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

Optimal. Leaf size=289 \[ \frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (4 a^2 C+14 a b B+7 A b^2+5 b^2 C\right )}{21 d}+\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)} \left (5 a^2 B+10 a A b+6 a b C+3 b^2 B\right )}{5 d}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (7 a^2 (3 A+C)+14 a b B+b^2 (7 A+5 C)\right )}{21 d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^2 B+10 a A b+6 a b C+3 b^2 B\right )}{5 d}+\frac {2 b (4 a C+7 b B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{35 d}+\frac {2 C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2}{7 d} \]

[Out]

2/21*(7*A*b^2+14*B*a*b+4*C*a^2+5*C*b^2)*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/35*b*(7*B*b+4*C*a)*sec(d*x+c)^(5/2)*si
n(d*x+c)/d+2/7*C*sec(d*x+c)^(3/2)*(a+b*sec(d*x+c))^2*sin(d*x+c)/d+2/5*(10*A*a*b+5*B*a^2+3*B*b^2+6*C*a*b)*sin(d
*x+c)*sec(d*x+c)^(1/2)/d-2/5*(10*A*a*b+5*B*a^2+3*B*b^2+6*C*a*b)*(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*(14*a*b*B+7*a^2*(3*A+C)+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)*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

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Rubi [A]  time = 0.53, antiderivative size = 289, normalized size of antiderivative = 1.00, 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 = {4096, 4076, 4047, 3768, 3771, 2639, 4046, 2641} \[ \frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (4 a^2 C+14 a b B+7 A b^2+5 b^2 C\right )}{21 d}+\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)} \left (5 a^2 B+10 a A b+6 a b C+3 b^2 B\right )}{5 d}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (7 a^2 (3 A+C)+14 a b B+b^2 (7 A+5 C)\right )}{21 d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^2 B+10 a A b+6 a b C+3 b^2 B\right )}{5 d}+\frac {2 b (4 a C+7 b B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{35 d}+\frac {2 C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(10*a*A*b + 5*a^2*B + 3*b^2*B + 6*a*b*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/
(5*d) + (2*(14*a*b*B + 7*a^2*(3*A + C) + b^2*(7*A + 5*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Se
c[c + d*x]])/(21*d) + (2*(10*a*A*b + 5*a^2*B + 3*b^2*B + 6*a*b*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(5*d) + (2*
(7*A*b^2 + 14*a*b*B + 4*a^2*C + 5*b^2*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(21*d) + (2*b*(7*b*B + 4*a*C)*Sec[c
+ d*x]^(5/2)*Sin[c + d*x])/(35*d) + (2*C*Sec[c + d*x]^(3/2)*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(7*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))^2 \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))^2 \sin (c+d x)}{7 d}+\frac {2}{7} \int \sqrt {\sec (c+d x)} (a+b \sec (c+d x)) \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+4 a C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {2 b (7 b B+4 a C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 C \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{7 d}+\frac {4}{35} \int \sqrt {\sec (c+d x)} \left (\frac {5}{4} a^2 (7 A+C)+\frac {7}{4} \left (10 a A b+5 a^2 B+3 b^2 B+6 a b C\right ) \sec (c+d x)+\frac {5}{4} \left (7 A b^2+14 a b B+4 a^2 C+5 b^2 C\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {2 b (7 b B+4 a C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 C \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{7 d}+\frac {4}{35} \int \sqrt {\sec (c+d x)} \left (\frac {5}{4} a^2 (7 A+C)+\frac {5}{4} \left (7 A b^2+14 a b B+4 a^2 C+5 b^2 C\right ) \sec ^2(c+d x)\right ) \, dx+\frac {1}{5} \left (10 a A b+5 a^2 B+3 b^2 B+6 a b C\right ) \int \sec ^{\frac {3}{2}}(c+d x) \, dx\\ &=\frac {2 \left (10 a A b+5 a^2 B+3 b^2 B+6 a b C\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 \left (7 A b^2+14 a b B+4 a^2 C+5 b^2 C\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {2 b (7 b B+4 a C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 C \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{7 d}+\frac {1}{5} \left (-10 a A b-5 a^2 B-3 b^2 B-6 a b C\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{21} \left (14 a b B+7 a^2 (3 A+C)+b^2 (7 A+5 C)\right ) \int \sqrt {\sec (c+d x)} \, dx\\ &=\frac {2 \left (10 a A b+5 a^2 B+3 b^2 B+6 a b C\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 \left (7 A b^2+14 a b B+4 a^2 C+5 b^2 C\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {2 b (7 b B+4 a C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 C \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{7 d}+\frac {1}{5} \left (\left (-10 a A b-5 a^2 B-3 b^2 B-6 a b C\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{21} \left (\left (14 a b B+7 a^2 (3 A+C)+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 {2 \left (10 a A b+5 a^2 B+3 b^2 B+6 a b 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 (14 a b B+7 a^2 (3 A+C)+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 (10 a A b+5 a^2 B+3 b^2 B+6 a b C\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 \left (7 A b^2+14 a b B+4 a^2 C+5 b^2 C\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {2 b (7 b B+4 a C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 C \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{7 d}\\ \end {align*}

