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

Optimal. Leaf size=29 \[ \frac{\sin (e+f x)}{a f \sqrt{a+b \sin ^2(e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.0439352, antiderivative size = 29, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {3190, 191} \[ \frac{\sin (e+f x)}{a f \sqrt{a+b \sin ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3190

Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e +
f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rubi steps

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

Mathematica [A]  time = 0.0302894, size = 29, normalized size = 1. \[ \frac{\sin (e+f x)}{a f \sqrt{a+b \sin ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.085, size = 28, normalized size = 1. \begin{align*}{\frac{\sin \left ( fx+e \right ) }{af}{\frac{1}{\sqrt{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.969579, size = 36, normalized size = 1.24 \begin{align*} \frac{\sin \left (f x + e\right )}{\sqrt{b \sin \left (f x + e\right )^{2} + a} a f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)/(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.43475, size = 116, normalized size = 4. \begin{align*} -\frac{\sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \sin \left (f x + e\right )}{a b f \cos \left (f x + e\right )^{2} -{\left (a^{2} + a b\right )} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Giac [A]  time = 1.36706, size = 39, normalized size = 1.34 \begin{align*} \frac{\sin \left (f x + e\right )}{\sqrt{b \sin \left (f x + e\right )^{2} + a} a f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)/(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

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