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

Optimal. Leaf size=115 \[ \frac{2 (9 A+7 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 b^2 d \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}}+\frac{2 (9 A+7 C) \sin (c+d x)}{45 b d (b \sec (c+d x))^{3/2}}+\frac{2 b^2 C \tan (c+d x)}{9 d (b \sec (c+d x))^{9/2}} \]

[Out]

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

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Rubi [A]  time = 0.127748, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3238, 4045, 3769, 3771, 2639} \[ \frac{2 (9 A+7 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 b^2 d \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}}+\frac{2 (9 A+7 C) \sin (c+d x)}{45 b d (b \sec (c+d x))^{3/2}}+\frac{2 b^2 C \tan (c+d x)}{9 d (b \sec (c+d x))^{9/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3238

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_.))^(p_.), x_Symbol] :> Dist
[d^(n*p), Int[(d*Csc[e + f*x])^(m - n*p)*(b + a*Csc[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x
] &&  !IntegerQ[m] && IntegersQ[n, p]

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

Mathematica [A]  time = 0.622843, size = 81, normalized size = 0.7 \[ \frac{4 \sin (2 (c+d x)) (18 A+5 C \cos (2 (c+d x))+19 C)+\frac{48 (9 A+7 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{\sqrt{\cos (c+d x)}}}{360 b^2 d \sqrt{b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((48*(9*A + 7*C)*EllipticE[(c + d*x)/2, 2])/Sqrt[Cos[c + d*x]] + 4*(18*A + 19*C + 5*C*Cos[2*(c + d*x)])*Sin[2*
(c + d*x)])/(360*b^2*d*Sqrt[b*Sec[c + d*x]])

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Maple [C]  time = 0.547, size = 636, normalized size = 5.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/45/d*(27*I*A*cos(d*x+c)*sin(d*x+c)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*EllipticE(I*(
-1+cos(d*x+c))/sin(d*x+c),I)-27*I*A*cos(d*x+c)*sin(d*x+c)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))
^(1/2)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)+5*C*cos(d*x+c)^6+21*I*C*cos(d*x+c)*sin(d*x+c)*EllipticE(I*(-1
+cos(d*x+c))/sin(d*x+c),I)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-21*I*C*cos(d*x+c)*sin(d*
x+c)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)+27*I
*A*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c
),I)-27*I*A*sin(d*x+c)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d
*x+c)))^(1/2)+21*I*C*sin(d*x+c)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)
/(1+cos(d*x+c)))^(1/2)-21*I*C*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)*EllipticF(
I*(-1+cos(d*x+c))/sin(d*x+c),I)+9*A*cos(d*x+c)^4+2*C*cos(d*x+c)^4+18*A*cos(d*x+c)^2+14*C*cos(d*x+c)^2-27*A*cos
(d*x+c)-21*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 \cos \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*cos(d*x+c)^2)/(b*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

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

[Out]

integral((C*cos(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*cos(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 \cos \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*cos(d*x+c)^2)/(b*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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