3.53 \(\int \frac {1}{(a \sec ^2(x))^{5/2}} \, dx\)

Optimal. Leaf size=55 \[ \frac {8 \tan (x)}{15 a^2 \sqrt {a \sec ^2(x)}}+\frac {4 \tan (x)}{15 a \left (a \sec ^2(x)\right )^{3/2}}+\frac {\tan (x)}{5 \left (a \sec ^2(x)\right )^{5/2}} \]

[Out]

1/5*tan(x)/(a*sec(x)^2)^(5/2)+4/15*tan(x)/a/(a*sec(x)^2)^(3/2)+8/15*tan(x)/a^2/(a*sec(x)^2)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {4122, 192, 191} \[ \frac {8 \tan (x)}{15 a^2 \sqrt {a \sec ^2(x)}}+\frac {4 \tan (x)}{15 a \left (a \sec ^2(x)\right )^{3/2}}+\frac {\tan (x)}{5 \left (a \sec ^2(x)\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Tan[x]/(5*(a*Sec[x]^2)^(5/2)) + (4*Tan[x])/(15*a*(a*Sec[x]^2)^(3/2)) + (8*Tan[x])/(15*a^2*Sqrt[a*Sec[x]^2])

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 192

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p + 1
], 0] && NeQ[p, -1]

Rule 4122

Int[((b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(b*ff)
/f, Subst[Int[(b + b*ff^2*x^2)^(p - 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{b, e, f, p}, x] &&  !IntegerQ[p
]

Rubi steps

\begin {align*} \int \frac {1}{\left (a \sec ^2(x)\right )^{5/2}} \, dx &=a \operatorname {Subst}\left (\int \frac {1}{\left (a+a x^2\right )^{7/2}} \, dx,x,\tan (x)\right )\\ &=\frac {\tan (x)}{5 \left (a \sec ^2(x)\right )^{5/2}}+\frac {4}{5} \operatorname {Subst}\left (\int \frac {1}{\left (a+a x^2\right )^{5/2}} \, dx,x,\tan (x)\right )\\ &=\frac {\tan (x)}{5 \left (a \sec ^2(x)\right )^{5/2}}+\frac {4 \tan (x)}{15 a \left (a \sec ^2(x)\right )^{3/2}}+\frac {8 \operatorname {Subst}\left (\int \frac {1}{\left (a+a x^2\right )^{3/2}} \, dx,x,\tan (x)\right )}{15 a}\\ &=\frac {\tan (x)}{5 \left (a \sec ^2(x)\right )^{5/2}}+\frac {4 \tan (x)}{15 a \left (a \sec ^2(x)\right )^{3/2}}+\frac {8 \tan (x)}{15 a^2 \sqrt {a \sec ^2(x)}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 36, normalized size = 0.65 \[ \frac {(150 \sin (x)+25 \sin (3 x)+3 \sin (5 x)) \cos (x) \sqrt {a \sec ^2(x)}}{240 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[x]*Sqrt[a*Sec[x]^2]*(150*Sin[x] + 25*Sin[3*x] + 3*Sin[5*x]))/(240*a^3)

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fricas [A]  time = 0.81, size = 32, normalized size = 0.58 \[ \frac {{\left (3 \, \cos \relax (x)^{5} + 4 \, \cos \relax (x)^{3} + 8 \, \cos \relax (x)\right )} \sqrt {\frac {a}{\cos \relax (x)^{2}}} \sin \relax (x)}{15 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*cos(x)^5 + 4*cos(x)^3 + 8*cos(x))*sqrt(a/cos(x)^2)*sin(x)/a^3

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giac [A]  time = 0.54, size = 84, normalized size = 1.53 \[ \frac {2 \, {\left (15 \, {\left (\frac {1}{\tan \left (\frac {1}{2} \, x\right )} + \tan \left (\frac {1}{2} \, x\right )\right )}^{4} \mathrm {sgn}\left (-\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right ) - 40 \, {\left (\frac {1}{\tan \left (\frac {1}{2} \, x\right )} + \tan \left (\frac {1}{2} \, x\right )\right )}^{2} \mathrm {sgn}\left (-\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right ) + 48 \, \mathrm {sgn}\left (-\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )\right )}}{15 \, a^{\frac {5}{2}} {\left (\frac {1}{\tan \left (\frac {1}{2} \, x\right )} + \tan \left (\frac {1}{2} \, x\right )\right )}^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(15*(1/tan(1/2*x) + tan(1/2*x))^4*sgn(-tan(1/2*x)^2 + 1) - 40*(1/tan(1/2*x) + tan(1/2*x))^2*sgn(-tan(1/2*
x)^2 + 1) + 48*sgn(-tan(1/2*x)^2 + 1))/(a^(5/2)*(1/tan(1/2*x) + tan(1/2*x))^5)

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maple [A]  time = 0.31, size = 31, normalized size = 0.56 \[ \frac {\sin \relax (x ) \left (3 \left (\cos ^{4}\relax (x )\right )+4 \left (\cos ^{2}\relax (x )\right )+8\right )}{15 \cos \relax (x )^{5} \left (\frac {a}{\cos \relax (x )^{2}}\right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*sec(x)^2)^(5/2),x)

[Out]

1/15*sin(x)*(3*cos(x)^4+4*cos(x)^2+8)/cos(x)^5/(a/cos(x)^2)^(5/2)

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maxima [A]  time = 0.68, size = 22, normalized size = 0.40 \[ \frac {3 \, \sin \left (5 \, x\right ) + 25 \, \sin \left (3 \, x\right ) + 150 \, \sin \relax (x)}{240 \, a^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(3*sin(5*x) + 25*sin(3*x) + 150*sin(x))/a^(5/2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{{\left (\frac {a}{{\cos \relax (x)}^2}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a/cos(x)^2)^(5/2),x)

[Out]

int(1/(a/cos(x)^2)^(5/2), x)

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sympy [A]  time = 10.57, size = 60, normalized size = 1.09 \[ \frac {8 \tan ^{5}{\relax (x )}}{15 a^{\frac {5}{2}} \left (\sec ^{2}{\relax (x )}\right )^{\frac {5}{2}}} + \frac {4 \tan ^{3}{\relax (x )}}{3 a^{\frac {5}{2}} \left (\sec ^{2}{\relax (x )}\right )^{\frac {5}{2}}} + \frac {\tan {\relax (x )}}{a^{\frac {5}{2}} \left (\sec ^{2}{\relax (x )}\right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sec(x)**2)**(5/2),x)

[Out]

8*tan(x)**5/(15*a**(5/2)*(sec(x)**2)**(5/2)) + 4*tan(x)**3/(3*a**(5/2)*(sec(x)**2)**(5/2)) + tan(x)/(a**(5/2)*
(sec(x)**2)**(5/2))

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