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

Optimal. Leaf size=397 \[ \frac {2 b \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (24 a^2 C+99 a b B+63 A b^2+49 b^2 C\right )}{315 d}+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (8 a^3 C+54 a^2 b B+9 a b^2 (7 A+5 C)+15 b^3 B\right )}{63 d}+\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)} \left (15 a^3 B+9 a^2 b (5 A+3 C)+27 a b^2 B+b^3 (9 A+7 C)\right )}{15 d}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (7 a^3 (3 A+C)+21 a^2 b B+3 a b^2 (7 A+5 C)+5 b^3 B\right )}{21 d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (15 a^3 B+9 a^2 b (5 A+3 C)+27 a b^2 B+b^3 (9 A+7 C)\right )}{15 d}+\frac {2 (2 a C+3 b B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2}{21 d}+\frac {2 C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^3}{9 d} \]

[Out]

2/63*(54*a^2*b*B+15*b^3*B+8*a^3*C+9*a*b^2*(7*A+5*C))*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/315*b*(63*A*b^2+99*B*a*b+
24*C*a^2+49*C*b^2)*sec(d*x+c)^(5/2)*sin(d*x+c)/d+2/21*(3*B*b+2*C*a)*sec(d*x+c)^(3/2)*(a+b*sec(d*x+c))^2*sin(d*
x+c)/d+2/9*C*sec(d*x+c)^(3/2)*(a+b*sec(d*x+c))^3*sin(d*x+c)/d+2/15*(15*a^3*B+27*a*b^2*B+9*a^2*b*(5*A+3*C)+b^3*
(9*A+7*C))*sin(d*x+c)*sec(d*x+c)^(1/2)/d-2/15*(15*a^3*B+27*a*b^2*B+9*a^2*b*(5*A+3*C)+b^3*(9*A+7*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*a^2*b*B+5*b^3*B+7*a^3*(3*A+C)+3*a*b^2*(7*A+5*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*El
lipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d

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Rubi [A]  time = 0.87, antiderivative size = 397, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.186, Rules used = {4096, 4076, 4047, 3768, 3771, 2639, 4046, 2641} \[ \frac {2 b \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (24 a^2 C+99 a b B+63 A b^2+49 b^2 C\right )}{315 d}+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (54 a^2 b B+8 a^3 C+9 a b^2 (7 A+5 C)+15 b^3 B\right )}{63 d}+\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)} \left (9 a^2 b (5 A+3 C)+15 a^3 B+27 a b^2 B+b^3 (9 A+7 C)\right )}{15 d}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (7 a^3 (3 A+C)+21 a^2 b B+3 a b^2 (7 A+5 C)+5 b^3 B\right )}{21 d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (9 a^2 b (5 A+3 C)+15 a^3 B+27 a b^2 B+b^3 (9 A+7 C)\right )}{15 d}+\frac {2 (2 a C+3 b B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2}{21 d}+\frac {2 C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^3}{9 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(15*a^3*B + 27*a*b^2*B + 9*a^2*b*(5*A + 3*C) + b^3*(9*A + 7*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2,
2]*Sqrt[Sec[c + d*x]])/(15*d) + (2*(21*a^2*b*B + 5*b^3*B + 7*a^3*(3*A + C) + 3*a*b^2*(7*A + 5*C))*Sqrt[Cos[c +
d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (2*(15*a^3*B + 27*a*b^2*B + 9*a^2*b*(5*A + 3*C)
+ b^3*(9*A + 7*C))*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(15*d) + (2*(54*a^2*b*B + 15*b^3*B + 8*a^3*C + 9*a*b^2*(7*
A + 5*C))*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(63*d) + (2*b*(63*A*b^2 + 99*a*b*B + 24*a^2*C + 49*b^2*C)*Sec[c + d
*x]^(5/2)*Sin[c + d*x])/(315*d) + (2*(3*b*B + 2*a*C)*Sec[c + d*x]^(3/2)*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(
21*d) + (2*C*Sec[c + d*x]^(3/2)*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/(9*d)

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]

Rule 3768

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 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 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 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 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 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]

