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

Optimal. Leaf size=150 \[ \frac{2 B \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \sqrt{b \sec (c+d x)}}{3 b^3 d}+\frac{2 (3 A+5 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 b^2 d \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}}+\frac{2 A \tan (c+d x)}{5 d (b \sec (c+d x))^{5/2}}+\frac{2 B \sin (c+d x)}{3 b^2 d \sqrt{b \sec (c+d x)}} \]

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

________________________________________________________________________________________

Rubi [A]  time = 0.154279, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {4047, 3769, 3771, 2641, 4045, 2639} \[ \frac{2 (3 A+5 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 b^2 d \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}}+\frac{2 A \tan (c+d x)}{5 d (b \sec (c+d x))^{5/2}}+\frac{2 B \sin (c+d x)}{3 b^2 d \sqrt{b \sec (c+d x)}}+\frac{2 B \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \sec (c+d x)}}{3 b^3 d} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sec[c + d*x] + 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*B*Sqrt[Cos[c
+ d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[b*Sec[c + d*x]])/(3*b^3*d) + (2*B*Sin[c + d*x])/(3*b^2*d*Sqrt[b*Sec[c +
d*x]]) + (2*A*Tan[c + d*x])/(5*d*(b*Sec[c + d*x])^(5/2))

Rule 4047

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

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]

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 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+B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx &=\frac{B \int \frac{1}{(b \sec (c+d x))^{3/2}} \, dx}{b}+\int \frac{A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx\\ &=\frac{2 B \sin (c+d x)}{3 b^2 d \sqrt{b \sec (c+d x)}}+\frac{2 A \tan (c+d x)}{5 d (b \sec (c+d x))^{5/2}}+\frac{B \int \sqrt{b \sec (c+d x)} \, dx}{3 b^3}+\frac{(3 A+5 C) \int \frac{1}{\sqrt{b \sec (c+d x)}} \, dx}{5 b^2}\\ &=\frac{2 B \sin (c+d x)}{3 b^2 d \sqrt{b \sec (c+d x)}}+\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{\left (B \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 (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 B \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 B \sin (c+d x)}{3 b^2 d \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 = 1.97786, size = 169, normalized size = 1.13 \[ \frac{e^{-i d x} (\cos (d x)+i \sin (d x)) \sqrt{b \sec (c+d x)} \left (-2 i (3 A+5 C) e^{i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )+10 B \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )+\cos (c+d x) (3 A \sin (2 (c+d x))+6 i (3 A+5 C)+10 B \sin (c+d x))\right )}{15 b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*Sec[c + d*x]]*(Cos[d*x] + I*Sin[d*x])*(10*B*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] - (2*I)*(3*A
+ 5*C)*E^(I*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))] +
Cos[c + d*x]*((6*I)*(3*A + 5*C) + 10*B*Sin[c + d*x] + 3*A*Sin[2*(c + d*x)])))/(15*b^3*d*E^(I*d*x))

________________________________________________________________________________________

Maple [C]  time = 0.229, size = 766, normalized size = 5.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15/d*(15*I*C*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(
d*x+c)+1))^(1/2)-15*I*C*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)*cos(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x
+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+5*I*B*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c)*(1/(cos(d*x+c)
+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-9*I*A*sin(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)
+1))^(1/2)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)+5*I*B*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*cos(d*x+c
)*sin(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+9*I*A*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*
x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c)+15*I*C*cos(d*x+c)*EllipticF(I*
(-1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-9*I*A*sin(
d*x+c)*cos(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(-1+cos(d*x+c))/sin(d
*x+c),I)-15*I*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)*sin(d*x+c)+9*I*A*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*cos(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x
+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-3*A*cos(d*x+c)^4-5*B*cos(d*x+c)^3-6*A*cos(d*x+c)^2-15*C*cos(d*x+c)^2+9*A*
cos(d*x+c)+5*B*cos(d*x+c)+15*C*cos(d*x+c))/sin(d*x+c)/cos(d*x+c)^3/(b/cos(d*x+c))^(5/2)

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + 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+B*sec(d*x+c)+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

integral((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sqrt(b*sec(d*x + c))/(b^3*sec(d*x + c)^3), 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((A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(b*sec(d*x+c))**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________

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

[Out]

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