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

Optimal. Leaf size=68 \[ 8 \tan ^{-1}\left (\frac{2 \tan (x)}{\sqrt{-5 \tan ^2(x)-1}}\right )-\frac{7}{2} \sqrt{5} \tan ^{-1}\left (\frac{\sqrt{5} \tan (x)}{\sqrt{-5 \tan ^2(x)-1}}\right )-\frac{5}{2} \tan (x) \sqrt{-5 \tan ^2(x)-1} \]

[Out]

8*ArcTan[(2*Tan[x])/Sqrt[-1 - 5*Tan[x]^2]] - (7*Sqrt[5]*ArcTan[(Sqrt[5]*Tan[x])/Sqrt[-1 - 5*Tan[x]^2]])/2 - (5
*Tan[x]*Sqrt[-1 - 5*Tan[x]^2])/2

________________________________________________________________________________________

Rubi [A]  time = 0.0554331, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {4128, 416, 523, 217, 203, 377} \[ 8 \tan ^{-1}\left (\frac{2 \tan (x)}{\sqrt{-5 \tan ^2(x)-1}}\right )-\frac{7}{2} \sqrt{5} \tan ^{-1}\left (\frac{\sqrt{5} \tan (x)}{\sqrt{-5 \tan ^2(x)-1}}\right )-\frac{5}{2} \tan (x) \sqrt{-5 \tan ^2(x)-1} \]

Antiderivative was successfully verified.

[In]

Int[(4 - 5*Sec[x]^2)^(3/2),x]

[Out]

8*ArcTan[(2*Tan[x])/Sqrt[-1 - 5*Tan[x]^2]] - (7*Sqrt[5]*ArcTan[(Sqrt[5]*Tan[x])/Sqrt[-1 - 5*Tan[x]^2]])/2 - (5
*Tan[x]*Sqrt[-1 - 5*Tan[x]^2])/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 416

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^(q - 1))/(b*(n*(p + q) + 1)), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, n, p, q, x]

Rule 523

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

Rule 217

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

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])

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]

Rubi steps

\begin{align*} \int \left (4-5 \sec ^2(x)\right )^{3/2} \, dx &=\operatorname{Subst}\left (\int \frac{\left (-1-5 x^2\right )^{3/2}}{1+x^2} \, dx,x,\tan (x)\right )\\ &=-\frac{5}{2} \tan (x) \sqrt{-1-5 \tan ^2(x)}+\frac{1}{2} \operatorname{Subst}\left (\int \frac{-3-35 x^2}{\sqrt{-1-5 x^2} \left (1+x^2\right )} \, dx,x,\tan (x)\right )\\ &=-\frac{5}{2} \tan (x) \sqrt{-1-5 \tan ^2(x)}+16 \operatorname{Subst}\left (\int \frac{1}{\sqrt{-1-5 x^2} \left (1+x^2\right )} \, dx,x,\tan (x)\right )-\frac{35}{2} \operatorname{Subst}\left (\int \frac{1}{\sqrt{-1-5 x^2}} \, dx,x,\tan (x)\right )\\ &=-\frac{5}{2} \tan (x) \sqrt{-1-5 \tan ^2(x)}+16 \operatorname{Subst}\left (\int \frac{1}{1+4 x^2} \, dx,x,\frac{\tan (x)}{\sqrt{-1-5 \tan ^2(x)}}\right )-\frac{35}{2} \operatorname{Subst}\left (\int \frac{1}{1+5 x^2} \, dx,x,\frac{\tan (x)}{\sqrt{-1-5 \tan ^2(x)}}\right )\\ &=8 \tan ^{-1}\left (\frac{2 \tan (x)}{\sqrt{-1-5 \tan ^2(x)}}\right )-\frac{7}{2} \sqrt{5} \tan ^{-1}\left (\frac{\sqrt{5} \tan (x)}{\sqrt{-1-5 \tan ^2(x)}}\right )-\frac{5}{2} \tan (x) \sqrt{-1-5 \tan ^2(x)}\\ \end{align*}

