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

Optimal result

Integrand size = 12, antiderivative size = 40 \[ \int \frac {1}{\left (4-5 \sec ^2(x)\right )^{3/2}} \, dx=\frac {1}{8} \arctan \left (\frac {2 \tan (x)}{\sqrt {-1-5 \tan ^2(x)}}\right )-\frac {5 \tan (x)}{4 \sqrt {-1-5 \tan ^2(x)}} \]

[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)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4213, 390, 385, 209} \[ \int \frac {1}{\left (4-5 \sec ^2(x)\right )^{3/2}} \, dx=\frac {1}{8} \arctan \left (\frac {2 \tan (x)}{\sqrt {-5 \tan ^2(x)-1}}\right )-\frac {5 \tan (x)}{4 \sqrt {-5 \tan ^2(x)-1}} \]

[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*} \text {integral}& = \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} \arctan \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] (verified)

Time = 0.14 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.98 \[ \int \frac {1}{\left (4-5 \sec ^2(x)\right )^{3/2}} \, dx=-\frac {(-3+2 \cos (2 x))^{3/2} \sec ^3(x) \left (\text {arcsinh}(2 \sin (x)) (-3+2 \cos (2 x))+10 \sqrt {3-2 \cos (2 x)} \sin (x)\right )}{8 \left (4-5 \sec ^2(x)\right )^{3/2} \sqrt {-\left (1+4 \sin ^2(x)\right )^2}} \]

[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])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(123\) vs. \(2(32)=64\).

Time = 1.36 (sec) , antiderivative size = 124, normalized size of antiderivative = 3.10

method	result	size
default	\(\frac {\left (\sec ^{3}\left (x \right )\right ) \left (4 \left (\cos ^{2}\left (x \right )\right )-5\right ) \left (\arctan \left (\frac {2 \sin \left (x \right )}{\left (\cos \left (x \right )+1\right ) \sqrt {\frac {4 \left (\cos ^{2}\left (x \right )\right )-5}{\left (\cos \left (x \right )+1\right )^{2}}}}\right ) \sqrt {\frac {4 \left (\cos ^{2}\left (x \right )\right )-5}{\left (\cos \left (x \right )+1\right )^{2}}}\, \cos \left (x \right )+\arctan \left (\frac {2 \sin \left (x \right )}{\left (\cos \left (x \right )+1\right ) \sqrt {\frac {4 \left (\cos ^{2}\left (x \right )\right )-5}{\left (\cos \left (x \right )+1\right )^{2}}}}\right ) \sqrt {\frac {4 \left (\cos ^{2}\left (x \right )\right )-5}{\left (\cos \left (x \right )+1\right )^{2}}}-10 \sin \left (x \right )\right )}{8 {\left (4-5 \left (\sec ^{2}\left (x \right )\right )\right )}^{\frac {3}{2}}}\)	\(124\)

[In]

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

[Out]

1/8*sec(x)^3*(4*cos(x)^2-5)*(arctan(2*sin(x)/(cos(x)+1)/((4*cos(x)^2-5)/(cos(x)+1)^2)^(1/2))*((4*cos(x)^2-5)/(
cos(x)+1)^2)^(1/2)*cos(x)+arctan(2*sin(x)/(cos(x)+1)/((4*cos(x)^2-5)/(cos(x)+1)^2)^(1/2))*((4*cos(x)^2-5)/(cos
(x)+1)^2)^(1/2)-10*sin(x))/(4-5*sec(x)^2)^(3/2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 115 vs. \(2 (32) = 64\).

Time = 0.28 (sec) , antiderivative size = 115, normalized size of antiderivative = 2.88 \[ \int \frac {1}{\left (4-5 \sec ^2(x)\right )^{3/2}} \, dx=-\frac {20 \, \sqrt {\frac {4 \, \cos \left (x\right )^{2} - 5}{\cos \left (x\right )^{2}}} \cos \left (x\right ) \sin \left (x\right ) - {\left (4 \, \cos \left (x\right )^{2} - 5\right )} \arctan \left (\frac {4 \, {\left (8 \, \cos \left (x\right )^{3} - 9 \, \cos \left (x\right )\right )} \sqrt {\frac {4 \, \cos \left (x\right )^{2} - 5}{\cos \left (x\right )^{2}}} \sin \left (x\right ) + \cos \left (x\right ) \sin \left (x\right )}{64 \, \cos \left (x\right )^{4} - 143 \, \cos \left (x\right )^{2} + 80}\right ) + {\left (4 \, \cos \left (x\right )^{2} - 5\right )} \arctan \left (\frac {\sin \left (x\right )}{\cos \left (x\right )}\right )}{16 \, {\left (4 \, \cos \left (x\right )^{2} - 5\right )}} \]

[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)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {1}{\left (4-5 \sec ^2(x)\right )^{3/2}} \, dx=\int { \frac {1}{{\left (-5 \, \sec \left (x\right )^{2} + 4\right )}^{\frac {3}{2}}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {1}{\left (4-5 \sec ^2(x)\right )^{3/2}} \, dx=\int { \frac {1}{{\left (-5 \, \sec \left (x\right )^{2} + 4\right )}^{\frac {3}{2}}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (4-5 \sec ^2(x)\right )^{3/2}} \, dx=\int \frac {1}{{\left (4-\frac {5}{{\cos \left (x\right )}^2}\right )}^{3/2}} \,d x \]

[In]

int(1/(4 - 5/cos(x)^2)^(3/2),x)

[Out]

int(1/(4 - 5/cos(x)^2)^(3/2), x)