3.435 \(\int \frac{1}{(4-5 \sec ^2(x))^{3/2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0301136, antiderivative size = 40, normalized size of antiderivative = 1., 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 = {4128, 382, 377, 203} \[ \frac{1}{8} \tan ^{-1}\left (\frac{2 \tan (x)}{\sqrt{-5 \tan ^2(x)-1}}\right )-\frac{5 \tan (x)}{4 \sqrt{-5 \tan ^2(x)-1}} \]

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 4128

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]

Rule 382

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] && EqQ[
n*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1]) && NeQ[p, -1]

Rule 377

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 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 (4-5 \sec ^2(x)\right )^{3/2}} \, dx &=\operatorname{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} \operatorname{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} \operatorname{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*}

Mathematica [A]  time = 0.124404, size = 79, normalized size = 1.98 \[ -\frac{(2 \cos (2 x)-3)^{3/2} \sec ^3(x) \left (10 \sin (x) \sqrt{3-2 \cos (2 x)}+(2 \cos (2 x)-3) \sinh ^{-1}(2 \sin (x))\right )}{8 \sqrt{-\left (4 \sin ^2(x)+1\right )^2} \left (4-5 \sec ^2(x)\right )^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(4 - 5*Sec[x]^2)^(-3/2),x]

[Out]

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

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Maple [C]  time = 0.211, size = 473, normalized size = 11.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)*5^(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)-2*5^(1
/2)+4*cos(x)-5)/(cos(x)+1))^(1/2)*((2*cos(x)*5^(1/2)-2*5^(1/2)-4*cos(x)+5)/(cos(x)+1))^(1/2)-I*sin(x)*2^(1/2)*
5^(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)-2*5^(1/2)+4*cos(x)-5)/(co
s(x)+1))^(1/2)*((2*cos(x)*5^(1/2)-2*5^(1/2)-4*cos(x)+5)/(cos(x)+1))^(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)-2*5^(1/2)+4*cos(x)-5)/(cos(x)+1))^(1/2)*((2*cos(x)*5^(1/2)-2*5^(1/2)-4*cos(x)+5)/(cos(x)+1))^(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)-2*5^(1/2)+4*cos(x)-5)
/(cos(x)+1))^(1/2)*((2*cos(x)*5^(1/2)-2*5^(1/2)-4*cos(x)+5)/(cos(x)+1))^(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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (-5 \, \sec \left (x\right )^{2} + 4\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

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]  time = 3.38956, size = 351, normalized size = 8.78 \begin{align*} -\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{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (4 - 5 \sec ^{2}{\left (x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (-5 \, \sec \left (x\right )^{2} + 4\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

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)