Mathematica [C]  time = 0.206649, size = 115, normalized size = 1.69 \[ -\frac{\left (4 \cos ^2(x)-5\right ) \sec (x) \sqrt{4-5 \sec ^2(x)} \left (5 \sin (x) \sqrt{2 \cos (2 x)-3}+16 i \cos ^2(x) \log \left (\sqrt{2 \cos (2 x)-3}+2 i \sin (x)\right )+7 \sqrt{5} \cos ^2(x) \tan ^{-1}\left (\frac{\sqrt{5} \sin (x)}{\sqrt{2 \cos (2 x)-3}}\right )\right )}{2 (2 \cos (2 x)-3)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(4 - 5*Sec[x]^2)^(3/2),x]

[Out]

-((-5 + 4*Cos[x]^2)*Sec[x]*Sqrt[4 - 5*Sec[x]^2]*(7*Sqrt[5]*ArcTan[(Sqrt[5]*Sin[x])/Sqrt[-3 + 2*Cos[2*x]]]*Cos[
x]^2 + (16*I)*Cos[x]^2*Log[Sqrt[-3 + 2*Cos[2*x]] + (2*I)*Sin[x]] + 5*Sqrt[-3 + 2*Cos[2*x]]*Sin[x]))/(2*(-3 + 2
*Cos[2*x])^(3/2))

________________________________________________________________________________________

Maple [C]  time = 0.264, size = 754, normalized size = 11.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*I/(-9-4*5^(1/2))^(1/2)/(2+5^(1/2))*(70*I*(-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)*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^(1/2)*sin(x)*cos(x)^2*5^(1/2)-64*I*(-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)*Ellipt
icPi((-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^(1/2)
*sin(x)*cos(x)^2*5^(1/2)-3*I*(-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)*EllipticF(I*(cos(x)-1)*(2+5^(1/2))/sin(x),9-4*5^(1/2))*2^(1/2)*sin(x)*
cos(x)^2*5^(1/2)+140*I*(-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)*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^(1/2)*sin(x)*cos(x)^2-128*I*(-2*(2*cos(x)*5^(1/2)-2*5^(1/2)+4*cos(x)-5)/(c
os(x)+1))^(1/2)*((2*cos(x)*5^(1/2)-2*5^(1/2)-4*cos(x)+5)/(cos(x)+1))^(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))*2^(1/2)*sin(x)*cos(x)^2-6*I*(-2*(2*c
os(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)*EllipticF(I*(cos(x)-1)*(2+5^(1/2))/sin(x),9-4*5^(1/2))*2^(1/2)*sin(x)*cos(x)^2-80*cos(x)^3*5^(1/2)-180*cos(
x)^3+80*cos(x)^2*5^(1/2)+180*cos(x)^2+100*cos(x)*5^(1/2)+225*cos(x)-100*5^(1/2)-225)*((4*cos(x)^2-5)/cos(x)^2)
^(3/2)*sin(x)*cos(x)/(cos(x)-1)/(4*cos(x)^2-5)^2

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\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((4-5*sec(x)^2)^(3/2),x, algorithm="maxima")

[Out]

integrate((-5*sec(x)^2 + 4)^(3/2), x)

________________________________________________________________________________________

Fricas [B]  time = 3.72658, size = 417, normalized size = 6.13 \begin{align*} \frac{7 \, \sqrt{5} \arctan \left (\frac{\sqrt{5} \sqrt{\frac{4 \, \cos \left (x\right )^{2} - 5}{\cos \left (x\right )^{2}}} \cos \left (x\right )}{5 \, \sin \left (x\right )}\right ) \cos \left (x\right ) + 8 \, \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 ) \cos \left (x\right ) - 8 \, \arctan \left (\frac{\sin \left (x\right )}{\cos \left (x\right )}\right ) \cos \left (x\right ) - 5 \, \sqrt{\frac{4 \, \cos \left (x\right )^{2} - 5}{\cos \left (x\right )^{2}}} \sin \left (x\right )}{2 \, \cos \left (x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(7*sqrt(5)*arctan(1/5*sqrt(5)*sqrt((4*cos(x)^2 - 5)/cos(x)^2)*cos(x)/sin(x))*cos(x) + 8*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))*cos
(x) - 8*arctan(sin(x)/cos(x))*cos(x) - 5*sqrt((4*cos(x)^2 - 5)/cos(x)^2)*sin(x))/cos(x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \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((4-5*sec(x)**2)**(3/2),x)

[Out]

Integral((4 - 5*sec(x)**2)**(3/2), x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\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((4-5*sec(x)^2)^(3/2),x, algorithm="giac")

[Out]

integrate((-5*sec(x)^2 + 4)^(3/2), x)