Rubi steps

\begin {align*} \int \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac {2 C \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{9 d}+\frac {2}{9} \int \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 \left (\frac {1}{2} a (9 A+C)+\frac {1}{2} (9 A b+9 a B+7 b C) \sec (c+d x)+\frac {3}{2} (3 b B+2 a C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {2 (3 b B+2 a C) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{21 d}+\frac {2 C \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{9 d}+\frac {4}{63} \int \sqrt {\sec (c+d x)} (a+b \sec (c+d x)) \left (\frac {1}{4} a (63 a A+9 b B+13 a C)+\frac {1}{4} \left (126 a A b+63 a^2 B+45 b^2 B+86 a b C\right ) \sec (c+d x)+\frac {1}{4} \left (63 A b^2+99 a b B+24 a^2 C+49 b^2 C\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {2 b \left (63 A b^2+99 a b B+24 a^2 C+49 b^2 C\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{315 d}+\frac {2 (3 b B+2 a C) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{21 d}+\frac {2 C \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{9 d}+\frac {8}{315} \int \sqrt {\sec (c+d x)} \left (\frac {5}{8} a^2 (63 a A+9 b B+13 a C)+\frac {21}{8} \left (15 a^3 B+27 a b^2 B+9 a^2 b (5 A+3 C)+b^3 (9 A+7 C)\right ) \sec (c+d x)+\frac {15}{8} \left (54 a^2 b B+15 b^3 B+8 a^3 C+9 a b^2 (7 A+5 C)\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {2 b \left (63 A b^2+99 a b B+24 a^2 C+49 b^2 C\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{315 d}+\frac {2 (3 b B+2 a C) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{21 d}+\frac {2 C \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{9 d}+\frac {8}{315} \int \sqrt {\sec (c+d x)} \left (\frac {5}{8} a^2 (63 a A+9 b B+13 a C)+\frac {15}{8} \left (54 a^2 b B+15 b^3 B+8 a^3 C+9 a b^2 (7 A+5 C)\right ) \sec ^2(c+d x)\right ) \, dx+\frac {1}{15} \left (15 a^3 B+27 a b^2 B+9 a^2 b (5 A+3 C)+b^3 (9 A+7 C)\right ) \int \sec ^{\frac {3}{2}}(c+d x) \, dx\\ &=\frac {2 \left (15 a^3 B+27 a b^2 B+9 a^2 b (5 A+3 C)+b^3 (9 A+7 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 \left (54 a^2 b B+15 b^3 B+8 a^3 C+9 a b^2 (7 A+5 C)\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 b \left (63 A b^2+99 a b B+24 a^2 C+49 b^2 C\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{315 d}+\frac {2 (3 b B+2 a C) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{21 d}+\frac {2 C \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{9 d}+\frac {1}{21} \left (21 a^2 b B+5 b^3 B+7 a^3 (3 A+C)+3 a b^2 (7 A+5 C)\right ) \int \sqrt {\sec (c+d x)} \, dx+\frac {1}{15} \left (-15 a^3 B-27 a b^2 B-9 a^2 b (5 A+3 C)-b^3 (9 A+7 C)\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx\\ &=\frac {2 \left (15 a^3 B+27 a b^2 B+9 a^2 b (5 A+3 C)+b^3 (9 A+7 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 \left (54 a^2 b B+15 b^3 B+8 a^3 C+9 a b^2 (7 A+5 C)\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 b \left (63 A b^2+99 a b B+24 a^2 C+49 b^2 C\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{315 d}+\frac {2 (3 b B+2 a C) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{21 d}+\frac {2 C \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{9 d}+\frac {1}{21} \left (\left (21 a^2 b B+5 b^3 B+7 a^3 (3 A+C)+3 a b^2 (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 {1}{15} \left (\left (-15 a^3 B-27 a b^2 B-9 a^2 b (5 A+3 C)-b^3 (9 A+7 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^3 B+27 a b^2 B+9 a^2 b (5 A+3 C)+b^3 (9 A+7 C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{15 d}+\frac {2 \left (21 a^2 b B+5 b^3 B+7 a^3 (3 A+C)+3 a b^2 (7 A+5 C)\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {2 \left (15 a^3 B+27 a b^2 B+9 a^2 b (5 A+3 C)+b^3 (9 A+7 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 \left (54 a^2 b B+15 b^3 B+8 a^3 C+9 a b^2 (7 A+5 C)\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 b \left (63 A b^2+99 a b B+24 a^2 C+49 b^2 C\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{315 d}+\frac {2 (3 b B+2 a C) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{21 d}+\frac {2 C \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{9 d}\\ \end {align*}

