3.39 \(\int \frac{\sec ^2(x)}{a+b \cos ^2(x)} \, dx\)

Optimal. Leaf size=37 \[ \frac{b \tan ^{-1}\left (\frac{\sqrt{a+b} \cot (x)}{\sqrt{a}}\right )}{a^{3/2} \sqrt{a+b}}+\frac{\tan (x)}{a} \]

[Out]

(b*ArcTan[(Sqrt[a + b]*Cot[x])/Sqrt[a]])/(a^(3/2)*Sqrt[a + b]) + Tan[x]/a

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Rubi [A]  time = 0.0605322, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3187, 453, 205} \[ \frac{b \tan ^{-1}\left (\frac{\sqrt{a+b} \cot (x)}{\sqrt{a}}\right )}{a^{3/2} \sqrt{a+b}}+\frac{\tan (x)}{a} \]

Antiderivative was successfully verified.

[In]

Int[Sec[x]^2/(a + b*Cos[x]^2),x]

[Out]

(b*ArcTan[(Sqrt[a + b]*Cot[x])/Sqrt[a]])/(a^(3/2)*Sqrt[a + b]) + Tan[x]/a

Rule 3187

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

Rule 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

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{\sec ^2(x)}{a+b \cos ^2(x)} \, dx &=-\operatorname{Subst}\left (\int \frac{1+x^2}{x^2 \left (a+(a+b) x^2\right )} \, dx,x,\cot (x)\right )\\ &=\frac{\tan (x)}{a}+\frac{b \operatorname{Subst}\left (\int \frac{1}{a+(a+b) x^2} \, dx,x,\cot (x)\right )}{a}\\ &=\frac{b \tan ^{-1}\left (\frac{\sqrt{a+b} \cot (x)}{\sqrt{a}}\right )}{a^{3/2} \sqrt{a+b}}+\frac{\tan (x)}{a}\\ \end{align*}

Mathematica [A]  time = 0.0743614, size = 38, normalized size = 1.03 \[ \frac{\tan (x)}{a}-\frac{b \tan ^{-1}\left (\frac{\sqrt{a} \tan (x)}{\sqrt{a+b}}\right )}{a^{3/2} \sqrt{a+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*ArcTan[(Sqrt[a]*Tan[x])/Sqrt[a + b]])/(a^(3/2)*Sqrt[a + b])) + Tan[x]/a

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Maple [A]  time = 0.028, size = 33, normalized size = 0.9 \begin{align*}{\frac{\tan \left ( x \right ) }{a}}-{\frac{b}{a}\arctan \left ({\tan \left ( x \right ) a{\frac{1}{\sqrt{ \left ( a+b \right ) a}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

tan(x)/a-b/a/((a+b)*a)^(1/2)*arctan(a*tan(x)/((a+b)*a)^(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(sec(x)^2/(a+b*cos(x)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.92837, size = 543, normalized size = 14.68 \begin{align*} \left [-\frac{\sqrt{-a^{2} - a b} b \cos \left (x\right ) \log \left (\frac{{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (x\right )^{4} - 2 \,{\left (4 \, a^{2} + 3 \, a b\right )} \cos \left (x\right )^{2} - 4 \,{\left ({\left (2 \, a + b\right )} \cos \left (x\right )^{3} - a \cos \left (x\right )\right )} \sqrt{-a^{2} - a b} \sin \left (x\right ) + a^{2}}{b^{2} \cos \left (x\right )^{4} + 2 \, a b \cos \left (x\right )^{2} + a^{2}}\right ) - 4 \,{\left (a^{2} + a b\right )} \sin \left (x\right )}{4 \,{\left (a^{3} + a^{2} b\right )} \cos \left (x\right )}, \frac{\sqrt{a^{2} + a b} b \arctan \left (\frac{{\left (2 \, a + b\right )} \cos \left (x\right )^{2} - a}{2 \, \sqrt{a^{2} + a b} \cos \left (x\right ) \sin \left (x\right )}\right ) \cos \left (x\right ) + 2 \,{\left (a^{2} + a b\right )} \sin \left (x\right )}{2 \,{\left (a^{3} + a^{2} b\right )} \cos \left (x\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.12358, size = 49, normalized size = 1.32 \begin{align*} -\frac{b \arctan \left (\frac{a \tan \left (x\right )}{\sqrt{a^{2} + a b}}\right )}{\sqrt{a^{2} + a b} a} + \frac{\tan \left (x\right )}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b*arctan(a*tan(x)/sqrt(a^2 + a*b))/(sqrt(a^2 + a*b)*a) + tan(x)/a