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

Optimal. Leaf size=343 \[ \frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \left (7 a^2 B+14 a A b+10 a b C+5 b^2 B\right )}{21 d}+\frac{2 \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x) \left (4 a^2 C+18 a b B+9 A b^2+7 b^2 C\right )}{45 d}+\frac{2 \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) \left (7 a^2 B+14 a A b+10 a b C+5 b^2 B\right )}{21 d}+\frac{2 \sin (c+d x) \sqrt{\sec (c+d x)} \left (3 a^2 (5 A+3 C)+18 a b B+b^2 (9 A+7 C)\right )}{15 d}-\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (3 a^2 (5 A+3 C)+18 a b B+b^2 (9 A+7 C)\right )}{15 d}+\frac{2 b (4 a C+9 b B) \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{63 d}+\frac{2 C \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x) (a+b \sec (c+d x))^2}{9 d} \]

[Out]

(-2*(18*a*b*B + 3*a^2*(5*A + 3*C) + b^2*(9*A + 7*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c +
d*x]])/(15*d) + (2*(14*a*A*b + 7*a^2*B + 5*b^2*B + 10*a*b*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqr
t[Sec[c + d*x]])/(21*d) + (2*(18*a*b*B + 3*a^2*(5*A + 3*C) + b^2*(9*A + 7*C))*Sqrt[Sec[c + d*x]]*Sin[c + d*x])
/(15*d) + (2*(14*a*A*b + 7*a^2*B + 5*b^2*B + 10*a*b*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(21*d) + (2*(9*A*b^2 +
18*a*b*B + 4*a^2*C + 7*b^2*C)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(45*d) + (2*b*(9*b*B + 4*a*C)*Sec[c + d*x]^(7/
2)*Sin[c + d*x])/(63*d) + (2*C*Sec[c + d*x]^(5/2)*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(9*d)

