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

Optimal result

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

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

Mathematica [A] (warning: unable to verify)

Time = 0.08 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.98 \[ \int \frac {1}{\left (4-5 \sec ^2(x)\right )^{3/2}} \, dx=-\frac {(-3+2 \cos (2 x))^{3/2} \sec ^3(x) \left (\text {arcsinh}(2 \sin (x)) (-3+2 \cos (2 x))+10 \sqrt {3-2 \cos (2 x)} \sin (x)\right )}{8 \left (4-5 \sec ^2(x)\right )^{3/2} \sqrt {-\left (1+4 \sin ^2(x)\right )^2}} \] Input:

Integrate[(4 - 5*Sec[x]^2)^(-3/2),x]
 

Output:

-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])
 

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {3042, 4616, 296, 291, 216}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4616

\(\displaystyle \int \frac {1}{\left (-5 \tan ^2(x)-1\right )^{3/2} \left (\tan ^2(x)+1\right )}d\tan (x)\)

\(\Big \downarrow \) 296

\(\displaystyle \frac {1}{4} \int \frac {1}{\sqrt {-5 \tan ^2(x)-1} \left (\tan ^2(x)+1\right )}d\tan (x)-\frac {5 \tan (x)}{4 \sqrt {-5 \tan ^2(x)-1}}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {1}{4} \int \frac {1}{\frac {4 \tan ^2(x)}{-5 \tan ^2(x)-1}+1}d\frac {\tan (x)}{\sqrt {-5 \tan ^2(x)-1}}-\frac {5 \tan (x)}{4 \sqrt {-5 \tan ^2(x)-1}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {1}{8} \arctan \left (\frac {2 \tan (x)}{\sqrt {-5 \tan ^2(x)-1}}\right )-\frac {5 \tan (x)}{4 \sqrt {-5 \tan ^2(x)-1}}\)

Input:

Int[(4 - 5*Sec[x]^2)^(-3/2),x]
 

Output:

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

Defintions of rubi rules used

rule 216

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

rule 291

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst 
[Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, 
d}, x] && NeQ[b*c - a*d, 0]
 

rule 296

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

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4616

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = 
FreeFactors[Tan[e + f*x], x]}, Simp[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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(88\) vs. \(2(32)=64\).

Time = 2.06 (sec) , antiderivative size = 89, normalized size of antiderivative = 2.22

method	result	size
default	\(\frac {-40 \tan \left (x \right )+50 \sec \left (x \right )^{2} \tan \left (x \right )+\sqrt {\frac {4 \cos \left (x \right )^{2}-5}{\left (1+\cos \left (x \right )\right )^{2}}}\, \arctan \left (\frac {2 \csc \left (x \right )-2 \cot \left (x \right )}{\sqrt {\frac {4 \cos \left (x \right )^{2}-5}{\left (1+\cos \left (x \right )\right )^{2}}}}\right ) \left (4+4 \sec \left (x \right )-5 \sec \left (x \right )^{2}-5 \sec \left (x \right )^{3}\right )}{8 \left (4-5 \sec \left (x \right )^{2}\right )^{\frac {3}{2}}}\)	\(89\)

Input:

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

Output:

1/8/(4-5*sec(x)^2)^(3/2)*(-40*tan(x)+50*sec(x)^2*tan(x)+((4*cos(x)^2-5)/(1 
+cos(x))^2)^(1/2)*arctan(2/((4*cos(x)^2-5)/(1+cos(x))^2)^(1/2)*(csc(x)-cot 
(x)))*(4+4*sec(x)-5*sec(x)^2-5*sec(x)^3))
 

Fricas [B] (verification not implemented)

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

Time = 0.08 (sec) , antiderivative size = 115, normalized size of antiderivative = 2.88 \[ \int \frac {1}{\left (4-5 \sec ^2(x)\right )^{3/2}} \, dx=-\frac {20 \, \sqrt {\frac {4 \, \cos \left (x\right )^{2} - 5}{\cos \left (x\right )^{2}}} \cos \left (x\right ) \sin \left (x\right ) - {\left (4 \, \cos \left (x\right )^{2} - 5\right )} \arctan \left (\frac {4 \, {\left (8 \, \cos \left (x\right )^{3} - 9 \, \cos \left (x\right )\right )} \sqrt {\frac {4 \, \cos \left (x\right )^{2} - 5}{\cos \left (x\right )^{2}}} \sin \left (x\right ) + \cos \left (x\right ) \sin \left (x\right )}{64 \, \cos \left (x\right )^{4} - 143 \, \cos \left (x\right )^{2} + 80}\right ) + {\left (4 \, \cos \left (x\right )^{2} - 5\right )} \arctan \left (\frac {\sin \left (x\right )}{\cos \left (x\right )}\right )}{16 \, {\left (4 \, \cos \left (x\right )^{2} - 5\right )}} \] Input:

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

Output:

-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))*sqrt((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)*arc 
tan(sin(x)/cos(x)))/(4*cos(x)^2 - 5)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\left (4-5 \sec ^2(x)\right )^{3/2}} \, dx=\int \frac {\sqrt {-5 \sec \left (x \right )^{2}+4}}{25 \sec \left (x \right )^{4}-40 \sec \left (x \right )^{2}+16}d x \] Input:

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

Output:

int(sqrt( - 5*sec(x)**2 + 4)/(25*sec(x)**4 - 40*sec(x)**2 + 16),x)