3.226 \(\int \frac{(a+a \sec (c+d x))^3 (A+C \sec ^2(c+d x))}{\sec ^{\frac{5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=253 \[ \frac{4 a^3 (3 A+5 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 d}-\frac{8 a^3 (3 A-10 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{15 d}-\frac{2 (9 A-5 C) \sin (c+d x) \sqrt{\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{15 d}+\frac{4 a^3 (9 A-5 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{4 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{5 a d \sqrt{\sec (c+d x)}}+\frac{2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{5 d \sec ^{\frac{3}{2}}(c+d x)} \]

[Out]

(4*a^3*(9*A - 5*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (4*a^3*(3*A + 5*C)
*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) - (8*a^3*(3*A - 10*C)*Sqrt[Sec[c + d*x
]]*Sin[c + d*x])/(15*d) + (2*A*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + (4*A*(a^2 + a^2
*Sec[c + d*x])^2*Sin[c + d*x])/(5*a*d*Sqrt[Sec[c + d*x]]) - (2*(9*A - 5*C)*Sqrt[Sec[c + d*x]]*(a^3 + a^3*Sec[c
+ d*x])*Sin[c + d*x])/(15*d)

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Rubi [A]  time = 0.569539, antiderivative size = 253, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.229, Rules used = {4087, 4017, 4018, 3997, 3787, 3771, 2639, 2641} \[ -\frac{8 a^3 (3 A-10 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{15 d}-\frac{2 (9 A-5 C) \sin (c+d x) \sqrt{\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{15 d}+\frac{4 a^3 (3 A+5 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{4 a^3 (9 A-5 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{4 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{5 a d \sqrt{\sec (c+d x)}}+\frac{2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{5 d \sec ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + a*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2))/Sec[c + d*x]^(5/2),x]

[Out]

(4*a^3*(9*A - 5*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (4*a^3*(3*A + 5*C)
*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) - (8*a^3*(3*A - 10*C)*Sqrt[Sec[c + d*x
]]*Sin[c + d*x])/(15*d) + (2*A*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + (4*A*(a^2 + a^2
*Sec[c + d*x])^2*Sin[c + d*x])/(5*a*d*Sqrt[Sec[c + d*x]]) - (2*(9*A - 5*C)*Sqrt[Sec[c + d*x]]*(a^3 + a^3*Sec[c
+ d*x])*Sin[c + d*x])/(15*d)

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 4018

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> -Simp[(b*B*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^n)/(f*(m + n
)), x] + Dist[1/(d*(m + n)), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^n*Simp[a*A*d*(m + n) + B*(b*d*n
) + (A*b*d*(m + n) + a*B*d*(2*m + n - 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && Ne
Q[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))^3 \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))^3 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 \int \frac{(a+a \sec (c+d x))^3 \left (3 a A-\frac{1}{2} a (3 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))^3 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{4 A \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{5 a d \sqrt{\sec (c+d x)}}+\frac{4 \int \frac{(a+a \sec (c+d x))^2 \left (\frac{3}{4} a^2 (11 A+5 C)-\frac{3}{4} a^2 (9 A-5 C) \sec (c+d x)\right )}{\sqrt{\sec (c+d x)}} \, dx}{15 a}\\ &=\frac{2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{4 A \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{5 a d \sqrt{\sec (c+d x)}}-\frac{2 (9 A-5 C) \sqrt{\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d}+\frac{8 \int \frac{(a+a \sec (c+d x)) \left (\frac{3}{4} a^3 (21 A+5 C)-\frac{3}{2} a^3 (3 A-10 C) \sec (c+d x)\right )}{\sqrt{\sec (c+d x)}} \, dx}{45 a}\\ &=-\frac{8 a^3 (3 A-10 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{4 A \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{5 a d \sqrt{\sec (c+d x)}}-\frac{2 (9 A-5 C) \sqrt{\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d}+\frac{16 \int \frac{\frac{9}{8} a^4 (9 A-5 C)+\frac{15}{8} a^4 (3 A+5 C) \sec (c+d x)}{\sqrt{\sec (c+d x)}} \, dx}{45 a}\\ &=-\frac{8 a^3 (3 A-10 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{4 A \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{5 a d \sqrt{\sec (c+d x)}}-\frac{2 (9 A-5 C) \sqrt{\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d}+\frac{1}{5} \left (2 a^3 (9 A-5 C)\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx+\frac{1}{3} \left (2 a^3 (3 A+5 C)\right ) \int \sqrt{\sec (c+d x)} \, dx\\ &=-\frac{8 a^3 (3 A-10 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{4 A \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{5 a d \sqrt{\sec (c+d x)}}-\frac{2 (9 A-5 C) \sqrt{\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d}+\frac{1}{5} \left (2 a^3 (9 A-5 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx+\frac{1}{3} \left (2 a^3 (3 A+5 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{4 a^3 (9 A-5 C) \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^3 (3 A+5 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{8 a^3 (3 A-10 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{4 A \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{5 a d \sqrt{\sec (c+d x)}}-\frac{2 (9 A-5 C) \sqrt{\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d}\\ \end{align*}

