3.501 \(\int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=229 \[ \frac {64 a^3 (165 A+143 B+125 C) \tan (c+d x)}{3465 d \sqrt {a \sec (c+d x)+a}}+\frac {16 a^2 (165 A+143 B+125 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3465 d}+\frac {2 (99 A-22 B+26 C) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{693 d}+\frac {2 a (165 A+143 B+125 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{1155 d}+\frac {2 (11 B+5 C) \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{99 a d}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d} \]

[Out]

2/1155*a*(165*A+143*B+125*C)*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d+2/693*(99*A-22*B+26*C)*(a+a*sec(d*x+c))^(5/2)
*tan(d*x+c)/d+2/11*C*sec(d*x+c)^2*(a+a*sec(d*x+c))^(5/2)*tan(d*x+c)/d+2/99*(11*B+5*C)*(a+a*sec(d*x+c))^(7/2)*t
an(d*x+c)/a/d+64/3465*a^3*(165*A+143*B+125*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+16/3465*a^2*(165*A+143*B+125
*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d

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Rubi [A]  time = 0.59, antiderivative size = 229, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.116, Rules used = {4088, 4010, 4001, 3793, 3792} \[ \frac {16 a^2 (165 A+143 B+125 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3465 d}+\frac {64 a^3 (165 A+143 B+125 C) \tan (c+d x)}{3465 d \sqrt {a \sec (c+d x)+a}}+\frac {2 (99 A-22 B+26 C) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{693 d}+\frac {2 a (165 A+143 B+125 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{1155 d}+\frac {2 (11 B+5 C) \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{99 a d}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(64*a^3*(165*A + 143*B + 125*C)*Tan[c + d*x])/(3465*d*Sqrt[a + a*Sec[c + d*x]]) + (16*a^2*(165*A + 143*B + 125
*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(3465*d) + (2*a*(165*A + 143*B + 125*C)*(a + a*Sec[c + d*x])^(3/2)*
Tan[c + d*x])/(1155*d) + (2*(99*A - 22*B + 26*C)*(a + a*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(693*d) + (2*C*Sec[c
+ d*x]^2*(a + a*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(11*d) + (2*(11*B + 5*C)*(a + a*Sec[c + d*x])^(7/2)*Tan[c +
d*x])/(99*a*d)

Rule 3792

Int[csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*b*Cot[e + f*x])/
(f*Sqrt[a + b*Csc[e + f*x]]), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rule 3793

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(b*Cot[e + f*x]*(a
+ b*Csc[e + f*x])^(m - 1))/(f*m), x] + Dist[(a*(2*m - 1))/m, Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m - 1), x
], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] && IntegerQ[2*m]

Rule 4001

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> -Simp[(B*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(a*B*m + A*b*(m + 1))/(b*(
m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] /; FreeQ[{a, b, A, B, e, f, m}, x] && NeQ[A*b - a*B,
0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b*(m + 1), 0] &&  !LtQ[m, -2^(-1)]

Rule 4010

Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_
)), x_Symbol] :> -Simp[(B*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), I
nt[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*B*(m + 1) + (A*b*(m + 2) - a*B)*Csc[e + f*x], x], x], x] /; Free
Q[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] &&  !LtQ[m, -1]

Rule 4088

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/(b*(m + n + 1)), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n*
Simp[A*b*(m + n + 1) + b*C*n + (a*C*m + b*B*(m + n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A,
B, C, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] &&  !LtQ[n, -2^(-1)] && NeQ[m + n + 1, 0]

Rubi steps

\begin {align*} \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac {2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{11 d}+\frac {2 \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (\frac {1}{2} a (11 A+4 C)+\frac {1}{2} a (11 B+5 C) \sec (c+d x)\right ) \, dx}{11 a}\\ &=\frac {2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{11 d}+\frac {2 (11 B+5 C) (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{99 a d}+\frac {4 \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} \left (\frac {7}{4} a^2 (11 B+5 C)+\frac {1}{4} a^2 (99 A-22 B+26 C) \sec (c+d x)\right ) \, dx}{99 a^2}\\ &=\frac {2 (99 A-22 B+26 C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{693 d}+\frac {2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{11 d}+\frac {2 (11 B+5 C) (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{99 a d}+\frac {1}{231} (165 A+143 B+125 C) \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} \, dx\\ &=\frac {2 a (165 A+143 B+125 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 d}+\frac {2 (99 A-22 B+26 C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{693 d}+\frac {2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{11 d}+\frac {2 (11 B+5 C) (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{99 a d}+\frac {(8 a (165 A+143 B+125 C)) \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \, dx}{1155}\\ &=\frac {16 a^2 (165 A+143 B+125 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{3465 d}+\frac {2 a (165 A+143 B+125 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 d}+\frac {2 (99 A-22 B+26 C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{693 d}+\frac {2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{11 d}+\frac {2 (11 B+5 C) (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{99 a d}+\frac {\left (32 a^2 (165 A+143 B+125 C)\right ) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx}{3465}\\ &=\frac {64 a^3 (165 A+143 B+125 C) \tan (c+d x)}{3465 d \sqrt {a+a \sec (c+d x)}}+\frac {16 a^2 (165 A+143 B+125 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{3465 d}+\frac {2 a (165 A+143 B+125 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 d}+\frac {2 (99 A-22 B+26 C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{693 d}+\frac {2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{11 d}+\frac {2 (11 B+5 C) (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{99 a d}\\ \end {align*}

