Integrand size = 35, antiderivative size = 275 \[ \int \frac {(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac {11}{2}}(c+d x)} \, dx=\frac {2 a^3 (194 A+209 B) \sin (c+d x)}{693 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (710 A+803 B) \sin (c+d x)}{1155 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {8 a^3 (710 A+803 B) \sin (c+d x)}{3465 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {16 a^3 (710 A+803 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3465 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (14 A+11 B) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{99 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 a A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)} \] Output:
2/693*a^3*(194*A+209*B)*sin(d*x+c)/d/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(1/ 2)+2/1155*a^3*(710*A+803*B)*sin(d*x+c)/d/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c)) ^(1/2)+8/3465*a^3*(710*A+803*B)*sin(d*x+c)/d/sec(d*x+c)^(1/2)/(a+a*sec(d*x +c))^(1/2)+16/3465*a^3*(710*A+803*B)*sec(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*se c(d*x+c))^(1/2)+2/99*a^2*(14*A+11*B)*(a+a*sec(d*x+c))^(1/2)*sin(d*x+c)/d/s ec(d*x+c)^(7/2)+2/11*a*A*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d/sec(d*x+c)^(9 /2)
Time = 2.85 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.46 \[ \int \frac {(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac {11}{2}}(c+d x)} \, dx=\frac {2 a^3 \left (315 A+35 (32 A+11 B) \sec (c+d x)+5 (355 A+286 B) \sec ^2(c+d x)+3 (710 A+803 B) \sec ^3(c+d x)+4 (710 A+803 B) \sec ^4(c+d x)+8 (710 A+803 B) \sec ^5(c+d x)\right ) \sin (c+d x)}{3465 d \sec ^{\frac {9}{2}}(c+d x) \sqrt {a (1+\sec (c+d x))}} \] Input:
Integrate[((a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]))/Sec[c + d*x]^( 11/2),x]
Output:
(2*a^3*(315*A + 35*(32*A + 11*B)*Sec[c + d*x] + 5*(355*A + 286*B)*Sec[c + d*x]^2 + 3*(710*A + 803*B)*Sec[c + d*x]^3 + 4*(710*A + 803*B)*Sec[c + d*x] ^4 + 8*(710*A + 803*B)*Sec[c + d*x]^5)*Sin[c + d*x])/(3465*d*Sec[c + d*x]^ (9/2)*Sqrt[a*(1 + Sec[c + d*x])])
Time = 1.58 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.01, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4505, 27, 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.
| \(\displaystyle \int \frac {(a \sec (c+d x)+a)^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac {11}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{11/2}}dx\) |
| \(\Big \downarrow \) 4505 |
| \(\displaystyle \frac {2}{11} \int \frac {(\sec (c+d x) a+a)^{3/2} (a (14 A+11 B)+a (6 A+11 B) \sec (c+d x))}{2 \sec ^{\frac {9}{2}}(c+d x)}dx+\frac {2 a A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{11} \int \frac {(\sec (c+d x) a+a)^{3/2} (a (14 A+11 B)+a (6 A+11 B) \sec (c+d x))}{\sec ^{\frac {9}{2}}(c+d x)}dx+\frac {2 a A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{11} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (a (14 A+11 B)+a (6 A+11 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{9/2}}dx+\frac {2 a A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4505 |
| \(\displaystyle \frac {1}{11} \left (\frac {2}{9} \int \frac {\sqrt {\sec (c+d x) a+a} \left ((194 A+209 B) a^2+3 (46 A+55 B) \sec (c+d x) a^2\right )}{2 \sec ^{\frac {7}{2}}(c+d x)}dx+\frac {2 a^2 (14 A+11 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{11} \left (\frac {1}{9} \int \frac {\sqrt {\sec (c+d x) a+a} \left ((194 A+209 B) a^2+3 (46 A+55 B) \sec (c+d x) a^2\right )}{\sec ^{\frac {7}{2}}(c+d x)}dx+\frac {2 a^2 (14 A+11 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{11} \left (\frac {1}{9} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((194 A+209 B) a^2+3 (46 A+55 B) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx+\frac {2 a^2 (14 A+11 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4503 |
| \(\displaystyle \frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^2 (710 A+803 B) \int \frac {\sqrt {\sec (c+d x) a+a}}{\sec ^{\frac {5}{2}}(c+d x)}dx+\frac {2 a^3 (194 A+209 B) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 (14 A+11 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^2 (710 A+803 B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 a^3 (194 A+209 B) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 (14 A+11 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4292 |
| \(\displaystyle \frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^2 (710 A+803 B) \left (\frac {4}{5} \int \frac {\sqrt {\sec (c+d x) a+a}}{\sec ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^3 (194 A+209 B) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 (14 A+11 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^2 (710 A+803 B) \left (\frac {4}{5} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 a \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^3 (194 A+209 B) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 (14 A+11 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4292 |