Mathematica [C]  time = 2.27082, size = 221, normalized size = 0.87 \[ \frac{a^3 e^{-i d x} \sec ^{\frac{3}{2}}(c+d x) (\cos (d x)+i \sin (d x)) \left (-8 i (9 A-5 C) \left (1+e^{2 i (c+d x)}\right )^{3/2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )+80 (3 A+5 C) \cos ^{\frac{3}{2}}(c+d x) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )+30 A \sin (c+d x)+6 A \sin (2 (c+d x))+30 A \sin (3 (c+d x))+3 A \sin (4 (c+d x))+216 i A \cos (2 (c+d x))+216 i A+40 C \sin (c+d x)+180 C \sin (2 (c+d x))-120 i C \cos (2 (c+d x))-120 i C\right )}{60 d} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + a*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2))/Sec[c + d*x]^(5/2),x]

[Out]

(a^3*Sec[c + d*x]^(3/2)*(Cos[d*x] + I*Sin[d*x])*((216*I)*A - (120*I)*C + (216*I)*A*Cos[2*(c + d*x)] - (120*I)*
C*Cos[2*(c + d*x)] + 80*(3*A + 5*C)*Cos[c + d*x]^(3/2)*EllipticF[(c + d*x)/2, 2] - (8*I)*(9*A - 5*C)*(1 + E^((
2*I)*(c + d*x)))^(3/2)*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))] + 30*A*Sin[c + d*x] + 40*C*Sin[c
+ d*x] + 6*A*Sin[2*(c + d*x)] + 180*C*Sin[2*(c + d*x)] + 30*A*Sin[3*(c + d*x)] + 3*A*Sin[4*(c + d*x)]))/(60*d
*E^(I*d*x))

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Maple [B]  time = 2.469, size = 704, normalized size = 2.8 \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))^3*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2),x)

[Out]

-4/15*(24*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)^8-96*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+6*(-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)*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-2*(-2*sin(1/2*d*x+1/
2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(9*A+25*C)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-2*(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)*(15*A*EllipticF(c
os(1/2*d*x+1/2*c),2^(1/2))-27*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+25*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2
))+15*C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*sin(1/2*d*x+1/2*c)^2+15*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(
1/2))-27*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))+25*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)+15*C*(-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)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2
)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*a^3/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d
*x+1/2*c)^2-1)^(3/2)/sin(1/2*d*x+1/2*c)/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))^3*(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^{3} \sec \left (d x + c\right )^{5} + 3 \, C a^{3} \sec \left (d x + c\right )^{4} +{\left (A + 3 \, C\right )} a^{3} \sec \left (d x + c\right )^{3} +{\left (3 \, A + C\right )} a^{3} \sec \left (d x + c\right )^{2} + 3 \, A a^{3} \sec \left (d x + c\right ) + A a^{3}}{\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))^3*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

integral((C*a^3*sec(d*x + c)^5 + 3*C*a^3*sec(d*x + c)^4 + (A + 3*C)*a^3*sec(d*x + c)^3 + (3*A + C)*a^3*sec(d*x
+ c)^2 + 3*A*a^3*sec(d*x + c) + A*a^3)/sec(d*x + c)^(5/2), 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+a*sec(d*x+c))**3*(A+C*sec(d*x+c)**2)/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{{\left (C \sec \left (d x + c\right )^{2} + A\right )}{\left (a \sec \left (d x + c\right ) + a\right )}^{3}}{\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))^3*(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)^3/sec(d*x + c)^(5/2), x)