\(\int \frac {(a+b \sec (c+d x))^3 (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sqrt {\sec (c+d x)}} \, dx\) [998]

Optimal result

Integrand size = 43, antiderivative size = 334 \[ \int \frac {(a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=-\frac {2 \left (15 a^2 b B+3 b^3 B-5 a^3 (A-C)+3 a b^2 (5 A+3 C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 \left (21 a^3 B+21 a b^2 B+21 a^2 b (3 A+C)+b^3 (7 A+5 C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {2 \left (98 a^2 b B+21 b^3 B+24 a^3 C+21 a b^2 (5 A+3 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{35 d}+\frac {2 b \left (35 A b^2+63 a b B+24 a^2 C+25 b^2 C\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac {2 (7 b B+6 a C) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{35 d}+\frac {2 C \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^3 \sin (c+d x)}{7 d} \]

[Out]

2/105*b*(35*A*b^2+63*B*a*b+24*C*a^2+25*C*b^2)*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/35*(98*B*a^2*b+21*B*b^3+24*a^3*C
+21*a*b^2*(5*A+3*C))*sin(d*x+c)*sec(d*x+c)^(1/2)/d+2/35*(7*B*b+6*C*a)*(a+b*sec(d*x+c))^2*sin(d*x+c)*sec(d*x+c)
^(1/2)/d+2/7*C*(a+b*sec(d*x+c))^3*sin(d*x+c)*sec(d*x+c)^(1/2)/d-2/5*(15*B*a^2*b+3*B*b^3-5*a^3*(A-C)+3*a*b^2*(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+2/21*(21*B*a^3+21*B*a*b^2+21*a^2*b*(3*A+C)+b^3*(7*A+5*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/co
s(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 = 0.87 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {4181, 4161, 4132, 3856, 2720, 4131, 2719} \[ \int \frac {(a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\frac {2 b \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (24 a^2 C+63 a b B+35 A b^2+25 b^2 C\right )}{105 d}+\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)} \left (24 a^3 C+98 a^2 b B+21 a b^2 (5 A+3 C)+21 b^3 B\right )}{35 d}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (21 a^3 B+21 a^2 b (3 A+C)+21 a b^2 B+b^3 (7 A+5 C)\right )}{21 d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-5 a^3 (A-C)+15 a^2 b B+3 a b^2 (5 A+3 C)+3 b^3 B\right )}{5 d}+\frac {2 (6 a C+7 b B) \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2}{35 d}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^3}{7 d} \]

[In]

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

[Out]

