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

Optimal. Leaf size=36 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{\sqrt {b} f \sqrt {a+b}} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 36, normalized size of antiderivative = 1.00, 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 = {4146, 205} \[ \frac {\tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{\sqrt {b} f \sqrt {a+b}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^2/(a + b*Sec[e + f*x]^2),x]

[Out]

ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]]/(Sqrt[b]*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 4146

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = Fre
eFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 + ff^2*x^2)^(m/2 - 1)*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/
2), x]^p, x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && IntegerQ[n/2]

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 36, normalized size = 1.00 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{\sqrt {b} f \sqrt {a+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.56, size = 209, normalized size = 5.81 \[ \left [-\frac {\sqrt {-a b - b^{2}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{3} - b \cos \left (f x + e\right )\right )} \sqrt {-a b - b^{2}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right )}{4 \, {\left (a b + b^{2}\right )} f}, -\frac {\arctan \left (\frac {{\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b}{2 \, \sqrt {a b + b^{2}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right )}{2 \, \sqrt {a b + b^{2}} f}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.22, size = 50, normalized size = 1.39 \[ \frac {\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )}{\sqrt {a b + b^{2}} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^2/(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)))/(sqrt(a*b + b^2)*f)

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maple [A]  time = 0.55, size = 28, normalized size = 0.78 \[ \frac {\arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{f \sqrt {\left (a +b \right ) b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.43, size = 27, normalized size = 0.75 \[ \frac {\arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.51, size = 31, normalized size = 0.86 \[ \frac {\mathrm {atan}\left (\frac {b\,\mathrm {tan}\left (e+f\,x\right )}{\sqrt {b^2+a\,b}}\right )}{f\,\sqrt {b^2+a\,b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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