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

Optimal. Leaf size=30 \[ -\frac{\cos (e+f x) \sqrt{a+b \sec ^2(e+f x)}}{a f} \]

[Out]

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

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Rubi [A]  time = 0.0420239, antiderivative size = 30, 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 = {4134, 264} \[ -\frac{\cos (e+f x) \sqrt{a+b \sec ^2(e+f x)}}{a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4134

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Cos[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^
n)^p)/x^(m + 1), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (Gt
Q[m, 0] || EqQ[n, 2] || EqQ[n, 4])

Rule 264

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

Rubi steps

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

Mathematica [A]  time = 0.114465, size = 48, normalized size = 1.6 \[ -\frac{\sec (e+f x) (a \cos (2 e+2 f x)+a+2 b)}{2 a f \sqrt{a+b \sec ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.076, size = 31, normalized size = 1. \begin{align*} -{\frac{1}{fa\sec \left ( fx+e \right ) }\sqrt{a+b \left ( \sec \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*sec(f*x+e)^2)^(1/2),x)

[Out]

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

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Maxima [A]  time = 0.995049, size = 38, normalized size = 1.27 \begin{align*} -\frac{\sqrt{a + \frac{b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.511495, size = 88, normalized size = 2.93 \begin{align*} -\frac{\sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/(a*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{a + b \sec ^{2}{\left (e + f x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.66257, size = 81, normalized size = 2.7 \begin{align*} \frac{\sqrt{b} \mathrm{sgn}\left (f\right ) \mathrm{sgn}\left (\cos \left (f x + e\right )\right )}{a{\left | f \right |}} - \frac{\sqrt{a \cos \left (f x + e\right )^{2} + b}}{a{\left | f \right |} \mathrm{sgn}\left (f\right ) \mathrm{sgn}\left (\cos \left (f x + e\right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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