3.5.35 \(\int \frac {1}{(4-5 \sec ^2(x))^{3/2}} \, dx\) [435]

Optimal. Leaf size=40 \[ \frac {1}{8} \tan ^{-1}\left (\frac {2 \tan (x)}{\sqrt {-1-5 \tan ^2(x)}}\right )-\frac {5 \tan (x)}{4 \sqrt {-1-5 \tan ^2(x)}} \]

[Out]

1/8*arctan(2*tan(x)/(-1-5*tan(x)^2)^(1/2))-5/4*tan(x)/(-1-5*tan(x)^2)^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4213, 390, 385, 209} \begin {gather*} \frac {1}{8} \text {ArcTan}\left (\frac {2 \tan (x)}{\sqrt {-5 \tan ^2(x)-1}}\right )-\frac {5 \tan (x)}{4 \sqrt {-5 \tan ^2(x)-1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4 - 5*Sec[x]^2)^(-3/2),x]

[Out]

ArcTan[(2*Tan[x])/Sqrt[-1 - 5*Tan[x]^2]]/8 - (5*Tan[x])/(4*Sqrt[-1 - 5*Tan[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 385

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

Rule 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-b)*x*(a + b*x^n)^(p + 1)*
((c + d*x^n)^(q + 1)/(a*n*(p + 1)*(b*c - a*d))), x] + Dist[(b*c + n*(p + 1)*(b*c - a*d))/(a*n*(p + 1)*(b*c - a
*d)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, q}, x] && NeQ[b*c - a*d, 0] && Eq
Q[n*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1]) && NeQ[p, -1]

Rule 4213

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff/f, Subst[Int[(a + b + b*ff^2*x^2)^p/(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p},
x] && NeQ[a + b, 0] && NeQ[p, -1]

Rubi steps

\begin {align*} \int \frac {1}{\left (4-5 \sec ^2(x)\right )^{3/2}} \, dx &=\text {Subst}\left (\int \frac {1}{\left (-1-5 x^2\right )^{3/2} \left (1+x^2\right )} \, dx,x,\tan (x)\right )\\ &=-\frac {5 \tan (x)}{4 \sqrt {-1-5 \tan ^2(x)}}+\frac {1}{4} \text {Subst}\left (\int \frac {1}{\sqrt {-1-5 x^2} \left (1+x^2\right )} \, dx,x,\tan (x)\right )\\ &=-\frac {5 \tan (x)}{4 \sqrt {-1-5 \tan ^2(x)}}+\frac {1}{4} \text {Subst}\left (\int \frac {1}{1+4 x^2} \, dx,x,\frac {\tan (x)}{\sqrt {-1-5 \tan ^2(x)}}\right )\\ &=\frac {1}{8} \tan ^{-1}\left (\frac {2 \tan (x)}{\sqrt {-1-5 \tan ^2(x)}}\right )-\frac {5 \tan (x)}{4 \sqrt {-1-5 \tan ^2(x)}}\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 79, normalized size = 1.98 \begin {gather*} -\frac {(-3+2 \cos (2 x))^{3/2} \sec ^3(x) \left (\sinh ^{-1}(2 \sin (x)) (-3+2 \cos (2 x))+10 \sqrt {3-2 \cos (2 x)} \sin (x)\right )}{8 \left (4-5 \sec ^2(x)\right )^{3/2} \sqrt {-\left (1+4 \sin ^2(x)\right )^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4 - 5*Sec[x]^2)^(-3/2),x]

[Out]

-1/8*((-3 + 2*Cos[2*x])^(3/2)*Sec[x]^3*(ArcSinh[2*Sin[x]]*(-3 + 2*Cos[2*x]) + 10*Sqrt[3 - 2*Cos[2*x]]*Sin[x]))
/((4 - 5*Sec[x]^2)^(3/2)*Sqrt[-(1 + 4*Sin[x]^2)^2])