| \(\displaystyle \frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^2 (710 A+803 B) \left (\frac {4}{5} \left (\frac {2}{3} \int \frac {\sqrt {\sec (c+d x) a+a}}{\sqrt {\sec (c+d x)}}dx+\frac {2 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^3 (194 A+209 B) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 (14 A+11 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^2 (710 A+803 B) \left (\frac {4}{5} \left (\frac {2}{3} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^3 (194 A+209 B) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 (14 A+11 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4291 |
| \(\displaystyle \frac {1}{11} \left (\frac {2 a^2 (14 A+11 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{9 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {1}{9} \left (\frac {2 a^3 (194 A+209 B) \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {3}{7} a^2 (710 A+803 B) \left (\frac {2 a \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {4}{5} \left (\frac {4 a \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )\right )\right )\right )+\frac {2 a A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\) |
Input:
Int[((a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]))/Sec[c + d*x]^(11/2), x]
Output:
(2*a*A*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(11*d*Sec[c + d*x]^(9/2)) + ((2*a^2*(14*A + 11*B)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(9*d*Sec[c + d*x]^(7/2)) + ((2*a^3*(194*A + 209*B)*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5 /2)*Sqrt[a + a*Sec[c + d*x]]) + (3*a^2*(710*A + 803*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*Sqrt[a + a*Sec[c + d*x]])))/5))/7)/9)/11
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 = 2.39 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.51
| method | result | size |
| default | \(\frac {2 a^{2} \left (\left (315 \cos \left (d x +c \right )^{5}+1120 \cos \left (d x +c \right )^{4}+1775 \cos \left (d x +c \right )^{3}+2130 \cos \left (d x +c \right )^{2}+2840 \cos \left (d x +c \right )+5680\right ) A +\left (385 \cos \left (d x +c \right )^{4}+1430 \cos \left (d x +c \right )^{3}+2409 \cos \left (d x +c \right )^{2}+3212 \cos \left (d x +c \right )+6424\right ) B \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{3465 d \left (1+\cos \left (d x +c \right )\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}\) | \(140\) |
| parts | \(\frac {2 A \,a^{2} \left (63 \cos \left (d x +c \right )^{5}+224 \cos \left (d x +c \right )^{4}+355 \cos \left (d x +c \right )^{3}+426 \cos \left (d x +c \right )^{2}+568 \cos \left (d x +c \right )+1136\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{693 d \left (1+\cos \left (d x +c \right )\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}+\frac {2 B \,a^{2} \left (35 \cos \left (d x +c \right )^{4}+130 \cos \left (d x +c \right )^{3}+219 \cos \left (d x +c \right )^{2}+292 \cos \left (d x +c \right )+584\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{315 d \left (1+\cos \left (d x +c \right )\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}\) | \(182\) |
Input:
int((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(11/2),x,method=_RE TURNVERBOSE)
Output:
2/3465/d*a^2*((315*cos(d*x+c)^5+1120*cos(d*x+c)^4+1775*cos(d*x+c)^3+2130*c os(d*x+c)^2+2840*cos(d*x+c)+5680)*A+(385*cos(d*x+c)^4+1430*cos(d*x+c)^3+24 09*cos(d*x+c)^2+3212*cos(d*x+c)+6424)*B)*(a*(1+sec(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 = 162, normalized size of antiderivative = 0.59 \[ \int \frac {(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac {11}{2}}(c+d x)} \, dx=\frac {2 \, {\left (315 \, A a^{2} \cos \left (d x + c\right )^{6} + 35 \, {\left (32 \, A + 11 \, B\right )} a^{2} \cos \left (d x + c\right )^{5} + 5 \, {\left (355 \, A + 286 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} + 3 \, {\left (710 \, A + 803 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 4 \, {\left (710 \, A + 803 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 8 \, {\left (710 \, A + 803 \, B\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3465 \, {\left (d \cos \left (d x + c\right ) + d\right )} \sqrt {\cos \left (d x + c\right )}} \] Input:
integrate((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(11/2),x, alg orithm="fricas")
Output:
2/3465*(315*A*a^2*cos(d*x + c)^6 + 35*(32*A + 11*B)*a^2*cos(d*x + c)^5 + 5 *(355*A + 286*B)*a^2*cos(d*x + c)^4 + 3*(710*A + 803*B)*a^2*cos(d*x + c)^3 + 4*(710*A + 803*B)*a^2*cos(d*x + c)^2 + 8*(710*A + 803*B)*a^2*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. \[ \int \frac {(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac {11}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:
integrate((a+a*sec(d*x+c))**(5/2)*(A+B*sec(d*x+c))/sec(d*x+c)**(11/2),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 945 vs. \(2 (239) = 478\).
