3.1.73 \(\int \frac {1}{1-\cos ^4(x)} \, dx\) [73]

Optimal. Leaf size=45 \[ \frac {x}{2 \sqrt {2}}-\frac {\text {ArcTan}\left (\frac {\cos (x) \sin (x)}{1+\sqrt {2}+\cos ^2(x)}\right )}{2 \sqrt {2}}-\frac {\cot (x)}{2} \]

[Out]

-1/2*cot(x)+1/4*x*2^(1/2)-1/4*arctan(cos(x)*sin(x)/(1+cos(x)^2+2^(1/2)))*2^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3288, 396, 209} \begin {gather*} -\frac {\text {ArcTan}\left (\frac {\sin (x) \cos (x)}{\cos ^2(x)+\sqrt {2}+1}\right )}{2 \sqrt {2}}+\frac {x}{2 \sqrt {2}}-\frac {\cot (x)}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - Cos[x]^4)^(-1),x]

[Out]

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

Rule 209

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

Rule 396

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*x*((a + b*x^n)^(p + 1)/(b*(n*(
p + 1) + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rule 3288

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dis
t[ff/f, Subst[Int[(a + 2*a*ff^2*x^2 + (a + b)*ff^4*x^4)^p/(1 + ff^2*x^2)^(2*p + 1), x], x, Tan[e + f*x]/ff], x
]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[p]

Rubi steps

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

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Mathematica [A]
time = 0.06, size = 24, normalized size = 0.53 \begin {gather*} \frac {1}{4} \left (\sqrt {2} \text {ArcTan}\left (\frac {\tan (x)}{\sqrt {2}}\right )-2 \cot (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 - Cos[x]^4)^(-1),x]

[Out]

(Sqrt[2]*ArcTan[Tan[x]/Sqrt[2]] - 2*Cot[x])/4

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Maple [A]
time = 0.06, size = 21, normalized size = 0.47

method	result	size
default	\(-\frac {1}{2 \tan \left (x \right )}+\frac {\arctan \left (\frac {\tan \left (x \right ) \sqrt {2}}{2}\right ) \sqrt {2}}{4}\)	\(21\)
risch	\(-\frac {i}{{\mathrm e}^{2 i x}-1}+\frac {i \sqrt {2}\, \ln \left ({\mathrm e}^{2 i x}+2 \sqrt {2}+3\right )}{8}-\frac {i \sqrt {2}\, \ln \left ({\mathrm e}^{2 i x}-2 \sqrt {2}+3\right )}{8}\)	\(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1-cos(x)^4),x,method=_RETURNVERBOSE)

[Out]

-1/2/tan(x)+1/4*arctan(1/2*tan(x)*2^(1/2))*2^(1/2)

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Maxima [A]
time = 0.48, size = 20, normalized size = 0.44 \begin {gather*} \frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} \tan \left (x\right )\right ) - \frac {1}{2 \, \tan \left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cos(x)^4),x, algorithm="maxima")

[Out]

1/4*sqrt(2)*arctan(1/2*sqrt(2)*tan(x)) - 1/2/tan(x)

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Fricas [A]
time = 0.41, size = 43, normalized size = 0.96 \begin {gather*} -\frac {\sqrt {2} \arctan \left (\frac {3 \, \sqrt {2} \cos \left (x\right )^{2} - \sqrt {2}}{4 \, \cos \left (x\right ) \sin \left (x\right )}\right ) \sin \left (x\right ) + 4 \, \cos \left (x\right )}{8 \, \sin \left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cos(x)^4),x, algorithm="fricas")

[Out]

-1/8*(sqrt(2)*arctan(1/4*(3*sqrt(2)*cos(x)^2 - sqrt(2))/(cos(x)*sin(x)))*sin(x) + 4*cos(x))/sin(x)

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Sympy [A]
time = 0.63, size = 78, normalized size = 1.73 \begin {gather*} \frac {\sqrt {2} \left (\operatorname {atan}{\left (\sqrt {2} \tan {\left (\frac {x}{2} \right )} - 1 \right )} + \pi \left \lfloor {\frac {\frac {x}{2} - \frac {\pi }{2}}{\pi }}\right \rfloor \right )}{4} + \frac {\sqrt {2} \left (\operatorname {atan}{\left (\sqrt {2} \tan {\left (\frac {x}{2} \right )} + 1 \right )} + \pi \left \lfloor {\frac {\frac {x}{2} - \frac {\pi }{2}}{\pi }}\right \rfloor \right )}{4} + \frac {\tan {\left (\frac {x}{2} \right )}}{4} - \frac {1}{4 \tan {\left (\frac {x}{2} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cos(x)**4),x)

[Out]

sqrt(2)*(atan(sqrt(2)*tan(x/2) - 1) + pi*floor((x/2 - pi/2)/pi))/4 + sqrt(2)*(atan(sqrt(2)*tan(x/2) + 1) + pi*
floor((x/2 - pi/2)/pi))/4 + tan(x/2)/4 - 1/(4*tan(x/2))

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Giac [A]
time = 0.40, size = 53, normalized size = 1.18 \begin {gather*} \frac {1}{4} \, \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 {1}{2 \, \tan \left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cos(x)^4),x, algorithm="giac")

[Out]

1/4*sqrt(2)*(x + arctan(-(sqrt(2)*sin(2*x) - sin(2*x))/(sqrt(2)*cos(2*x) + sqrt(2) - cos(2*x) + 1))) - 1/2/tan
(x)

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Mupad [B]
time = 2.16, size = 20, normalized size = 0.44 \begin {gather*} \frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\mathrm {tan}\left (x\right )}{2}\right )}{4}-\frac {1}{2\,\mathrm {tan}\left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/(cos(x)^4 - 1),x)

[Out]

(2^(1/2)*atan((2^(1/2)*tan(x))/2))/4 - 1/(2*tan(x))

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