3.56 \(\int (a \sec ^3(x))^{3/2} \, dx\)

Optimal. Leaf size=65 \[ \frac {10}{21} a \sin (x) \sqrt {a \sec ^3(x)}+\frac {2}{7} a \tan (x) \sec (x) \sqrt {a \sec ^3(x)}+\frac {10}{21} a \cos ^{\frac {3}{2}}(x) F\left (\left .\frac {x}{2}\right |2\right ) \sqrt {a \sec ^3(x)} \]

[Out]

10/21*a*cos(x)^(3/2)*(cos(1/2*x)^2)^(1/2)/cos(1/2*x)*EllipticF(sin(1/2*x),2^(1/2))*(a*sec(x)^3)^(1/2)+10/21*a*
sin(x)*(a*sec(x)^3)^(1/2)+2/7*a*sec(x)*(a*sec(x)^3)^(1/2)*tan(x)

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Rubi [A]  time = 0.04, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4123, 3768, 3771, 2641} \[ \frac {10}{21} a \sin (x) \sqrt {a \sec ^3(x)}+\frac {2}{7} a \tan (x) \sec (x) \sqrt {a \sec ^3(x)}+\frac {10}{21} a \cos ^{\frac {3}{2}}(x) F\left (\left .\frac {x}{2}\right |2\right ) \sqrt {a \sec ^3(x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(10*a*Cos[x]^(3/2)*EllipticF[x/2, 2]*Sqrt[a*Sec[x]^3])/21 + (10*a*Sqrt[a*Sec[x]^3]*Sin[x])/21 + (2*a*Sec[x]*Sq
rt[a*Sec[x]^3]*Tan[x])/7

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 3768

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

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

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]

Rubi steps

\begin {align*} \int \left (a \sec ^3(x)\right )^{3/2} \, dx &=\frac {\left (a \sqrt {a \sec ^3(x)}\right ) \int \sec ^{\frac {9}{2}}(x) \, dx}{\sec ^{\frac {3}{2}}(x)}\\ &=\frac {2}{7} a \sec (x) \sqrt {a \sec ^3(x)} \tan (x)+\frac {\left (5 a \sqrt {a \sec ^3(x)}\right ) \int \sec ^{\frac {5}{2}}(x) \, dx}{7 \sec ^{\frac {3}{2}}(x)}\\ &=\frac {10}{21} a \sqrt {a \sec ^3(x)} \sin (x)+\frac {2}{7} a \sec (x) \sqrt {a \sec ^3(x)} \tan (x)+\frac {\left (5 a \sqrt {a \sec ^3(x)}\right ) \int \sqrt {\sec (x)} \, dx}{21 \sec ^{\frac {3}{2}}(x)}\\ &=\frac {10}{21} a \sqrt {a \sec ^3(x)} \sin (x)+\frac {2}{7} a \sec (x) \sqrt {a \sec ^3(x)} \tan (x)+\frac {1}{21} \left (5 a \cos ^{\frac {3}{2}}(x) \sqrt {a \sec ^3(x)}\right ) \int \frac {1}{\sqrt {\cos (x)}} \, dx\\ &=\frac {10}{21} a \cos ^{\frac {3}{2}}(x) F\left (\left .\frac {x}{2}\right |2\right ) \sqrt {a \sec ^3(x)}+\frac {10}{21} a \sqrt {a \sec ^3(x)} \sin (x)+\frac {2}{7} a \sec (x) \sqrt {a \sec ^3(x)} \tan (x)\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 43, normalized size = 0.66 \[ \frac {2}{21} a \sec (x) \sqrt {a \sec ^3(x)} \left (3 \tan (x)+5 \cos ^{\frac {5}{2}}(x) F\left (\left .\frac {x}{2}\right |2\right )+5 \sin (x) \cos (x)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*Sec[x]*Sqrt[a*Sec[x]^3]*(5*Cos[x]^(5/2)*EllipticF[x/2, 2] + 5*Cos[x]*Sin[x] + 3*Tan[x]))/21

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fricas [F]  time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {a \sec \relax (x)^{3}} a \sec \relax (x)^{3}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(a*sec(x)^3)*a*sec(x)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*sec(x)^3)^(3/2), x)

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maple [C]  time = 0.55, size = 87, normalized size = 1.34 \[ -\frac {2 \left (\cos \relax (x )+1\right )^{2} \left (-1+\cos \relax (x )\right ) \left (5 i \left (\cos ^{3}\relax (x )\right ) \sin \relax (x ) \sqrt {\frac {1}{\cos \relax (x )+1}}\, \sqrt {\frac {\cos \relax (x )}{\cos \relax (x )+1}}\, \EllipticF \left (\frac {i \left (-1+\cos \relax (x )\right )}{\sin \relax (x )}, i\right )-5 \left (\cos ^{3}\relax (x )\right )+5 \left (\cos ^{2}\relax (x )\right )-3 \cos \relax (x )+3\right ) \cos \relax (x ) \left (\frac {a}{\cos \relax (x )^{3}}\right )^{\frac {3}{2}}}{21 \sin \relax (x )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/21*(cos(x)+1)^2*(-1+cos(x))*(5*I*cos(x)^3*sin(x)*(1/(cos(x)+1))^(1/2)*(cos(x)/(cos(x)+1))^(1/2)*EllipticF(I
*(-1+cos(x))/sin(x),I)-5*cos(x)^3+5*cos(x)^2-3*cos(x)+3)*cos(x)*(a/cos(x)^3)^(3/2)/sin(x)^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*sec(x)^3)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int {\left (\frac {a}{{\cos \relax (x)}^3}\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((a/cos(x)^3)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*sec(x)**3)**(3/2), x)

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