3.1.42 \(\int \frac {1}{(a+b \cos ^2(x))^2} \, dx\) [42]

Optimal. Leaf size=65 \[ -\frac {(2 a+b) \text {ArcTan}\left (\frac {\sqrt {a+b} \cot (x)}{\sqrt {a}}\right )}{2 a^{3/2} (a+b)^{3/2}}-\frac {b \cos (x) \sin (x)}{2 a (a+b) \left (a+b \cos ^2(x)\right )} \]

[Out]

-1/2*(2*a+b)*arctan(cot(x)*(a+b)^(1/2)/a^(1/2))/a^(3/2)/(a+b)^(3/2)-1/2*b*cos(x)*sin(x)/a/(a+b)/(a+b*cos(x)^2)

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Rubi [A]
time = 0.04, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3263, 12, 3260, 211} \begin {gather*} -\frac {(2 a+b) \text {ArcTan}\left (\frac {\sqrt {a+b} \cot (x)}{\sqrt {a}}\right )}{2 a^{3/2} (a+b)^{3/2}}-\frac {b \sin (x) \cos (x)}{2 a (a+b) \left (a+b \cos ^2(x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[x]^2)^(-2),x]

[Out]

-1/2*((2*a + b)*ArcTan[(Sqrt[a + b]*Cot[x])/Sqrt[a]])/(a^(3/2)*(a + b)^(3/2)) - (b*Cos[x]*Sin[x])/(2*a*(a + b)
*(a + b*Cos[x]^2))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 3260

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(-1), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff/f, Subst[Int[1/(a + (a + b)*ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x]

Rule 3263

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Simp[(-b)*Cos[e + f*x]*Sin[e + f*x]*((a + b*Si
n[e + f*x]^2)^(p + 1)/(2*a*f*(p + 1)*(a + b))), x] + Dist[1/(2*a*(p + 1)*(a + b)), Int[(a + b*Sin[e + f*x]^2)^
(p + 1)*Simp[2*a*(p + 1) + b*(2*p + 3) - 2*b*(p + 2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f}, x] && N
eQ[a + b, 0] && LtQ[p, -1]

Rubi steps

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

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Mathematica [A]
time = 0.26, size = 70, normalized size = 1.08 \begin {gather*} -\frac {(-2 a-b) \text {ArcTan}\left (\frac {\sqrt {a} \tan (x)}{\sqrt {a+b}}\right )}{2 a^{3/2} (a+b)^{3/2}}-\frac {b \sin (2 x)}{2 a (a+b) (2 a+b+b \cos (2 x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[x]^2)^(-2),x]

[Out]

-1/2*((-2*a - b)*ArcTan[(Sqrt[a]*Tan[x])/Sqrt[a + b]])/(a^(3/2)*(a + b)^(3/2)) - (b*Sin[2*x])/(2*a*(a + b)*(2*
a + b + b*Cos[2*x]))

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Maple [A]
time = 0.13, size = 60, normalized size = 0.92

