3.43 \(\int \frac{1}{(a+b \cos ^2(x))^3} \, dx\)

Optimal. Leaf size=107 \[ -\frac{\left (8 a^2+8 a b+3 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a+b} \cot (x)}{\sqrt{a}}\right )}{8 a^{5/2} (a+b)^{5/2}}-\frac{3 b (2 a+b) \sin (x) \cos (x)}{8 a^2 (a+b)^2 \left (a+b \cos ^2(x)\right )}-\frac{b \sin (x) \cos (x)}{4 a (a+b) \left (a+b \cos ^2(x)\right )^2} \]

[Out]

-((8*a^2 + 8*a*b + 3*b^2)*ArcTan[(Sqrt[a + b]*Cot[x])/Sqrt[a]])/(8*a^(5/2)*(a + b)^(5/2)) - (b*Cos[x]*Sin[x])/
(4*a*(a + b)*(a + b*Cos[x]^2)^2) - (3*b*(2*a + b)*Cos[x]*Sin[x])/(8*a^2*(a + b)^2*(a + b*Cos[x]^2))

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Rubi [A]  time = 0.117157, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {3184, 3173, 12, 3181, 205} \[ -\frac{\left (8 a^2+8 a b+3 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a+b} \cot (x)}{\sqrt{a}}\right )}{8 a^{5/2} (a+b)^{5/2}}-\frac{3 b (2 a+b) \sin (x) \cos (x)}{8 a^2 (a+b)^2 \left (a+b \cos ^2(x)\right )}-\frac{b \sin (x) \cos (x)}{4 a (a+b) \left (a+b \cos ^2(x)\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[x]^2)^(-3),x]

[Out]

-((8*a^2 + 8*a*b + 3*b^2)*ArcTan[(Sqrt[a + b]*Cot[x])/Sqrt[a]])/(8*a^(5/2)*(a + b)^(5/2)) - (b*Cos[x]*Sin[x])/
(4*a*(a + b)*(a + b*Cos[x]^2)^2) - (3*b*(2*a + b)*Cos[x]*Sin[x])/(8*a^2*(a + b)^2*(a + b*Cos[x]^2))

Rule 3184

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> -Simp[(b*Cos[e + f*x]*Sin[e + f*x]*(a + b*Sin[
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] && NeQ
[a + b, 0] && LtQ[p, -1]

Rule 3173

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

Rule 12

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

Rule 3181

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 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{1}{\left (a+b \cos ^2(x)\right )^3} \, dx &=-\frac{b \cos (x) \sin (x)}{4 a (a+b) \left (a+b \cos ^2(x)\right )^2}-\frac{\int \frac{-4 a-3 b+2 b \cos ^2(x)}{\left (a+b \cos ^2(x)\right )^2} \, dx}{4 a (a+b)}\\ &=-\frac{b \cos (x) \sin (x)}{4 a (a+b) \left (a+b \cos ^2(x)\right )^2}-\frac{3 b (2 a+b) \cos (x) \sin (x)}{8 a^2 (a+b)^2 \left (a+b \cos ^2(x)\right )}-\frac{\int \frac{-8 a^2-8 a b-3 b^2}{a+b \cos ^2(x)} \, dx}{8 a^2 (a+b)^2}\\ &=-\frac{b \cos (x) \sin (x)}{4 a (a+b) \left (a+b \cos ^2(x)\right )^2}-\frac{3 b (2 a+b) \cos (x) \sin (x)}{8 a^2 (a+b)^2 \left (a+b \cos ^2(x)\right )}+\frac{\left (8 a^2+8 a b+3 b^2\right ) \int \frac{1}{a+b \cos ^2(x)} \, dx}{8 a^2 (a+b)^2}\\ &=-\frac{b \cos (x) \sin (x)}{4 a (a+b) \left (a+b \cos ^2(x)\right )^2}-\frac{3 b (2 a+b) \cos (x) \sin (x)}{8 a^2 (a+b)^2 \left (a+b \cos ^2(x)\right )}-\frac{\left (8 a^2+8 a b+3 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{a+(a+b) x^2} \, dx,x,\cot (x)\right )}{8 a^2 (a+b)^2}\\ &=-\frac{\left (8 a^2+8 a b+3 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a+b} \cot (x)}{\sqrt{a}}\right )}{8 a^{5/2} (a+b)^{5/2}}-\frac{b \cos (x) \sin (x)}{4 a (a+b) \left (a+b \cos ^2(x)\right )^2}-\frac{3 b (2 a+b) \cos (x) \sin (x)}{8 a^2 (a+b)^2 \left (a+b \cos ^2(x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.615142, size = 106, normalized size = 0.99 \[ \frac{\frac{\left (8 a^2+8 a b+3 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a} \tan (x)}{\sqrt{a+b}}\right )}{(a+b)^{5/2}}-\frac{\sqrt{a} b \sin (2 x) \left (16 a^2+3 b (2 a+b) \cos (2 x)+16 a b+3 b^2\right )}{(a+b)^2 (2 a+b \cos (2 x)+b)^2}}{8 a^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[x]^2)^(-3),x]

[Out]

(((8*a^2 + 8*a*b + 3*b^2)*ArcTan[(Sqrt[a]*Tan[x])/Sqrt[a + b]])/(a + b)^(5/2) - (Sqrt[a]*b*(16*a^2 + 16*a*b +
3*b^2 + 3*b*(2*a + b)*Cos[2*x])*Sin[2*x])/((a + b)^2*(2*a + b + b*Cos[2*x])^2))/(8*a^(5/2))

