3.42 \(\int \frac {\sec (x)}{\sqrt {-1+\sec ^4(x)}} \, dx\)

Optimal. Leaf size=28 \[ -\frac {\tanh ^{-1}\left (\frac {\cos (x) \cot (x) \sqrt {\sec ^4(x)-1}}{\sqrt {2}}\right )}{\sqrt {2}} \]

[Out]

-1/2*arctanh(1/2*cos(x)*cot(x)*(-1+sec(x)^4)^(1/2)*2^(1/2))*2^(1/2)

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Rubi [B]  time = 0.18, antiderivative size = 59, normalized size of antiderivative = 2.11, number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {4148, 6722, 1988, 2008, 206} \[ -\frac {\sqrt {1-\cos ^4(x)} \sec ^2(x) \tanh ^{-1}\left (\frac {\sqrt {2} \sin (x)}{\sqrt {2 \sin ^2(x)-\sin ^4(x)}}\right )}{\sqrt {2} \sqrt {\sec ^4(x)-1}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[x]/Sqrt[-1 + Sec[x]^4],x]

[Out]

-((ArcTanh[(Sqrt[2]*Sin[x])/Sqrt[2*Sin[x]^2 - Sin[x]^4]]*Sqrt[1 - Cos[x]^4]*Sec[x]^2)/(Sqrt[2]*Sqrt[-1 + Sec[x
]^4]))

Rule 206

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

Rule 1988

Int[(u_)^(p_), x_Symbol] :> Int[ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && TrinomialQ[u, x] &&  !TrinomialMatch
Q[u, x]

Rule 2008

Int[1/Sqrt[(a_.)*(x_)^2 + (b_.)*(x_)^(n_.)], x_Symbol] :> Dist[2/(2 - n), Subst[Int[1/(1 - a*x^2), x], x, x/Sq
rt[a*x^2 + b*x^n]], x] /; FreeQ[{a, b, n}, x] && NeQ[n, 2]

Rule 4148

Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = Fr
eeFactors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[(a + b/(1 - ff^2*x^2)^(n/2))^p/(1 - ff^2*x^2)^((m + 1)/2), x
], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2] && IntegerQ[n/2] &&  !IntegerQ
[p]

Rule 6722

Int[(u_.)*((a_.) + (b_.)*(v_)^(n_))^(p_), x_Symbol] :> Dist[(a + b*v^n)^FracPart[p]/(v^(n*FracPart[p])*(b + a/
v^n)^FracPart[p]), Int[u*v^(n*p)*(b + a/v^n)^p, x], x] /; FreeQ[{a, b, p}, x] &&  !IntegerQ[p] && ILtQ[n, 0] &
& BinomialQ[v, x] &&  !LinearQ[v, x]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 45, normalized size = 1.61 \[ -\frac {\sqrt {\cos (2 x)+3} \tan (x) \sec (x) \tanh ^{-1}\left (\frac {1}{2} \sqrt {4-2 \sin ^2(x)}\right )}{2 \sqrt {\sec ^4(x)-1}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[x]/Sqrt[-1 + Sec[x]^4],x]

[Out]

-1/2*(ArcTanh[Sqrt[4 - 2*Sin[x]^2]/2]*Sqrt[3 + Cos[2*x]]*Sec[x]*Tan[x])/Sqrt[-1 + Sec[x]^4]

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fricas [B]  time = 0.49, size = 54, normalized size = 1.93 \[ \frac {1}{4} \, \sqrt {2} \log \left (-\frac {2 \, {\left (2 \, \sqrt {2} \sqrt {-\frac {\cos \relax (x)^{4} - 1}{\cos \relax (x)^{4}}} \cos \relax (x)^{2} - {\left (\cos \relax (x)^{2} + 3\right )} \sin \relax (x)\right )}}{{\left (\cos \relax (x)^{2} - 1\right )} \sin \relax (x)}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)/(-1+sec(x)^4)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 2.22, size = 92, normalized size = 3.29 \[ \frac {\sqrt {2} {\left (\log \left (\tan \left (\frac {1}{2} \, x\right )^{2} - \sqrt {\tan \left (\frac {1}{2} \, x\right )^{4} + 1} + 1\right ) - \log \left (-\tan \left (\frac {1}{2} \, x\right )^{2} + \sqrt {\tan \left (\frac {1}{2} \, x\right )^{4} + 1} + 1\right ) + \log \left (-\tan \left (\frac {1}{2} \, x\right )^{2} + \sqrt {\tan \left (\frac {1}{2} \, x\right )^{4} + 1}\right )\right )}}{4 \, \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, x\right )^{5} - 2 \, \tan \left (\frac {1}{2} \, x\right )^{3} + \tan \left (\frac {1}{2} \, x\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)/(-1+sec(x)^4)^(1/2),x, algorithm="giac")

[Out]

1/4*sqrt(2)*(log(tan(1/2*x)^2 - sqrt(tan(1/2*x)^4 + 1) + 1) - log(-tan(1/2*x)^2 + sqrt(tan(1/2*x)^4 + 1) + 1)
+ log(-tan(1/2*x)^2 + sqrt(tan(1/2*x)^4 + 1)))/sgn(tan(1/2*x)^5 - 2*tan(1/2*x)^3 + tan(1/2*x))

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maple [B]  time = 0.25, size = 91, normalized size = 3.25 \[ \frac {\sqrt {8}\, \sqrt {2}\, \left (-\arcsinh \left (\frac {\cos \relax (x )-1}{\cos \relax (x )+1}\right )+\arctanh \left (\frac {\sqrt {2}\, \sqrt {4}}{4 \sqrt {\frac {\cos ^{2}\relax (x )+1}{\left (\cos \relax (x )+1\right )^{2}}}}\right )\right ) \sqrt {\frac {\cos ^{2}\relax (x )+1}{\left (\cos \relax (x )+1\right )^{2}}}\, \left (\sin ^{3}\relax (x )\right )}{8 \left (\cos \relax (x )-1\right ) \sqrt {-\frac {2 \left (\cos ^{4}\relax (x )-1\right )}{\cos \relax (x )^{4}}}\, \cos \relax (x )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(x)/(-1+sec(x)^4)^(1/2),x)

[Out]

1/8*8^(1/2)*2^(1/2)*(arctanh(1/4*2^(1/2)*4^(1/2)/((1+cos(x)^2)/(1+cos(x))^2)^(1/2))-arcsinh((cos(x)-1)/(1+cos(
x))))*sin(x)^3*((1+cos(x)^2)/(1+cos(x))^2)^(1/2)/(cos(x)-1)/cos(x)^2/(-2*(cos(x)^4-1)/cos(x)^4)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \relax (x)}{\sqrt {\sec \relax (x)^{4} - 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)/(-1+sec(x)^4)^(1/2),x, algorithm="maxima")

[Out]

integrate(sec(x)/sqrt(sec(x)^4 - 1), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \[ \int \frac {1}{\cos \relax (x)\,\sqrt {\frac {1}{{\cos \relax (x)}^4}-1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(x)*(1/cos(x)^4 - 1)^(1/2)),x)

[Out]

int(1/(cos(x)*(1/cos(x)^4 - 1)^(1/2)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)/(-1+sec(x)**4)**(1/2),x)

[Out]

Integral(sec(x)/sqrt((sec(x) - 1)*(sec(x) + 1)*(sec(x)**2 + 1)), x)

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