Integrand size = 45, antiderivative size = 181 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {8 a^2 (19 A+21 B+35 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (19 A+21 B+35 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{105 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 (3 A+7 B) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)} \] Output:
8/105*a^2*(19*A+21*B+35*C)*sec(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^ (1/2)+2/105*a*(19*A+21*B+35*C)*(a+a*sec(d*x+c))^(1/2)*sin(d*x+c)/d/sec(d*x +c)^(1/2)+2/7*A*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d/sec(d*x+c)^(5/2)+2/35* (3*A+7*B)*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d/sec(d*x+c)^(3/2)
Time = 6.57 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.55 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {a (494 A+546 B+700 C+(253 A+28 (9 B+5 C)) \cos (c+d x)+6 (13 A+7 B) \cos (2 (c+d x))+15 A \cos (3 (c+d x))) \sqrt {a (1+\sec (c+d x))} \tan \left (\frac {1}{2} (c+d x)\right )}{210 d \sqrt {\sec (c+d x)}} \] Input:
Integrate[((a + a*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x] ^2))/Sec[c + d*x]^(7/2),x]
Output:
(a*(494*A + 546*B + 700*C + (253*A + 28*(9*B + 5*C))*Cos[c + d*x] + 6*(13* A + 7*B)*Cos[2*(c + d*x)] + 15*A*Cos[3*(c + d*x)])*Sqrt[a*(1 + Sec[c + d*x ])]*Tan[(c + d*x)/2])/(210*d*Sqrt[Sec[c + d*x]])
Time = 0.98 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4574, 27, 3042, 4501, 3042, 4296, 3042, 4291}
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.
| \(\displaystyle \int \frac {(a \sec (c+d x)+a)^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx\) |
| \(\Big \downarrow \) 4574 |
| \(\displaystyle \frac {2 \int \frac {(\sec (c+d x) a+a)^{3/2} (a (3 A+7 B)+a (2 A+7 C) \sec (c+d x))}{2 \sec ^{\frac {5}{2}}(c+d x)}dx}{7 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {(\sec (c+d x) a+a)^{3/2} (a (3 A+7 B)+a (2 A+7 C) \sec (c+d x))}{\sec ^{\frac {5}{2}}(c+d x)}dx}{7 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (a (3 A+7 B)+a (2 A+7 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx}{7 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4501 |
| \(\displaystyle \frac {\frac {1}{5} a (19 A+21 B+35 C) \int \frac {(\sec (c+d x) a+a)^{3/2}}{\sec ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a (3 A+7 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} a (19 A+21 B+35 C) \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 a (3 A+7 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4296 |
| \(\displaystyle \frac {\frac {1}{5} a (19 A+21 B+35 C) \left (\frac {4}{3} a \int \frac {\sqrt {\sec (c+d x) a+a}}{\sqrt {\sec (c+d x)}}dx+\frac {2 a \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a (3 A+7 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} a (19 A+21 B+35 C) \left (\frac {4}{3} a \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a (3 A+7 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4291 |
| \(\displaystyle \frac {\frac {1}{5} a (19 A+21 B+35 C) \left (\frac {8 a^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a (3 A+7 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\) |
Input:
Int[((a + a*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/S ec[c + d*x]^(7/2),x]
Output:
(2*A*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2)) + ( (2*a*(3*A + 7*B)*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(5*d*Sec[c + d*x ]^(3/2)) + (a*(19*A + 21*B + 35*C)*((8*a^2*Sqrt[Sec[c + d*x]]*Sin[c + d*x] )/(3*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d *x])/(3*d*Sqrt[Sec[c + d*x]])))/5)/(7*a)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)] *(d_.)], x_Symbol] :> Simp[-2*a*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]]*S qrt[d*Csc[e + f*x]])), x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_), x_Symbol] :> Simp[(-a)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1) *((d*Csc[e + f*x])^n/(f*m)), x] + Simp[b*((2*m - 1)/(d*m)) Int[(a + b*Csc [e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f , m, n}, x] && EqQ[a^2 - b^2, 0] && EqQ[m + n, 0] && GtQ[m, 1/2] && Integer Q[2*m]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[(a*A*m - b*B*n)/(b*d*n) Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1), x] , x] /; FreeQ[{a, b, d, e, f, A, B, m, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a ^2 - b^2, 0] && EqQ[m + n + 1, 0] && !LeQ[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[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[1/(b*d*n) Int[(a + b*Csc[e + f*x])^m*(d*Csc[ e + f*x])^(n + 1)*Simp[a*A*m - b*B*n - b*(A*(m + n + 1) + C*n)*Csc[e + f*x] , x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || EqQ[m + n + 1, 0])
