Integrand size = 35, antiderivative size = 228 \[ \int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sec ^{\frac {9}{2}}(c+d x)} \, dx=\frac {2 a^2 (10 A+9 B) \sin (c+d x)}{63 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (34 A+39 B) \sin (c+d x)}{105 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {8 a^2 (34 A+39 B) \sin (c+d x)}{315 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {16 a^2 (34 A+39 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{315 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a A \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x)} \] Output:
2/63*a^2*(10*A+9*B)*sin(d*x+c)/d/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(1/2)+2 /105*a^2*(34*A+39*B)*sin(d*x+c)/d/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(1/2)+ 8/315*a^2*(34*A+39*B)*sin(d*x+c)/d/sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2) +16/315*a^2*(34*A+39*B)*sec(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/ 2)+2/9*a*A*(a+a*sec(d*x+c))^(1/2)*sin(d*x+c)/d/sec(d*x+c)^(7/2)
Time = 6.58 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.50 \[ \int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sec ^{\frac {9}{2}}(c+d x)} \, dx=\frac {a (2689 A+2964 B+2 (799 A+759 B) \cos (c+d x)+(548 A+468 B) \cos (2 (c+d x))+170 A \cos (3 (c+d x))+90 B \cos (3 (c+d x))+35 A \cos (4 (c+d x))) \sqrt {a (1+\sec (c+d x))} \tan \left (\frac {1}{2} (c+d x)\right )}{1260 d \sqrt {\sec (c+d x)}} \] Input:
Integrate[((a + a*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x]))/Sec[c + d*x]^( 9/2),x]
Output:
(a*(2689*A + 2964*B + 2*(799*A + 759*B)*Cos[c + d*x] + (548*A + 468*B)*Cos [2*(c + d*x)] + 170*A*Cos[3*(c + d*x)] + 90*B*Cos[3*(c + d*x)] + 35*A*Cos[ 4*(c + d*x)])*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d*x)/2])/(1260*d*Sqrt[Se c[c + d*x]])
Time = 1.17 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.98, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.314, Rules used = {3042, 4505, 27, 3042, 4503, 3042, 4292, 3042, 4292, 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.
| \(\displaystyle \int \frac {(a \sec (c+d x)+a)^{3/2} (A+B \sec (c+d x))}{\sec ^{\frac {9}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{9/2}}dx\) |
| \(\Big \downarrow \) 4505 |
| \(\displaystyle \frac {2}{9} \int \frac {\sqrt {\sec (c+d x) a+a} (a (10 A+9 B)+3 a (2 A+3 B) \sec (c+d x))}{2 \sec ^{\frac {7}{2}}(c+d x)}dx+\frac {2 a A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{9} \int \frac {\sqrt {\sec (c+d x) a+a} (a (10 A+9 B)+3 a (2 A+3 B) \sec (c+d x))}{\sec ^{\frac {7}{2}}(c+d x)}dx+\frac {2 a A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{9} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (a (10 A+9 B)+3 a (2 A+3 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx+\frac {2 a A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4503 |
| \(\displaystyle \frac {1}{9} \left (\frac {3}{7} a (34 A+39 B) \int \frac {\sqrt {\sec (c+d x) a+a}}{\sec ^{\frac {5}{2}}(c+d x)}dx+\frac {2 a^2 (10 A+9 B) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{9} \left (\frac {3}{7} a (34 A+39 B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 a^2 (10 A+9 B) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4292 |
| \(\displaystyle \frac {1}{9} \left (\frac {3}{7} a (34 A+39 B) \left (\frac {4}{5} \int \frac {\sqrt {\sec (c+d x) a+a}}{\sec ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 (10 A+9 B) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{9} \left (\frac {3}{7} a (34 A+39 B) \left (\frac {4}{5} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 a \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 (10 A+9 B) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4292 |
| \(\displaystyle \frac {1}{9} \left (\frac {3}{7} a (34 A+39 B) \left (\frac {4}{5} \left (\frac {2}{3} \int \frac {\sqrt {\sec (c+d x) a+a}}{\sqrt {\sec (c+d x)}}dx+\frac {2 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 (10 A+9 B) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{9} \left (\frac {3}{7} a (34 A+39 B) \left (\frac {4}{5} \left (\frac {2}{3} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 (10 A+9 B) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4291 |
| \(\displaystyle \frac {1}{9} \left (\frac {2 a^2 (10 A+9 B) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {3}{7} a (34 A+39 B) \left (\frac {2 a \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {4}{5} \left (\frac {4 a \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )\right )\right )+\frac {2 a A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\) |
