3.453 \(\int \frac{\sqrt{b \sec (e+f x)}}{(a \sin (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=33 \[ -\frac{2 b}{a f \sqrt{a \sin (e+f x)} \sqrt{b \sec (e+f x)}} \]

[Out]

(-2*b)/(a*f*Sqrt[b*Sec[e + f*x]]*Sqrt[a*Sin[e + f*x]])

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Rubi [A]  time = 0.0516478, antiderivative size = 33, 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 = {2578} \[ -\frac{2 b}{a f \sqrt{a \sin (e+f x)} \sqrt{b \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b)/(a*f*Sqrt[b*Sec[e + f*x]]*Sqrt[a*Sin[e + f*x]])

Rule 2578

Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Simp[(b*(a*Sin[e
 + f*x])^(m + 1)*(b*Sec[e + f*x])^(n - 1))/(a*f*(m + 1)), x] /; FreeQ[{a, b, e, f, m, n}, x] && EqQ[m - n + 2,
 0] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \frac{\sqrt{b \sec (e+f x)}}{(a \sin (e+f x))^{3/2}} \, dx &=-\frac{2 b}{a f \sqrt{b \sec (e+f x)} \sqrt{a \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.0740362, size = 37, normalized size = 1.12 \[ -\frac{\sin (2 (e+f x)) \sqrt{b \sec (e+f x)}}{f (a \sin (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[b*Sec[e + f*x]]*Sin[2*(e + f*x)])/(f*(a*Sin[e + f*x])^(3/2)))

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Maple [A]  time = 0.123, size = 40, normalized size = 1.2 \begin{align*} -2\,{\frac{\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) }{f \left ( a\sin \left ( fx+e \right ) \right ) ^{3/2}}\sqrt{{\frac{b}{\cos \left ( fx+e \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Fricas [A]  time = 3.85517, size = 108, normalized size = 3.27 \begin{align*} -\frac{2 \, \sqrt{a \sin \left (f x + e\right )} \sqrt{\frac{b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{a^{2} f \sin \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*sqrt(a*sin(f*x + e))*sqrt(b/cos(f*x + e))*cos(f*x + e)/(a^2*f*sin(f*x + e))

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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((b*sec(f*x+e))**(1/2)/(a*sin(f*x+e))**(3/2),x)

[Out]

Timed out

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

[Out]

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