3.269 \(\int \frac{1}{(a \sec (x)+b \tan (x))^2} \, dx\)

Optimal. Leaf size=66 \[ \frac{2 a \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{b^2 \sqrt{a^2-b^2}}-\frac{\cos (x)}{b (a+b \sin (x))}-\frac{x}{b^2} \]

[Out]

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

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Rubi [A]  time = 0.133339, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.546, Rules used = {4391, 2693, 2735, 2660, 618, 204} \[ \frac{2 a \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{b^2 \sqrt{a^2-b^2}}-\frac{\cos (x)}{b (a+b \sin (x))}-\frac{x}{b^2} \]

Antiderivative was successfully verified.

[In]

Int[(a*Sec[x] + b*Tan[x])^(-2),x]

[Out]

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

Rule 4391

Int[(u_.)*((b_.)*sec[(c_.) + (d_.)*(x_)]^(n_.) + (a_.)*tan[(c_.) + (d_.)*(x_)]^(n_.))^(p_), x_Symbol] :> Int[A
ctivateTrig[u]*Sec[c + d*x]^(n*p)*(b + a*Sin[c + d*x]^n)^p, x] /; FreeQ[{a, b, c, d}, x] && IntegersQ[n, p]

Rule 2693

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*(g*
Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Dist[(g^2*(p - 1))/(b*(m + 1)), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1)*Sin[e + f*x], x], x] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a
^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && IntegersQ[2*m, 2*p]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rubi steps

\begin{align*} \int \frac{1}{(a \sec (x)+b \tan (x))^2} \, dx &=\int \frac{\cos ^2(x)}{(a+b \sin (x))^2} \, dx\\ &=-\frac{\cos (x)}{b (a+b \sin (x))}-\frac{\int \frac{\sin (x)}{a+b \sin (x)} \, dx}{b}\\ &=-\frac{x}{b^2}-\frac{\cos (x)}{b (a+b \sin (x))}+\frac{a \int \frac{1}{a+b \sin (x)} \, dx}{b^2}\\ &=-\frac{x}{b^2}-\frac{\cos (x)}{b (a+b \sin (x))}+\frac{(2 a) \operatorname{Subst}\left (\int \frac{1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )}{b^2}\\ &=-\frac{x}{b^2}-\frac{\cos (x)}{b (a+b \sin (x))}-\frac{(4 a) \operatorname{Subst}\left (\int \frac{1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac{x}{2}\right )\right )}{b^2}\\ &=-\frac{x}{b^2}+\frac{2 a \tan ^{-1}\left (\frac{b+a \tan \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{b^2 \sqrt{a^2-b^2}}-\frac{\cos (x)}{b (a+b \sin (x))}\\ \end{align*}

Mathematica [B]  time = 3.75459, size = 422, normalized size = 6.39 \[ \frac{\cos (x) \left (\sqrt{a+b} \left (\sqrt{a-b} \sqrt{1-\sin (x)} \left (b \left (b^2-a^2\right ) \sqrt{\frac{b (\sin (x)+1)}{b-a}} \sqrt{\frac{b-b \sin (x)}{a+b}}+2 a \left (a \sqrt{-b^2}+\sqrt{-b} b^{3/2} \sin (x)\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\frac{b (\sin (x)+1)}{b-a}}}{\sqrt{-b} \sqrt{\frac{b-b \sin (x)}{a+b}}}\right )\right )+2 \sqrt{\frac{b-b \sin (x)}{a+b}} \left (a^2 \sqrt{-b} \sqrt{-b^2}-a b^{5/2} \sin (x)+a b^{5/2}+(-b)^{3/2} \sqrt{-b^2} b \sin (x)\right ) \sinh ^{-1}\left (\frac{\sqrt{a-b} \sqrt{-\frac{b (\sin (x)+1)}{a-b}}}{\sqrt{2} \sqrt{b}}\right )\right )-2 a b (a-b) \sqrt{1-\sin (x)} (a+b \sin (x)) \tanh ^{-1}\left (\frac{\sqrt{a-b} \sqrt{-\frac{b (\sin (x)+1)}{a-b}}}{\sqrt{a+b} \sqrt{-\frac{b (\sin (x)-1)}{a+b}}}\right )\right )}{b^2 (a-b)^{3/2} (a+b)^{3/2} \sqrt{1-\sin (x)} \sqrt{-\frac{b (\sin (x)+1)}{a-b}} \sqrt{\frac{b-b \sin (x)}{a+b}} (a+b \sin (x))} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a*Sec[x] + b*Tan[x])^(-2),x]

[Out]