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.44, size = 473, normalized size = 11.82

method	result	size
default	\(-\frac {i \left (4 \left (\cos ^{2}\left (x \right )\right )-5\right ) \left (2 i \sin \left (x \right ) \sqrt {2}\, \EllipticPi \left (\frac {\sqrt {-9-4 \sqrt {5}}\, \left (\cos \left (x \right )-1\right )}{\sin \left (x \right )}, \frac {1}{9+4 \sqrt {5}}, \frac {\sqrt {-9+4 \sqrt {5}}}{\sqrt {-9-4 \sqrt {5}}}\right ) \sqrt {5}\, \sqrt {-\frac {2 \left (2 \cos \left (x \right ) \sqrt {5}+4 \cos \left (x \right )-2 \sqrt {5}-5\right )}{1+\cos \left (x \right )}}\, \sqrt {\frac {2 \cos \left (x \right ) \sqrt {5}-4 \cos \left (x \right )-2 \sqrt {5}+5}{1+\cos \left (x \right )}}-i \sin \left (x \right ) \sqrt {2}\, \EllipticF \left (\frac {i \left (\cos \left (x \right )-1\right ) \left (2+\sqrt {5}\right )}{\sin \left (x \right )}, 9-4 \sqrt {5}\right ) \sqrt {5}\, \sqrt {-\frac {2 \left (2 \cos \left (x \right ) \sqrt {5}+4 \cos \left (x \right )-2 \sqrt {5}-5\right )}{1+\cos \left (x \right )}}\, \sqrt {\frac {2 \cos \left (x \right ) \sqrt {5}-4 \cos \left (x \right )-2 \sqrt {5}+5}{1+\cos \left (x \right )}}+4 i \sin \left (x \right ) \sqrt {2}\, \EllipticPi \left (\frac {\sqrt {-9-4 \sqrt {5}}\, \left (\cos \left (x \right )-1\right )}{\sin \left (x \right )}, \frac {1}{9+4 \sqrt {5}}, \frac {\sqrt {-9+4 \sqrt {5}}}{\sqrt {-9-4 \sqrt {5}}}\right ) \sqrt {-\frac {2 \left (2 \cos \left (x \right ) \sqrt {5}+4 \cos \left (x \right )-2 \sqrt {5}-5\right )}{1+\cos \left (x \right )}}\, \sqrt {\frac {2 \cos \left (x \right ) \sqrt {5}-4 \cos \left (x \right )-2 \sqrt {5}+5}{1+\cos \left (x \right )}}-2 i \sin \left (x \right ) \sqrt {2}\, \EllipticF \left (\frac {i \left (\cos \left (x \right )-1\right ) \left (2+\sqrt {5}\right )}{\sin \left (x \right )}, 9-4 \sqrt {5}\right ) \sqrt {-\frac {2 \left (2 \cos \left (x \right ) \sqrt {5}+4 \cos \left (x \right )-2 \sqrt {5}-5\right )}{1+\cos \left (x \right )}}\, \sqrt {\frac {2 \cos \left (x \right ) \sqrt {5}-4 \cos \left (x \right )-2 \sqrt {5}+5}{1+\cos \left (x \right )}}+20 \cos \left (x \right ) \sqrt {5}+45 \cos \left (x \right )-20 \sqrt {5}-45\right ) \sin \left (x \right )}{4 \sqrt {-9-4 \sqrt {5}}\, \left (2+\sqrt {5}\right ) \left (\cos \left (x \right )-1\right ) \cos \left (x \right )^{3} \left (\frac {4 \left (\cos ^{2}\left (x \right )\right )-5}{\cos \left (x \right )^{2}}\right )^{\frac {3}{2}}}\)	\(473\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(4-5*sec(x)^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/4*I/(-9-4*5^(1/2))^(1/2)/(2+5^(1/2))*(4*cos(x)^2-5)*(2*I*sin(x)*2^(1/2)*EllipticPi((-9-4*5^(1/2))^(1/2)*(co
s(x)-1)/sin(x),1/(9+4*5^(1/2)),(-9+4*5^(1/2))^(1/2)/(-9-4*5^(1/2))^(1/2))*5^(1/2)*(-2*(2*cos(x)*5^(1/2)+4*cos(
x)-2*5^(1/2)-5)/(1+cos(x)))^(1/2)*((2*cos(x)*5^(1/2)-4*cos(x)-2*5^(1/2)+5)/(1+cos(x)))^(1/2)-I*sin(x)*2^(1/2)*
EllipticF(I*(cos(x)-1)*(2+5^(1/2))/sin(x),9-4*5^(1/2))*5^(1/2)*(-2*(2*cos(x)*5^(1/2)+4*cos(x)-2*5^(1/2)-5)/(1+
cos(x)))^(1/2)*((2*cos(x)*5^(1/2)-4*cos(x)-2*5^(1/2)+5)/(1+cos(x)))^(1/2)+4*I*sin(x)*2^(1/2)*EllipticPi((-9-4*
5^(1/2))^(1/2)*(cos(x)-1)/sin(x),1/(9+4*5^(1/2)),(-9+4*5^(1/2))^(1/2)/(-9-4*5^(1/2))^(1/2))*(-2*(2*cos(x)*5^(1
/2)+4*cos(x)-2*5^(1/2)-5)/(1+cos(x)))^(1/2)*((2*cos(x)*5^(1/2)-4*cos(x)-2*5^(1/2)+5)/(1+cos(x)))^(1/2)-2*I*sin
(x)*2^(1/2)*EllipticF(I*(cos(x)-1)*(2+5^(1/2))/sin(x),9-4*5^(1/2))*(-2*(2*cos(x)*5^(1/2)+4*cos(x)-2*5^(1/2)-5)
/(1+cos(x)))^(1/2)*((2*cos(x)*5^(1/2)-4*cos(x)-2*5^(1/2)+5)/(1+cos(x)))^(1/2)+20*cos(x)*5^(1/2)+45*cos(x)-20*5
^(1/2)-45)*sin(x)/(cos(x)-1)/cos(x)^3/((4*cos(x)^2-5)/cos(x)^2)^(3/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(4-5*sec(x)^2)^(3/2),x, algorithm="maxima")