method	result	size
default	\(-\frac {b \tan \left (x \right )}{2 \left (a +b \right ) a \left (a \left (\tan ^{2}\left (x \right )\right )+a +b \right )}+\frac {\left (2 a +b \right ) \arctan \left (\frac {a \tan \left (x \right )}{\sqrt {\left (a +b \right ) a}}\right )}{2 \left (a +b \right ) a \sqrt {\left (a +b \right ) a}}\)	\(60\)
risch	\(-\frac {i \left (2 a \,{\mathrm e}^{2 i x}+b \,{\mathrm e}^{2 i x}+b \right )}{\left (a +b \right ) a \left (b \,{\mathrm e}^{4 i x}+4 a \,{\mathrm e}^{2 i x}+2 b \,{\mathrm e}^{2 i x}+b \right )}-\frac {\ln \left ({\mathrm e}^{2 i x}+\frac {2 i a^{2}+2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{2 \sqrt {-a^{2}-a b}\, \left (a +b \right )}-\frac {\ln \left ({\mathrm e}^{2 i x}+\frac {2 i a^{2}+2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right ) b}{4 \sqrt {-a^{2}-a b}\, \left (a +b \right ) a}+\frac {\ln \left ({\mathrm e}^{2 i x}+\frac {-2 i a^{2}-2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{2 \sqrt {-a^{2}-a b}\, \left (a +b \right )}+\frac {\ln \left ({\mathrm e}^{2 i x}+\frac {-2 i a^{2}-2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right ) b}{4 \sqrt {-a^{2}-a b}\, \left (a +b \right ) a}\)	\(397\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*cos(x)^2)^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.47, size = 72, normalized size = 1.11 \begin {gather*} -\frac {b \tan \left (x\right )}{2 \, {\left (a^{3} + 2 \, a^{2} b + a b^{2} + {\left (a^{3} + a^{2} b\right )} \tan \left (x\right )^{2}\right )}} + \frac {{\left (2 \, a + b\right )} \arctan \left (\frac {a \tan \left (x\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{2 \, \sqrt {{\left (a + b\right )} a} {\left (a^{2} + a b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 131 vs. \(2 (53) = 106\).
time = 0.45, size = 326, normalized size = 5.02 \begin {gather*} \left [-\frac {4 \, {\left (a^{2} b + a b^{2}\right )} \cos \left (x\right ) \sin \left (x\right ) + {\left ({\left (2 \, a b + b^{2}\right )} \cos \left (x\right )^{2} + 2 \, a^{2} + a b\right )} \sqrt {-a^{2} - a b} \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 )}{8 \, {\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2} + {\left (a^{4} b + 2 \, a^{3} b^{2} + a^{2} b^{3}\right )} \cos \left (x\right )^{2}\right )}}, -\frac {2 \, {\left (a^{2} b + a b^{2}\right )} \cos \left (x\right ) \sin \left (x\right ) + {\left ({\left (2 \, a b + b^{2}\right )} \cos \left (x\right )^{2} + 2 \, a^{2} + a b\right )} \sqrt {a^{2} + a 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 )}{4 \, {\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2} + {\left (a^{4} b + 2 \, a^{3} b^{2} + a^{2} b^{3}\right )} \cos \left (x\right )^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*(4*(a^2*b + a*b^2)*cos(x)*sin(x) + ((2*a*b + b^2)*cos(x)^2 + 2*a^2 + a*b)*sqrt(-a^2 - a*b)*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)))/(a^5 + 2*a^4*b + a^3*b^2 + (a^4*b + 2*a^3*b^2 + a^2*b^3)*cos(
x)^2), -1/4*(2*(a^2*b + a*b^2)*cos(x)*sin(x) + ((2*a*b + b^2)*cos(x)^2 + 2*a^2 + a*b)*sqrt(a^2 + a*b)*arctan(1
/2*((2*a + b)*cos(x)^2 - a)/(sqrt(a^2 + a*b)*cos(x)*sin(x))))/(a^5 + 2*a^4*b + a^3*b^2 + (a^4*b + 2*a^3*b^2 +
a^2*b^3)*cos(x)^2)]

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]
time = 0.40, size = 69, normalized size = 1.06 \begin {gather*} \frac {{\left (\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (x\right )}{\sqrt {a^{2} + a b}}\right )\right )} {\left (2 \, a + b\right )}}{2 \, {\left (a^{2} + a b\right )}^{\frac {3}{2}}} - \frac {b \tan \left (x\right )}{2 \, {\left (a \tan \left (x\right )^{2} + a + b\right )} {\left (a^{2} + a b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(pi*floor(x/pi + 1/2)*sgn(a) + arctan(a*tan(x)/sqrt(a^2 + a*b)))*(2*a + b)/(a^2 + a*b)^(3/2) - 1/2*b*tan(x
)/((a*tan(x)^2 + a + b)*(a^2 + a*b))

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Mupad [B]
time = 2.34, size = 52, normalized size = 0.80 \begin {gather*} \frac {\mathrm {atan}\left (\frac {\sqrt {a}\,\mathrm {tan}\left (x\right )}{\sqrt {a+b}}\right )\,\left (2\,a+b\right )}{2\,a^{3/2}\,{\left (a+b\right )}^{3/2}}-\frac {b\,\mathrm {tan}\left (x\right )}{2\,a\,\left (a+b\right )\,\left (a\,{\mathrm {tan}\left (x\right )}^2+a+b\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*cos(x)^2)^2,x)

[Out]

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

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