1.56 \begin{align*} \begin{cases} \frac{2 \sqrt{b} \tan ^{\frac{3}{2}}{\left (e + f x \right )}}{3 d^{\frac{3}{2}} f \sec ^{\frac{3}{2}}{\left (e + f x \right )}} & \text{for}\: f \neq 0 \\\frac{x \sqrt{b \tan{\left (e \right )}}}{\left (d \sec{\left (e \right )}\right )^{\frac{3}{2}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e))**(1/2)/(d*sec(f*x+e))**(3/2),x)

[Out]

Piecewise((2*sqrt(b)*tan(e + f*x)**(3/2)/(3*d**(3/2)*f*sec(e + f*x)**(3/2)), Ne(f, 0)), (x*sqrt(b*tan(e))/(d*s
ec(e))**(3/2), True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{b \tan \left (f x + e\right )}}{\left (d \sec \left (f x + e\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e))^(1/2)/(d*sec(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate(sqrt(b*tan(f*x + e))/(d*sec(f*x + e))^(3/2), x)