Input:
Int[((a + a*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x]))/Sec[c + d*x]^(9/2),x ]
Output:
(2*a*A*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(9*d*Sec[c + d*x]^(7/2)) + ( (2*a^2*(10*A + 9*B)*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2)*Sqrt[a + a*Sec[c + d*x]]) + (3*a*(34*A + 39*B)*((2*a*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2) *Sqrt[a + a*Sec[c + d*x]]) + (4*((2*a*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x] ]*Sqrt[a + a*Sec[c + d*x]]) + (4*a*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(3*d*S qrt[a + a*Sec[c + d*x]])))/5))/7)/9
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_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[a*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*n*Sqrt[a + b*Csc[e + f*x]])), x] + Simp[a*((2*n + 1)/(2*b*d*n)) Int[Sqrt[a + b*Csc [e + f*x]]*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[n, -2^(-1)] && IntegerQ[2*n]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*b^2*Co t[e + f*x]*((d*Csc[e + f*x])^n/(a*f*n*Sqrt[a + b*Csc[e + f*x]])), x] + Simp [(A*b*(2*n + 1) + 2*a*B*n)/(2*a*d*n) Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[ e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, A, B}, 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[a*A*Cot [e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*n)), x] - Sim p[b/(a*d*n) Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Sim p[a*A*(m - n - 1) - b*B*n - (a*B*n + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0 ] && GtQ[m, 1/2] && LtQ[n, -1]
Time = 2.31 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.52
| method | result | size |
| default | \(\frac {2 a \left (\left (35 \cos \left (d x +c \right )^{4}+85 \cos \left (d x +c \right )^{3}+102 \cos \left (d x +c \right )^{2}+136 \cos \left (d x +c \right )+272\right ) A +\left (45 \cos \left (d x +c \right )^{3}+117 \cos \left (d x +c \right )^{2}+156 \cos \left (d x +c \right )+312\right ) B \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{315 d \left (1+\cos \left (d x +c \right )\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}\) | \(118\) |
| parts | \(\frac {2 A a \left (35 \cos \left (d x +c \right )^{4}+85 \cos \left (d x +c \right )^{3}+102 \cos \left (d x +c \right )^{2}+136 \cos \left (d x +c \right )+272\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{315 d \left (1+\cos \left (d x +c \right )\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}+\frac {2 B a \left (15 \cos \left (d x +c \right )^{3}+39 \cos \left (d x +c \right )^{2}+52 \cos \left (d x +c \right )+104\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{105 d \left (1+\cos \left (d x +c \right )\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}\) | \(158\) |
Input:
int((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(9/2),x,method=_RET URNVERBOSE)
Output:
2/315/d*a*((35*cos(d*x+c)^4+85*cos(d*x+c)^3+102*cos(d*x+c)^2+136*cos(d*x+c )+272)*A+(45*cos(d*x+c)^3+117*cos(d*x+c)^2+156*cos(d*x+c)+312)*B)*(a*(1+se c(d*x+c)))^(1/2)/(1+cos(d*x+c))/sec(d*x+c)^(3/2)*tan(d*x+c)
Time = 0.09 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.58 \[ \int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sec ^{\frac {9}{2}}(c+d x)} \, dx=\frac {2 \, {\left (35 \, A a \cos \left (d x + c\right )^{5} + 5 \, {\left (17 \, A + 9 \, B\right )} a \cos \left (d x + c\right )^{4} + 3 \, {\left (34 \, A + 39 \, B\right )} a \cos \left (d x + c\right )^{3} + 4 \, {\left (34 \, A + 39 \, B\right )} a \cos \left (d x + c\right )^{2} + 8 \, {\left (34 \, A + 39 \, B\right )} a \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right ) + d\right )} \sqrt {\cos \left (d x + c\right )}} \] Input:
integrate((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(9/2),x, algo rithm="fricas")
Output:
2/315*(35*A*a*cos(d*x + c)^5 + 5*(17*A + 9*B)*a*cos(d*x + c)^4 + 3*(34*A + 39*B)*a*cos(d*x + c)^3 + 4*(34*A + 39*B)*a*cos(d*x + c)^2 + 8*(34*A + 39* B)*a*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/(( d*cos(d*x + c) + d)*sqrt(cos(d*x + c)))
Timed out. \[ \int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sec ^{\frac {9}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:
integrate((a+a*sec(d*x+c))**(3/2)*(A+B*sec(d*x+c))/sec(d*x+c)**(9/2),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 700 vs. \(2 (198) = 396\).