(-2*(15*a^2*b*B + 3*b^3*B - 5*a^3*(A - C) + 3*a*b^2*(5*A + 3*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*
Sqrt[Sec[c + d*x]])/(5*d) + (2*(21*a^3*B + 21*a*b^2*B + 21*a^2*b*(3*A + C) + b^3*(7*A + 5*C))*Sqrt[Cos[c + d*x
]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (2*(98*a^2*b*B + 21*b^3*B + 24*a^3*C + 21*a*b^2*(5*A
+ 3*C))*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(35*d) + (2*b*(35*A*b^2 + 63*a*b*B + 24*a^2*C + 25*b^2*C)*Sec[c + d*
x]^(3/2)*Sin[c + d*x])/(105*d) + (2*(7*b*B + 6*a*C)*Sqrt[Sec[c + d*x]]*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(3
5*d) + (2*C*Sqrt[Sec[c + d*x]]*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/(7*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 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 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 C \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^3 \sin (c+d x)}{7 d}+\frac {2}{7} \int \frac {(a+b \sec (c+d x))^2 \left (\frac {1}{2} a (7 A-C)+\frac {1}{2} (7 A b+7 a B+5 b C) \sec (c+d x)+\frac {1}{2} (7 b B+6 a C) \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx \\ & = \frac {2 (7 b B+6 a C) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{35 d}+\frac {2 C \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^3 \sin (c+d x)}{7 d}+\frac {4}{35} \int \frac {(a+b \sec (c+d x)) \left (\frac {1}{4} a (35 a A-7 b B-11 a C)+\frac {1}{4} \left (70 a A b+35 a^2 B+21 b^2 B+38 a b C\right ) \sec (c+d x)+\frac {1}{4} \left (35 A b^2+63 a b B+24 a^2 C+25 b^2 C\right ) \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx \\ & = \frac {2 b \left (35 A b^2+63 a b B+24 a^2 C+25 b^2 C\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac {2 (7 b B+6 a C) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{35 d}+\frac {2 C \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^3 \sin (c+d x)}{7 d}+\frac {8}{105} \int \frac {\frac {3}{8} a^2 (35 a A-7 b B-11 a C)+\frac {5}{8} \left (21 a^3 B+21 a b^2 B+21 a^2 b (3 A+C)+b^3 (7 A+5 C)\right ) \sec (c+d x)+\frac {3}{8} \left (98 a^2 b B+21 b^3 B+24 a^3 C+21 a b^2 (5 A+3 C)\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)}} \, dx \\ & = \frac {2 b \left (35 A b^2+63 a b B+24 a^2 C+25 b^2 C\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac {2 (7 b B+6 a C) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{35 d}+\frac {2 C \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^3 \sin (c+d x)}{7 d}+\frac {8}{105} \int \frac {\frac {3}{8} a^2 (35 a A-7 b B-11 a C)+\frac {3}{8} \left (98 a^2 b B+21 b^3 B+24 a^3 C+21 a b^2 (5 A+3 C)\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{21} \left (21 a^3 B+21 a b^2 B+21 a^2 b (3 A+C)+b^3 (7 A+5 C)\right ) \int \sqrt {\sec (c+d x)} \, dx \\ & = \frac {2 \left (98 a^2 b B+21 b^3 B+24 a^3 C+21 a b^2 (5 A+3 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{35 d}+\frac {2 b \left (35 A b^2+63 a b B+24 a^2 C+25 b^2 C\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac {2 (7 b B+6 a C) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{35 d}+\frac {2 C \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^3 \sin (c+d x)}{7 d}+\frac {1}{5} \left (-15 a^2 b B-3 b^3 B+5 a^3 (A-C)-3 a b^2 (5 A+3 C)\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{21} \left (\left (21 a^3 B+21 a b^2 B+21 a^2 b (3 A+C)+b^3 (7 A+5 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 (21 a^3 B+21 a b^2 B+21 a^2 b (3 A+C)+b^3 (7 A+5 C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {2 \left (98 a^2 b B+21 b^3 B+24 a^3 C+21 a b^2 (5 A+3 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{35 d}+\frac {2 b \left (35 A b^2+63 a b B+24 a^2 C+25 b^2 C\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac {2 (7 b B+6 a C) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{35 d}+\frac {2 C \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^3 \sin (c+d x)}{7 d}+\frac {1}{5} \left (\left (-15 a^2 b B-3 b^3 B+5 a^3 (A-C)-3 a b^2 (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 (15 a^2 b B+3 b^3 B-5 a^3 (A-C)+3 a b^2 (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 (21 a^3 B+21 a b^2 B+21 a^2 b (3 A+C)+b^3 (7 A+5 C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {2 \left (98 a^2 b B+21 b^3 B+24 a^3 C+21 a b^2 (5 A+3 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{35 d}+\frac {2 b \left (35 A b^2+63 a b B+24 a^2 C+25 b^2 C\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac {2 (7 b B+6 a C) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{35 d}+\frac {2 C \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^3 \sin (c+d x)}{7 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 7.98 (sec) , antiderivative size = 377, normalized size of antiderivative = 1.13 \[ \int \frac {(a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\frac {4 (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (21 \left (-15 a^2 b B-3 b^3 B+5 a^3 (A-C)-3 a b^2 (5 A+3 C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+5 \left (21 a^3 B+21 a b^2 B+21 a^2 b (3 A+C)+b^3 (7 A+5 C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+315 a A b^2 \sin (c+d x)+315 a^2 b B \sin (c+d x)+63 b^3 B \sin (c+d x)+105 a^3 C \sin (c+d x)+189 a b^2 C \sin (c+d x)+35 A b^3 \tan (c+d x)+105 a b^2 B \tan (c+d x)+105 a^2 b C \tan (c+d x)+25 b^3 C \tan (c+d x)+21 b^3 B \sec (c+d x) \tan (c+d x)+63 a b^2 C \sec (c+d x) \tan (c+d x)+15 b^3 C \sec ^2(c+d x) \tan (c+d x)\right )}{105 d (b+a \cos (c+d x))^3 (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x))) \sec ^{\frac {9}{2}}(c+d x)} \]

[In]

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

[Out]

