3.1.11 \(\int \sec ^{\frac {3}{2}}(a+b x) \, dx\) [11]

Optimal. Leaf size=58 \[ -\frac {2 \sqrt {\cos (a+b x)} E\left (\left .\frac {1}{2} (a+b x)\right |2\right ) \sqrt {\sec (a+b x)}}{b}+\frac {2 \sqrt {\sec (a+b x)} \sin (a+b x)}{b} \]

[Out]

2*sin(b*x+a)*sec(b*x+a)^(1/2)/b-2*(cos(1/2*a+1/2*b*x)^2)^(1/2)/cos(1/2*a+1/2*b*x)*EllipticE(sin(1/2*a+1/2*b*x)
,2^(1/2))*cos(b*x+a)^(1/2)*sec(b*x+a)^(1/2)/b

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Rubi [A]
time = 0.02, antiderivative size = 58, 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 = {3853, 3856, 2719} \begin {gather*} \frac {2 \sin (a+b x) \sqrt {\sec (a+b x)}}{b}-\frac {2 \sqrt {\cos (a+b x)} \sqrt {\sec (a+b x)} E\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]^(3/2),x]

[Out]

(-2*Sqrt[Cos[a + b*x]]*EllipticE[(a + b*x)/2, 2]*Sqrt[Sec[a + b*x]])/b + (2*Sqrt[Sec[a + b*x]]*Sin[a + b*x])/b

Rule 2719

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

Rule 3853

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 3856

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

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Mathematica [A]
time = 0.04, size = 45, normalized size = 0.78 \begin {gather*} \frac {2 \sqrt {\sec (a+b x)} \left (-\sqrt {\cos (a+b x)} E\left (\left .\frac {1}{2} (a+b x)\right |2\right )+\sin (a+b x)\right )}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[a + b*x]^(3/2),x]

[Out]

(2*Sqrt[Sec[a + b*x]]*(-(Sqrt[Cos[a + b*x]]*EllipticE[(a + b*x)/2, 2]) + Sin[a + b*x]))/b

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(181\) vs. \(2(78)=156\).
time = 2.85, size = 182, normalized size = 3.14

method	result	size
default	\(-\frac {2 \left (-2 \cos \left (\frac {b x}{2}+\frac {a}{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )}\, \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1}\, \sqrt {-2 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )}\, \EllipticE \left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right )\right )}{\sqrt {-2 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )}\, \sin \left (\frac {b x}{2}+\frac {a}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1}\, b}\)	\(182\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-2*(-2*cos(1/2*b*x+1/2*a)*(-2*sin(1/2*b*x+1/2*a)^4+sin(1/2*b*x+1/2*a)^2)^(1/2)*sin(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)*(-2*sin(1/2*b*x+1/2*a)^4+sin(1/2*b*x+1/2*a)^2)^(1/2)*Ellip
ticE(cos(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)/sin(1/2*b*x+1/2*a)/(2*c
os(1/2*b*x+1/2*a)^2-1)^(1/2)/b

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sec(b*x + a)^(3/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.41, size = 73, normalized size = 1.26 \begin {gather*} \frac {-i \, \sqrt {2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\right ) + i \, \sqrt {2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\right ) + \frac {2 \, \sin \left (b x + a\right )}{\sqrt {\cos \left (b x + a\right )}}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-I*sqrt(2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(b*x + a) + I*sin(b*x + a))) + I*sqrt(2)*weie
rstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(b*x + a) - I*sin(b*x + a))) + 2*sin(b*x + a)/sqrt(cos(b*x +
a)))/b

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sec ^{\frac {3}{2}}{\left (a + b x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sec(a + b*x)**(3/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sec(b*x + a)^(3/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int {\left (\frac {1}{\cos \left (a+b\,x\right )}\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/cos(a + b*x))^(3/2),x)

[Out]

int((1/cos(a + b*x))^(3/2), x)

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