Integrand size = 43, antiderivative size = 229 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {64 a^3 (165 A+143 B+125 C) \tan (c+d x)}{3465 d \sqrt {a+a \sec (c+d x)}}+\frac {16 a^2 (165 A+143 B+125 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{3465 d}+\frac {2 a (165 A+143 B+125 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 d}+\frac {2 (99 A-22 B+26 C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{693 d}+\frac {2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{11 d}+\frac {2 (11 B+5 C) (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{99 a d} \] Output:
64/3465*a^3*(165*A+143*B+125*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+16/346 5*a^2*(165*A+143*B+125*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d+2/1155*a*(16 5*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)*tan(d*x+c)/a/d
Time = 2.43 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.80 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^3 (13365 A+15356 B+18140 C+(49830 A+49654 B+50140 C) \cos (c+d x)+4 (4290 A+4642 B+4615 C) \cos (2 (c+d x))+22935 A \cos (3 (c+d x))+20878 B \cos (3 (c+d x))+18460 C \cos (3 (c+d x))+3795 A \cos (4 (c+d x))+3212 B \cos (4 (c+d x))+2840 C \cos (4 (c+d x))+3795 A \cos (5 (c+d x))+3212 B \cos (5 (c+d x))+2840 C \cos (5 (c+d x))) \sec ^5(c+d x) \tan (c+d x)}{13860 d \sqrt {a (1+\sec (c+d x))}} \] Input:
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]
Output:
(a^3*(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*C os[5*(c + d*x)] + 3212*B*Cos[5*(c + d*x)] + 2840*C*Cos[5*(c + d*x)])*Sec[c + d*x]^5*Tan[c + d*x])/(13860*d*Sqrt[a*(1 + Sec[c + d*x])])
Time = 1.37 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.03, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.326, Rules used = {3042, 4576, 27, 3042, 4498, 27, 3042, 4489, 3042, 4280, 3042, 4280, 3042, 4279}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \sec ^2(c+d x) (a \sec (c+d x)+a)^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 4576 |
| \(\displaystyle \frac {2 \int \frac {1}{2} \sec ^2(c+d x) (\sec (c+d x) a+a)^{5/2} (a (11 A+4 C)+a (11 B+5 C) \sec (c+d x))dx}{11 a}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \sec ^2(c+d x) (\sec (c+d x) a+a)^{5/2} (a (11 A+4 C)+a (11 B+5 C) \sec (c+d x))dx}{11 a}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (a (11 A+4 C)+a (11 B+5 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{11 a}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d}\) |
| \(\Big \downarrow \) 4498 |
| \(\displaystyle \frac {\frac {2 \int \frac {1}{2} \sec (c+d x) (\sec (c+d x) a+a)^{5/2} \left (7 (11 B+5 C) a^2+(99 A-22 B+26 C) \sec (c+d x) a^2\right )dx}{9 a}+\frac {2 (11 B+5 C) \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{9 d}}{11 a}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \sec (c+d x) (\sec (c+d x) a+a)^{5/2} \left (7 (11 B+5 C) a^2+(99 A-22 B+26 C) \sec (c+d x) a^2\right )dx}{9 a}+\frac {2 (11 B+5 C) \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{9 d}}{11 a}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (7 (11 B+5 C) a^2+(99 A-22 B+26 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )dx}{9 a}+\frac {2 (11 B+5 C) \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{9 d}}{11 a}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d}\) |
| \(\Big \downarrow \) 4489 |
| \(\displaystyle \frac {\frac {\frac {3}{7} a^2 (165 A+143 B+125 C) \int \sec (c+d x) (\sec (c+d x) a+a)^{5/2}dx+\frac {2 a^2 (99 A-22 B+26 C) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}}{9 a}+\frac {2 (11 B+5 C) \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{9 d}}{11 a}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {3}{7} a^2 (165 A+143 B+125 C) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2}dx+\frac {2 a^2 (99 A-22 B+26 C) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}}{9 a}+\frac {2 (11 B+5 C) \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{9 d}}{11 a}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d}\) |
| \(\Big \downarrow \) 4280 |
