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

Optimal. Leaf size=319 \[ \frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \left (a^3 (A+3 C)+9 a^2 b B+3 a b^2 (3 A+C)+b^3 B\right )}{3 d}+\frac{2 b \sin (c+d x) \sqrt{\sec (c+d x)} \left (a^2 (-(10 A-42 C))+45 a b B+3 b^2 (5 A+3 C)\right )}{15 d}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (15 a^2 b (A-C)+5 a^3 B-15 a b^2 B-b^3 (5 A+3 C)\right )}{5 d}-\frac{2 b^2 \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) (5 a A-9 a C-5 b B)}{15 d}-\frac{2 b (5 A-3 C) \sin (c+d x) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^2}{15 d}+\frac{2 A \sin (c+d x) (a+b \sec (c+d x))^3}{3 d \sqrt{\sec (c+d x)}} \]

[Out]

(2*(5*a^3*B - 15*a*b^2*B + 15*a^2*b*(A - C) - b^3*(5*A + 3*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sq
rt[Sec[c + d*x]])/(5*d) + (2*(9*a^2*b*B + b^3*B + 3*a*b^2*(3*A + C) + a^3*(A + 3*C))*Sqrt[Cos[c + d*x]]*Ellipt
icF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (2*b*(45*a*b*B - a^2*(10*A - 42*C) + 3*b^2*(5*A + 3*C))*Sqrt[S
ec[c + d*x]]*Sin[c + d*x])/(15*d) - (2*b^2*(5*a*A - 5*b*B - 9*a*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(15*d) - (
2*b*(5*A - 3*C)*Sqrt[Sec[c + d*x]]*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(15*d) + (2*A*(a + b*Sec[c + d*x])^3*S
in[c + d*x])/(3*d*Sqrt[Sec[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.829917, antiderivative size = 319, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.186, Rules used = {4094, 4096, 4076, 4047, 3771, 2641, 4046, 2639} \[ \frac{2 b \sin (c+d x) \sqrt{\sec (c+d x)} \left (a^2 (-(10 A-42 C))+45 a b B+3 b^2 (5 A+3 C)\right )}{15 d}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^3 (A+3 C)+9 a^2 b B+3 a b^2 (3 A+C)+b^3 B\right )}{3 d}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (15 a^2 b (A-C)+5 a^3 B-15 a b^2 B-b^3 (5 A+3 C)\right )}{5 d}-\frac{2 b^2 \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) (5 a A-9 a C-5 b B)}{15 d}-\frac{2 b (5 A-3 C) \sin (c+d x) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^2}{15 d}+\frac{2 A \sin (c+d x) (a+b \sec (c+d x))^3}{3 d \sqrt{\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(5*a^3*B - 15*a*b^2*B + 15*a^2*b*(A - C) - b^3*(5*A + 3*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sq
rt[Sec[c + d*x]])/(5*d) + (2*(9*a^2*b*B + b^3*B + 3*a*b^2*(3*A + C) + a^3*(A + 3*C))*Sqrt[Cos[c + d*x]]*Ellipt
icF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (2*b*(45*a*b*B - a^2*(10*A - 42*C) + 3*b^2*(5*A + 3*C))*Sqrt[S
ec[c + d*x]]*Sin[c + d*x])/(15*d) - (2*b^2*(5*a*A - 5*b*B - 9*a*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(15*d) - (
2*b*(5*A - 3*C)*Sqrt[Sec[c + d*x]]*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(15*d) + (2*A*(a + b*Sec[c + d*x])^3*S
in[c + d*x])/(3*d*Sqrt[Sec[c + d*x]])

Rule 4094

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + 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] - Dist[1/(d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[A*b*
m - a*B*n - (b*B*n + a*(C*n + A*(n + 1)))*Csc[e + f*x] - b*(C*n + A*(m + n + 1))*Csc[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && LeQ[n, -1]

Rule 4096

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

Rule 4076

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

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 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 4046