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Mathematica [A]  time = 2.51, size = 333, normalized size = 1.15 \[ \frac {4 (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (5 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (7 a^2 (3 A+C)+14 a b B+b^2 (7 A+5 C)\right )-21 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^2 B+2 a b (5 A+3 C)+3 b^2 B\right )+105 a^2 B \sin (c+d x)+35 a^2 C \tan (c+d x)+210 a A b \sin (c+d x)+70 a b B \tan (c+d x)+126 a b C \sin (c+d x)+42 a b C \tan (c+d x) \sec (c+d x)+35 A b^2 \tan (c+d x)+63 b^2 B \sin (c+d x)+21 b^2 B \tan (c+d x) \sec (c+d x)+25 b^2 C \tan (c+d x)+15 b^2 C \tan (c+d x) \sec ^2(c+d x)\right )}{105 d \sec ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+b)^2 (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(a + b*Sec[c + d*x])^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(-21*(5*a^2*B + 3*b^2*B + 2*a*b*(5*A + 3*C))
*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 5*(14*a*b*B + 7*a^2*(3*A + C) + b^2*(7*A + 5*C))*Sqrt[Cos[c +
d*x]]*EllipticF[(c + d*x)/2, 2] + 210*a*A*b*Sin[c + d*x] + 105*a^2*B*Sin[c + d*x] + 63*b^2*B*Sin[c + d*x] + 12
6*a*b*C*Sin[c + d*x] + 35*A*b^2*Tan[c + d*x] + 70*a*b*B*Tan[c + d*x] + 35*a^2*C*Tan[c + d*x] + 25*b^2*C*Tan[c
+ d*x] + 21*b^2*B*Sec[c + d*x]*Tan[c + d*x] + 42*a*b*C*Sec[c + d*x]*Tan[c + d*x] + 15*b^2*C*Sec[c + d*x]^2*Tan
[c + d*x]))/(105*d*(b + a*Cos[c + d*x])^2*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)])*Sec[c + d*x]^(7/2)
)

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fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C b^{2} \sec \left (d x + c\right )^{4} + {\left (2 \, C a b + B b^{2}\right )} \sec \left (d x + c\right )^{3} + A a^{2} + {\left (C a^{2} + 2 \, B a b + A b^{2}\right )} \sec \left (d x + c\right )^{2} + {\left (B a^{2} + 2 \, A a b\right )} \sec \left (d x + c\right )\right )} \sqrt {\sec \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*b^2*sec(d*x + c)^4 + (2*C*a*b + B*b^2)*sec(d*x + c)^3 + A*a^2 + (C*a^2 + 2*B*a*b + A*b^2)*sec(d*x
+ c)^2 + (B*a^2 + 2*A*a*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 \[ \int {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{2} \sqrt {\sec \left (d x + c\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [B]  time = 19.39, size = 947, normalized size = 3.28 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^(1/2)*(a+b*sec(d*x+c))^2*(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^2*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*b*(B*b+2*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*(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)*s
in(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*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+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*C*(-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*(si
n(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*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*s
in(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*b^2+2*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))^2*(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 \[ \int {\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^2\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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