\(\int \frac {(a+b \sec (c+d x))^3 (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\) [999]

Optimal result

Integrand size = 43, antiderivative size = 319 \[ \int \frac {(a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2 \left (5 a^3 B-15 a b^2 B+15 a^2 b (A-C)-b^3 (5 A+3 C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 \left (9 a^2 b B+b^3 B+3 a b^2 (3 A+C)+a^3 (A+3 C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 b \left (45 a b B-a^2 (10 A-42 C)+3 b^2 (5 A+3 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}-\frac {2 b^2 (5 a A-5 b B-9 a C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}-\frac {2 b (5 A-3 C) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{15 d}+\frac {2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}} \]

[Out]

-2/15*b^2*(5*A*a-5*B*b-9*C*a)*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/3*A*(a+b*sec(d*x+c))^3*sin(d*x+c)/d/sec(d*x+c)^(
1/2)+2/15*b*(45*B*a*b-a^2*(10*A-42*C)+3*b^2*(5*A+3*C))*sin(d*x+c)*sec(d*x+c)^(1/2)/d-2/15*b*(5*A-3*C)*(a+b*sec
(d*x+c))^2*sin(d*x+c)*sec(d*x+c)^(1/2)/d+2/5*(5*B*a^3-15*B*a*b^2+15*a^2*b*(A-C)-b^3*(5*A+3*C))*(cos(1/2*d*x+1/
2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+2/3
*(9*B*a^2*b+B*b^3+3*a*b^2*(3*A+C)+a^3*(A+3*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1
/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d

Rubi [A] (verified)

Time = 0.89 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.186, Rules used = {4179, 4181, 4161, 4132, 3856, 2720, 4131, 2719} \[ \int \frac {(a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2 b \sin (c+d x) \sqrt {\sec (c+d x)} \left (-\left (a^2 (10 A-42 C)\right )+45 a b B+3 b^2 (5 A+3 C)\right )}{15 d}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (a^3 (A+3 C)+9 a^2 b B+3 a b^2 (3 A+C)+b^3 B\right )}{3 d}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^3 B+15 a^2 b (A-C)-15 a b^2 B-b^3 (5 A+3 C)\right )}{5 d}-\frac {2 b^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (5 a A-9 a C-5 b B)}{15 d}-\frac {2 b (5 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2}{15 d}+\frac {2 A \sin (c+d x) (a+b \sec (c+d x))^3}{3 d \sqrt {\sec (c+d x)}} \]

[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 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 3856

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 4131

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 4132

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 4161

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 4179

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 4181

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]

Rubi steps \begin{align*} \text {integral}& = \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)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),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)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),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] (verified)

Time = 12.47 (sec) , antiderivative size = 311, normalized size of antiderivative = 0.97 \[ \int \frac {(a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2 (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (6 \left (5 a^3 B-15 a b^2 B+15 a^2 b (A-C)-b^3 (5 A+3 C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+10 \left (9 a^2 b B+b^3 B+3 a b^2 (3 A+C)+a^3 (A+3 C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+30 A b^3 \sin (c+d x)+90 a b^2 B \sin (c+d x)+90 a^2 b C \sin (c+d x)+18 b^3 C \sin (c+d x)+5 a^3 A \sin (2 (c+d x))+10 b^3 B \tan (c+d x)+30 a b^2 C \tan (c+d x)+6 b^3 C \sec (c+d x) \tan (c+d x)\right )}{15 d (b+a \cos (c+d x))^3 (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x))) \sec ^{\frac {9}{2}}(c+d x)} \]

[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] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1267\) vs. \(2(343)=686\).

