3.380 \(\int \frac {1}{(2 \sec (x)+\sin (x))^2} \, dx\)

Optimal. Leaf size=67 \[ \frac {8 x}{15 \sqrt {15}}+\frac {4 \tan (x)+1}{15 \left (2 \tan ^2(x)+\tan (x)+2\right )}-\frac {8 \tan ^{-1}\left (\frac {1-2 \cos ^2(x)}{2 \sin (x) \cos (x)+\sqrt {15}+4}\right )}{15 \sqrt {15}} \]

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Rubi [A]  time = 0.05, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {614, 618, 204} \[ \frac {8 x}{15 \sqrt {15}}+\frac {4 \tan (x)+1}{15 \left (2 \tan ^2(x)+\tan (x)+2\right )}-\frac {8 \tan ^{-1}\left (\frac {1-2 \cos ^2(x)}{2 \sin (x) \cos (x)+\sqrt {15}+4}\right )}{15 \sqrt {15}} \]

Antiderivative was successfully verified.

[In]

Int[(2*Sec[x] + Sin[x])^(-2),x]

[Out]

(8*x)/(15*Sqrt[15]) - (8*ArcTan[(1 - 2*Cos[x]^2)/(4 + Sqrt[15] + 2*Cos[x]*Sin[x])])/(15*Sqrt[15]) + (1 + 4*Tan
[x])/(15*(2 + Tan[x] + 2*Tan[x]^2))

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])

Rule 614

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^(p + 1))/((p +
1)*(b^2 - 4*a*c)), x] - Dist[(2*c*(2*p + 3))/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2] && IntegerQ[4*p]

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]

Rubi steps

\begin {align*} \int \frac {1}{(2 \sec (x)+\sin (x))^2} \, dx &=\operatorname {Subst}\left (\int \frac {1}{\left (2+x+2 x^2\right )^2} \, dx,x,\tan (x)\right )\\ &=\frac {1+4 \tan (x)}{15 \left (2+\tan (x)+2 \tan ^2(x)\right )}+\frac {4}{15} \operatorname {Subst}\left (\int \frac {1}{2+x+2 x^2} \, dx,x,\tan (x)\right )\\ &=\frac {1+4 \tan (x)}{15 \left (2+\tan (x)+2 \tan ^2(x)\right )}-\frac {8}{15} \operatorname {Subst}\left (\int \frac {1}{-15-x^2} \, dx,x,1+4 \tan (x)\right )\\ &=\frac {8 x}{15 \sqrt {15}}-\frac {8 \tan ^{-1}\left (\frac {1-2 \cos ^2(x)}{4+\sqrt {15}+2 \cos (x) \sin (x)}\right )}{15 \sqrt {15}}+\frac {1+4 \tan (x)}{15 \left (2+\tan (x)+2 \tan ^2(x)\right )}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 58, normalized size = 0.87 \[ \frac {(\sin (2 x)+4) \sec ^2(x) \left (15 (\cos (2 x)-15)+8 \sqrt {15} (\sin (2 x)+4) \tan ^{-1}\left (\frac {4 \tan (x)+1}{\sqrt {15}}\right )\right )}{900 (\sin (x)+2 \sec (x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(2*Sec[x] + Sin[x])^(-2),x]

[Out]

(Sec[x]^2*(4 + Sin[2*x])*(15*(-15 + Cos[2*x]) + 8*Sqrt[15]*ArcTan[(1 + 4*Tan[x])/Sqrt[15]]*(4 + Sin[2*x])))/(9
00*(2*Sec[x] + Sin[x])^2)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{(2 \sec (x)+\sin (x))^2} \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[(2*Sec[x] + Sin[x])^(-2),x]

[Out]

Could not integrate

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fricas [A]  time = 0.88, size = 61, normalized size = 0.91 \[ \frac {4 \, {\left (\sqrt {15} \cos \relax (x) \sin \relax (x) + 2 \, \sqrt {15}\right )} \arctan \left (\frac {8 \, \sqrt {15} \cos \relax (x) \sin \relax (x) + \sqrt {15}}{15 \, {\left (2 \, \cos \relax (x)^{2} - 1\right )}}\right ) + 15 \, \cos \relax (x)^{2} - 120}{225 \, {\left (\cos \relax (x) \sin \relax (x) + 2\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/225*(4*(sqrt(15)*cos(x)*sin(x) + 2*sqrt(15))*arctan(1/15*(8*sqrt(15)*cos(x)*sin(x) + sqrt(15))/(2*cos(x)^2 -
1)) + 15*cos(x)^2 - 120)/(cos(x)*sin(x) + 2)

