Integrand size = 42, antiderivative size = 237 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 a^3 (803 B+710 C) \tan (c+d x)}{495 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (209 B+194 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}-\frac {4 a^2 (803 B+710 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{3465 d}+\frac {2 a^2 (11 B+14 C) \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{99 d}+\frac {2 a (803 B+710 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 d}+\frac {2 a C \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{11 d} \]
2/1155*a*(803*B+710*C)*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d+2/11*a*C*sec(d* x+c)^3*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d+2/495*a^3*(803*B+710*C)*tan(d*x +c)/d/(a+a*sec(d*x+c))^(1/2)+2/693*a^3*(209*B+194*C)*sec(d*x+c)^3*tan(d*x+ c)/d/(a+a*sec(d*x+c))^(1/2)-4/3465*a^2*(803*B+710*C)*(a+a*sec(d*x+c))^(1/2 )*tan(d*x+c)/d+2/99*a^2*(11*B+14*C)*sec(d*x+c)^3*(a+a*sec(d*x+c))^(1/2)*ta n(d*x+c)/d
Time = 4.80 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.49 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 a^3 \left (8 (803 B+710 C)+4 (803 B+710 C) \sec (c+d x)+3 (803 B+710 C) \sec ^2(c+d x)+5 (286 B+355 C) \sec ^3(c+d x)+35 (11 B+32 C) \sec ^4(c+d x)+315 C \sec ^5(c+d x)\right ) \tan (c+d x)}{3465 d \sqrt {a (1+\sec (c+d x))}} \]
(2*a^3*(8*(803*B + 710*C) + 4*(803*B + 710*C)*Sec[c + d*x] + 3*(803*B + 71 0*C)*Sec[c + d*x]^2 + 5*(286*B + 355*C)*Sec[c + d*x]^3 + 35*(11*B + 32*C)* Sec[c + d*x]^4 + 315*C*Sec[c + d*x]^5)*Tan[c + d*x])/(3465*d*Sqrt[a*(1 + S ec[c + d*x])])
Time = 1.58 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.05, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.405, Rules used = {3042, 4560, 3042, 4506, 27, 3042, 4506, 27, 3042, 4504, 3042, 4287, 27, 3042, 4489, 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.
| \(\displaystyle \int \sec ^2(c+d x) (a \sec (c+d x)+a)^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 4560 |
| \(\displaystyle \int \sec ^3(c+d x) (a \sec (c+d x)+a)^{5/2} (B+C \sec (c+d x))dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (B+C \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \frac {2}{11} \int \frac {1}{2} \sec ^3(c+d x) (\sec (c+d x) a+a)^{3/2} (a (11 B+6 C)+a (11 B+14 C) \sec (c+d x))dx+\frac {2 a C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{11} \int \sec ^3(c+d x) (\sec (c+d x) a+a)^{3/2} (a (11 B+6 C)+a (11 B+14 C) \sec (c+d x))dx+\frac {2 a C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{11} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (a (11 B+6 C)+a (11 B+14 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {2 a C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \frac {1}{11} \left (\frac {2}{9} \int \frac {1}{2} \sec ^3(c+d x) \sqrt {\sec (c+d x) a+a} \left (3 (55 B+46 C) a^2+(209 B+194 C) \sec (c+d x) a^2\right )dx+\frac {2 a^2 (11 B+14 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\right )+\frac {2 a C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{11} \left (\frac {1}{9} \int \sec ^3(c+d x) \sqrt {\sec (c+d x) a+a} \left (3 (55 B+46 C) a^2+(209 B+194 C) \sec (c+d x) a^2\right )dx+\frac {2 a^2 (11 B+14 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\right )+\frac {2 a C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{11} \left (\frac {1}{9} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (3 (55 B+46 C) a^2+(209 B+194 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )dx+\frac {2 a^2 (11 B+14 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\right )+\frac {2 a C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}\) |
| \(\Big \downarrow \) 4504 |
| \(\displaystyle \frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^2 (803 B+710 C) \int \sec ^3(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {2 a^3 (209 B+194 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 (11 B+14 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\right )+\frac {2 a C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^2 (803 B+710 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a^3 (209 B+194 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 (11 B+14 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\right )+\frac {2 a C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}\) |
| \(\Big \downarrow \) 4287 |
