3.66 \(\int \frac {1}{(a \sec ^4(x))^{3/2}} \, dx\)

Optimal. Leaf size=86 \[ \frac {5 x \sec ^2(x)}{16 a \sqrt {a \sec ^4(x)}}+\frac {5 \tan (x)}{16 a \sqrt {a \sec ^4(x)}}+\frac {\sin (x) \cos ^3(x)}{6 a \sqrt {a \sec ^4(x)}}+\frac {5 \sin (x) \cos (x)}{24 a \sqrt {a \sec ^4(x)}} \]

[Out]

5/16*x*sec(x)^2/a/(a*sec(x)^4)^(1/2)+5/24*cos(x)*sin(x)/a/(a*sec(x)^4)^(1/2)+1/6*cos(x)^3*sin(x)/a/(a*sec(x)^4
)^(1/2)+5/16*tan(x)/a/(a*sec(x)^4)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {4123, 2635, 8} \[ \frac {5 x \sec ^2(x)}{16 a \sqrt {a \sec ^4(x)}}+\frac {5 \tan (x)}{16 a \sqrt {a \sec ^4(x)}}+\frac {\sin (x) \cos ^3(x)}{6 a \sqrt {a \sec ^4(x)}}+\frac {5 \sin (x) \cos (x)}{24 a \sqrt {a \sec ^4(x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*x*Sec[x]^2)/(16*a*Sqrt[a*Sec[x]^4]) + (5*Cos[x]*Sin[x])/(24*a*Sqrt[a*Sec[x]^4]) + (Cos[x]^3*Sin[x])/(6*a*Sq
rt[a*Sec[x]^4]) + (5*Tan[x])/(16*a*Sqrt[a*Sec[x]^4])

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 4123

Int[((b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> Dist[(b^IntPart[p]*(b*(c*Sec[e + f*x])^n)^
FracPart[p])/(c*Sec[e + f*x])^(n*FracPart[p]), Int[(c*Sec[e + f*x])^(n*p), x], x] /; FreeQ[{b, c, e, f, n, p},
x] &&  !IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {1}{\left (a \sec ^4(x)\right )^{3/2}} \, dx &=\frac {\sec ^2(x) \int \cos ^6(x) \, dx}{a \sqrt {a \sec ^4(x)}}\\ &=\frac {\cos ^3(x) \sin (x)}{6 a \sqrt {a \sec ^4(x)}}+\frac {\left (5 \sec ^2(x)\right ) \int \cos ^4(x) \, dx}{6 a \sqrt {a \sec ^4(x)}}\\ &=\frac {5 \cos (x) \sin (x)}{24 a \sqrt {a \sec ^4(x)}}+\frac {\cos ^3(x) \sin (x)}{6 a \sqrt {a \sec ^4(x)}}+\frac {\left (5 \sec ^2(x)\right ) \int \cos ^2(x) \, dx}{8 a \sqrt {a \sec ^4(x)}}\\ &=\frac {5 \cos (x) \sin (x)}{24 a \sqrt {a \sec ^4(x)}}+\frac {\cos ^3(x) \sin (x)}{6 a \sqrt {a \sec ^4(x)}}+\frac {5 \tan (x)}{16 a \sqrt {a \sec ^4(x)}}+\frac {\left (5 \sec ^2(x)\right ) \int 1 \, dx}{16 a \sqrt {a \sec ^4(x)}}\\ &=\frac {5 x \sec ^2(x)}{16 a \sqrt {a \sec ^4(x)}}+\frac {5 \cos (x) \sin (x)}{24 a \sqrt {a \sec ^4(x)}}+\frac {\cos ^3(x) \sin (x)}{6 a \sqrt {a \sec ^4(x)}}+\frac {5 \tan (x)}{16 a \sqrt {a \sec ^4(x)}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 38, normalized size = 0.44 \[ \frac {(60 x+45 \sin (2 x)+9 \sin (4 x)+\sin (6 x)) \sec ^6(x)}{192 \left (a \sec ^4(x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[x]^6*(60*x + 45*Sin[2*x] + 9*Sin[4*x] + Sin[6*x]))/(192*(a*Sec[x]^4)^(3/2))

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fricas [A]  time = 0.67, size = 43, normalized size = 0.50 \[ \frac {{\left (15 \, x \cos \relax (x)^{2} + {\left (8 \, \cos \relax (x)^{7} + 10 \, \cos \relax (x)^{5} + 15 \, \cos \relax (x)^{3}\right )} \sin \relax (x)\right )} \sqrt {\frac {a}{\cos \relax (x)^{4}}}}{48 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(15*x*cos(x)^2 + (8*cos(x)^7 + 10*cos(x)^5 + 15*cos(x)^3)*sin(x))*sqrt(a/cos(x)^4)/a^2

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.38, size = 41, normalized size = 0.48 \[ \frac {8 \sin \relax (x ) \left (\cos ^{5}\relax (x )\right )+10 \left (\cos ^{3}\relax (x )\right ) \sin \relax (x )+15 \cos \relax (x ) \sin \relax (x )+15 x}{48 \cos \relax (x )^{6} \left (\frac {a}{\cos \relax (x )^{4}}\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(8*sin(x)*cos(x)^5+10*cos(x)^3*sin(x)+15*cos(x)*sin(x)+15*x)/cos(x)^6/(a/cos(x)^4)^(3/2)

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maxima [A]  time = 0.69, size = 58, normalized size = 0.67 \[ \frac {15 \, \tan \relax (x)^{5} + 40 \, \tan \relax (x)^{3} + 33 \, \tan \relax (x)}{48 \, {\left (a^{\frac {3}{2}} \tan \relax (x)^{6} + 3 \, a^{\frac {3}{2}} \tan \relax (x)^{4} + 3 \, a^{\frac {3}{2}} \tan \relax (x)^{2} + a^{\frac {3}{2}}\right )}} + \frac {5 \, x}{16 \, a^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(15*tan(x)^5 + 40*tan(x)^3 + 33*tan(x))/(a^(3/2)*tan(x)^6 + 3*a^(3/2)*tan(x)^4 + 3*a^(3/2)*tan(x)^2 + a^(
3/2)) + 5/16*x/a^(3/2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (\frac {a}{{\cos \relax (x)}^4}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/(a/cos(x)^4)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a \sec ^{4}{\relax (x )}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*sec(x)**4)**(-3/2), x)

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