3.127 \(\int \frac{\sec (c+d x)}{(b \sec (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=72 \[ \frac{2 \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \sqrt{b \sec (c+d x)}}{3 b^3 d}+\frac{2 \sin (c+d x)}{3 b^2 d \sqrt{b \sec (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.0400432, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {16, 3769, 3771, 2641} \[ \frac{2 \sin (c+d x)}{3 b^2 d \sqrt{b \sec (c+d x)}}+\frac{2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \sec (c+d x)}}{3 b^3 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]/(b*Sec[c + d*x])^(5/2),x]

[Out]

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

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 3769

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Csc[c + d*x])^(n + 1))/(b*d*n), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[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 \frac{\sec (c+d x)}{(b \sec (c+d x))^{5/2}} \, dx &=\frac{\int \frac{1}{(b \sec (c+d x))^{3/2}} \, dx}{b}\\ &=\frac{2 \sin (c+d x)}{3 b^2 d \sqrt{b \sec (c+d x)}}+\frac{\int \sqrt{b \sec (c+d x)} \, dx}{3 b^3}\\ &=\frac{2 \sin (c+d x)}{3 b^2 d \sqrt{b \sec (c+d x)}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{3 b^3}\\ &=\frac{2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \sec (c+d x)}}{3 b^3 d}+\frac{2 \sin (c+d x)}{3 b^2 d \sqrt{b \sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.0306355, size = 62, normalized size = 0.86 \[ \frac{\sec ^2(c+d x) \left (2 \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )+\sin (2 (c+d x))\right )}{3 b d (b \sec (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]/(b*Sec[c + d*x])^(5/2),x]

[Out]

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

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Maple [C]  time = 0.147, size = 131, normalized size = 1.8 \begin{align*} -{\frac{ \left ( -2+2\,\cos \left ( dx+c \right ) \right ) \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2}}{3\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}} \left ( i{\it EllipticF} \left ({\frac{i \left ( -1+\cos \left ( dx+c \right ) \right ) }{\sin \left ( dx+c \right ) }},i \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\sin \left ( dx+c \right ) - \left ( \cos \left ( dx+c \right ) \right ) ^{2}+\cos \left ( dx+c \right ) \right ) \left ({\frac{b}{\cos \left ( dx+c \right ) }} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3/d*(-1+cos(d*x+c))*(I*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(
d*x+c)+1))^(1/2)*sin(d*x+c)-cos(d*x+c)^2+cos(d*x+c))*(cos(d*x+c)+1)^2/sin(d*x+c)^3/cos(d*x+c)^3/(b/cos(d*x+c))
^(5/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b \sec \left (d x + c\right )}}{b^{3} \sec \left (d x + c\right )^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*sec(d*x + c))/(b^3*sec(d*x + c)^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sec(c + d*x)/(b*sec(c + d*x))**(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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