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Mathematica [A]  time = 1.67, size = 188, normalized size = 0.82 \[ \frac {a^2 \tan \left (\frac {1}{2} (c+d x)\right ) \sec ^5(c+d x) \sqrt {a (\sec (c+d x)+1)} ((49830 A+49654 B+50140 C) \cos (c+d x)+4 (4290 A+4642 B+4615 C) \cos (2 (c+d x))+22935 A \cos (3 (c+d x))+3795 A \cos (4 (c+d x))+3795 A \cos (5 (c+d x))+13365 A+20878 B \cos (3 (c+d x))+3212 B \cos (4 (c+d x))+3212 B \cos (5 (c+d x))+15356 B+18460 C \cos (3 (c+d x))+2840 C \cos (4 (c+d x))+2840 C \cos (5 (c+d x))+18140 C)}{13860 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^2*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(a^2*(13365*A + 15356*B + 18140*C + (49830*A + 49654*B + 50140*C)*Cos[c + d*x] + 4*(4290*A + 4642*B + 4615*C)*
Cos[2*(c + d*x)] + 22935*A*Cos[3*(c + d*x)] + 20878*B*Cos[3*(c + d*x)] + 18460*C*Cos[3*(c + d*x)] + 3795*A*Cos
[4*(c + d*x)] + 3212*B*Cos[4*(c + d*x)] + 2840*C*Cos[4*(c + d*x)] + 3795*A*Cos[5*(c + d*x)] + 3212*B*Cos[5*(c
+ d*x)] + 2840*C*Cos[5*(c + d*x)])*Sec[c + d*x]^5*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d*x)/2])/(13860*d)

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fricas [A]  time = 0.45, size = 168, normalized size = 0.73 \[ \frac {2 \, {\left (2 \, {\left (3795 \, A + 3212 \, B + 2840 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} + {\left (3795 \, A + 3212 \, B + 2840 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 3 \, {\left (660 \, A + 803 \, B + 710 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 5 \, {\left (99 \, A + 286 \, B + 355 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 35 \, {\left (11 \, B + 32 \, C\right )} a^{2} \cos \left (d x + c\right ) + 315 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3465 \, {\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

2/3465*(2*(3795*A + 3212*B + 2840*C)*a^2*cos(d*x + c)^5 + (3795*A + 3212*B + 2840*C)*a^2*cos(d*x + c)^4 + 3*(6
60*A + 803*B + 710*C)*a^2*cos(d*x + c)^3 + 5*(99*A + 286*B + 355*C)*a^2*cos(d*x + c)^2 + 35*(11*B + 32*C)*a^2*
cos(d*x + c) + 315*C*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c)^6 + d*cos(d*x +
c)^5)

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giac [A]  time = 2.90, size = 383, normalized size = 1.67 \[ \frac {8 \, {\left ({\left ({\left ({\left (4 \, {\left (2 \, \sqrt {2} {\left (165 \, A a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 143 \, B a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 125 \, C a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 11 \, \sqrt {2} {\left (165 \, A a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 143 \, B a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 125 \, C a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 99 \, \sqrt {2} {\left (165 \, A a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 143 \, B a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 125 \, C a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 231 \, \sqrt {2} {\left (85 \, A a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 69 \, B a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 65 \, C a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1155 \, \sqrt {2} {\left (11 \, A a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 9 \, B a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 7 \, C a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3465 \, \sqrt {2} {\left (A a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + B a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + C a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{3465 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{5} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

8/3465*((((4*(2*sqrt(2)*(165*A*a^8*sgn(cos(d*x + c)) + 143*B*a^8*sgn(cos(d*x + c)) + 125*C*a^8*sgn(cos(d*x + c
)))*tan(1/2*d*x + 1/2*c)^2 - 11*sqrt(2)*(165*A*a^8*sgn(cos(d*x + c)) + 143*B*a^8*sgn(cos(d*x + c)) + 125*C*a^8
*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 99*sqrt(2)*(165*A*a^8*sgn(cos(d*x + c)) + 143*B*a^8*sgn(cos(d*x
+ c)) + 125*C*a^8*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 - 231*sqrt(2)*(85*A*a^8*sgn(cos(d*x + c)) + 69*B*
a^8*sgn(cos(d*x + c)) + 65*C*a^8*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 1155*sqrt(2)*(11*A*a^8*sgn(cos(d
*x + c)) + 9*B*a^8*sgn(cos(d*x + c)) + 7*C*a^8*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 - 3465*sqrt(2)*(A*a^
8*sgn(cos(d*x + c)) + B*a^8*sgn(cos(d*x + c)) + C*a^8*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x
+ 1/2*c)^2 - a)^5*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*d)