Time = 5.72 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.61
| method | result | size |
| default | \(\frac {2 a \left (\left (15 \cos \left (d x +c \right )^{3}+39 \cos \left (d x +c \right )^{2}+52 \cos \left (d x +c \right )+104\right ) A +\left (21 \cos \left (d x +c \right )^{2}+63 \cos \left (d x +c \right )+126\right ) B +\left (35 \cos \left (d x +c \right )+175\right ) C \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{105 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}\) | \(110\) |
| parts | \(\frac {2 A a \left (15 \cos \left (d x +c \right )^{3}+39 \cos \left (d x +c \right )^{2}+52 \cos \left (d x +c \right )+104\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{105 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}+\frac {B \left (2 \sin \left (d x +c \right )+6 \tan \left (d x +c \right )+12 \sec \left (d x +c \right ) \tan \left (d x +c \right )\right ) a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d \left (5 \cos \left (d x +c \right )+5\right ) \sec \left (d x +c \right )^{\frac {5}{2}}}+\frac {C \left (2 \sin \left (d x +c \right )+10 \tan \left (d x +c \right )\right ) a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d \left (3 \cos \left (d x +c \right )+3\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}\) | \(199\) |
Input:
int((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(7/2 ),x,method=_RETURNVERBOSE)
Output:
2/105/d*a*((15*cos(d*x+c)^3+39*cos(d*x+c)^2+52*cos(d*x+c)+104)*A+(21*cos(d *x+c)^2+63*cos(d*x+c)+126)*B+(35*cos(d*x+c)+175)*C)*(a*(1+sec(d*x+c)))^(1/ 2)/(cos(d*x+c)+1)/sec(d*x+c)^(3/2)*tan(d*x+c)
Time = 0.08 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.65 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 \, {\left (15 \, A a \cos \left (d x + c\right )^{4} + 3 \, {\left (13 \, A + 7 \, B\right )} a \cos \left (d x + c\right )^{3} + {\left (52 \, A + 63 \, B + 35 \, C\right )} a \cos \left (d x + c\right )^{2} + {\left (104 \, A + 126 \, B + 175 \, C\right )} a \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right ) + d\right )} \sqrt {\cos \left (d x + c\right )}} \] Input:
integrate((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c )^(7/2),x, algorithm="fricas")
Output:
2/105*(15*A*a*cos(d*x + c)^4 + 3*(13*A + 7*B)*a*cos(d*x + c)^3 + (52*A + 6 3*B + 35*C)*a*cos(d*x + c)^2 + (104*A + 126*B + 175*C)*a*cos(d*x + c))*sqr t((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/((d*cos(d*x + c) + d)*sq rt(cos(d*x + c)))
Timed out. \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:
integrate((a+a*sec(d*x+c))**(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/sec(d*x +c)**(7/2),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 550 vs. \(2 (157) = 314\).
Time = 0.33 (sec) , antiderivative size = 550, normalized size of antiderivative = 3.04 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:
integrate((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c )^(7/2),x, algorithm="maxima")
Output:
1/840*(sqrt(2)*(735*a*cos(6/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))*sin(7/2*d*x + 7/2*c) + 175*a*cos(4/7*arctan2(sin(7/2*d*x + 7/2*c) , cos(7/2*d*x + 7/2*c)))*sin(7/2*d*x + 7/2*c) + 63*a*cos(2/7*arctan2(sin(7 /2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))*sin(7/2*d*x + 7/2*c) - 735*a*cos(7 /2*d*x + 7/2*c)*sin(6/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c) )) - 175*a*cos(7/2*d*x + 7/2*c)*sin(4/7*arctan2(sin(7/2*d*x + 7/2*c), cos( 7/2*d*x + 7/2*c))) - 63*a*cos(7/2*d*x + 7/2*c)*sin(2/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 30*a*sin(7/2*d*x + 7/2*c) + 63*a*sin(5 /7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 175*a*sin(3/7*ar ctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 735*a*sin(1/7*arctan2 (sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))))*A*sqrt(a) + 42*sqrt(2)*(20* a*cos(4/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))*sin(5/2*d*x + 5/2*c) + 5*a*cos(2/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c) ))*sin(5/2*d*x + 5/2*c) - 20*a*cos(5/2*d*x + 5/2*c)*sin(4/5*arctan2(sin(5/ 2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) - 5*a*cos(5/2*d*x + 5/2*c)*sin(2/5* arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 2*a*sin(5/2*d*x + 5 /2*c) + 5*a*sin(3/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 20*a*sin(1/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))))*B*sqrt (a) + 280*(sqrt(2)*a*sin(3/2*d*x + 3/2*c) + 9*sqrt(2)*a*sin(1/2*d*x + 1/2* c))*C*sqrt(a))/d
