3.413 \(\int \frac{\sin (e+f x)}{\sqrt{b \sec (e+f x)}} \, dx\)

Optimal. Leaf size=20 \[ -\frac{2 b}{3 f (b \sec (e+f x))^{3/2}} \]

[Out]

(-2*b)/(3*f*(b*Sec[e + f*x])^(3/2))

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Rubi [A]  time = 0.0307073, antiderivative size = 20, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2622, 30} \[ -\frac{2 b}{3 f (b \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]/Sqrt[b*Sec[e + f*x]],x]

[Out]

(-2*b)/(3*f*(b*Sec[e + f*x])^(3/2))

Rule 2622

Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[1/(f*a^n), Subst[Int
[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n
+ 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \frac{\sin (e+f x)}{\sqrt{b \sec (e+f x)}} \, dx &=\frac{b \operatorname{Subst}\left (\int \frac{1}{x^{5/2}} \, dx,x,b \sec (e+f x)\right )}{f}\\ &=-\frac{2 b}{3 f (b \sec (e+f x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0450973, size = 20, normalized size = 1. \[ -\frac{2 b}{3 f (b \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]/Sqrt[b*Sec[e + f*x]],x]

[Out]

(-2*b)/(3*f*(b*Sec[e + f*x])^(3/2))

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Maple [A]  time = 0.019, size = 17, normalized size = 0.9 \begin{align*} -{\frac{2\,b}{3\,f} \left ( b\sec \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*b/f/(b*sec(f*x+e))^(3/2)

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Maxima [A]  time = 1.0032, size = 31, normalized size = 1.55 \begin{align*} -\frac{2 \, \cos \left (f x + e\right )}{3 \, f \sqrt{\frac{b}{\cos \left (f x + e\right )}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*cos(f*x + e)/(f*sqrt(b/cos(f*x + e)))

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Fricas [A]  time = 2.35535, size = 65, normalized size = 3.25 \begin{align*} -\frac{2 \, \sqrt{\frac{b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )^{2}}{3 \, b f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.43903, size = 51, normalized size = 2.55 \begin{align*} -\frac{2 \, \sqrt{b \cos \left (f x + e\right )}{\left | f \right |} \cos \left (f x + e\right ) \mathrm{sgn}\left (f\right ) \mathrm{sgn}\left (\cos \left (f x + e\right )\right )}{3 \, b f^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*sqrt(b*cos(f*x + e))*abs(f)*cos(f*x + e)*sgn(f)*sgn(cos(f*x + e))/(b*f^2)