| \(\displaystyle \frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^2 (803 B+710 C) \left (\frac {2 \int \frac {1}{2} \sec (c+d x) (3 a-2 a \sec (c+d x)) \sqrt {\sec (c+d x) a+a}dx}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a^3 (209 B+194 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 (11 B+14 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\right )+\frac {2 a C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^2 (803 B+710 C) \left (\frac {\int \sec (c+d x) (3 a-2 a \sec (c+d x)) \sqrt {\sec (c+d x) a+a}dx}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a^3 (209 B+194 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 (11 B+14 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\right )+\frac {2 a C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^2 (803 B+710 C) \left (\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (3 a-2 a \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a^3 (209 B+194 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 (11 B+14 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\right )+\frac {2 a C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}\) |
| \(\Big \downarrow \) 4489 |
| \(\displaystyle \frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^2 (803 B+710 C) \left (\frac {\frac {7}{3} a \int \sec (c+d x) \sqrt {\sec (c+d x) a+a}dx-\frac {4 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a^3 (209 B+194 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 (11 B+14 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\right )+\frac {2 a C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^2 (803 B+710 C) \left (\frac {\frac {7}{3} a \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-\frac {4 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a^3 (209 B+194 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 (11 B+14 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\right )+\frac {2 a C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}\) |
| \(\Big \downarrow \) 4279 |
| \(\displaystyle \frac {1}{11} \left (\frac {2 a^2 (11 B+14 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}+\frac {1}{9} \left (\frac {2 a^3 (209 B+194 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}+\frac {3}{7} a^2 (803 B+710 C) \left (\frac {\frac {14 a^2 \tan (c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {4 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )\right )\right )+\frac {2 a C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}\) |
(2*a*C*Sec[c + d*x]^3*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(11*d) + (( 2*a^2*(11*B + 14*C)*Sec[c + d*x]^3*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/ (9*d) + ((2*a^3*(209*B + 194*C)*Sec[c + d*x]^3*Tan[c + d*x])/(7*d*Sqrt[a + a*Sec[c + d*x]]) + (3*a^2*(803*B + 710*C)*((2*(a + a*Sec[c + d*x])^(3/2)* Tan[c + d*x])/(5*a*d) + ((14*a^2*Tan[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x ]]) - (4*a*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(3*d))/(5*a)))/7)/9)/11
3.4.75.3.1 Defintions of rubi rules used
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_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-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*(b*( m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)]
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_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[-2*b*B*C ot[e + f*x]*((d*Csc[e + f*x])^n/(f*(2*n + 1)*Sqrt[a + b*Csc[e + f*x]])), x] + Simp[(A*b*(2*n + 1) + 2*a*B*n)/(b*(2*n + 1)) Int[Sqrt[a + b*Csc[e + f* x]]*(d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ [A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[A*b*(2*n + 1) + 2*a*B*n, 0] && !LtQ[n, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B* Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*(m + n))), x] + Simp[1/(d*(m + n)) Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x] )^n*Simp[a*A*d*(m + n) + B*(b*d*n) + (A*b*d*(m + n) + a*B*d*(2*m + n - 1))* Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] && !LtQ[n, -1]
Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_. )*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.) *(x_)]*(d_.))^(n_.), x_Symbol] :> Simp[1/b^2 Int[(a + b*Csc[e + f*x])^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Time = 198.82 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.65
| method | result | size |
