\(\int \sqrt {\sec (c+d x)} (a+b \sec (c+d x)) (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [983]

Optimal result

Integrand size = 41, antiderivative size = 192 \[ \int \sqrt {\sec (c+d x)} (a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {2 (5 A b+5 a B+3 b C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 (b B+a (3 A+C)) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 (5 A b+5 a B+3 b C) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 (b B+a C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {2 b C \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d} \]

[Out]

2/3*(B*b+C*a)*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/5*b*C*sec(d*x+c)^(5/2)*sin(d*x+c)/d+2/5*(5*A*b+5*B*a+3*C*b)*sin(
d*x+c)*sec(d*x+c)^(1/2)/d-2/5*(5*A*b+5*B*a+3*C*b)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(si
n(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*(B*b+a*(3*A+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 = 0.27 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {4161, 4132, 3853, 3856, 2719, 4131, 2720} \[ \int \sqrt {\sec (c+d x)} (a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)} (5 a B+5 A b+3 b C)}{5 d}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) (a (3 A+C)+b B)}{3 d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) (5 a B+5 A b+3 b C)}{5 d}+\frac {2 (a C+b B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 b C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d} \]

[In]

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

[Out]

(-2*(5*A*b + 5*a*B + 3*b*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (2*(b*B +
a*(3*A + C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (2*(5*A*b + 5*a*B + 3*b
*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(5*d) + (2*(b*B + a*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*d) + (2*b*C*Se
c[c + d*x]^(5/2)*Sin[c + d*x])/(5*d)

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 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

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]

Rubi steps \begin{align*} \text {integral}& = \frac {2 b C \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2}{5} \int \sqrt {\sec (c+d x)} \left (\frac {5 a A}{2}+\frac {1}{2} (5 A b+5 a B+3 b C) \sec (c+d x)+\frac {5}{2} (b B+a C) \sec ^2(c+d x)\right ) \, dx \\ & = \frac {2 b C \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2}{5} \int \sqrt {\sec (c+d x)} \left (\frac {5 a A}{2}+\frac {5}{2} (b B+a C) \sec ^2(c+d x)\right ) \, dx+\frac {1}{5} (5 A b+5 a B+3 b C) \int \sec ^{\frac {3}{2}}(c+d x) \, dx \\ & = \frac {2 (5 A b+5 a B+3 b C) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 (b B+a C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {2 b C \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {1}{5} (-5 A b-5 a B-3 b C) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{3} (b B+a (3 A+C)) \int \sqrt {\sec (c+d x)} \, dx \\ & = \frac {2 (5 A b+5 a B+3 b C) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 (b B+a C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {2 b C \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {1}{5} \left ((-5 A b-5 a B-3 b C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{3} \left ((b B+a (3 A+C)) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx \\ & = -\frac {2 (5 A b+5 a B+3 b 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 {2 (b B+a (3 A+C)) \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 (5 A b+5 a B+3 b C) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 (b B+a C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {2 b C \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 8.65 (sec) , antiderivative size = 1140, normalized size of antiderivative = 5.94 \[ \int \sqrt {\sec (c+d x)} (a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \sqrt {2} A b e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \cos ^3(c+d x) \csc (c) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right ) (a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{3 d (b+a \cos (c+d x)) (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x))}+\frac {2 \sqrt {2} a B e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \cos ^3(c+d x) \csc (c) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right ) (a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{3 d (b+a \cos (c+d x)) (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x))}+\frac {2 \sqrt {2} b C e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \cos ^3(c+d x) \csc (c) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right ) (a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{5 d (b+a \cos (c+d x)) (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x))}+\frac {4 a A \cos ^{\frac {7}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)} (a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{d (b+a \cos (c+d x)) (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x))}+\frac {4 b B \cos ^{\frac {7}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)} (a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{3 d (b+a \cos (c+d x)) (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x))}+\frac {4 a C \cos ^{\frac {7}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)} (a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{3 d (b+a \cos (c+d x)) (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x))}+\frac {(a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac {4 (5 A b+5 a B+3 b C) \cos (d x) \csc (c)}{5 d}+\frac {4 b C \sec (c) \sec ^2(c+d x) \sin (d x)}{5 d}+\frac {4 \sec (c) \sec (c+d x) (3 b C \sin (c)+5 b B \sin (d x)+5 a C \sin (d x))}{15 d}+\frac {4 (b B+a C) \tan (c)}{3 d}\right )}{(b+a \cos (c+d x)) (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) \sec ^{\frac {5}{2}}(c+d x)} \]

