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

Optimal. Leaf size=40 \[ \frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \sin (x)}{\sqrt{a}}\right )}{\sqrt{a} (a+b)}+\frac{\tanh ^{-1}(\sin (x))}{a+b} \]

[Out]

(Sqrt[b]*ArcTan[(Sqrt[b]*Sin[x])/Sqrt[a]])/(Sqrt[a]*(a + b)) + ArcTanh[Sin[x]]/(a + b)

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Rubi [A]  time = 0.0509571, antiderivative size = 40, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {3190, 391, 206, 205} \[ \frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \sin (x)}{\sqrt{a}}\right )}{\sqrt{a} (a+b)}+\frac{\tanh ^{-1}(\sin (x))}{a+b} \]

Antiderivative was successfully verified.

[In]

Int[Sec[x]/(a + b*Sin[x]^2),x]

[Out]

(Sqrt[b]*ArcTan[(Sqrt[b]*Sin[x])/Sqrt[a]])/(Sqrt[a]*(a + b)) + ArcTanh[Sin[x]]/(a + b)

Rule 3190

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

Rule 391

Int[1/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x^n),
x], x] - Dist[d/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

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

Mathematica [B]  time = 0.124296, size = 96, normalized size = 2.4 \[ \frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \sin (x)}{\sqrt{a}}\right )-\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{a} \csc (x)}{\sqrt{b}}\right )+2 \sqrt{a} \left (\log \left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right )-\log \left (\cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )\right )\right )}{2 \sqrt{a} (a+b)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[x]/(a + b*Sin[x]^2),x]

[Out]

(-(Sqrt[b]*ArcTan[(Sqrt[a]*Csc[x])/Sqrt[b]]) + Sqrt[b]*ArcTan[(Sqrt[b]*Sin[x])/Sqrt[a]] + 2*Sqrt[a]*(-Log[Cos[
x/2] - Sin[x/2]] + Log[Cos[x/2] + Sin[x/2]]))/(2*Sqrt[a]*(a + b))

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Maple [A]  time = 0.055, size = 55, normalized size = 1.4 \begin{align*} -{\frac{\ln \left ( -1+\sin \left ( x \right ) \right ) }{2\,a+2\,b}}+{\frac{\ln \left ( 1+\sin \left ( x \right ) \right ) }{2\,a+2\,b}}+{\frac{b}{a+b}\arctan \left ({\sin \left ( x \right ) b{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.28405, size = 308, normalized size = 7.7 \begin{align*} \left [\frac{\sqrt{-\frac{b}{a}} \log \left (-\frac{b \cos \left (x\right )^{2} - 2 \, a \sqrt{-\frac{b}{a}} \sin \left (x\right ) + a - b}{b \cos \left (x\right )^{2} - a - b}\right ) + \log \left (\sin \left (x\right ) + 1\right ) - \log \left (-\sin \left (x\right ) + 1\right )}{2 \,{\left (a + b\right )}}, \frac{2 \, \sqrt{\frac{b}{a}} \arctan \left (\sqrt{\frac{b}{a}} \sin \left (x\right )\right ) + \log \left (\sin \left (x\right ) + 1\right ) - \log \left (-\sin \left (x\right ) + 1\right )}{2 \,{\left (a + b\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(sqrt(-b/a)*log(-(b*cos(x)^2 - 2*a*sqrt(-b/a)*sin(x) + a - b)/(b*cos(x)^2 - a - b)) + log(sin(x) + 1) - l
og(-sin(x) + 1))/(a + b), 1/2*(2*sqrt(b/a)*arctan(sqrt(b/a)*sin(x)) + log(sin(x) + 1) - log(-sin(x) + 1))/(a +
b)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sec(x)/(a + b*sin(x)**2), x)

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Giac [A]  time = 1.09957, size = 66, normalized size = 1.65 \begin{align*} \frac{b \arctan \left (\frac{b \sin \left (x\right )}{\sqrt{a b}}\right )}{\sqrt{a b}{\left (a + b\right )}} + \frac{\log \left (\sin \left (x\right ) + 1\right )}{2 \,{\left (a + b\right )}} - \frac{\log \left (-\sin \left (x\right ) + 1\right )}{2 \,{\left (a + b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*arctan(b*sin(x)/sqrt(a*b))/(sqrt(a*b)*(a + b)) + 1/2*log(sin(x) + 1)/(a + b) - 1/2*log(-sin(x) + 1)/(a + b)