3.10 \(\int \sec ^{\frac {5}{2}}(a+b x) \, dx\)

Optimal. Leaf size=62 \[ \frac {2 \sin (a+b x) \sec ^{\frac {3}{2}}(a+b x)}{3 b}+\frac {2 \sqrt {\cos (a+b x)} \sqrt {\sec (a+b x)} F\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{3 b} \]

[Out]

2/3*sec(b*x+a)^(3/2)*sin(b*x+a)/b+2/3*(cos(1/2*b*x+1/2*a)^2)^(1/2)/cos(1/2*b*x+1/2*a)*EllipticF(sin(1/2*b*x+1/
2*a),2^(1/2))*cos(b*x+a)^(1/2)*sec(b*x+a)^(1/2)/b

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Rubi [A]  time = 0.03, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3768, 3771, 2641} \[ \frac {2 \sin (a+b x) \sec ^{\frac {3}{2}}(a+b x)}{3 b}+\frac {2 \sqrt {\cos (a+b x)} \sqrt {\sec (a+b x)} F\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]^(5/2),x]

[Out]

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 46, normalized size = 0.74 \[ \frac {2 \sec ^{\frac {3}{2}}(a+b x) \left (\sin (a+b x)+\cos ^{\frac {3}{2}}(a+b x) F\left (\left .\frac {1}{2} (a+b x)\right |2\right )\right )}{3 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[a + b*x]^(5/2),x]

[Out]

(2*Sec[a + b*x]^(3/2)*(Cos[a + b*x]^(3/2)*EllipticF[(a + b*x)/2, 2] + Sin[a + b*x]))/(3*b)

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fricas [F]  time = 0.80, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sec \left (b x + a\right )^{\frac {5}{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sec(b*x + a)^(5/2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sec \left (b x + a\right )^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sec(b*x + a)^(5/2), x)

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maple [B]  time = 3.07, size = 213, normalized size = 3.44 \[ -\frac {2 \left (-2 \sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right ) \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right )-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \cos \left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}}{3 \sqrt {-2 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )}\, \left (2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right )^{\frac {3}{2}} \sin \left (\frac {b x}{2}+\frac {a}{2}\right ) b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(-2*(sin(1/2*b*x+1/2*a)^2)^(1/2)*(2*sin(1/2*b*x+1/2*a)^2-1)^(1/2)*EllipticF(cos(1/2*b*x+1/2*a),2^(1/2))*s
in(1/2*b*x+1/2*a)^2+(sin(1/2*b*x+1/2*a)^2)^(1/2)*(2*sin(1/2*b*x+1/2*a)^2-1)^(1/2)*EllipticF(cos(1/2*b*x+1/2*a)
,2^(1/2))-2*sin(1/2*b*x+1/2*a)^2*cos(1/2*b*x+1/2*a))*((2*cos(1/2*b*x+1/2*a)^2-1)*sin(1/2*b*x+1/2*a)^2)^(1/2)/(
-2*sin(1/2*b*x+1/2*a)^4+sin(1/2*b*x+1/2*a)^2)^(1/2)/(2*cos(1/2*b*x+1/2*a)^2-1)^(3/2)/sin(1/2*b*x+1/2*a)/b

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sec \left (b x + a\right )^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sec(b*x + a)^(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/cos(a + b*x))^(5/2),x)

[Out]

int((1/cos(a + b*x))^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sec ^{\frac {5}{2}}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)**(5/2),x)

[Out]

Integral(sec(a + b*x)**(5/2), x)

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