3.218 \(\int \frac{(a+a \sec (c+d x))^2 (A+C \sec ^2(c+d x))}{\sec ^{\frac{5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=196 \[ \frac{4 a^2 (A+3 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 d}-\frac{2 a^2 (7 A-15 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{15 d}+\frac{8 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{15 d \sqrt{\sec (c+d x)}}+\frac{16 a^2 A \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 A \sin (c+d x) (a \sec (c+d x)+a)^2}{5 d \sec ^{\frac{3}{2}}(c+d x)} \]

[Out]

(16*a^2*A*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (4*a^2*(A + 3*C)*Sqrt[Cos[c
+ d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) - (2*a^2*(7*A - 15*C)*Sqrt[Sec[c + d*x]]*Sin[c +
d*x])/(15*d) + (2*A*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + (8*A*(a^2 + a^2*Sec[c + d*
x])*Sin[c + d*x])/(15*d*Sqrt[Sec[c + d*x]])

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Rubi [A]  time = 0.39229, antiderivative size = 196, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4087, 4017, 3997, 3787, 3771, 2639, 2641} \[ -\frac{2 a^2 (7 A-15 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{15 d}+\frac{4 a^2 (A+3 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{8 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{15 d \sqrt{\sec (c+d x)}}+\frac{16 a^2 A \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 A \sin (c+d x) (a \sec (c+d x)+a)^2}{5 d \sec ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(16*a^2*A*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (4*a^2*(A + 3*C)*Sqrt[Cos[c
+ d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) - (2*a^2*(7*A - 15*C)*Sqrt[Sec[c + d*x]]*Sin[c +
d*x])/(15*d) + (2*A*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + (8*A*(a^2 + a^2*Sec[c + d*
x])*Sin[c + d*x])/(15*d*Sqrt[Sec[c + d*x]])

Rule 4087

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b
_.) + (a_))^(m_), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(f*n), x] - Dis
t[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*(A*(m + n + 1) + C*n)*Csc[e +
f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2
^(-1)] || EqQ[m + n + 1, 0])

Rule 4017

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(a*A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^n)/(f*n), x]
- Dist[b/(a*d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*(m - n - 1) - b*B*n - (a*
B*n + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2
- b^2, 0] && GtQ[m, 1/2] && LtQ[n, -1]

Rule 3997

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.
) + (A_)), x_Symbol] :> -Simp[(b*B*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*(n + 1)), x] + Dist[1/(n + 1), Int[(d*C
sc[e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*a)*(n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f
, A, B}, x] && NeQ[A*b - a*B, 0] &&  !LeQ[n, -1]

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

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]