| \(\displaystyle \frac {\frac {\frac {3}{7} a^2 (165 A+143 B+125 C) \left (\frac {8}{5} a \int \sec (c+d x) (\sec (c+d x) a+a)^{3/2}dx+\frac {2 a \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\right )+\frac {2 a^2 (99 A-22 B+26 C) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}}{9 a}+\frac {2 (11 B+5 C) \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{9 d}}{11 a}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {3}{7} a^2 (165 A+143 B+125 C) \left (\frac {8}{5} a \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}dx+\frac {2 a \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\right )+\frac {2 a^2 (99 A-22 B+26 C) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}}{9 a}+\frac {2 (11 B+5 C) \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{9 d}}{11 a}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d}\) |
| \(\Big \downarrow \) 4280 |
| \(\displaystyle \frac {\frac {\frac {3}{7} a^2 (165 A+143 B+125 C) \left (\frac {8}{5} a \left (\frac {4}{3} a \int \sec (c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {2 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {2 a \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\right )+\frac {2 a^2 (99 A-22 B+26 C) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}}{9 a}+\frac {2 (11 B+5 C) \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{9 d}}{11 a}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {3}{7} a^2 (165 A+143 B+125 C) \left (\frac {8}{5} a \left (\frac {4}{3} a \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {2 a \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\right )+\frac {2 a^2 (99 A-22 B+26 C) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}}{9 a}+\frac {2 (11 B+5 C) \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{9 d}}{11 a}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d}\) |
| \(\Big \downarrow \) 4279 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (99 A-22 B+26 C) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}+\frac {3}{7} a^2 (165 A+143 B+125 C) \left (\frac {8}{5} a \left (\frac {8 a^2 \tan (c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {2 a \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\right )}{9 a}+\frac {2 (11 B+5 C) \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{9 d}}{11 a}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d}\) |
Input:
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]
Output:
(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])/(9*d) + ((2*a^2*(99* A - 22*B + 26*C)*(a + a*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(7*d) + (3*a^2*( 165*A + 143*B + 125*C)*((2*a*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d ) + (8*a*((8*a^2*Tan[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*Sqrt[ a + a*Sec[c + d*x]]*Tan[c + d*x])/(3*d)))/5))/7)/(9*a))/(11*a)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S ymbol] :> Simp[-2*b*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]])), x] /; Free Q[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]
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] + Simp[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] && Intege rQ[2*m]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]*(( a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[(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)]
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] + Simp[1/(b*(m + 2)) Int [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] /; FreeQ[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a *B, 0] && !LtQ[m, -1]
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*Cs c[e + f*x])^n/(f*(m + n + 1))), x] + Simp[1/(b*(m + n + 1)) Int[(a + b*Cs c[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]
Time = 231.09 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.81
| method | result | size |
| default | \(\frac {2 a^{2} \left (\cos \left (d x +c \right )^{2} \left (7590 \cos \left (d x +c \right )^{3}+3795 \cos \left (d x +c \right )^{2}+1980 \cos \left (d x +c \right )+495\right ) A +\cos \left (d x +c \right ) \left (6424 \cos \left (d x +c \right )^{4}+3212 \cos \left (d x +c \right )^{3}+2409 \cos \left (d x +c \right )^{2}+1430 \cos \left (d x +c \right )+385\right ) B +\left (5680 \cos \left (d x +c \right )^{5}+2840 \cos \left (d x +c \right )^{4}+2130 \cos \left (d x +c \right )^{3}+1775 \cos \left (d x +c \right )^{2}+1120 \cos \left (d x +c \right )+315\right ) C \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{4}}{3465 d \left (\cos \left (d x +c \right )+1\right )}\) | \(186\) |