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maple [A]  time = 1.76, size = 207, normalized size = 0.90 \[ -\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (7590 A \left (\cos ^{5}\left (d x +c \right )\right )+6424 B \left (\cos ^{5}\left (d x +c \right )\right )+5680 C \left (\cos ^{5}\left (d x +c \right )\right )+3795 A \left (\cos ^{4}\left (d x +c \right )\right )+3212 B \left (\cos ^{4}\left (d x +c \right )\right )+2840 C \left (\cos ^{4}\left (d x +c \right )\right )+1980 A \left (\cos ^{3}\left (d x +c \right )\right )+2409 B \left (\cos ^{3}\left (d x +c \right )\right )+2130 C \left (\cos ^{3}\left (d x +c \right )\right )+495 A \left (\cos ^{2}\left (d x +c \right )\right )+1430 B \left (\cos ^{2}\left (d x +c \right )\right )+1775 C \left (\cos ^{2}\left (d x +c \right )\right )+385 B \cos \left (d x +c \right )+1120 C \cos \left (d x +c \right )+315 C \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a^{2}}{3465 d \cos \left (d x +c \right )^{5} \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

-2/3465/d*(-1+cos(d*x+c))*(7590*A*cos(d*x+c)^5+6424*B*cos(d*x+c)^5+5680*C*cos(d*x+c)^5+3795*A*cos(d*x+c)^4+321
2*B*cos(d*x+c)^4+2840*C*cos(d*x+c)^4+1980*A*cos(d*x+c)^3+2409*B*cos(d*x+c)^3+2130*C*cos(d*x+c)^3+495*A*cos(d*x
+c)^2+1430*B*cos(d*x+c)^2+1775*C*cos(d*x+c)^2+385*B*cos(d*x+c)+1120*C*cos(d*x+c)+315*C)*(a*(1+cos(d*x+c))/cos(
d*x+c))^(1/2)/cos(d*x+c)^5/sin(d*x+c)*a^2

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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)^2*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

Timed out

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mupad [B]  time = 17.05, size = 1034, normalized size = 4.52 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + a/cos(c + d*x))^(5/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/cos(c + d*x)^2,x)

[Out]

((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*((a^2*(9*A + 10*B + 4*C)*4i
)/(5*d) - (A*a^2*4i)/(5*d) + (a^2*(33*A + 44*B - 31*C)*16i)/(1155*d)) + (a^2*(5*A + 16*B + 20*C)*4i)/(5*d) - (
a^2*(5*A + 2*B)*4i)/(5*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^2) + ((a + a/(exp(- c*1i - d*x*
1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*((A*a^2*4i)/(11*d) - (a^2*(6*A + 5*B + 2*C)*8i)/(11*d
) - (a^2*(10*A + 11*B + 10*C)*8i)/(11*d) + (a^2*(13*A + 15*B + 20*C)*8i)/(11*d) + (a^2*(5*A + 2*B)*4i)/(11*d))
+ (A*a^2*4i)/(11*d) - (a^2*(6*A + 5*B + 2*C)*8i)/(11*d) - (a^2*(10*A + 11*B + 10*C)*8i)/(11*d) + (a^2*(13*A +
15*B + 20*C)*8i)/(11*d) + (a^2*(5*A + 2*B)*4i)/(11*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^5)
- ((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*((A*a^2*4i)/(9*d) + (C*a
^2*64i)/(99*d) - (a^2*(5*A + 2*B - 16*C)*4i)/(9*d) - (a^2*(11*A + 10*B + 4*C)*4i)/(9*d) + (a^2*(15*A + 20*B +
36*C)*4i)/(9*d)) - (a^2*(A - 16*C)*4i)/(9*d) - (a^2*(3*A + 4*B + 4*C)*20i)/(9*d) + (a^2*(11*A + 10*B + 20*C)*4
i)/(9*d) + (a^2*(5*A + 2*B)*4i)/(9*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^4) + ((a + a/(exp(-
c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*((A*a^2*4i)/(7*d) - (a^2*(5*A + 5*B + 2*C
)*8i)/(7*d) + (a^2*(5*A + 10*B + 32*C)*4i)/(7*d) + (a^2*(11*B + 50*C)*32i)/(693*d)) + (a^2*(A - 8*B)*4i)/(7*d)
- (a^2*(5*A + 9*B + 10*C)*8i)/(7*d) + (a^2*(5*A + 2*B)*4i)/(7*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*
2i) + 1)^3) + ((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*((A*a^2*4i)/(
3*d) - (a^2*(660*A + 803*B + 710*C)*8i)/(3465*d)) + (a^2*(5*A + 2*B)*4i)/(3*d)))/((exp(c*1i + d*x*1i) + 1)*(ex
p(c*2i + d*x*2i) + 1)) - (a^2*exp(c*1i + d*x*1i)*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)
*(3795*A + 3212*B + 2840*C)*4i)/(3465*d*(exp(c*1i + d*x*1i) + 1))

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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)**2*(a+a*sec(d*x+c))**(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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