3.5.34 \(\int (4-5 \sec ^2(x))^{3/2} \, dx\) [434]

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

[Out]

8*arctan(2*tan(x)/(-1-5*tan(x)^2)^(1/2))-7/2*arctan(5^(1/2)*tan(x)/(-1-5*tan(x)^2)^(1/2))*5^(1/2)-5/2*(-1-5*ta
n(x)^2)^(1/2)*tan(x)

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Rubi [A]
time = 0.08, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4213, 427, 537, 223, 209, 385} \begin {gather*} 8 \text {ArcTan}\left (\frac {2 \tan (x)}{\sqrt {-5 \tan ^2(x)-1}}\right )-\frac {7}{2} \sqrt {5} \text {ArcTan}\left (\frac {\sqrt {5} \tan (x)}{\sqrt {-5 \tan ^2(x)-1}}\right )-\frac {5}{2} \tan (x) \sqrt {-5 \tan ^2(x)-1} \end {gather*}

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 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 223

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 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 427

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 537

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 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 \left (4-5 \sec ^2(x)\right )^{3/2} \, dx &=\text {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} \text {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 \text {Subst}\left (\int \frac {1}{\sqrt {-1-5 x^2} \left (1+x^2\right )} \, dx,x,\tan (x)\right )-\frac {35}{2} \text {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 \text {Subst}\left (\int \frac {1}{1+4 x^2} \, dx,x,\frac {\tan (x)}{\sqrt {-1-5 \tan ^2(x)}}\right )-\frac {35}{2} \text {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*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.15, size = 115, normalized size = 1.69 \begin {gather*} -\frac {\left (-5+4 \cos ^2(x)\right ) \sec (x) \sqrt {4-5 \sec ^2(x)} \left (7 \sqrt {5} \tan ^{-1}\left (\frac {\sqrt {5} \sin (x)}{\sqrt {-3+2 \cos (2 x)}}\right ) \cos ^2(x)+16 i \cos ^2(x) \log \left (\sqrt {-3+2 \cos (2 x)}+2 i \sin (x)\right )+5 \sqrt {-3+2 \cos (2 x)} \sin (x)\right )}{2 (-3+2 \cos (2 x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4 - 5*Sec[x]^2)^(3/2),x]

[Out]

-1/2*((-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]))/(-3 +
2*Cos[2*x])^(3/2)

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.46, size = 754, normalized size = 11.09

method	result	size
default	\(\text {Expression too large to display}\)	\(754\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*I/(-9-4*5^(1/2))^(1/2)/(2+5^(1/2))*(64*I*cos(x)^2*sin(x)*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)*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))*5^(1/2)+3*I*cos(x)^2*sin(x)*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)*EllipticF(I*(cos(x)-1)*(2+5^(1/2))/sin(x),9-4*5^(1/2))*5^(1/2)-70*I*cos(x)^2*sin(x)*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)*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))*5^(1/2)+128*I*cos(x)^2*sin(x)*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)*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))+6*I*cos(x)^2*sin(x)*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)*EllipticF(I*(
cos(x)-1)*(2+5^(1/2))/sin(x),9-4*5^(1/2))-140*I*cos(x)^2*sin(x)*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)*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))+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)*cos(x)*sin(x)*((4*cos(x)
^2-5)/cos(x)^2)^(3/2)/(cos(x)-1)/(4*cos(x)^2-5)^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((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. 130 vs. \(2 (54) = 108\).
time = 1.19, size = 130, normalized size = 1.91 \begin {gather*} \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 {gather*}

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)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \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((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((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 = -0.01 \begin {gather*} \int {\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((4 - 5/cos(x)^2)^(3/2),x)

[Out]

int((4 - 5/cos(x)^2)^(3/2), x)

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