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

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

[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)

________________________________________________________________________________________

Rubi [A]  time = 0.785919, antiderivative size = 334, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.163, Rules used = {4096, 4076, 4047, 3771, 2641, 4046, 2639} \[ \frac{2 b \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) \left (24 a^2 C+63 a b B+35 A b^2+25 b^2 C\right )}{105 d}+\frac{2 \sin (c+d x) \sqrt{\sec (c+d x)} \left (98 a^2 b B+24 a^3 C+21 a b^2 (5 A+3 C)+21 b^3 B\right )}{35 d}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (21 a^2 b (3 A+C)+21 a^3 B+21 a b^2 B+b^3 (7 A+5 C)\right )}{21 d}-\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (-5 a^3 (A-C)+15 a^2 b B+3 a b^2 (5 A+3 C)+3 b^3 B\right )}{5 d}+\frac{2 (6 a C+7 b B) \sin (c+d x) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^2}{35 d}+\frac{2 C \sin (c+d x) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^3}{7 d} \]

Antiderivative was successfully verified.

[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 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]

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 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 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 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 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 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]

Rubi steps

\begin{align*} \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 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)} F\left (\left .\frac{1}{2} (c+d x)\right |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)} F\left (\left .\frac{1}{2} (c+d x)\right |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]  time = 3.77841, size = 377, normalized size = 1.13 \[ \frac{4 (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (5 \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \left (21 a^2 b (3 A+C)+21 a^3 B+21 a b^2 B+b^3 (7 A+5 C)\right )+21 \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (5 a^3 (A-C)-15 a^2 b B-3 a b^2 (5 A+3 C)-3 b^3 B\right )+315 a^2 b B \sin (c+d x)+105 a^2 b C \tan (c+d x)+105 a^3 C \sin (c+d x)+315 a A b^2 \sin (c+d x)+105 a b^2 B \tan (c+d x)+189 a b^2 C \sin (c+d x)+63 a b^2 C \tan (c+d x) \sec (c+d x)+35 A b^3 \tan (c+d x)+63 b^3 B \sin (c+d x)+21 b^3 B \tan (c+d x) \sec (c+d x)+25 b^3 C \tan (c+d x)+15 b^3 C \tan (c+d x) \sec ^2(c+d x)\right )}{105 d \sec ^{\frac{9}{2}}(c+d x) (a \cos (c+d x)+b)^3 (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[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))

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Maple [B]  time = 9.369, size = 1205, normalized size = 3.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

[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
))-EllipticE(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*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))-2/5*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*(12*(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))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^4-
24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-12*(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))*(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*(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))*(sin(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*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)*EllipticF(cos(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)^(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(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))/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., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{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 )}{\sqrt{\sec \left (d x + c\right )}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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]

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)