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

Optimal. Leaf size=77 \[ \frac{2 (3 A+5 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 b^2 d \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}}+\frac{2 A \tan (c+d x)}{5 d (b \sec (c+d x))^{5/2}} \]

[Out]

(2*(3*A + 5*C)*EllipticE[(c + d*x)/2, 2])/(5*b^2*d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d*x]]) + (2*A*Tan[c + d*x
])/(5*d*(b*Sec[c + d*x])^(5/2))

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Rubi [A]  time = 0.0636336, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {4045, 3771, 2639} \[ \frac{2 (3 A+5 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 b^2 d \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}}+\frac{2 A \tan (c+d x)}{5 d (b \sec (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(3*A + 5*C)*EllipticE[(c + d*x)/2, 2])/(5*b^2*d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d*x]]) + (2*A*Tan[c + d*x
])/(5*d*(b*Sec[c + d*x])^(5/2))

Rule 4045

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e
+ f*x]*(b*Csc[e + f*x])^m)/(f*m), x] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]

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 2639

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

Rubi steps

\begin{align*} \int \frac{A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx &=\frac{2 A \tan (c+d x)}{5 d (b \sec (c+d x))^{5/2}}+\frac{(3 A+5 C) \int \frac{1}{\sqrt{b \sec (c+d x)}} \, dx}{5 b^2}\\ &=\frac{2 A \tan (c+d x)}{5 d (b \sec (c+d x))^{5/2}}+\frac{(3 A+5 C) \int \sqrt{\cos (c+d x)} \, dx}{5 b^2 \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}}\\ &=\frac{2 (3 A+5 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 b^2 d \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}}+\frac{2 A \tan (c+d x)}{5 d (b \sec (c+d x))^{5/2}}\\ \end{align*}

Mathematica [C]  time = 1.15229, size = 133, normalized size = 1.73 \[ \frac{e^{-i d x} \sec ^2(c+d x) (\cos (d x)+i \sin (d x)) \left (-\frac{8 i (3 A+5 C) e^{2 i (c+d x)} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )}{\sqrt{1+e^{2 i (c+d x)}}}+6 A \sin (2 (c+d x))+12 i (3 A+5 C)\right )}{30 d (b \sec (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^2*(Cos[d*x] + I*Sin[d*x])*((12*I)*(3*A + 5*C) - ((8*I)*(3*A + 5*C)*E^((2*I)*(c + d*x))*Hypergeom
etric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])/Sqrt[1 + E^((2*I)*(c + d*x))] + 6*A*Sin[2*(c + d*x)]))/(30*d*E^
(I*d*x)*(b*Sec[c + d*x])^(5/2))

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Maple [C]  time = 0.217, size = 613, normalized size = 8. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5/d*(3*I*A*sin(d*x+c)*cos(d*x+c)*EllipticE(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)-3*I*A*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)*sin(d*x
+c)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)+5*I*C*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)
*cos(d*x+c)*sin(d*x+c)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)-5*I*C*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(c
os(d*x+c)+1))^(1/2)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*cos(d*x+c)*sin(d*x+c)+3*I*A*sin(d*x+c)*EllipticE
(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)-3*I*A*sin(d*x+c)*E
llipticF(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)+5*I*C*(cos
(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)-5*
I*C*sin(d*x+c)*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)+A*cos(d*x+c)^4+2*A*cos(d*x+c)^2+5*C*cos(d*x+c)^2-3*A*cos(d*x+c)-5*C*cos(d*x+c))/cos(d*x+c)^3/(b/cos(d*x+
c))^(5/2)/sin(d*x+c)

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

[Out]

integrate((C*sec(d*x + c)^2 + A)/(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{{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sqrt{b \sec \left (d x + c\right )}}{b^{3} \sec \left (d x + c\right )^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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

[Out]

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