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

Optimal. Leaf size=37 \[ -\frac{\cos (e+f x) \sqrt{a+b \sec ^2(e+f x)-b}}{f (a-b)} \]

[Out]

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

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Rubi [A]  time = 0.0465261, antiderivative size = 37, 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 = {3664, 264} \[ -\frac{\cos (e+f x) \sqrt{a+b \sec ^2(e+f x)-b}}{f (a-b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3664

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

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

Mathematica [A]  time = 0.603409, size = 52, normalized size = 1.41 \[ \frac{\cos (e+f x) \sqrt{\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}}{\sqrt{2} f (b-a)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.069, size = 78, normalized size = 2.1 \begin{align*} -{\frac{a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b}{f\cos \left ( fx+e \right ) \left ( a-b \right ) }{\frac{1}{\sqrt{{\frac{a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b}{ \left ( \cos \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*tan(f*x+e)^2)^(1/2),x)

[Out]

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

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Maxima [A]  time = 0.98396, size = 47, normalized size = 1.27 \begin{align*} -\frac{\sqrt{a - b + \frac{b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\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*tan(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.76052, size = 104, normalized size = 2.81 \begin{align*} -\frac{\sqrt{\frac{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\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*tan(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

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

[Out]

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

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Giac [B]  time = 1.7781, size = 119, normalized size = 3.22 \begin{align*} \frac{\sqrt{b} \mathrm{sgn}\left (f\right ) \mathrm{sgn}\left (\cos \left (f x + e\right )\right )}{a{\left | f \right |} - b{\left | f \right |}} - \frac{\sqrt{a \cos \left (f x + e\right )^{2} - b \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 ) - b{\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*tan(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

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