3.1.18 \(\int (c \sec (a+b x))^{5/2} \, dx\) [18]

Optimal. Leaf size=70 \[ \frac {2 c^2 \sqrt {\cos (a+b x)} F\left (\left .\frac {1}{2} (a+b x)\right |2\right ) \sqrt {c \sec (a+b x)}}{3 b}+\frac {2 c (c \sec (a+b x))^{3/2} \sin (a+b x)}{3 b} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3853, 3856, 2720} \begin {gather*} \frac {2 c^2 \sqrt {\cos (a+b x)} F\left (\left .\frac {1}{2} (a+b x)\right |2\right ) \sqrt {c \sec (a+b x)}}{3 b}+\frac {2 c \sin (a+b x) (c \sec (a+b x))^{3/2}}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c*Sec[a + b*x])^(5/2),x]

[Out]

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

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(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 (c \sec (a+b x))^{5/2} \, dx &=\frac {2 c (c \sec (a+b x))^{3/2} \sin (a+b x)}{3 b}+\frac {1}{3} c^2 \int \sqrt {c \sec (a+b x)} \, dx\\ &=\frac {2 c (c \sec (a+b x))^{3/2} \sin (a+b x)}{3 b}+\frac {1}{3} \left (c^2 \sqrt {\cos (a+b x)} \sqrt {c \sec (a+b x)}\right ) \int \frac {1}{\sqrt {\cos (a+b x)}} \, dx\\ &=\frac {2 c^2 \sqrt {\cos (a+b x)} F\left (\left .\frac {1}{2} (a+b x)\right |2\right ) \sqrt {c \sec (a+b x)}}{3 b}+\frac {2 c (c \sec (a+b x))^{3/2} \sin (a+b x)}{3 b}\\ \end {align*}

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Mathematica [A]
time = 0.08, size = 51, normalized size = 0.73 \begin {gather*} \frac {2 c^2 \sqrt {c \sec (a+b x)} \left (\sqrt {\cos (a+b x)} F\left (\left .\frac {1}{2} (a+b x)\right |2\right )+\tan (a+b x)\right )}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c*Sec[a + b*x])^(5/2),x]

[Out]

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

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Maple [C] Result contains complex when optimal does not.
time = 41.75, size = 128, normalized size = 1.83

method	result	size
default	\(-\frac {2 \left (-1+\cos \left (b x +a \right )\right ) \left (i \sin \left (b x +a \right ) \cos \left (b x +a \right ) \sqrt {\frac {1}{\cos \left (b x +a \right )+1}}\, \sqrt {\frac {\cos \left (b x +a \right )}{\cos \left (b x +a \right )+1}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (b x +a \right )\right )}{\sin \left (b x +a \right )}, i\right )-\cos \left (b x +a \right )+1\right ) \cos \left (b x +a \right ) \left (\cos \left (b x +a \right )+1\right )^{2} \left (\frac {c}{\cos \left (b x +a \right )}\right )^{\frac {5}{2}}}{3 b \sin \left (b x +a \right )^{3}}\)	\(128\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*sec(b*x+a))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-2/3/b*(-1+cos(b*x+a))*(I*sin(b*x+a)*cos(b*x+a)*(1/(cos(b*x+a)+1))^(1/2)*(cos(b*x+a)/(cos(b*x+a)+1))^(1/2)*Ell
ipticF(I*(-1+cos(b*x+a))/sin(b*x+a),I)-cos(b*x+a)+1)*cos(b*x+a)*(cos(b*x+a)+1)^2*(c/cos(b*x+a))^(5/2)/sin(b*x+
a)^3

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

[Out]

integrate((c*sec(b*x + a))^(5/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.65, size = 101, normalized size = 1.44 \begin {gather*} \frac {-i \, \sqrt {2} c^{\frac {5}{2}} \cos \left (b x + a\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + i \, \sqrt {2} c^{\frac {5}{2}} \cos \left (b x + a\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) + 2 \, c^{2} \sqrt {\frac {c}{\cos \left (b x + a\right )}} \sin \left (b x + a\right )}{3 \, b \cos \left (b x + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(-I*sqrt(2)*c^(5/2)*cos(b*x + a)*weierstrassPInverse(-4, 0, cos(b*x + a) + I*sin(b*x + a)) + I*sqrt(2)*c^(
5/2)*cos(b*x + a)*weierstrassPInverse(-4, 0, cos(b*x + a) - I*sin(b*x + a)) + 2*c^2*sqrt(c/cos(b*x + a))*sin(b
*x + a))/(b*cos(b*x + a))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate((c*sec(b*x + a))^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (\frac {c}{\cos \left (a+b\,x\right )}\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c/cos(a + b*x))^(5/2),x)

[Out]

int((c/cos(a + b*x))^(5/2), x)

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