\(\int \frac {\sec ^2(x)}{a+b \sin ^2(x)} \, dx\) [309]

Optimal result

Integrand size = 15, antiderivative size = 39 \[ \int \frac {\sec ^2(x)}{a+b \sin ^2(x)} \, dx=\frac {b \arctan \left (\frac {\sqrt {a+b} \tan (x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)^{3/2}}+\frac {\tan (x)}{a+b} \]

[Out]

b*arctan((a+b)^(1/2)*tan(x)/a^(1/2))/(a+b)^(3/2)/a^(1/2)+tan(x)/(a+b)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3270, 396, 211} \[ \int \frac {\sec ^2(x)}{a+b \sin ^2(x)} \, dx=\frac {b \arctan \left (\frac {\sqrt {a+b} \tan (x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)^{3/2}}+\frac {\tan (x)}{a+b} \]

[In]

Int[Sec[x]^2/(a + b*Sin[x]^2),x]

[Out]

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

Rule 211

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

Rule 396

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

Rule 3270

Int[cos[(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/f, Subst[Int[(a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p + 1), x], x, T
an[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {1+x^2}{a+(a+b) x^2} \, dx,x,\tan (x)\right ) \\ & = \frac {\tan (x)}{a+b}+\frac {b \text {Subst}\left (\int \frac {1}{a+(a+b) x^2} \, dx,x,\tan (x)\right )}{a+b} \\ & = \frac {b \arctan \left (\frac {\sqrt {a+b} \tan (x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)^{3/2}}+\frac {\tan (x)}{a+b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00 \[ \int \frac {\sec ^2(x)}{a+b \sin ^2(x)} \, dx=\frac {b \arctan \left (\frac {\sqrt {a+b} \tan (x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)^{3/2}}+\frac {\tan (x)}{a+b} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.60 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.97

method	result	size
default	\(\frac {\tan \left (x \right )}{a +b}+\frac {b \arctan \left (\frac {\left (a +b \right ) \tan \left (x \right )}{\sqrt {a \left (a +b \right )}}\right )}{\left (a +b \right ) \sqrt {a \left (a +b \right )}}\)	\(38\)
risch	\(\frac {2 i}{\left ({\mathrm e}^{2 i x}+1\right ) \left (a +b \right )}-\frac {b \ln \left ({\mathrm e}^{2 i x}-\frac {2 i a^{2}+2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{2 \sqrt {-a^{2}-a b}\, \left (a +b \right )}+\frac {b \ln \left ({\mathrm e}^{2 i x}+\frac {2 i a^{2}+2 i a b -2 a \sqrt {-a^{2}-a b}-b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{2 \sqrt {-a^{2}-a b}\, \left (a +b \right )}\)	\(189\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 85 vs. \(2 (31) = 62\).

Time = 0.32 (sec) , antiderivative size = 255, normalized size of antiderivative = 6.54 \[ \int \frac {\sec ^2(x)}{a+b \sin ^2(x)} \, dx=\left [-\frac {\sqrt {-a^{2} - a b} b \cos \left (x\right ) \log \left (\frac {{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (x\right )^{4} - 2 \, {\left (4 \, a^{2} + 5 \, a b + b^{2}\right )} \cos \left (x\right )^{2} + 4 \, {\left ({\left (2 \, a + b\right )} \cos \left (x\right )^{3} - {\left (a + b\right )} \cos \left (x\right )\right )} \sqrt {-a^{2} - a b} \sin \left (x\right ) + a^{2} + 2 \, a b + b^{2}}{b^{2} \cos \left (x\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (x\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right ) - 4 \, {\left (a^{2} + a b\right )} \sin \left (x\right )}{4 \, {\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \cos \left (x\right )}, -\frac {\sqrt {a^{2} + a b} b \arctan \left (\frac {{\left (2 \, a + b\right )} \cos \left (x\right )^{2} - a - b}{2 \, \sqrt {a^{2} + a b} \cos \left (x\right ) \sin \left (x\right )}\right ) \cos \left (x\right ) - 2 \, {\left (a^{2} + a b\right )} \sin \left (x\right )}{2 \, {\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \cos \left (x\right )}\right ] \]

[In]

integrate(sec(x)^2/(a+b*sin(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 + 5*a*b + b^2)*cos(x)^2 + 4*((
2*a + b)*cos(x)^3 - (a + b)*cos(x))*sqrt(-a^2 - a*b)*sin(x) + a^2 + 2*a*b + b^2)/(b^2*cos(x)^4 - 2*(a*b + b^2)
*cos(x)^2 + a^2 + 2*a*b + b^2)) - 4*(a^2 + a*b)*sin(x))/((a^3 + 2*a^2*b + a*b^2)*cos(x)), -1/2*(sqrt(a^2 + a*b
)*b*arctan(1/2*((2*a + b)*cos(x)^2 - a - b)/(sqrt(a^2 + a*b)*cos(x)*sin(x)))*cos(x) - 2*(a^2 + a*b)*sin(x))/((
a^3 + 2*a^2*b + a*b^2)*cos(x))]

Sympy [F]

\[ \int \frac {\sec ^2(x)}{a+b \sin ^2(x)} \, dx=\int \frac {\sec ^{2}{\left (x \right )}}{a + b \sin ^{2}{\left (x \right )}}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.95 \[ \int \frac {\sec ^2(x)}{a+b \sin ^2(x)} \, dx=\frac {b \arctan \left (\frac {{\left (a + b\right )} \tan \left (x\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{\sqrt {{\left (a + b\right )} a} {\left (a + b\right )}} + \frac {\tan \left (x\right )}{a + b} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.15 \[ \int \frac {\sec ^2(x)}{a+b \sin ^2(x)} \, dx=\frac {b \arctan \left (\frac {a \tan \left (x\right ) + b \tan \left (x\right )}{\sqrt {a^{2} + a b}}\right )}{\sqrt {a^{2} + a b} {\left (a + b\right )}} + \frac {\tan \left (x\right )}{a + b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 14.10 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00 \[ \int \frac {\sec ^2(x)}{a+b \sin ^2(x)} \, dx=\frac {\mathrm {tan}\left (x\right )}{a+b}+\frac {b\,\mathrm {atan}\left (\frac {\mathrm {tan}\left (x\right )\,\left (2\,a+2\,b\right )}{2\,\sqrt {a}\,\sqrt {a+b}}\right )}{\sqrt {a}\,{\left (a+b\right )}^{3/2}} \]

[In]

int(1/(cos(x)^2*(a + b*sin(x)^2)),x)

[Out]

tan(x)/(a + b) + (b*atan((tan(x)*(2*a + 2*b))/(2*a^(1/2)*(a + b)^(1/2))))/(a^(1/2)*(a + b)^(3/2))