(Cos[x]*(-2*a*(a - b)*b*ArcTanh[(Sqrt[a - b]*Sqrt[-((b*(1 + Sin[x]))/(a - b))])/(Sqrt[a + b]*Sqrt[-((b*(-1 + S
in[x]))/(a + b))])]*Sqrt[1 - Sin[x]]*(a + b*Sin[x]) + Sqrt[a + b]*(2*ArcSinh[(Sqrt[a - b]*Sqrt[-((b*(1 + Sin[x
]))/(a - b))])/(Sqrt[2]*Sqrt[b])]*Sqrt[(b - b*Sin[x])/(a + b)]*(a*b^(5/2) + a^2*Sqrt[-b]*Sqrt[-b^2] - a*b^(5/2
)*Sin[x] + (-b)^(3/2)*b*Sqrt[-b^2]*Sin[x]) + Sqrt[a - b]*Sqrt[1 - Sin[x]]*(b*(-a^2 + b^2)*Sqrt[(b*(1 + Sin[x])
)/(-a + b)]*Sqrt[(b - b*Sin[x])/(a + b)] + 2*a*ArcTan[(Sqrt[b]*Sqrt[(b*(1 + Sin[x]))/(-a + b)])/(Sqrt[-b]*Sqrt
[(b - b*Sin[x])/(a + b)])]*(a*Sqrt[-b^2] + Sqrt[-b]*b^(3/2)*Sin[x])))))/((a - b)^(3/2)*b^2*(a + b)^(3/2)*Sqrt[
1 - Sin[x]]*Sqrt[-((b*(1 + Sin[x]))/(a - b))]*Sqrt[(b - b*Sin[x])/(a + b)]*(a + b*Sin[x]))

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Maple [A]  time = 0.079, size = 106, normalized size = 1.6 \begin{align*} -2\,{\frac{\tan \left ( x/2 \right ) }{ \left ( a \left ( \tan \left ( x/2 \right ) \right ) ^{2}+2\,b\tan \left ( x/2 \right ) +a \right ) a}}-2\,{\frac{1}{b \left ( a \left ( \tan \left ( x/2 \right ) \right ) ^{2}+2\,b\tan \left ( x/2 \right ) +a \right ) }}+2\,{\frac{a}{{b}^{2}\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( x/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }-2\,{\frac{\arctan \left ( \tan \left ( x/2 \right ) \right ) }{{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*sec(x)+b*tan(x))^2,x)

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.39972, size = 687, normalized size = 10.41 \begin{align*} \left [-\frac{2 \,{\left (a^{2} b - b^{3}\right )} x \sin \left (x\right ) +{\left (a b \sin \left (x\right ) + a^{2}\right )} \sqrt{-a^{2} + b^{2}} \log \left (\frac{{\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2} + 2 \,{\left (a \cos \left (x\right ) \sin \left (x\right ) + b \cos \left (x\right )\right )} \sqrt{-a^{2} + b^{2}}}{b^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2}}\right ) + 2 \,{\left (a^{3} - a b^{2}\right )} x + 2 \,{\left (a^{2} b - b^{3}\right )} \cos \left (x\right )}{2 \,{\left (a^{3} b^{2} - a b^{4} +{\left (a^{2} b^{3} - b^{5}\right )} \sin \left (x\right )\right )}}, -\frac{{\left (a^{2} b - b^{3}\right )} x \sin \left (x\right ) +{\left (a b \sin \left (x\right ) + a^{2}\right )} \sqrt{a^{2} - b^{2}} \arctan \left (-\frac{a \sin \left (x\right ) + b}{\sqrt{a^{2} - b^{2}} \cos \left (x\right )}\right ) +{\left (a^{3} - a b^{2}\right )} x +{\left (a^{2} b - b^{3}\right )} \cos \left (x\right )}{a^{3} b^{2} - a b^{4} +{\left (a^{2} b^{3} - b^{5}\right )} \sin \left (x\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sec(x)+b*tan(x))^2,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a \sec{\left (x \right )} + b \tan{\left (x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sec(x)+b*tan(x))**2,x)

[Out]

Integral((a*sec(x) + b*tan(x))**(-2), x)

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Giac [A]  time = 1.14163, size = 127, normalized size = 1.92 \begin{align*} \frac{2 \,{\left (\pi \left \lfloor \frac{x}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, x\right ) + b}{\sqrt{a^{2} - b^{2}}}\right )\right )} a}{\sqrt{a^{2} - b^{2}} b^{2}} - \frac{x}{b^{2}} - \frac{2 \,{\left (b \tan \left (\frac{1}{2} \, x\right ) + a\right )}}{{\left (a \tan \left (\frac{1}{2} \, x\right )^{2} + 2 \, b \tan \left (\frac{1}{2} \, x\right ) + a\right )} a b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sec(x)+b*tan(x))^2,x, algorithm="giac")

[Out]

2*(pi*floor(1/2*x/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*x) + b)/sqrt(a^2 - b^2)))*a/(sqrt(a^2 - b^2)*b^2) - x/b
^2 - 2*(b*tan(1/2*x) + a)/((a*tan(1/2*x)^2 + 2*b*tan(1/2*x) + a)*a*b)