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

Optimal. Leaf size=71 \[ \frac{19 x}{32 \sqrt{2}}-\frac{9 \sin (x) \cos (x)}{32 \left (\cos ^2(x)+1\right )}-\frac{\sin (x) \cos (x)}{8 \left (\cos ^2(x)+1\right )^2}-\frac{19 \tan ^{-1}\left (\frac{\sin (x) \cos (x)}{\cos ^2(x)+\sqrt{2}+1}\right )}{32 \sqrt{2}} \]

[Out]

(19*x)/(32*Sqrt[2]) - (19*ArcTan[(Cos[x]*Sin[x])/(1 + Sqrt[2] + Cos[x]^2)])/(32*Sqrt[2]) - (Cos[x]*Sin[x])/(8*
(1 + Cos[x]^2)^2) - (9*Cos[x]*Sin[x])/(32*(1 + Cos[x]^2))

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Rubi [A]  time = 0.0523739, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.625, Rules used = {3184, 3173, 12, 3181, 203} \[ \frac{19 x}{32 \sqrt{2}}-\frac{9 \sin (x) \cos (x)}{32 \left (\cos ^2(x)+1\right )}-\frac{\sin (x) \cos (x)}{8 \left (\cos ^2(x)+1\right )^2}-\frac{19 \tan ^{-1}\left (\frac{\sin (x) \cos (x)}{\cos ^2(x)+\sqrt{2}+1}\right )}{32 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Int[(1 + Cos[x]^2)^(-3),x]

[Out]

(19*x)/(32*Sqrt[2]) - (19*ArcTan[(Cos[x]*Sin[x])/(1 + Sqrt[2] + Cos[x]^2)])/(32*Sqrt[2]) - (Cos[x]*Sin[x])/(8*
(1 + Cos[x]^2)^2) - (9*Cos[x]*Sin[x])/(32*(1 + 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 203

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

Rubi steps

\begin{align*} \int \frac{1}{\left (1+\cos ^2(x)\right )^3} \, dx &=-\frac{\cos (x) \sin (x)}{8 \left (1+\cos ^2(x)\right )^2}-\frac{1}{8} \int \frac{-7+2 \cos ^2(x)}{\left (1+\cos ^2(x)\right )^2} \, dx\\ &=-\frac{\cos (x) \sin (x)}{8 \left (1+\cos ^2(x)\right )^2}-\frac{9 \cos (x) \sin (x)}{32 \left (1+\cos ^2(x)\right )}-\frac{1}{32} \int -\frac{19}{1+\cos ^2(x)} \, dx\\ &=-\frac{\cos (x) \sin (x)}{8 \left (1+\cos ^2(x)\right )^2}-\frac{9 \cos (x) \sin (x)}{32 \left (1+\cos ^2(x)\right )}+\frac{19}{32} \int \frac{1}{1+\cos ^2(x)} \, dx\\ &=-\frac{\cos (x) \sin (x)}{8 \left (1+\cos ^2(x)\right )^2}-\frac{9 \cos (x) \sin (x)}{32 \left (1+\cos ^2(x)\right )}-\frac{19}{32} \operatorname{Subst}\left (\int \frac{1}{1+2 x^2} \, dx,x,\cot (x)\right )\\ &=\frac{19 x}{32 \sqrt{2}}-\frac{19 \tan ^{-1}\left (\frac{\cos (x) \sin (x)}{1+\sqrt{2}+\cos ^2(x)}\right )}{32 \sqrt{2}}-\frac{\cos (x) \sin (x)}{8 \left (1+\cos ^2(x)\right )^2}-\frac{9 \cos (x) \sin (x)}{32 \left (1+\cos ^2(x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.123027, size = 51, normalized size = 0.72 \[ \frac{19 \tan ^{-1}\left (\frac{\tan (x)}{\sqrt{2}}\right )}{32 \sqrt{2}}-\frac{9 \sin (2 x)}{32 (\cos (2 x)+3)}-\frac{\sin (2 x)}{4 (\cos (2 x)+3)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + Cos[x]^2)^(-3),x]

[Out]

(19*ArcTan[Tan[x]/Sqrt[2]])/(32*Sqrt[2]) - Sin[2*x]/(4*(3 + Cos[2*x])^2) - (9*Sin[2*x])/(32*(3 + Cos[2*x]))

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Maple [A]  time = 0.016, size = 35, normalized size = 0.5 \begin{align*}{\frac{1}{ \left ( \left ( \tan \left ( x \right ) \right ) ^{2}+2 \right ) ^{2}} \left ( -{\frac{13\, \left ( \tan \left ( x \right ) \right ) ^{3}}{32}}-{\frac{11\,\tan \left ( x \right ) }{16}} \right ) }+{\frac{19\,\sqrt{2}}{64}\arctan \left ({\frac{\tan \left ( x \right ) \sqrt{2}}{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-13/32*tan(x)^3-11/16*tan(x))/(tan(x)^2+2)^2+19/64*arctan(1/2*tan(x)*2^(1/2))*2^(1/2)

