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

Optimal. Leaf size=34 \[ -\frac{\cos (e+f x)}{f (a+b) \sqrt{a-b \cos ^2(e+f x)+b}} \]

[Out]

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

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Rubi [A]  time = 0.045294, antiderivative size = 34, 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 = {3186, 191} \[ -\frac{\cos (e+f x)}{f (a+b) \sqrt{a-b \cos ^2(e+f x)+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3186

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Cos[e + f*x], x]}, -Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p, x], x, Cos
[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{\sin (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-b x^2\right )^{3/2}} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac{\cos (e+f x)}{(a+b) f \sqrt{a+b-b \cos ^2(e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.114126, size = 41, normalized size = 1.21 \[ -\frac{\sqrt{2} \cos (e+f x)}{f (a+b) \sqrt{2 a-b \cos (2 (e+f x))+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.839, size = 31, normalized size = 0.9 \begin{align*} -{\frac{\cos \left ( fx+e \right ) }{ \left ( a+b \right ) f}{\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(sin(f*x+e)/(a+b*sin(f*x+e)^2)^(3/2),x)

[Out]

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

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Maxima [A]  time = 0.945119, size = 43, normalized size = 1.26 \begin{align*} -\frac{\cos \left (f x + e\right )}{\sqrt{-b \cos \left (f x + e\right )^{2} + a + b}{\left (a + b\right )} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.55985, size = 136, normalized size = 4. \begin{align*} \frac{\sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \cos \left (f x + e\right )}{{\left (a b + b^{2}\right )} f \cos \left (f x + e\right )^{2} -{\left (a^{2} + 2 \, a b + b^{2}\right )} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.3584, size = 72, normalized size = 2.12 \begin{align*} \frac{\sqrt{-{\left (\cos \left (f x + e\right )^{2} - 1\right )} b + a} \cos \left (f x + e\right )}{{\left ({\left (\cos \left (f x + e\right )^{2} - 1\right )} b - a\right )}{\left (a + b\right )} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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