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giac [A]  time = 0.64, size = 78, normalized size = 1.16 \[ \frac {8}{225} \, \sqrt {15} {\left (x + \arctan \left (-\frac {\sqrt {15} \sin \left (2 \, x\right ) - \cos \left (2 \, x\right ) - 4 \, \sin \left (2 \, x\right ) - 1}{\sqrt {15} \cos \left (2 \, x\right ) + \sqrt {15} - 4 \, \cos \left (2 \, x\right ) + \sin \left (2 \, x\right ) + 4}\right )\right )} + \frac {4 \, \tan \relax (x) + 1}{15 \, {\left (2 \, \tan \relax (x)^{2} + \tan \relax (x) + 2\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8/225*sqrt(15)*(x + arctan(-(sqrt(15)*sin(2*x) - cos(2*x) - 4*sin(2*x) - 1)/(sqrt(15)*cos(2*x) + sqrt(15) - 4*
cos(2*x) + sin(2*x) + 4))) + 1/15*(4*tan(x) + 1)/(2*tan(x)^2 + tan(x) + 2)

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maple [A]  time = 0.22, size = 39, normalized size = 0.58

method	result	size
default	\(\frac {1+4 \tan \relax (x )}{30+15 \tan \relax (x )+30 \left (\tan ^{2}\relax (x )\right )}+\frac {8 \sqrt {15}\, \arctan \left (\frac {\left (1+4 \tan \relax (x )\right ) \sqrt {15}}{15}\right )}{225}\)	\(39\)
risch	\(\frac {\left (\frac {8}{3615}-\frac {2 i}{241}\right ) \left (241 \,{\mathrm e}^{2 i x}-15+4 i\right )}{{\mathrm e}^{4 i x}+8 i {\mathrm e}^{2 i x}-1}+\frac {4 i \sqrt {15}\, \ln \left ({\mathrm e}^{2 i x}+i \sqrt {15}+4 i\right )}{225}-\frac {4 i \sqrt {15}\, \ln \left ({\mathrm e}^{2 i x}-i \sqrt {15}+4 i\right )}{225}\)	\(76\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*sec(x)+sin(x))^2,x,method=_RETURNVERBOSE)

[Out]

1/15*(1+4*tan(x))/(2+tan(x)+2*tan(x)^2)+8/225*15^(1/2)*arctan(1/15*(1+4*tan(x))*15^(1/2))

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maxima [A]  time = 0.96, size = 38, normalized size = 0.57 \[ \frac {8}{225} \, \sqrt {15} \arctan \left (\frac {1}{15} \, \sqrt {15} {\left (4 \, \tan \relax (x) + 1\right )}\right ) + \frac {4 \, \tan \relax (x) + 1}{15 \, {\left (2 \, \tan \relax (x)^{2} + \tan \relax (x) + 2\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*sec(x)+sin(x))^2,x, algorithm="maxima")

[Out]

8/225*sqrt(15)*arctan(1/15*sqrt(15)*(4*tan(x) + 1)) + 1/15*(4*tan(x) + 1)/(2*tan(x)^2 + tan(x) + 2)

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mupad [B]  time = 0.43, size = 120, normalized size = 1.79 \[ \frac {4\,\sqrt {15}\,\left (2\,\mathrm {atan}\left (\frac {2\,\sqrt {15}\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3}{15}-\frac {2\,\sqrt {15}\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{15}+\frac {2\,\sqrt {15}\,\mathrm {tan}\left (\frac {x}{2}\right )}{5}+\frac {\sqrt {15}}{15}\right )-2\,\mathrm {atan}\left (\frac {\sqrt {15}}{15}-\frac {2\,\sqrt {15}\,\mathrm {tan}\left (\frac {x}{2}\right )}{15}\right )\right )}{225}-\frac {\frac {7\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3}{30}+\frac {2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{15}-\frac {7\,\mathrm {tan}\left (\frac {x}{2}\right )}{30}}{{\mathrm {tan}\left (\frac {x}{2}\right )}^4-{\mathrm {tan}\left (\frac {x}{2}\right )}^3+2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+\mathrm {tan}\left (\frac {x}{2}\right )+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(x) + 2/cos(x))^2,x)

[Out]

(4*15^(1/2)*(2*atan((2*15^(1/2)*tan(x/2))/5 + 15^(1/2)/15 - (2*15^(1/2)*tan(x/2)^2)/15 + (2*15^(1/2)*tan(x/2)^
3)/15) - 2*atan(15^(1/2)/15 - (2*15^(1/2)*tan(x/2))/15)))/225 - ((2*tan(x/2)^2)/15 - (7*tan(x/2))/30 + (7*tan(
x/2)^3)/30)/(tan(x/2) + 2*tan(x/2)^2 - tan(x/2)^3 + tan(x/2)^4 + 1)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (\sin {\relax (x )} + 2 \sec {\relax (x )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*sec(x)+sin(x))**2,x)

[Out]

Integral((sin(x) + 2*sec(x))**(-2), x)

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