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{(a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac{5}{2}}(c+d x)} \, dx &=\frac{2 A (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 \int \frac{(a+a \sec (c+d x))^2 \left (2 a A-\frac{1}{2} a (A-5 C) \sec (c+d x)\right )}{\sec ^{\frac{3}{2}}(c+d x)} \, dx}{5 a}\\ &=\frac{2 A (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{8 A \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{4 \int \frac{(a+a \sec (c+d x)) \left (\frac{1}{4} a^2 (17 A+15 C)-\frac{1}{4} a^2 (7 A-15 C) \sec (c+d x)\right )}{\sqrt{\sec (c+d x)}} \, dx}{15 a}\\ &=-\frac{2 a^2 (7 A-15 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 A (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{8 A \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{8 \int \frac{3 a^3 A+\frac{5}{4} a^3 (A+3 C) \sec (c+d x)}{\sqrt{\sec (c+d x)}} \, dx}{15 a}\\ &=-\frac{2 a^2 (7 A-15 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 A (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{8 A \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{1}{5} \left (8 a^2 A\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx+\frac{1}{3} \left (2 a^2 (A+3 C)\right ) \int \sqrt{\sec (c+d x)} \, dx\\ &=-\frac{2 a^2 (7 A-15 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 A (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{8 A \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{1}{5} \left (8 a^2 A \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx+\frac{1}{3} \left (2 a^2 (A+3 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{16 a^2 A \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{5 d}+\frac{4 a^2 (A+3 C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{3 d}-\frac{2 a^2 (7 A-15 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 A (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{8 A \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 4.64055, size = 318, normalized size = 1.62 \[ \frac{a^2 \sec ^4\left (\frac{1}{2} (c+d x)\right ) (\sec (c+d x)+1)^2 \left (A+C \sec ^2(c+d x)\right ) \left (-\frac{\csc (c) ((99 A+60 C) \cos (2 c+d x)-2 A \sin (c) (20 \sin (2 (c+d x))+3 \sin (3 (c+d x)))+(93 A-60 C) \cos (d x))}{8 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 i \sqrt{2} \cos ^4(c+d x) \left (-5 \left (-1+e^{2 i c}\right ) (A+3 C) e^{i (c+d x)} \text{Hypergeometric2F1}\left (\frac{1}{4},\frac{1}{2},\frac{5}{4},-e^{2 i (c+d x)}\right )+12 A \left (-1+e^{2 i c}\right ) \text{Hypergeometric2F1}\left (-\frac{1}{4},\frac{1}{2},\frac{3}{4},-e^{2 i (c+d x)}\right )+12 A \sqrt{1+e^{2 i (c+d x)}}\right )}{\left (-1+e^{2 i c}\right ) d \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}}}\right )}{15 (A \cos (2 (c+d x))+A+2 C)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a^2*Sec[(c + d*x)/2]^4*(1 + Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2)*(((2*I)*Sqrt[2]*Cos[c + d*x]^4*(12*A*Sqrt[
1 + E^((2*I)*(c + d*x))] + 12*A*(-1 + E^((2*I)*c))*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))] - 5
*(A + 3*C)*E^(I*(c + d*x))*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/4, 1/2, 5/4, -E^((2*I)*(c + d*x))]))/(d*(-1
+ E^((2*I)*c))*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]) - (Csc[c]*((93*A
- 60*C)*Cos[d*x] + (99*A + 60*C)*Cos[2*c + d*x] - 2*A*Sin[c]*(20*Sin[2*(c + d*x)] + 3*Sin[3*(c + d*x)])))/(8*
d*Sec[c + d*x]^(7/2))))/(15*(A + 2*C + A*Cos[2*(c + d*x)]))

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Maple [A]  time = 2.163, size = 440, normalized size = 2.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/15*a^2*(-12*A*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+
32*A*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-(-2*sin(1/2*
d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(13*A+15*C)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+5*A*(sin(1/2*d*x+
1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*Elliptic
F(cos(1/2*d*x+1/2*c),2^(1/2))-12*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(-2*sin(1/2*d
*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+15*C*(sin(1/2*d*x+1/2*c)^2)^(1/2
)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+
1/2*c)^2)^(1/2))/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)
^2-1)^(1/2)/d

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Maxima [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+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{C a^{2} \sec \left (d x + c\right )^{4} + 2 \, C a^{2} \sec \left (d x + c\right )^{3} +{\left (A + C\right )} a^{2} \sec \left (d x + c\right )^{2} + 2 \, A a^{2} \sec \left (d x + c\right ) + A a^{2}}{\sec \left (d x + c\right )^{\frac{5}{2}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*a^2*sec(d*x + c)^4 + 2*C*a^2*sec(d*x + c)^3 + (A + C)*a^2*sec(d*x + c)^2 + 2*A*a^2*sec(d*x + c) +
A*a^2)/sec(d*x + c)^(5/2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{2} \left (\int \frac{A}{\sec ^{\frac{5}{2}}{\left (c + d x \right )}}\, dx + \int \frac{2 A}{\sec ^{\frac{3}{2}}{\left (c + d x \right )}}\, dx + \int \frac{A}{\sqrt{\sec{\left (c + d x \right )}}}\, dx + \int \frac{C}{\sqrt{\sec{\left (c + d x \right )}}}\, dx + \int 2 C \sqrt{\sec{\left (c + d x \right )}}\, dx + \int C \sec ^{\frac{3}{2}}{\left (c + d x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*(Integral(A/sec(c + d*x)**(5/2), x) + Integral(2*A/sec(c + d*x)**(3/2), x) + Integral(A/sqrt(sec(c + d*x)
), x) + Integral(C/sqrt(sec(c + d*x)), x) + Integral(2*C*sqrt(sec(c + d*x)), x) + Integral(C*sec(c + d*x)**(3/
2), x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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