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

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Rubi [A]  time = 0.04, antiderivative size = 45, normalized size of antiderivative = 1.00, 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 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]

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 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]

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*}

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Mathematica [C]  time = 0.29, 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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fricas [A]  time = 0.63, size = 231, normalized size = 5.13 \[ \left [\frac {4 \, f x + \sqrt {-\frac {b}{a + b}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-\frac {b}{a + b}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right )}{4 \, a f}, \frac {2 \, f x + \sqrt {\frac {b}{a + b}} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {b}{a + b}}}{2 \, b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right )}{2 \, a f}\right ] \]

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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giac [A]  time = 0.21, size = 68, normalized size = 1.51 \[ -\frac {\frac {{\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )} b}{\sqrt {a b + b^{2}} a} - \frac {f x + e}{a}}{f} \]

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

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maple [A]  time = 0.78, size = 48, normalized size = 1.07 \[ -\frac {b \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{f a \sqrt {\left (a +b \right ) b}}+\frac {\arctan \left (\tan \left (f x +e \right )\right )}{f a} \]

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

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maxima [A]  time = 0.43, size = 44, normalized size = 0.98 \[ -\frac {\frac {b \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} a} - \frac {f x + e}{a}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b*arctan(b*tan(f*x + e)/sqrt((a + b)*b))/(sqrt((a + b)*b)*a) - (f*x + e)/a)/f

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mupad [B]  time = 4.71, size = 460, normalized size = 10.22 \[ \frac {x}{a}-\frac {\mathrm {atan}\left (\frac {\frac {\left (2\,b^3\,\mathrm {tan}\left (e+f\,x\right )-\frac {\left (2\,a^2\,b^2-\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (8\,a^3\,b^2+16\,a^2\,b^3\right )\,\sqrt {-b\,\left (a+b\right )}}{4\,\left (a^2+b\,a\right )}\right )\,\sqrt {-b\,\left (a+b\right )}}{2\,\left (a^2+b\,a\right )}\right )\,\sqrt {-b\,\left (a+b\right )}\,1{}\mathrm {i}}{a^2+b\,a}+\frac {\left (2\,b^3\,\mathrm {tan}\left (e+f\,x\right )+\frac {\left (2\,a^2\,b^2+\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (8\,a^3\,b^2+16\,a^2\,b^3\right )\,\sqrt {-b\,\left (a+b\right )}}{4\,\left (a^2+b\,a\right )}\right )\,\sqrt {-b\,\left (a+b\right )}}{2\,\left (a^2+b\,a\right )}\right )\,\sqrt {-b\,\left (a+b\right )}\,1{}\mathrm {i}}{a^2+b\,a}}{\frac {\left (2\,b^3\,\mathrm {tan}\left (e+f\,x\right )-\frac {\left (2\,a^2\,b^2-\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (8\,a^3\,b^2+16\,a^2\,b^3\right )\,\sqrt {-b\,\left (a+b\right )}}{4\,\left (a^2+b\,a\right )}\right )\,\sqrt {-b\,\left (a+b\right )}}{2\,\left (a^2+b\,a\right )}\right )\,\sqrt {-b\,\left (a+b\right )}}{a^2+b\,a}-\frac {\left (2\,b^3\,\mathrm {tan}\left (e+f\,x\right )+\frac {\left (2\,a^2\,b^2+\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (8\,a^3\,b^2+16\,a^2\,b^3\right )\,\sqrt {-b\,\left (a+b\right )}}{4\,\left (a^2+b\,a\right )}\right )\,\sqrt {-b\,\left (a+b\right )}}{2\,\left (a^2+b\,a\right )}\right )\,\sqrt {-b\,\left (a+b\right )}}{a^2+b\,a}}\right )\,\sqrt {-b\,\left (a+b\right )}\,1{}\mathrm {i}}{f\,\left (a^2+b\,a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b/cos(e + f*x)^2),x)

[Out]

x/a - (atan((((2*b^3*tan(e + f*x) - ((2*a^2*b^2 - (tan(e + f*x)*(16*a^2*b^3 + 8*a^3*b^2)*(-b*(a + b))^(1/2))/(
4*(a*b + a^2)))*(-b*(a + b))^(1/2))/(2*(a*b + a^2)))*(-b*(a + b))^(1/2)*1i)/(a*b + a^2) + ((2*b^3*tan(e + f*x)
+ ((2*a^2*b^2 + (tan(e + f*x)*(16*a^2*b^3 + 8*a^3*b^2)*(-b*(a + b))^(1/2))/(4*(a*b + a^2)))*(-b*(a + b))^(1/2
))/(2*(a*b + a^2)))*(-b*(a + b))^(1/2)*1i)/(a*b + a^2))/(((2*b^3*tan(e + f*x) - ((2*a^2*b^2 - (tan(e + f*x)*(1
6*a^2*b^3 + 8*a^3*b^2)*(-b*(a + b))^(1/2))/(4*(a*b + a^2)))*(-b*(a + b))^(1/2))/(2*(a*b + a^2)))*(-b*(a + b))^
(1/2))/(a*b + a^2) - ((2*b^3*tan(e + f*x) + ((2*a^2*b^2 + (tan(e + f*x)*(16*a^2*b^3 + 8*a^3*b^2)*(-b*(a + b))^
(1/2))/(4*(a*b + a^2)))*(-b*(a + b))^(1/2))/(2*(a*b + a^2)))*(-b*(a + b))^(1/2))/(a*b + a^2)))*(-b*(a + b))^(1
/2)*1i)/(f*(a*b + a^2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{a + b \sec ^{2}{\left (e + f x \right )}}\, dx \]

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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