[Out]

integrate((-5*sec(x)^2 + 4)^(-3/2), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 115 vs. \(2 (32) = 64\).
time = 1.15, size = 115, normalized size = 2.88 \begin {gather*} -\frac {20 \, \sqrt {\frac {4 \, \cos \left (x\right )^{2} - 5}{\cos \left (x\right )^{2}}} \cos \left (x\right ) \sin \left (x\right ) - {\left (4 \, \cos \left (x\right )^{2} - 5\right )} \arctan \left (\frac {4 \, {\left (8 \, \cos \left (x\right )^{3} - 9 \, \cos \left (x\right )\right )} \sqrt {\frac {4 \, \cos \left (x\right )^{2} - 5}{\cos \left (x\right )^{2}}} \sin \left (x\right ) + \cos \left (x\right ) \sin \left (x\right )}{64 \, \cos \left (x\right )^{4} - 143 \, \cos \left (x\right )^{2} + 80}\right ) + {\left (4 \, \cos \left (x\right )^{2} - 5\right )} \arctan \left (\frac {\sin \left (x\right )}{\cos \left (x\right )}\right )}{16 \, {\left (4 \, \cos \left (x\right )^{2} - 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(4-5*sec(x)^2)^(3/2),x, algorithm="fricas")

[Out]

-1/16*(20*sqrt((4*cos(x)^2 - 5)/cos(x)^2)*cos(x)*sin(x) - (4*cos(x)^2 - 5)*arctan((4*(8*cos(x)^3 - 9*cos(x))*s
qrt((4*cos(x)^2 - 5)/cos(x)^2)*sin(x) + cos(x)*sin(x))/(64*cos(x)^4 - 143*cos(x)^2 + 80)) + (4*cos(x)^2 - 5)*a
rctan(sin(x)/cos(x)))/(4*cos(x)^2 - 5)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (4 - 5 \sec ^{2}{\left (x \right )}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(4-5*sec(x)**2)**(3/2),x)

[Out]

Integral((4 - 5*sec(x)**2)**(-3/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(4-5*sec(x)^2)^(3/2),x, algorithm="giac")

[Out]

integrate((-5*sec(x)^2 + 4)^(-3/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{{\left (4-\frac {5}{{\cos \left (x\right )}^2}\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(4 - 5/cos(x)^2)^(3/2),x)

[Out]

int(1/(4 - 5/cos(x)^2)^(3/2), x)

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