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> -Simp[(C*Cot[
e + f*x]*(b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x]
/; FreeQ[{b, e, f, A, C, m}, 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))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac{3}{2}}(c+d x)} \, dx &=\frac{2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{2}{3} \int \frac{(a+b \sec (c+d x))^2 \left (\frac{3}{2} (2 A b+a B)+\frac{1}{2} (3 b B+a (A+3 C)) \sec (c+d x)-\frac{1}{2} b (5 A-3 C) \sec ^2(c+d x)\right )}{\sqrt{\sec (c+d x)}} \, dx\\ &=-\frac{2 b (5 A-3 C) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{15 d}+\frac{2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{4}{15} \int \frac{(a+b \sec (c+d x)) \left (\frac{1}{4} a (35 A b+15 a B-3 b C)+\frac{1}{4} \left (30 a b B+5 a^2 (A+3 C)+3 b^2 (5 A+3 C)\right ) \sec (c+d x)-\frac{3}{4} b (5 a A-5 b B-9 a C) \sec ^2(c+d x)\right )}{\sqrt{\sec (c+d x)}} \, dx\\ &=-\frac{2 b^2 (5 a A-5 b B-9 a C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 d}-\frac{2 b (5 A-3 C) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{15 d}+\frac{2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{8}{45} \int \frac{\frac{3}{8} a^2 (35 A b+15 a B-3 b C)+\frac{15}{8} \left (9 a^2 b B+b^3 B+3 a b^2 (3 A+C)+a^3 (A+3 C)\right ) \sec (c+d x)+\frac{3}{8} b \left (45 a b B-a^2 (10 A-42 C)+3 b^2 (5 A+3 C)\right ) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)}} \, dx\\ &=-\frac{2 b^2 (5 a A-5 b B-9 a C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 d}-\frac{2 b (5 A-3 C) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{15 d}+\frac{2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{8}{45} \int \frac{\frac{3}{8} a^2 (35 A b+15 a B-3 b C)+\frac{3}{8} b \left (45 a b B-a^2 (10 A-42 C)+3 b^2 (5 A+3 C)\right ) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)}} \, dx+\frac{1}{3} \left (9 a^2 b B+b^3 B+3 a b^2 (3 A+C)+a^3 (A+3 C)\right ) \int \sqrt{\sec (c+d x)} \, dx\\ &=\frac{2 b \left (45 a b B-a^2 (10 A-42 C)+3 b^2 (5 A+3 C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}-\frac{2 b^2 (5 a A-5 b B-9 a C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 d}-\frac{2 b (5 A-3 C) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{15 d}+\frac{2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{1}{5} \left (5 a^3 B-15 a b^2 B+15 a^2 b (A-C)-b^3 (5 A+3 C)\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx+\frac{1}{3} \left (\left (9 a^2 b B+b^3 B+3 a b^2 (3 A+C)+a^3 (A+3 C)\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 \left (9 a^2 b B+b^3 B+3 a b^2 (3 A+C)+a^3 (A+3 C)\right ) \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 b \left (45 a b B-a^2 (10 A-42 C)+3 b^2 (5 A+3 C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}-\frac{2 b^2 (5 a A-5 b B-9 a C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 d}-\frac{2 b (5 A-3 C) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{15 d}+\frac{2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{1}{5} \left (\left (5 a^3 B-15 a b^2 B+15 a^2 b (A-C)-b^3 (5 A+3 C)\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=\frac{2 \left (5 a^3 B-15 a b^2 B+15 a^2 b (A-C)-b^3 (5 A+3 C)\right ) \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{2 \left (9 a^2 b B+b^3 B+3 a b^2 (3 A+C)+a^3 (A+3 C)\right ) \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 b \left (45 a b B-a^2 (10 A-42 C)+3 b^2 (5 A+3 C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}-\frac{2 b^2 (5 a A-5 b B-9 a C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 d}-\frac{2 b (5 A-3 C) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{15 d}+\frac{2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 3.5338, size = 311, normalized size = 0.97 \[ \frac{2 (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (10 \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \left (a^3 (A+3 C)+9 a^2 b B+3 a b^2 (3 A+C)+b^3 B\right )+6 \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (15 a^2 b (A-C)+5 a^3 B-15 a b^2 B-b^3 (5 A+3 C)\right )+5 a^3 A \sin (2 (c+d x))+90 a^2 b C \sin (c+d x)+90 a b^2 B \sin (c+d x)+30 a b^2 C \tan (c+d x)+30 A b^3 \sin (c+d x)+10 b^3 B \tan (c+d x)+18 b^3 C \sin (c+d x)+6 b^3 C \tan (c+d x) \sec (c+d x)\right )}{15 d \sec ^{\frac{9}{2}}(c+d x) (a \cos (c+d x)+b)^3 (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(6*(5*a^3*B - 15*a*b^2*B + 15*a^2*b*(A - C)
- b^3*(5*A + 3*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 10*(9*a^2*b*B + b^3*B + 3*a*b^2*(3*A + C) +
a^3*(A + 3*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + 30*A*b^3*Sin[c + d*x] + 90*a*b^2*B*Sin[c + d*x]
+ 90*a^2*b*C*Sin[c + d*x] + 18*b^3*C*Sin[c + d*x] + 5*a^3*A*Sin[2*(c + d*x)] + 10*b^3*B*Tan[c + d*x] + 30*a*b^
2*C*Tan[c + d*x] + 6*b^3*C*Sec[c + d*x]*Tan[c + d*x]))/(15*d*(b + a*Cos[c + d*x])^3*(A + 2*C + 2*B*Cos[c + d*x
] + A*Cos[2*(c + d*x)])*Sec[c + d*x]^(9/2))

________________________________________________________________________________________

Maple [B]  time = 8.263, size = 1419, normalized size = 4.5 \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))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2),x)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4/3*A*a^3*(2*sin(1/2*d*x+1/2*c)^4*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*(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))*(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*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+(-4*A*a^3+6*A*a^2*b+2*B*a^3)*
(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(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))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+2*A*a^3*(sin(1/2*d*x+1/2*c
)^2)^(1/2)*(-2*cos(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(co
s(1/2*d*x+1/2*c),2^(1/2))-6*A*a^2*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(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))+6*A*a*b^2*(sin(1/2*d*x+1/2*c)^
2)^(1/2)*(-2*cos(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))-2*B*a^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(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))+6*B*a^2*b*(sin(1/2*d*x+1/2*c)^2)^(
1/2)*(-2*cos(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))+2*a^3*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(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))-2/5*C*b^3/(8*sin(1/2*d*x+1/2*c)^6-12*s
in(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1/2*d*x+1/2*c)^2*(12*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ellipt
icE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^4-24*sin(1/2*d*x+1/2*c)^6*cos(
1/2*d*x+1/2*c)-12*(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))*(sin(1/2*d*x+1/2*c)^2
)^(1/2)*sin(1/2*d*x+1/2*c)^2+24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+3*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ell
ipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)-8*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)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+2*b^2*(B*b+3*C*a)*(-1/6*cos(1/2*d*x+1/2*c)*(-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)^2-1/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(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
)))+2*b*(A*b^2+3*B*a*b+3*C*a^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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+2*(-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)^2)/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)
)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

________________________________________________________________________________________

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+b*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^3/sec(d*x + c)^(3/2), x)