3.145 \(\int \frac{\sin (e+f x)}{\sqrt{a+b \sin ^2(e+f x)}} \, dx\)

Optimal. Leaf size=41 \[ -\frac{\tan ^{-1}\left (\frac{\sqrt{b} \cos (e+f x)}{\sqrt{a-b \cos ^2(e+f x)+b}}\right )}{\sqrt{b} f} \]

[Out]

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

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Rubi [A]  time = 0.0472215, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3186, 217, 203} \[ -\frac{\tan ^{-1}\left (\frac{\sqrt{b} \cos (e+f x)}{\sqrt{a-b \cos ^2(e+f x)+b}}\right )}{\sqrt{b} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.113035, size = 53, normalized size = 1.29 \[ -\frac{\log \left (\sqrt{2 a-b \cos (2 (e+f x))+b}+\sqrt{2} \sqrt{-b} \cos (e+f x)\right )}{\sqrt{-b} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.861, size = 99, normalized size = 2.4 \begin{align*}{\frac{1}{2\,f\cos \left ( fx+e \right ) }\sqrt{ \left ( \cos \left ( fx+e \right ) \right ) ^{2} \left ( a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) }\arctan \left ({\frac{2\,b \left ( \sin \left ( fx+e \right ) \right ) ^{2}+a-b}{2}{\frac{1}{\sqrt{b}}}{\frac{1}{\sqrt{ \left ( \cos \left ( fx+e \right ) \right ) ^{2} \left ( a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) }}}} \right ){\frac{1}{\sqrt{b}}}{\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)^(1/2),x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.94815, size = 900, normalized size = 21.95 \begin{align*} \left [-\frac{\sqrt{-b} \log \left (128 \, b^{4} \cos \left (f x + e\right )^{8} - 256 \,{\left (a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{6} + 160 \,{\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{4} + a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4} - 32 \,{\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{2} + 8 \,{\left (16 \, b^{3} \cos \left (f x + e\right )^{7} - 24 \,{\left (a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{5} + 10 \,{\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{3} -{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \sqrt{-b}\right )}{8 \, b f}, \frac{\arctan \left (\frac{{\left (8 \, b^{2} \cos \left (f x + e\right )^{4} - 8 \,{\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \sqrt{b}}{4 \,{\left (2 \, b^{3} \cos \left (f x + e\right )^{5} - 3 \,{\left (a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{3} +{\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \cos \left (f x + e\right )\right )}}\right )}{4 \, \sqrt{b} f}\right ] \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)^(1/2),x, algorithm="fricas")

[Out]

[-1/8*sqrt(-b)*log(128*b^4*cos(f*x + e)^8 - 256*(a*b^3 + b^4)*cos(f*x + e)^6 + 160*(a^2*b^2 + 2*a*b^3 + b^4)*c
os(f*x + e)^4 + a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4 - 32*(a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*cos(f*x + e
)^2 + 8*(16*b^3*cos(f*x + e)^7 - 24*(a*b^2 + b^3)*cos(f*x + e)^5 + 10*(a^2*b + 2*a*b^2 + b^3)*cos(f*x + e)^3 -
(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(f*x + e))*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-b))/(b*f), 1/4*arctan(1/4
*(8*b^2*cos(f*x + e)^4 - 8*(a*b + b^2)*cos(f*x + e)^2 + a^2 + 2*a*b + b^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqr
t(b)/(2*b^3*cos(f*x + e)^5 - 3*(a*b^2 + b^3)*cos(f*x + e)^3 + (a^2*b + 2*a*b^2 + b^3)*cos(f*x + e)))/(sqrt(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(sin(f*x+e)/(a+b*sin(f*x+e)**2)**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.76948, size = 68, normalized size = 1.66 \begin{align*} -\frac{\log \left ({\left | \sqrt{-b \cos \left (f x + e\right )^{2} + a + b} + \frac{\sqrt{-b f^{2}} \cos \left (f x + e\right )}{f} \right |}\right )}{\sqrt{-b}{\left | f \right |}} \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)^(1/2),x, algorithm="giac")

[Out]

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