Time = 2.34 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.46 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {4 \, {\left (105 \, \sqrt {2} A a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 105 \, \sqrt {2} B a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 105 \, \sqrt {2} C a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + {\left (140 \, \sqrt {2} A a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 210 \, \sqrt {2} B a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 280 \, \sqrt {2} C a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + {\left (133 \, \sqrt {2} A a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 147 \, \sqrt {2} B a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 245 \, \sqrt {2} C a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 2 \, {\left (19 \, \sqrt {2} A a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 21 \, \sqrt {2} B a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 35 \, \sqrt {2} C a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{105 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {7}{2}} d} \] Input:
integrate((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c )^(7/2),x, algorithm="giac")
Output:
4/105*(105*sqrt(2)*A*a^5*sgn(cos(d*x + c)) + 105*sqrt(2)*B*a^5*sgn(cos(d*x + c)) + 105*sqrt(2)*C*a^5*sgn(cos(d*x + c)) + (140*sqrt(2)*A*a^5*sgn(cos( d*x + c)) + 210*sqrt(2)*B*a^5*sgn(cos(d*x + c)) + 280*sqrt(2)*C*a^5*sgn(co s(d*x + c)) + (133*sqrt(2)*A*a^5*sgn(cos(d*x + c)) + 147*sqrt(2)*B*a^5*sgn (cos(d*x + c)) + 245*sqrt(2)*C*a^5*sgn(cos(d*x + c)) + 2*(19*sqrt(2)*A*a^5 *sgn(cos(d*x + c)) + 21*sqrt(2)*B*a^5*sgn(cos(d*x + c)) + 35*sqrt(2)*C*a^5 *sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*tan(1/ 2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 + a)^(7/ 2)*d)
Time = 15.15 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.84 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {a\,\cos \left (c+d\,x\right )\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\sqrt {\frac {a\,\left (\cos \left (c+d\,x\right )+1\right )}{\cos \left (c+d\,x\right )}}\,\left (910\,A\,\sin \left (c+d\,x\right )+1050\,B\,\sin \left (c+d\,x\right )+1400\,C\,\sin \left (c+d\,x\right )+238\,A\,\sin \left (2\,c+2\,d\,x\right )+78\,A\,\sin \left (3\,c+3\,d\,x\right )+15\,A\,\sin \left (4\,c+4\,d\,x\right )+252\,B\,\sin \left (2\,c+2\,d\,x\right )+42\,B\,\sin \left (3\,c+3\,d\,x\right )+140\,C\,\sin \left (2\,c+2\,d\,x\right )\right )}{420\,d\,\left (\cos \left (c+d\,x\right )+1\right )} \] Input:
int(((a + a/cos(c + d*x))^(3/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/( 1/cos(c + d*x))^(7/2),x)
Output:
(a*cos(c + d*x)*(1/cos(c + d*x))^(1/2)*((a*(cos(c + d*x) + 1))/cos(c + d*x ))^(1/2)*(910*A*sin(c + d*x) + 1050*B*sin(c + d*x) + 1400*C*sin(c + d*x) + 238*A*sin(2*c + 2*d*x) + 78*A*sin(3*c + 3*d*x) + 15*A*sin(4*c + 4*d*x) + 252*B*sin(2*c + 2*d*x) + 42*B*sin(3*c + 3*d*x) + 140*C*sin(2*c + 2*d*x)))/ (420*d*(cos(c + d*x) + 1))
\[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\sqrt {a}\, a \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{4}}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{3}}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{3}}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{2}}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{2}}d x \right ) c +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )}d x \right ) c \right ) \] Input:
int((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(7/2 ),x)
Output:
sqrt(a)*a*(int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1))/sec(c + d*x)**4 ,x)*a + int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1))/sec(c + d*x)**3,x) *a + int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1))/sec(c + d*x)**3,x)*b + int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1))/sec(c + d*x)**2,x)*b + i nt((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1))/sec(c + d*x)**2,x)*c + int( (sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1))/sec(c + d*x),x)*c)