3.23 \(\int \frac{A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{9/2}} \, dx\)

Optimal. Leaf size=112 \[ \frac{2 (7 A+9 C) \sin (c+d x)}{45 b^3 d (b \sec (c+d x))^{3/2}}+\frac{2 (7 A+9 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 b^4 d \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}}+\frac{2 A \tan (c+d x)}{9 d (b \sec (c+d x))^{9/2}} \]

[Out]

(2*(7*A + 9*C)*EllipticE[(c + d*x)/2, 2])/(15*b^4*d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d*x]]) + (2*(7*A + 9*C)*
Sin[c + d*x])/(45*b^3*d*(b*Sec[c + d*x])^(3/2)) + (2*A*Tan[c + d*x])/(9*d*(b*Sec[c + d*x])^(9/2))

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Rubi [A]  time = 0.0864181, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {4045, 3769, 3771, 2639} \[ \frac{2 (7 A+9 C) \sin (c+d x)}{45 b^3 d (b \sec (c+d x))^{3/2}}+\frac{2 (7 A+9 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 b^4 d \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}}+\frac{2 A \tan (c+d x)}{9 d (b \sec (c+d x))^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + C*Sec[c + d*x]^2)/(b*Sec[c + d*x])^(9/2),x]

[Out]

(2*(7*A + 9*C)*EllipticE[(c + d*x)/2, 2])/(15*b^4*d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d*x]]) + (2*(7*A + 9*C)*
Sin[c + d*x])/(45*b^3*d*(b*Sec[c + d*x])^(3/2)) + (2*A*Tan[c + d*x])/(9*d*(b*Sec[c + d*x])^(9/2))

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 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 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))^{9/2}} \, dx &=\frac{2 A \tan (c+d x)}{9 d (b \sec (c+d x))^{9/2}}+\frac{(7 A+9 C) \int \frac{1}{(b \sec (c+d x))^{5/2}} \, dx}{9 b^2}\\ &=\frac{2 (7 A+9 C) \sin (c+d x)}{45 b^3 d (b \sec (c+d x))^{3/2}}+\frac{2 A \tan (c+d x)}{9 d (b \sec (c+d x))^{9/2}}+\frac{(7 A+9 C) \int \frac{1}{\sqrt{b \sec (c+d x)}} \, dx}{15 b^4}\\ &=\frac{2 (7 A+9 C) \sin (c+d x)}{45 b^3 d (b \sec (c+d x))^{3/2}}+\frac{2 A \tan (c+d x)}{9 d (b \sec (c+d x))^{9/2}}+\frac{(7 A+9 C) \int \sqrt{\cos (c+d x)} \, dx}{15 b^4 \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}}\\ &=\frac{2 (7 A+9 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 b^4 d \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}}+\frac{2 (7 A+9 C) \sin (c+d x)}{45 b^3 d (b \sec (c+d x))^{3/2}}+\frac{2 A \tan (c+d x)}{9 d (b \sec (c+d x))^{9/2}}\\ \end{align*}

Mathematica [C]  time = 1.41676, size = 143, normalized size = 1.28 \[ \frac{e^{-i d x} (\cos (d x)+i \sin (d x)) \left (-\frac{32 i (7 A+9 C) e^{2 i (c+d x)} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )}{\sqrt{1+e^{2 i (c+d x)}}}+(76 A+72 C) \sin (2 (c+d x))+10 A \sin (4 (c+d x))+336 i A+432 i C\right )}{360 b^4 d \sqrt{b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + C*Sec[c + d*x]^2)/(b*Sec[c + d*x])^(9/2),x]

[Out]

((Cos[d*x] + I*Sin[d*x])*((336*I)*A + (432*I)*C - ((32*I)*(7*A + 9*C)*E^((2*I)*(c + d*x))*Hypergeometric2F1[1/
2, 3/4, 7/4, -E^((2*I)*(c + d*x))])/Sqrt[1 + E^((2*I)*(c + d*x))] + (76*A + 72*C)*Sin[2*(c + d*x)] + 10*A*Sin[
4*(c + d*x)]))/(360*b^4*d*E^(I*d*x)*Sqrt[b*Sec[c + d*x]])

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Maple [C]  time = 0.261, size = 636, normalized size = 5.7 \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))^(9/2),x)

[Out]

-2/45/d*(21*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)-21*I*A*sin(d*x+c)*cos(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)+5*A*cos(d*x+c)^6+27*I*C*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)-27*I*C*(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)*cos(d*x+c)*sin(d*x+c)+21*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)-21*I*A*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)+27*I*C*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)-27*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)*(c
os(d*x+c)/(cos(d*x+c)+1))^(1/2)+2*A*cos(d*x+c)^4+9*C*cos(d*x+c)^4+14*A*cos(d*x+c)^2+18*C*cos(d*x+c)^2-21*A*cos
(d*x+c)-27*C*cos(d*x+c))/sin(d*x+c)/cos(d*x+c)^5/(b/cos(d*x+c))^(9/2)

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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{9}{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))^(9/2),x, algorithm="maxima")

[Out]

integrate((C*sec(d*x + c)^2 + A)/(b*sec(d*x + c))^(9/2), 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^{5} \sec \left (d x + c\right )^{5}}, 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))^(9/2),x, algorithm="fricas")

[Out]

integral((C*sec(d*x + c)^2 + A)*sqrt(b*sec(d*x + c))/(b^5*sec(d*x + c)^5), 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))**(9/2),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{9}{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))^(9/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)/(b*sec(d*x + c))^(9/2), x)