(4*(a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(21*(-15*a^2*b*B - 3*b^3*B + 5*a^3*(A - C) -
3*a*b^2*(5*A + 3*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 5*(21*a^3*B + 21*a*b^2*B + 21*a^2*b*(3*A
+ C) + b^3*(7*A + 5*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + 315*a*A*b^2*Sin[c + d*x] + 315*a^2*b*B*
Sin[c + d*x] + 63*b^3*B*Sin[c + d*x] + 105*a^3*C*Sin[c + d*x] + 189*a*b^2*C*Sin[c + d*x] + 35*A*b^3*Tan[c + d*
x] + 105*a*b^2*B*Tan[c + d*x] + 105*a^2*b*C*Tan[c + d*x] + 25*b^3*C*Tan[c + d*x] + 21*b^3*B*Sec[c + d*x]*Tan[c
+ d*x] + 63*a*b^2*C*Sec[c + d*x]*Tan[c + d*x] + 15*b^3*C*Sec[c + d*x]^2*Tan[c + d*x]))/(105*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. \(1177\) vs. \(2(358)=716\).

Time = 7.37 (sec) , antiderivative size = 1178, normalized size of antiderivative = 3.53

method	result	size
default	\(\text {Expression too large to display}\)	\(1178\)
parts	\(\text {Expression too large to display}\)	\(1332\)

[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)^(1/2),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*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)
)-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(cos(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))+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(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+2*C*b^3*(-1/56*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)^4-5/42*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+5/21*(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)*Ellipti
cF(cos(1/2*d*x+1/2*c),2^(1/2)))+2/5*b^2*(B*b+3*C*a)/(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*(A*b^2+3*B*a*b+3*C*a^2)*(-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*a*(3*A*b^2+3*B
*a*b+C*a^2)/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 = 0.15 (sec) , antiderivative size = 415, normalized size of antiderivative = 1.24 \[ \int \frac {(a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=-\frac {5 \, \sqrt {2} {\left (21 i \, B a^{3} + 21 i \, {\left (3 \, A + C\right )} a^{2} b + 21 i \, B a b^{2} + i \, {\left (7 \, A + 5 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-21 i \, B a^{3} - 21 i \, {\left (3 \, A + C\right )} a^{2} b - 21 i \, B a b^{2} - i \, {\left (7 \, A + 5 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 21 \, \sqrt {2} {\left (-5 i \, {\left (A - C\right )} a^{3} + 15 i \, B a^{2} b + 3 i \, {\left (5 \, A + 3 \, C\right )} a b^{2} + 3 i \, B b^{3}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 21 \, \sqrt {2} {\left (5 i \, {\left (A - C\right )} a^{3} - 15 i \, B a^{2} b - 3 i \, {\left (5 \, A + 3 \, C\right )} a b^{2} - 3 i \, B b^{3}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (15 \, C b^{3} + 21 \, {\left (5 \, C a^{3} + 15 \, B a^{2} b + 3 \, {\left (5 \, A + 3 \, C\right )} a b^{2} + 3 \, B b^{3}\right )} \cos \left (d x + c\right )^{3} + 5 \, {\left (21 \, C a^{2} b + 21 \, B a b^{2} + {\left (7 \, A + 5 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{2} + 21 \, {\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{105 \, d \cos \left (d x + c\right )^{3}} \]

[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)^(1/2),x, algorithm="fricas")

[Out]

-1/105*(5*sqrt(2)*(21*I*B*a^3 + 21*I*(3*A + C)*a^2*b + 21*I*B*a*b^2 + I*(7*A + 5*C)*b^3)*cos(d*x + c)^3*weiers
trassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*sqrt(2)*(-21*I*B*a^3 - 21*I*(3*A + C)*a^2*b - 21*I*B*a
*b^2 - I*(7*A + 5*C)*b^3)*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 21*sqrt(2
)*(-5*I*(A - C)*a^3 + 15*I*B*a^2*b + 3*I*(5*A + 3*C)*a*b^2 + 3*I*B*b^3)*cos(d*x + c)^3*weierstrassZeta(-4, 0,
weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 21*sqrt(2)*(5*I*(A - C)*a^3 - 15*I*B*a^2*b - 3*I*
(5*A + 3*C)*a*b^2 - 3*I*B*b^3)*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) -
I*sin(d*x + c))) - 2*(15*C*b^3 + 21*(5*C*a^3 + 15*B*a^2*b + 3*(5*A + 3*C)*a*b^2 + 3*B*b^3)*cos(d*x + c)^3 + 5
*(21*C*a^2*b + 21*B*a*b^2 + (7*A + 5*C)*b^3)*cos(d*x + c)^2 + 21*(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)^3)

Sympy [F(-1)]

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

[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)**(1/2),x)

[Out]

Timed out

Maxima [F]

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

[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)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[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)^(1/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

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

[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))^(1/2),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))^(1/2), x)