| default | \(\frac {2 a^{2} \left (6424 B \cos \left (d x +c \right )^{5}+5680 C \cos \left (d x +c \right )^{5}+3212 B \cos \left (d x +c \right )^{4}+2840 C \cos \left (d x +c \right )^{4}+2409 B \cos \left (d x +c \right )^{3}+2130 C \cos \left (d x +c \right )^{3}+1430 B \cos \left (d x +c \right )^{2}+1775 C \cos \left (d x +c \right )^{2}+385 B \cos \left (d x +c \right )+1120 C \cos \left (d x +c \right )+315 C \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{4}}{3465 d \left (\cos \left (d x +c \right )+1\right )}\) | \(155\) |
| parts | \(\frac {2 B \,a^{2} \left (584 \cos \left (d x +c \right )^{4}+292 \cos \left (d x +c \right )^{3}+219 \cos \left (d x +c \right )^{2}+130 \cos \left (d x +c \right )+35\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{3}}{315 d \left (\cos \left (d x +c \right )+1\right )}+\frac {2 C \,a^{2} \left (1136 \cos \left (d x +c \right )^{5}+568 \cos \left (d x +c \right )^{4}+426 \cos \left (d x +c \right )^{3}+355 \cos \left (d x +c \right )^{2}+224 \cos \left (d x +c \right )+63\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{4}}{693 d \left (\cos \left (d x +c \right )+1\right )}\) | \(182\) |
int(sec(d*x+c)^2*(a+a*sec(d*x+c))^(5/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x,me thod=_RETURNVERBOSE)
2/3465*a^2/d*(6424*B*cos(d*x+c)^5+5680*C*cos(d*x+c)^5+3212*B*cos(d*x+c)^4+ 2840*C*cos(d*x+c)^4+2409*B*cos(d*x+c)^3+2130*C*cos(d*x+c)^3+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+ sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)*tan(d*x+c)*sec(d*x+c)^4
Time = 0.29 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.66 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (8 \, {\left (803 \, B + 710 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} + 4 \, {\left (803 \, B + 710 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 3 \, {\left (803 \, B + 710 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 5 \, {\left (286 \, B + 355 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 35 \, {\left (11 \, B + 32 \, C\right )} a^{2} \cos \left (d x + c\right ) + 315 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3465 \, {\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\right )}} \]
integrate(sec(d*x+c)^2*(a+a*sec(d*x+c))^(5/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2 ),x, algorithm="fricas")
2/3465*(8*(803*B + 710*C)*a^2*cos(d*x + c)^5 + 4*(803*B + 710*C)*a^2*cos(d *x + c)^4 + 3*(803*B + 710*C)*a^2*cos(d*x + c)^3 + 5*(286*B + 355*C)*a^2*c os(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)
Timed out. \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
Timed out. \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
integrate(sec(d*x+c)^2*(a+a*sec(d*x+c))^(5/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2 ),x, algorithm="maxima")
\[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{2} \,d x } \]
integrate(sec(d*x+c)^2*(a+a*sec(d*x+c))^(5/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2 ),x, algorithm="giac")
Time = 26.59 (sec) , antiderivative size = 855, normalized size of antiderivative = 3.61 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \]
((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*((B*a^2*8i) /(3*d) - (a^2*exp(c*1i + d*x*1i)*(803*B + 710*C)*8i)/(3465*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)) - ((a + a/(exp(- c*1i - d*x*1i)/ 2 + exp(c*1i + d*x*1i)/2))^(1/2)*((B*a^2*24i)/(7*d) - exp(c*1i + d*x*1i)*( (a^2*(5*B + 16*C)*8i)/(7*d) - (a^2*(5*B + 2*C)*8i)/(7*d) + (a^2*(11*B + 50 *C)*32i)/(693*d)) + (a^2*(9*B + 10*C)*8i)/(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)*((B*a^2*8i)/(11*d) + (a^2*(3*B + 4*C)*40i)/(11*d) - (a^2*(5*B + 2*C)*8i)/(11*d) - (a^2*(11*B + 10*C)*8i)/(1 1*d)) + (B*a^2*8i)/(11*d) + (a^2*(3*B + 4*C)*40i)/(11*d) - (a^2*(5*B + 2*C )*8i)/(11*d) - (a^2*(11*B + 10*C)*8i)/(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)*((C*a^2*64i)/(99*d) - (a^2*(B - 8*C) *8i)/(9*d) - (a^2*(5*B + 2*C)*8i)/(9*d) + (a^2*(5*B + 9*C)*16i)/(9*d)) + ( B*a^2*8i)/(9*d) + (C*a^2*64i)/(9*d) + (a^2*(B + 2*C)*40i)/(9*d) - (a^2*(B + C)*80i)/(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^2*(5*B + 2*C)*8i)/(5*d) + (a^2*(44*B - 31*C)*16i)/(1155*d)) - (B*a^2*8i)/(5*d) + (a^2*(4*B + 5*C)*16i)/(5*d)))/((exp(c*1i + d*x*1i) + 1 )*(exp(c*2i + d*x*2i) + 1)^2) - (a^2*exp(c*1i + d*x*1i)*(a + a/(exp(- c...