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

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Rubi [A]  time = 0.05, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, 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*}

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Mathematica [A]  time = 0.08, 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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fricas [A]  time = 1.13, size = 44, normalized size = 1.33 \[ -\frac {2 \, \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{a^{2} f \sin \left (f x + e\right )} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {b \sec \left (f x + e\right )}}{\left (a \sin \left (f x + e\right )\right )^{\frac {3}{2}}}\,{d x} \]

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)

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maple [A]  time = 0.17, size = 40, normalized size = 1.21 \[ -\frac {2 \sin \left (f x +e \right ) \cos \left (f x +e \right ) \sqrt {\frac {b}{\cos \left (f x +e \right )}}}{f \left (a \sin \left (f x +e \right )\right )^{\frac {3}{2}}} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {b \sec \left (f x + e\right )}}{\left (a \sin \left (f x + e\right )\right )^{\frac {3}{2}}}\,{d x} \]

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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mupad [B]  time = 0.87, size = 36, normalized size = 1.09 \[ -\frac {2\,\cos \left (e+f\,x\right )\,\sqrt {\frac {b}{\cos \left (e+f\,x\right )}}}{a\,f\,\sqrt {a\,\sin \left (e+f\,x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {b \sec {\left (e + f x \right )}}}{\left (a \sin {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]

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]

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

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