Time = 0.29 (sec) , antiderivative size = 700, normalized size of antiderivative = 3.07 \[ \int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sec ^{\frac {9}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:
integrate((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(9/2),x, algo rithm="maxima")
Output:
1/5040*(sqrt(2)*(3780*a*cos(8/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c)))*sin(9/2*d*x + 9/2*c) + 1050*a*cos(2/3*arctan2(sin(9/2*d*x + 9/2 *c), cos(9/2*d*x + 9/2*c)))*sin(9/2*d*x + 9/2*c) + 378*a*cos(4/9*arctan2(s in(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c)))*sin(9/2*d*x + 9/2*c) + 135*a*c os(2/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c)))*sin(9/2*d*x + 9/2*c) - 3780*a*cos(9/2*d*x + 9/2*c)*sin(8/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) - 1050*a*cos(9/2*d*x + 9/2*c)*sin(2/3*arctan2(sin( 9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) - 378*a*cos(9/2*d*x + 9/2*c)*sin( 4/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) - 135*a*cos(9/2*d *x + 9/2*c)*sin(2/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 70*a*sin(9/2*d*x + 9/2*c) + 135*a*sin(7/9*arctan2(sin(9/2*d*x + 9/2*c), c os(9/2*d*x + 9/2*c))) + 378*a*sin(5/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/ 2*d*x + 9/2*c))) + 1050*a*sin(1/3*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d* x + 9/2*c))) + 3780*a*sin(1/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))))*A*sqrt(a) + 6*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/...
Time = 1.90 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.04 \[ \int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sec ^{\frac {9}{2}}(c+d x)} \, dx=\frac {4 \, {\left ({\left ({\left ({\left (2 \, \sqrt {2} {\left (47 \, A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 57 \, B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 9 \, \sqrt {2} {\left (47 \, A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 57 \, B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 819 \, \sqrt {2} {\left (A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 105 \, \sqrt {2} {\left (5 \, A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 7 \, B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 315 \, \sqrt {2} {\left (A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{315 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {9}{2}} d} \] Input:
integrate((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(9/2),x, algo rithm="giac")
Output:
4/315*((((2*sqrt(2)*(47*A*a^6*sgn(cos(d*x + c)) + 57*B*a^6*sgn(cos(d*x + c )))*tan(1/2*d*x + 1/2*c)^2 + 9*sqrt(2)*(47*A*a^6*sgn(cos(d*x + c)) + 57*B* a^6*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 819*sqrt(2)*(A*a^6*sgn(co s(d*x + c)) + B*a^6*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 105*sqrt( 2)*(5*A*a^6*sgn(cos(d*x + c)) + 7*B*a^6*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 315*sqrt(2)*(A*a^6*sgn(cos(d*x + c)) + B*a^6*sgn(cos(d*x + c))) )*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 + a)^(9/2)*d)
Time = 14.65 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.68 \[ \int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sec ^{\frac {9}{2}}(c+d x)} \, dx=\frac {a\,\cos \left (c+d\,x\right )\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\sqrt {\frac {a\,\left (\cos \left (c+d\,x\right )+1\right )}{\cos \left (c+d\,x\right )}}\,\left (4830\,A\,\sin \left (c+d\,x\right )+5460\,B\,\sin \left (c+d\,x\right )+1428\,A\,\sin \left (2\,c+2\,d\,x\right )+513\,A\,\sin \left (3\,c+3\,d\,x\right )+170\,A\,\sin \left (4\,c+4\,d\,x\right )+35\,A\,\sin \left (5\,c+5\,d\,x\right )+1428\,B\,\sin \left (2\,c+2\,d\,x\right )+468\,B\,\sin \left (3\,c+3\,d\,x\right )+90\,B\,\sin \left (4\,c+4\,d\,x\right )\right )}{2520\,d\,\left (\cos \left (c+d\,x\right )+1\right )} \] Input:
int(((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^(3/2))/(1/cos(c + d*x))^(9/ 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)*(4830*A*sin(c + d*x) + 5460*B*sin(c + d*x) + 1428*A*sin(2*c + 2*d *x) + 513*A*sin(3*c + 3*d*x) + 170*A*sin(4*c + 4*d*x) + 35*A*sin(5*c + 5*d *x) + 1428*B*sin(2*c + 2*d*x) + 468*B*sin(3*c + 3*d*x) + 90*B*sin(4*c + 4* d*x)))/(2520*d*(cos(c + d*x) + 1))
\[ \int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sec ^{\frac {9}{2}}(c+d x)} \, dx=\sqrt {a}\, a \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{5}}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{4}}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{4}}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{3}}d x \right ) b \right ) \] Input:
int((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(9/2),x)
Output:
sqrt(a)*a*(int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1))/sec(c + d*x)**5 ,x)*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)**4,x)*b + int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1))/sec(c + d*x)**3,x)*b)