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Maxima [A]  time = 1.45831, size = 55, normalized size = 0.77 \begin{align*} \frac{19}{64} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2} \tan \left (x\right )\right ) - \frac{13 \, \tan \left (x\right )^{3} + 22 \, \tan \left (x\right )}{32 \,{\left (\tan \left (x\right )^{4} + 4 \, \tan \left (x\right )^{2} + 4\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+cos(x)^2)^3,x, algorithm="maxima")

[Out]

19/64*sqrt(2)*arctan(1/2*sqrt(2)*tan(x)) - 1/32*(13*tan(x)^3 + 22*tan(x))/(tan(x)^4 + 4*tan(x)^2 + 4)

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Fricas [A]  time = 1.6348, size = 251, normalized size = 3.54 \begin{align*} -\frac{19 \,{\left (\sqrt{2} \cos \left (x\right )^{4} + 2 \, \sqrt{2} \cos \left (x\right )^{2} + \sqrt{2}\right )} \arctan \left (\frac{3 \, \sqrt{2} \cos \left (x\right )^{2} - \sqrt{2}}{4 \, \cos \left (x\right ) \sin \left (x\right )}\right ) + 4 \,{\left (9 \, \cos \left (x\right )^{3} + 13 \, \cos \left (x\right )\right )} \sin \left (x\right )}{128 \,{\left (\cos \left (x\right )^{4} + 2 \, \cos \left (x\right )^{2} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/128*(19*(sqrt(2)*cos(x)^4 + 2*sqrt(2)*cos(x)^2 + sqrt(2))*arctan(1/4*(3*sqrt(2)*cos(x)^2 - sqrt(2))/(cos(x)
*sin(x))) + 4*(9*cos(x)^3 + 13*cos(x))*sin(x))/(cos(x)^4 + 2*cos(x)^2 + 1)

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Sympy [B]  time = 34.9375, size = 439, normalized size = 6.18 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

19*sqrt(2)*(atan(sqrt(2)*tan(x/2) - 1) + pi*floor((x/2 - pi/2)/pi))*tan(x/2)**8/(64*tan(x/2)**8 + 128*tan(x/2)
**4 + 64) + 38*sqrt(2)*(atan(sqrt(2)*tan(x/2) - 1) + pi*floor((x/2 - pi/2)/pi))*tan(x/2)**4/(64*tan(x/2)**8 +
128*tan(x/2)**4 + 64) + 19*sqrt(2)*(atan(sqrt(2)*tan(x/2) - 1) + pi*floor((x/2 - pi/2)/pi))/(64*tan(x/2)**8 +
128*tan(x/2)**4 + 64) + 19*sqrt(2)*(atan(sqrt(2)*tan(x/2) + 1) + pi*floor((x/2 - pi/2)/pi))*tan(x/2)**8/(64*ta
n(x/2)**8 + 128*tan(x/2)**4 + 64) + 38*sqrt(2)*(atan(sqrt(2)*tan(x/2) + 1) + pi*floor((x/2 - pi/2)/pi))*tan(x/
2)**4/(64*tan(x/2)**8 + 128*tan(x/2)**4 + 64) + 19*sqrt(2)*(atan(sqrt(2)*tan(x/2) + 1) + pi*floor((x/2 - pi/2)
/pi))/(64*tan(x/2)**8 + 128*tan(x/2)**4 + 64) + 22*tan(x/2)**7/(64*tan(x/2)**8 + 128*tan(x/2)**4 + 64) - 14*ta
n(x/2)**5/(64*tan(x/2)**8 + 128*tan(x/2)**4 + 64) + 14*tan(x/2)**3/(64*tan(x/2)**8 + 128*tan(x/2)**4 + 64) - 2
2*tan(x/2)/(64*tan(x/2)**8 + 128*tan(x/2)**4 + 64)

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Giac [A]  time = 1.22654, size = 92, normalized size = 1.3 \begin{align*} \frac{19}{64} \, \sqrt{2}{\left (x + \arctan \left (-\frac{\sqrt{2} \sin \left (2 \, x\right ) - \sin \left (2 \, x\right )}{\sqrt{2} \cos \left (2 \, x\right ) + \sqrt{2} - \cos \left (2 \, x\right ) + 1}\right )\right )} - \frac{13 \, \tan \left (x\right )^{3} + 22 \, \tan \left (x\right )}{32 \,{\left (\tan \left (x\right )^{2} + 2\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

19/64*sqrt(2)*(x + arctan(-(sqrt(2)*sin(2*x) - sin(2*x))/(sqrt(2)*cos(2*x) + sqrt(2) - cos(2*x) + 1))) - 1/32*
(13*tan(x)^3 + 22*tan(x))/(tan(x)^2 + 2)^2