[In]

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

[Out]

(2*Sqrt[2]*A*b*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c + d*x]^3*Cs
c[c]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^
((2*I)*(c + d*x))])*(a + b*Sec[c + d*x])*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(3*d*E^(I*d*x)*(b + a*Cos[c
+ d*x])*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + (2*Sqrt[2]*a*B*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)
*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c + d*x]^3*Csc[c]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)
*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])*(a + b*Sec[c + d*x])*(A + B*S
ec[c + d*x] + C*Sec[c + d*x]^2))/(3*d*E^(I*d*x)*(b + a*Cos[c + d*x])*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c +
2*d*x])) + (2*Sqrt[2]*b*C*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c
+ d*x]^3*Csc[c]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3
/4, 7/4, -E^((2*I)*(c + d*x))])*(a + b*Sec[c + d*x])*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(5*d*E^(I*d*x)*(
b + a*Cos[c + d*x])*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + (4*a*A*Cos[c + d*x]^(7/2)*EllipticF[(
c + d*x)/2, 2]*Sqrt[Sec[c + d*x]]*(a + b*Sec[c + d*x])*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(d*(b + a*Cos[
c + d*x])*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + (4*b*B*Cos[c + d*x]^(7/2)*EllipticF[(c + d*x)/2
, 2]*Sqrt[Sec[c + d*x]]*(a + b*Sec[c + d*x])*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(3*d*(b + a*Cos[c + d*x]
)*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + (4*a*C*Cos[c + d*x]^(7/2)*EllipticF[(c + d*x)/2, 2]*Sqr
t[Sec[c + d*x]]*(a + b*Sec[c + d*x])*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(3*d*(b + a*Cos[c + d*x])*(A + 2
*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + ((a + b*Sec[c + d*x])*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(
(4*(5*A*b + 5*a*B + 3*b*C)*Cos[d*x]*Csc[c])/(5*d) + (4*b*C*Sec[c]*Sec[c + d*x]^2*Sin[d*x])/(5*d) + (4*Sec[c]*S
ec[c + d*x]*(3*b*C*Sin[c] + 5*b*B*Sin[d*x] + 5*a*C*Sin[d*x]))/(15*d) + (4*(b*B + a*C)*Tan[c])/(3*d)))/((b + a*
Cos[c + d*x])*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sec[c + d*x]^(5/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(714\) vs. \(2(220)=440\).

Time = 4.74 (sec) , antiderivative size = 715, normalized size of antiderivative = 3.72

method	result	size
default	\(\text {Expression too large to display}\)	\(715\)
parts	\(\text {Expression too large to display}\)	\(902\)

[In]

int(sec(d*x+c)^(1/2)*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*a*A*(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/(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)^2*(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)*sin(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*(B*b+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)*Ellip
ticF(cos(1/2*d*x+1/2*c),2^(1/2)))+2*(A*b+B*a)/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)*(-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)^2*cos(1/2*d*x+1/2*c)-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*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 = 0.12 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.35 \[ \int \sqrt {\sec (c+d x)} (a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {5 \, \sqrt {2} {\left (i \, {\left (3 \, A + C\right )} a + i \, B b\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-i \, {\left (3 \, A + C\right )} a - i \, B b\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, \sqrt {2} {\left (5 i \, B a + i \, {\left (5 \, A + 3 \, C\right )} b\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {2} {\left (-5 i \, B a - i \, {\left (5 \, A + 3 \, C\right )} b\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (3 \, {\left (5 \, B a + {\left (5 \, A + 3 \, C\right )} b\right )} \cos \left (d x + c\right )^{2} + 3 \, C b + 5 \, {\left (C a + B b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{15 \, d \cos \left (d x + c\right )^{2}} \]

[In]

integrate(sec(d*x+c)^(1/2)*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

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

[In]

integrate(sec(d*x+c)**(1/2)*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

Maxima [F]

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

[In]

integrate(sec(d*x+c)^(1/2)*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate(sec(d*x+c)^(1/2)*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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