________________________________________________________________________________________

Rubi [A]  time = 0.587621, antiderivative size = 343, normalized size of antiderivative = 1., 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, 2641, 4046, 2639} \[ \frac{2 \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x) \left (4 a^2 C+18 a b B+9 A b^2+7 b^2 C\right )}{45 d}+\frac{2 \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) \left (7 a^2 B+14 a A b+10 a b C+5 b^2 B\right )}{21 d}+\frac{2 \sin (c+d x) \sqrt{\sec (c+d x)} \left (3 a^2 (5 A+3 C)+18 a b B+b^2 (9 A+7 C)\right )}{15 d}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (7 a^2 B+14 a A b+10 a b C+5 b^2 B\right )}{21 d}-\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (3 a^2 (5 A+3 C)+18 a b B+b^2 (9 A+7 C)\right )}{15 d}+\frac{2 b (4 a C+9 b B) \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{63 d}+\frac{2 C \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x) (a+b \sec (c+d x))^2}{9 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(18*a*b*B + 3*a^2*(5*A + 3*C) + b^2*(9*A + 7*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c +
d*x]])/(15*d) + (2*(14*a*A*b + 7*a^2*B + 5*b^2*B + 10*a*b*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqr
t[Sec[c + d*x]])/(21*d) + (2*(18*a*b*B + 3*a^2*(5*A + 3*C) + b^2*(9*A + 7*C))*Sqrt[Sec[c + d*x]]*Sin[c + d*x])
/(15*d) + (2*(14*a*A*b + 7*a^2*B + 5*b^2*B + 10*a*b*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(21*d) + (2*(9*A*b^2 +
18*a*b*B + 4*a^2*C + 7*b^2*C)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(45*d) + (2*b*(9*b*B + 4*a*C)*Sec[c + d*x]^(7/
2)*Sin[c + d*x])/(63*d) + (2*C*Sec[c + d*x]^(5/2)*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(9*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 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 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 \sec ^{\frac{3}{2}}(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{5}{2}}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{9 d}+\frac{2}{9} \int \sec ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x)) \left (\frac{3}{2} a (3 A+C)+\frac{1}{2} (9 A b+9 a B+7 b C) \sec (c+d x)+\frac{1}{2} (9 b B+4 a C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{2 b (9 b B+4 a C) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac{2 C \sec ^{\frac{5}{2}}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{9 d}+\frac{4}{63} \int \sec ^{\frac{3}{2}}(c+d x) \left (\frac{21}{4} a^2 (3 A+C)+\frac{9}{4} \left (14 a A b+7 a^2 B+5 b^2 B+10 a b C\right ) \sec (c+d x)+\frac{7}{4} \left (9 A b^2+18 a b B+4 a^2 C+7 b^2 C\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{2 b (9 b B+4 a C) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac{2 C \sec ^{\frac{5}{2}}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{9 d}+\frac{4}{63} \int \sec ^{\frac{3}{2}}(c+d x) \left (\frac{21}{4} a^2 (3 A+C)+\frac{7}{4} \left (9 A b^2+18 a b B+4 a^2 C+7 b^2 C\right ) \sec ^2(c+d x)\right ) \, dx+\frac{1}{7} \left (14 a A b+7 a^2 B+5 b^2 B+10 a b C\right ) \int \sec ^{\frac{5}{2}}(c+d x) \, dx\\ &=\frac{2 \left (14 a A b+7 a^2 B+5 b^2 B+10 a b C\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac{2 \left (9 A b^2+18 a b B+4 a^2 C+7 b^2 C\right ) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac{2 b (9 b B+4 a C) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac{2 C \sec ^{\frac{5}{2}}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{9 d}+\frac{1}{21} \left (14 a A b+7 a^2 B+5 b^2 B+10 a b C\right ) \int \sqrt{\sec (c+d x)} \, dx+\frac{1}{15} \left (18 a b B+3 a^2 (5 A+3 C)+b^2 (9 A+7 C)\right ) \int \sec ^{\frac{3}{2}}(c+d x) \, dx\\ &=\frac{2 \left (18 a b B+3 a^2 (5 A+3 C)+b^2 (9 A+7 C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 \left (14 a A b+7 a^2 B+5 b^2 B+10 a b C\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac{2 \left (9 A b^2+18 a b B+4 a^2 C+7 b^2 C\right ) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac{2 b (9 b B+4 a C) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac{2 C \sec ^{\frac{5}{2}}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{9 d}+\frac{1}{15} \left (-18 a b B-3 a^2 (5 A+3 C)-b^2 (9 A+7 C)\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx+\frac{1}{21} \left (\left (14 a A b+7 a^2 B+5 b^2 B+10 a b 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 (14 a A b+7 a^2 B+5 b^2 B+10 a b 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 (18 a b B+3 a^2 (5 A+3 C)+b^2 (9 A+7 C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 \left (14 a A b+7 a^2 B+5 b^2 B+10 a b C\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac{2 \left (9 A b^2+18 a b B+4 a^2 C+7 b^2 C\right ) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac{2 b (9 b B+4 a C) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac{2 C \sec ^{\frac{5}{2}}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{9 d}+\frac{1}{15} \left (\left (-18 a b B-3 a^2 (5 A+3 C)-b^2 (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 (18 a b B+3 a^2 (5 A+3 C)+b^2 (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 (14 a A b+7 a^2 B+5 b^2 B+10 a b 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 (18 a b B+3 a^2 (5 A+3 C)+b^2 (9 A+7 C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 \left (14 a A b+7 a^2 B+5 b^2 B+10 a b C\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac{2 \left (9 A b^2+18 a b B+4 a^2 C+7 b^2 C\right ) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac{2 b (9 b B+4 a C) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac{2 C \sec ^{\frac{5}{2}}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{9 d}\\ \end{align*}

