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*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])

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Rubi [A]  time = 0.0319133, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, 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 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]

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 8

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

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*}

Mathematica [A]  time = 0.0388284, 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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Maple [A]  time = 0.104, size = 41, normalized size = 0.5 \begin{align*}{\frac{8\, \left ( \cos \left ( x \right ) \right ) ^{5}\sin \left ( x \right ) +10\, \left ( \cos \left ( x \right ) \right ) ^{3}\sin \left ( x \right ) +15\,\cos \left ( x \right ) \sin \left ( x \right ) +15\,x}{48\, \left ( \cos \left ( x \right ) \right ) ^{6}} \left ({\frac{a}{ \left ( \cos \left ( x \right ) \right ) ^{4}}} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(8*cos(x)^5*sin(x)+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 = 1.64573, size = 78, normalized size = 0.91 \begin{align*} \frac{15 \, \tan \left (x\right )^{5} + 40 \, \tan \left (x\right )^{3} + 33 \, \tan \left (x\right )}{48 \,{\left (a^{\frac{3}{2}} \tan \left (x\right )^{6} + 3 \, a^{\frac{3}{2}} \tan \left (x\right )^{4} + 3 \, a^{\frac{3}{2}} \tan \left (x\right )^{2} + a^{\frac{3}{2}}\right )}} + \frac{5 \, x}{16 \, a^{\frac{3}{2}}} \end{align*}

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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Fricas [A]  time = 1.46067, size = 126, normalized size = 1.47 \begin{align*} \frac{{\left (15 \, x \cos \left (x\right )^{2} +{\left (8 \, \cos \left (x\right )^{7} + 10 \, \cos \left (x\right )^{5} + 15 \, \cos \left (x\right )^{3}\right )} \sin \left (x\right )\right )} \sqrt{\frac{a}{\cos \left (x\right )^{4}}}}{48 \, a^{2}} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a \sec ^{4}{\left (x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

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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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

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