3.189 \(\int \frac{1}{a+b \sec ^2(e+f x)} \, dx\)

Optimal. Leaf size=45 \[ \frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{a+b} \cot (e+f x)}{\sqrt{b}}\right )}{a f \sqrt{a+b}}+\frac{x}{a} \]

[Out]

x/a + (Sqrt[b]*ArcTan[(Sqrt[a + b]*Cot[e + f*x])/Sqrt[b]])/(a*Sqrt[a + b]*f)

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Rubi [A]  time = 0.0420105, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {4127, 3181, 205} \[ \frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{a+b} \cot (e+f x)}{\sqrt{b}}\right )}{a f \sqrt{a+b}}+\frac{x}{a} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[e + f*x]^2)^(-1),x]

[Out]

x/a + (Sqrt[b]*ArcTan[(Sqrt[a + b]*Cot[e + f*x])/Sqrt[b]])/(a*Sqrt[a + b]*f)

Rule 4127

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(-1), x_Symbol] :> Simp[x/a, x] - Dist[b/a, Int[1/(b + a*Cos[e +
f*x]^2), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a + b, 0]

Rule 3181

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(-1), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff/f, Subst[Int[1/(a + (a + b)*ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x]

Rule 205

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

Rubi steps

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

Mathematica [C]  time = 0.318623, size = 182, normalized size = 4.04 \[ \frac{\sec ^2(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (f x \sqrt{a+b} \sqrt{b (\cos (e)-i \sin (e))^4}+b (\cos (2 e)-i \sin (2 e)) \tan ^{-1}\left (\frac{(\cos (2 e)-i \sin (2 e)) \sec (f x) (a \sin (2 e+f x)-(a+2 b) \sin (f x))}{2 \sqrt{a+b} \sqrt{b (\cos (e)-i \sin (e))^4}}\right )\right )}{2 a f \sqrt{a+b} \sqrt{b (\cos (e)-i \sin (e))^4} \left (a+b \sec ^2(e+f x)\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sec[e + f*x]^2)^(-1),x]

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^2*(Sqrt[a + b]*f*x*Sqrt[b*(Cos[e] - I*Sin[e])^4] + b*ArcTan[(Sec[
f*x]*(Cos[2*e] - I*Sin[2*e])*(-((a + 2*b)*Sin[f*x]) + a*Sin[2*e + f*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin
[e])^4])]*(Cos[2*e] - I*Sin[2*e])))/(2*a*Sqrt[a + b]*f*(a + b*Sec[e + f*x]^2)*Sqrt[b*(Cos[e] - I*Sin[e])^4])

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Maple [A]  time = 0.069, size = 48, normalized size = 1.1 \begin{align*}{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) }{fa}}-{\frac{b}{fa}\arctan \left ({\tan \left ( fx+e \right ) b{\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.554304, size = 544, normalized size = 12.09 \begin{align*} \left [\frac{4 \, f x + \sqrt{-\frac{b}{a + b}} \log \left (\frac{{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \,{\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \,{\left ({\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} -{\left (a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt{-\frac{b}{a + b}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right )}{4 \, a f}, \frac{2 \, f x + \sqrt{\frac{b}{a + b}} \arctan \left (\frac{{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt{\frac{b}{a + b}}}{2 \, b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right )}{2 \, a f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(4*f*x + sqrt(-b/(a + b))*log(((a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^4 - 2*(3*a*b + 4*b^2)*cos(f*x + e)^2 +
4*((a^2 + 3*a*b + 2*b^2)*cos(f*x + e)^3 - (a*b + b^2)*cos(f*x + e))*sqrt(-b/(a + b))*sin(f*x + e) + b^2)/(a^2*
cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2)))/(a*f), 1/2*(2*f*x + sqrt(b/(a + b))*arctan(1/2*((a + 2*b)*cos(f
*x + e)^2 - b)*sqrt(b/(a + b))/(b*cos(f*x + e)*sin(f*x + e))))/(a*f)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{a + b \sec ^{2}{\left (e + f x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.34351, size = 92, normalized size = 2.04 \begin{align*} -\frac{\frac{{\left (\pi \left \lfloor \frac{f x + e}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (b\right ) + \arctan \left (\frac{b \tan \left (f x + e\right )}{\sqrt{a b + b^{2}}}\right )\right )} b}{\sqrt{a b + b^{2}} a} - \frac{f x + e}{a}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b + b^2)))*b/(sqrt(a*b + b^2)*a) - (f*x
+ e)/a)/f