Mathematica [A]  time = 6.74075, size = 507, normalized size = 1.48 \[ \frac{(a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac{4}{15} \sin (c+d x) \left (15 a^2 A+9 a^2 C+18 a b B+9 A b^2+7 b^2 C\right )+\frac{4}{45} \sec ^2(c+d x) \left (9 a^2 C \sin (c+d x)+18 a b B \sin (c+d x)+9 A b^2 \sin (c+d x)+7 b^2 C \sin (c+d x)\right )+\frac{4}{21} \sec (c+d x) \left (7 a^2 B \sin (c+d x)+14 a A b \sin (c+d x)+10 a b C \sin (c+d x)+5 b^2 B \sin (c+d x)\right )+\frac{4}{7} \sec ^3(c+d x) \left (2 a b C \sin (c+d x)+b^2 B \sin (c+d x)\right )+\frac{4}{9} b^2 C \tan (c+d x) \sec ^3(c+d x)\right )}{d \sec ^{\frac{7}{2}}(c+d x) (a \cos (c+d x)+b)^2 (A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C)}-\frac{2 \cos ^4(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \left (-35 a^2 B-70 a A b-50 a b C-25 b^2 B\right )+\frac{2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (105 a^2 A+63 a^2 C+126 a b B+63 A b^2+49 b^2 C\right )}{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}}\right )}{105 d (a \cos (c+d x)+b)^2 (A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Cos[c + d*x]^4*((2*(105*a^2*A + 63*A*b^2 + 126*a*b*B + 63*a^2*C + 49*b^2*C)*EllipticE[(c + d*x)/2, 2])/(Sq
rt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + 2*(-70*a*A*b - 35*a^2*B - 25*b^2*B - 50*a*b*C)*Sqrt[Cos[c + d*x]]*Ellip
ticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])*(a + b*Sec[c + d*x])^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(105*
d*(b + a*Cos[c + d*x])^2*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + ((a + b*Sec[c + d*x])^2*(A + B*S
ec[c + d*x] + C*Sec[c + d*x]^2)*((4*(15*a^2*A + 9*A*b^2 + 18*a*b*B + 9*a^2*C + 7*b^2*C)*Sin[c + d*x])/15 + (4*
Sec[c + d*x]^3*(b^2*B*Sin[c + d*x] + 2*a*b*C*Sin[c + d*x]))/7 + (4*Sec[c + d*x]*(14*a*A*b*Sin[c + d*x] + 7*a^2
*B*Sin[c + d*x] + 5*b^2*B*Sin[c + d*x] + 10*a*b*C*Sin[c + d*x]))/21 + (4*Sec[c + d*x]^2*(9*A*b^2*Sin[c + d*x]
+ 18*a*b*B*Sin[c + d*x] + 9*a^2*C*Sin[c + d*x] + 7*b^2*C*Sin[c + d*x]))/45 + (4*b^2*C*Sec[c + d*x]^3*Tan[c + d
*x])/9))/(d*(b + a*Cos[c + d*x])^2*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sec[c + d*x]^(7/2))

________________________________________________________________________________________

Maple [B]  time = 11.647, size = 1196, normalized size = 3.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^(3/2)*(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*b+B*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)*EllipticF(cos(1/2*d*x+1/2*c
),2^(1/2)))+2*b*(B*b+2*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)/(co
s(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^2*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)/(cos(1/2*d*x+1/2*c)^2-1/2)^5-7/180*cos(1/2*d*x+1/2*c)*(-2*si
n(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)^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*co
s(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*(A*b^2+
2*B*a*b+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*(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*s
in(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*a^2*A*(-(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)*El
lipticE(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(sec(d*x+c)^(3/2)*(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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C b^{2} \sec \left (d x + c\right )^{5} +{\left (2 \, C a b + B b^{2}\right )} \sec \left (d x + c\right )^{4} + A a^{2} \sec \left (d x + c\right ) +{\left (C a^{2} + 2 \, B a b + A b^{2}\right )} \sec \left (d x + c\right )^{3} +{\left (B a^{2} + 2 \, A a b\right )} \sec \left (d x + c\right )^{2}\right )} \sqrt{\sec \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(3/2)*(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)^5 + (2*C*a*b + B*b^2)*sec(d*x + c)^4 + A*a^2*sec(d*x + c) + (C*a^2 + 2*B*a*b + A*
b^2)*sec(d*x + c)^3 + (B*a^2 + 2*A*a*b)*sec(d*x + c)^2)*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(sec(d*x+c)**(3/2)*(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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \sec \left (d x + c\right )^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(3/2)*(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*sec(d*x + c)^(3/2), x)