3.18 \(\int (c \sec (a+b x))^{5/2} \, dx\)

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

[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)

________________________________________________________________________________________

Rubi [A]  time = 0.0336493, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3768, 3771, 2641} \[ \frac{2 c^2 \sqrt{\cos (a+b x)} F\left (\left .\frac{1}{2} (a+b x)\right |2\right ) \sqrt{c \sec (a+b x)}}{3 b}+\frac{2 c \sin (a+b x) (c \sec (a+b x))^{3/2}}{3 b} \]

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 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 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]

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*}

Mathematica [A]  time = 0.0658061, size = 51, normalized size = 0.73 \[ \frac{2 c^2 \sqrt{c \sec (a+b x)} \left (\sqrt{\cos (a+b x)} \text{EllipticF}\left (\frac{1}{2} (a+b x),2\right )+\tan (a+b x)\right )}{3 b} \]

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)

________________________________________________________________________________________

Maple [C]  time = 0.198, size = 128, normalized size = 1.8 \begin{align*} -{\frac{ \left ( -2+2\,\cos \left ( bx+a \right ) \right ) \cos \left ( bx+a \right ) \left ( \cos \left ( bx+a \right ) +1 \right ) ^{2}}{3\,b \left ( \sin \left ( bx+a \right ) \right ) ^{3}} \left ( i\sqrt{ \left ( \cos \left ( bx+a \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( bx+a \right ) }{\cos \left ( bx+a \right ) +1}}}{\it EllipticF} \left ({\frac{i \left ( -1+\cos \left ( bx+a \right ) \right ) }{\sin \left ( bx+a \right ) }},i \right ) \cos \left ( bx+a \right ) \sin \left ( bx+a \right ) -\cos \left ( bx+a \right ) +1 \right ) \left ({\frac{c}{\cos \left ( bx+a \right ) }} \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c \sec \left (b x + a\right )\right )^{\frac{5}{2}}\,{d x} \end{align*}

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)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{c \sec \left (b x + a\right )} c^{2} \sec \left (b x + a\right )^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c*sec(b*x + a))*c^2*sec(b*x + a)^2, x)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c \sec \left (b x + a\right )\right )^{\frac{5}{2}}\,{d x} \end{align*}

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)