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

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

[Out]

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

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Rubi [A]  time = 0.0503519, 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} \[ -\frac{2 \sqrt{d \sec (e+f x)}}{b f \sqrt{b \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.123912, size = 32, normalized size = 1. \[ -\frac{2 \sqrt{d \sec (e+f x)}}{b f \sqrt{b \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.154, size = 50, normalized size = 1.6 \begin{align*} -2\,{\frac{\sin \left ( fx+e \right ) }{f\cos \left ( fx+e \right ) }\sqrt{{\frac{d}{\cos \left ( fx+e \right ) }}} \left ({\frac{b\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }} \right ) ^{-3/2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{d \sec \left (f x + e\right )}}{\left (b \tan \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((d*sec(f*x+e))^(1/2)/(b*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.65273, size = 126, normalized size = 3.94 \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^{2} f \sin \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [A]  time = 24.0339, size = 53, normalized size = 1.66 \begin{align*} \begin{cases} - \frac{2 \sqrt{d} \sqrt{\sec{\left (e + f x \right )}}}{b^{\frac{3}{2}} f \sqrt{\tan{\left (e + f x \right )}}} & \text{for}\: f \neq 0 \\\frac{x \sqrt{d \sec{\left (e \right )}}}{\left (b \tan{\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((d*sec(f*x+e))**(1/2)/(b*tan(f*x+e))**(3/2),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{d \sec \left (f x + e\right )}}{\left (b \tan \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((d*sec(f*x+e))^(1/2)/(b*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

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