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Mathematica [A]  time = 7.14, size = 566, normalized size = 1.43 \[ \frac {2 \cos ^5(c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (105 a^3 A+35 a^3 C+105 a^2 b B+105 a A b^2+75 a b^2 C+25 b^3 B\right )+\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-105 a^3 B-315 a^2 A b-189 a^2 b C-189 a b^2 B-63 A b^3-49 b^3 C\right )}{\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}\right )}{105 d (a \cos (c+d x)+b)^3 (A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C)}+\frac {(a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac {4}{45} \sec ^2(c+d x) \left (27 a^2 b C \sin (c+d x)+27 a b^2 B \sin (c+d x)+9 A b^3 \sin (c+d x)+7 b^3 C \sin (c+d x)\right )+\frac {4}{15} \sin (c+d x) \left (15 a^3 B+45 a^2 A b+27 a^2 b C+27 a b^2 B+9 A b^3+7 b^3 C\right )+\frac {4}{21} \sec (c+d x) \left (7 a^3 C \sin (c+d x)+21 a^2 b B \sin (c+d x)+21 a A b^2 \sin (c+d x)+15 a b^2 C \sin (c+d x)+5 b^3 B \sin (c+d x)\right )+\frac {4}{7} \sec ^3(c+d x) \left (3 a b^2 C \sin (c+d x)+b^3 B \sin (c+d x)\right )+\frac {4}{9} b^3 C \tan (c+d x) \sec ^3(c+d x)\right )}{d \sec ^{\frac {9}{2}}(c+d x) (a \cos (c+d x)+b)^3 (A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Cos[c + d*x]^5*((2*(-315*a^2*A*b - 63*A*b^3 - 105*a^3*B - 189*a*b^2*B - 189*a^2*b*C - 49*b^3*C)*EllipticE[(
c + d*x)/2, 2])/(Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + 2*(105*a^3*A + 105*a*A*b^2 + 105*a^2*b*B + 25*b^3*B
+ 35*a^3*C + 75*a*b^2*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])*(a + b*Sec[c + d*x])
^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(105*d*(b + a*Cos[c + d*x])^3*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[
2*c + 2*d*x])) + ((a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((4*(45*a^2*A*b + 9*A*b^3 + 1
5*a^3*B + 27*a*b^2*B + 27*a^2*b*C + 7*b^3*C)*Sin[c + d*x])/15 + (4*Sec[c + d*x]^3*(b^3*B*Sin[c + d*x] + 3*a*b^
2*C*Sin[c + d*x]))/7 + (4*Sec[c + d*x]*(21*a*A*b^2*Sin[c + d*x] + 21*a^2*b*B*Sin[c + d*x] + 5*b^3*B*Sin[c + d*
x] + 7*a^3*C*Sin[c + d*x] + 15*a*b^2*C*Sin[c + d*x]))/21 + (4*Sec[c + d*x]^2*(9*A*b^3*Sin[c + d*x] + 27*a*b^2*
B*Sin[c + d*x] + 27*a^2*b*C*Sin[c + d*x] + 7*b^3*C*Sin[c + d*x]))/45 + (4*b^3*C*Sec[c + d*x]^3*Tan[c + d*x])/9
))/(d*(b + a*Cos[c + d*x])^3*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sec[c + d*x]^(9/2))

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C b^{3} \sec \left (d x + c\right )^{5} + {\left (3 \, C a b^{2} + B b^{3}\right )} \sec \left (d x + c\right )^{4} + A a^{3} + {\left (3 \, C a^{2} b + 3 \, B a b^{2} + A b^{3}\right )} \sec \left (d x + c\right )^{3} + {\left (C a^{3} + 3 \, B a^{2} b + 3 \, A a b^{2}\right )} \sec \left (d x + c\right )^{2} + {\left (B a^{3} + 3 \, A a^{2} b\right )} \sec \left (d x + c\right )\right )} \sqrt {\sec \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(1/2)*(a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^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))*sqrt(sec(d
*x + c)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{3} \sqrt {\sec \left (d x + c\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(1/2)*(a+b*sec(d*x+c))^3*(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)^3*sqrt(sec(d*x + c)), x)

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maple [B]  time = 24.60, size = 1292, normalized size = 3.25 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^(1/2)*(a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^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)*(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)
)+2*b^3*C*(-1/144*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)/(-1/2+cos(1/2*d*x+1/
2*c)^2)^5-7/180*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)/(-1/2+cos(1/2*d*x+1/2*
c)^2)^3-14/15*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)/(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2
)+7/15*(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))-7/15*(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))))-2/5*b*(A*b^2+3*B*a*b+3*C*a^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)^2*(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)^6-12*Ell
ipticE(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+24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+3*(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))-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)*(-1/56*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)/(-1/2+cos(1/2*d*x+1/2*c)^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)/(-1/2+cos(1/2*d*x+1/2*c)^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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+2*a^2*(3*A*b+B*a)
*(-(-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))+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)+2*a*(3*A*b^2+3*B*a*b+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)/(-1/2+cos(1/2*d*x+1/2*c)^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))))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(1/2)*(a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^3\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**(1/2)*(a+b*sec(d*x+c))**3*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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