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Maple [A]  time = 0.022, size = 175, normalized size = 1.6 \begin{align*}{\frac{1}{ \left ( \left ( \tan \left ( x \right ) \right ) ^{2}a+a+b \right ) ^{2}} \left ( -{\frac{b \left ( 8\,a+5\,b \right ) \left ( \tan \left ( x \right ) \right ) ^{3}}{8\,a \left ({a}^{2}+2\,ab+{b}^{2} \right ) }}-{\frac{ \left ( 8\,a+3\,b \right ) b\tan \left ( x \right ) }{8\,{a}^{2} \left ( a+b \right ) }} \right ) }+{\frac{1}{{a}^{2}+2\,ab+{b}^{2}}\arctan \left ({a\tan \left ( x \right ){\frac{1}{\sqrt{ \left ( a+b \right ) a}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) a}}}}+{\frac{b}{a \left ({a}^{2}+2\,ab+{b}^{2} \right ) }\arctan \left ({a\tan \left ( x \right ){\frac{1}{\sqrt{ \left ( a+b \right ) a}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) a}}}}+{\frac{3\,{b}^{2}}{ \left ( 8\,{a}^{2}+16\,ab+8\,{b}^{2} \right ){a}^{2}}\arctan \left ({a\tan \left ( x \right ){\frac{1}{\sqrt{ \left ( a+b \right ) a}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*cos(x)^2)^3,x)

[Out]

(-1/8*b*(8*a+5*b)/a/(a^2+2*a*b+b^2)*tan(x)^3-1/8*(8*a+3*b)/a^2*b/(a+b)*tan(x))/(tan(x)^2*a+a+b)^2+1/(a^2+2*a*b
+b^2)/((a+b)*a)^(1/2)*arctan(a*tan(x)/((a+b)*a)^(1/2))+1/(a^2+2*a*b+b^2)/a/((a+b)*a)^(1/2)*arctan(a*tan(x)/((a
+b)*a)^(1/2))*b+3/8/(a^2+2*a*b+b^2)/a^2/((a+b)*a)^(1/2)*arctan(a*tan(x)/((a+b)*a)^(1/2))*b^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(1/(a+b*cos(x)^2)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.18925, size = 1381, normalized size = 12.91 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(x)^2)^3,x, algorithm="fricas")

[Out]

[-1/32*(((8*a^2*b^2 + 8*a*b^3 + 3*b^4)*cos(x)^4 + 8*a^4 + 8*a^3*b + 3*a^2*b^2 + 2*(8*a^3*b + 8*a^2*b^2 + 3*a*b
^3)*cos(x)^2)*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)) + 4*(3*(2*a^3*b^2
+ 3*a^2*b^3 + a*b^4)*cos(x)^3 + (8*a^4*b + 13*a^3*b^2 + 5*a^2*b^3)*cos(x))*sin(x))/(a^8 + 3*a^7*b + 3*a^6*b^2
+ a^5*b^3 + (a^6*b^2 + 3*a^5*b^3 + 3*a^4*b^4 + a^3*b^5)*cos(x)^4 + 2*(a^7*b + 3*a^6*b^2 + 3*a^5*b^3 + a^4*b^4)
*cos(x)^2), -1/16*(((8*a^2*b^2 + 8*a*b^3 + 3*b^4)*cos(x)^4 + 8*a^4 + 8*a^3*b + 3*a^2*b^2 + 2*(8*a^3*b + 8*a^2*
b^2 + 3*a*b^3)*cos(x)^2)*sqrt(a^2 + a*b)*arctan(1/2*((2*a + b)*cos(x)^2 - a)/(sqrt(a^2 + a*b)*cos(x)*sin(x)))
+ 2*(3*(2*a^3*b^2 + 3*a^2*b^3 + a*b^4)*cos(x)^3 + (8*a^4*b + 13*a^3*b^2 + 5*a^2*b^3)*cos(x))*sin(x))/(a^8 + 3*
a^7*b + 3*a^6*b^2 + a^5*b^3 + (a^6*b^2 + 3*a^5*b^3 + 3*a^4*b^4 + a^3*b^5)*cos(x)^4 + 2*(a^7*b + 3*a^6*b^2 + 3*
a^5*b^3 + a^4*b^4)*cos(x)^2)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(x)**2)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.15086, size = 201, normalized size = 1.88 \begin{align*} \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 (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )}}{8 \,{\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} \sqrt{a^{2} + a b}} - \frac{8 \, a^{2} b \tan \left (x\right )^{3} + 5 \, a b^{2} \tan \left (x\right )^{3} + 8 \, a^{2} b \tan \left (x\right ) + 11 \, a b^{2} \tan \left (x\right ) + 3 \, b^{3} \tan \left (x\right )}{8 \,{\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )}{\left (a \tan \left (x\right )^{2} + a + b\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(x)^2)^3,x, algorithm="giac")

[Out]

1/8*(pi*floor(x/pi + 1/2)*sgn(a) + arctan(a*tan(x)/sqrt(a^2 + a*b)))*(8*a^2 + 8*a*b + 3*b^2)/((a^4 + 2*a^3*b +
a^2*b^2)*sqrt(a^2 + a*b)) - 1/8*(8*a^2*b*tan(x)^3 + 5*a*b^2*tan(x)^3 + 8*a^2*b*tan(x) + 11*a*b^2*tan(x) + 3*b
^3*tan(x))/((a^4 + 2*a^3*b + a^2*b^2)*(a*tan(x)^2 + a + b)^2)