| parts | \(\frac {2 A \,a^{2} \left (46 \cos \left (d x +c \right )^{3}+23 \cos \left (d x +c \right )^{2}+12 \cos \left (d x +c \right )+3\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}}{21 d \left (\cos \left (d x +c \right )+1\right )}+\frac {2 B \,a^{2} \left (584 \cos \left (d x +c \right )^{4}+292 \cos \left (d x +c \right )^{3}+219 \cos \left (d x +c \right )^{2}+130 \cos \left (d x +c \right )+35\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{3}}{315 d \left (\cos \left (d x +c \right )+1\right )}+\frac {2 C \,a^{2} \left (1136 \cos \left (d x +c \right )^{5}+568 \cos \left (d x +c \right )^{4}+426 \cos \left (d x +c \right )^{3}+355 \cos \left (d x +c \right )^{2}+224 \cos \left (d x +c \right )+63\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{4}}{693 d \left (\cos \left (d x +c \right )+1\right )}\) | \(257\) |
Input:
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, method=_RETURNVERBOSE)
Output:
2/3465/d*a^2*(cos(d*x+c)^2*(7590*cos(d*x+c)^3+3795*cos(d*x+c)^2+1980*cos(d *x+c)+495)*A+cos(d*x+c)*(6424*cos(d*x+c)^4+3212*cos(d*x+c)^3+2409*cos(d*x+ c)^2+1430*cos(d*x+c)+385)*B+(5680*cos(d*x+c)^5+2840*cos(d*x+c)^4+2130*cos( d*x+c)^3+1775*cos(d*x+c)^2+1120*cos(d*x+c)+315)*C)*(a*(1+sec(d*x+c)))^(1/2 )/(cos(d*x+c)+1)*tan(d*x+c)*sec(d*x+c)^4
Time = 0.09 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.73 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (2 \, {\left (3795 \, A + 3212 \, B + 2840 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} + {\left (3795 \, A + 3212 \, B + 2840 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 3 \, {\left (660 \, A + 803 \, B + 710 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 5 \, {\left (99 \, A + 286 \, B + 355 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 35 \, {\left (11 \, B + 32 \, C\right )} a^{2} \cos \left (d x + c\right ) + 315 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3465 \, {\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\right )}} \] Input:
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")
Output:
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*(660*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*co s(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)
Timed out. \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:
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)
Output:
Timed out
Timed out. \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:
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")
Output:
Timed out
Time = 1.23 (sec) , antiderivative size = 383, normalized size of antiderivative = 1.67 \[ \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 {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} \] Input:
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")
Output:
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*s gn(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))))*ta n(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)
Time = 23.07 (sec) , antiderivative size = 1034, normalized size of antiderivative = 4.52 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \] Input:
int(((a + a/cos(c + d*x))^(5/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/c os(c + d*x)^2,x)
Output:
((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)*(ex p(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*1 i + 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)*4i)/(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*...
\[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\sqrt {a}\, a^{2} \left (\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{6}d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{5}d x \right ) b +2 \left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{5}d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{4}d x \right ) a +2 \left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{4}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{4}d x \right ) c +2 \left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{3}d x \right ) a +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{3}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{2}d x \right ) a \right ) \] Input:
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)
Output:
sqrt(a)*a**2*(int(sqrt(sec(c + d*x) + 1)*sec(c + d*x)**6,x)*c + int(sqrt(s ec(c + d*x) + 1)*sec(c + d*x)**5,x)*b + 2*int(sqrt(sec(c + d*x) + 1)*sec(c + d*x)**5,x)*c + int(sqrt(sec(c + d*x) + 1)*sec(c + d*x)**4,x)*a + 2*int( sqrt(sec(c + d*x) + 1)*sec(c + d*x)**4,x)*b + int(sqrt(sec(c + d*x) + 1)*s ec(c + d*x)**4,x)*c + 2*int(sqrt(sec(c + d*x) + 1)*sec(c + d*x)**3,x)*a + int(sqrt(sec(c + d*x) + 1)*sec(c + d*x)**3,x)*b + int(sqrt(sec(c + d*x) + 1)*sec(c + d*x)**2,x)*a)