Time = 5.61 (sec) , antiderivative size = 1268, normalized size of antiderivative = 3.97

method	result	size
parts	\(\text {Expression too large to display}\)	\(1268\)
default	\(\text {Expression too large to display}\)	\(1392\)

[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,method=_RETURNVERBOSE)

[Out]

2*(3*A*a^2*b+B*a^3)*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*c
os(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE(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)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d-2/3*(B*b^3+3*C*a*b^2)*(-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))*sin(1/2*d*x+1/2*c)^2-2*sin(1/2
*d*x+1/2*c)^2*cos(1/2*d*x+1/2*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*cos(1/2*d*x+1/2*c)^2-1)*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)/(2*cos(1/2*d*x+1/2*c)^2-1)^(3/2)/sin(1/2*d*x+1/2*c)/d-2*(A*b^3+3*B*a*b^2+3*C*a^2*b)*(-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)^(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(c
os(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-2*(3*A*a*b^2+3*B*a^2*b+C*a^3)*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)
*(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))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d-2/3*A*a^3*(
(2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4*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)^2*cos(1/2*d*x+1/2*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)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/
2*c)^2-1)^(1/2)/d-2/5*C*b^3*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(8*sin(1/2*d*x+1/2*c)^6-
12*sin(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)^3*(24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/
2*c)-12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*si
n(1/2*d*x+1/2*c)^4-24*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(
1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2+8*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x
+1/2*c)-3*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(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)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.14 (sec) , antiderivative size = 382, normalized size of antiderivative = 1.20 \[ \int \frac {(a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {5 \, \sqrt {2} {\left (i \, {\left (A + 3 \, C\right )} a^{3} + 9 i \, B a^{2} b + 3 i \, {\left (3 \, A + C\right )} a b^{2} + i \, B b^{3}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-i \, {\left (A + 3 \, C\right )} a^{3} - 9 i \, B a^{2} b - 3 i \, {\left (3 \, A + C\right )} a b^{2} - i \, B b^{3}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, \sqrt {2} {\left (-5 i \, B a^{3} - 15 i \, {\left (A - C\right )} a^{2} b + 15 i \, B a b^{2} + i \, {\left (5 \, A + 3 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {2} {\left (5 i \, B a^{3} + 15 i \, {\left (A - C\right )} a^{2} b - 15 i \, B a b^{2} - i \, {\left (5 \, A + 3 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (5 \, A a^{3} \cos \left (d x + c\right )^{3} + 3 \, C b^{3} + 3 \, {\left (15 \, C a^{2} b + 15 \, B a b^{2} + {\left (5 \, A + 3 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{2} + 5 \, {\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{15 \, d \cos \left (d x + c\right )^{2}} \]

[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]

-1/15*(5*sqrt(2)*(I*(A + 3*C)*a^3 + 9*I*B*a^2*b + 3*I*(3*A + C)*a*b^2 + I*B*b^3)*cos(d*x + c)^2*weierstrassPIn
verse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*sqrt(2)*(-I*(A + 3*C)*a^3 - 9*I*B*a^2*b - 3*I*(3*A + C)*a*b^2
- I*B*b^3)*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*sqrt(2)*(-5*I*B*a^3 -
15*I*(A - C)*a^2*b + 15*I*B*a*b^2 + I*(5*A + 3*C)*b^3)*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInver
se(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*sqrt(2)*(5*I*B*a^3 + 15*I*(A - C)*a^2*b - 15*I*B*a*b^2 - I*(5*A
+ 3*C)*b^3)*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) -
2*(5*A*a^3*cos(d*x + c)^3 + 3*C*b^3 + 3*(15*C*a^2*b + 15*B*a*b^2 + (5*A + 3*C)*b^3)*cos(d*x + c)^2 + 5*(3*C*a
*b^2 + B*b^3)*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^2)

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \]

[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

Maxima [F(-1)]

Timed out. \[ \int \frac {(a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \]

[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

Giac [F]

\[ \int \frac {(a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{3}}{\sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^3\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

[In]

int(((a + b/cos(c + d*x))^3*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(1/cos(c + d*x))^(3/2),x)

[Out]

int(((a + b/cos(c + d*x))^3*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(1/cos(c + d*x))^(3/2), x)