Time = 0.28 (sec) , antiderivative size = 945, normalized size of antiderivative = 3.44 \[ \int \frac {(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac {11}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:
integrate((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(11/2),x, alg orithm="maxima")
Output:
1/110880*(5*sqrt(2)*(31878*a^2*cos(10/11*arctan2(sin(11/2*d*x + 11/2*c), c os(11/2*d*x + 11/2*c)))*sin(11/2*d*x + 11/2*c) + 8778*a^2*cos(8/11*arctan2 (sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c)))*sin(11/2*d*x + 11/2*c) + 3465*a^2*cos(6/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c)) )*sin(11/2*d*x + 11/2*c) + 1287*a^2*cos(4/11*arctan2(sin(11/2*d*x + 11/2*c ), cos(11/2*d*x + 11/2*c)))*sin(11/2*d*x + 11/2*c) + 385*a^2*cos(2/11*arct an2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c)))*sin(11/2*d*x + 11/2*c ) - 31878*a^2*cos(11/2*d*x + 11/2*c)*sin(10/11*arctan2(sin(11/2*d*x + 11/2 *c), cos(11/2*d*x + 11/2*c))) - 8778*a^2*cos(11/2*d*x + 11/2*c)*sin(8/11*a rctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) - 3465*a^2*cos(11/ 2*d*x + 11/2*c)*sin(6/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11 /2*c))) - 1287*a^2*cos(11/2*d*x + 11/2*c)*sin(4/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) - 385*a^2*cos(11/2*d*x + 11/2*c)*sin(2/1 1*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 126*a^2*sin(1 1/2*d*x + 11/2*c) + 385*a^2*sin(9/11*arctan2(sin(11/2*d*x + 11/2*c), cos(1 1/2*d*x + 11/2*c))) + 1287*a^2*sin(7/11*arctan2(sin(11/2*d*x + 11/2*c), co s(11/2*d*x + 11/2*c))) + 3465*a^2*sin(5/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 8778*a^2*sin(3/11*arctan2(sin(11/2*d*x + 11/2* c), cos(11/2*d*x + 11/2*c))) + 31878*a^2*sin(1/11*arctan2(sin(11/2*d*x + 1 1/2*c), cos(11/2*d*x + 11/2*c))))*A*sqrt(a) + 22*sqrt(2)*(8190*a^2*cos(...
Time = 2.21 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.04 \[ \int \frac {(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac {11}{2}}(c+d x)} \, dx=\frac {8 \, {\left ({\left ({\left ({\left (4 \, {\left (2 \, \sqrt {2} {\left (125 \, A a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 143 \, B a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 11 \, \sqrt {2} {\left (125 \, A a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 143 \, B a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 99 \, \sqrt {2} {\left (125 \, A a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 143 \, B a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 231 \, \sqrt {2} {\left (65 \, A a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 69 \, B a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1155 \, \sqrt {2} {\left (7 \, A a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 9 \, B a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3465 \, \sqrt {2} {\left (A a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + B a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{3465 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {11}{2}} d} \] Input:
integrate((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(11/2),x, alg orithm="giac")
Output:
8/3465*((((4*(2*sqrt(2)*(125*A*a^8*sgn(cos(d*x + c)) + 143*B*a^8*sgn(cos(d *x + c)))*tan(1/2*d*x + 1/2*c)^2 + 11*sqrt(2)*(125*A*a^8*sgn(cos(d*x + c)) + 143*B*a^8*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 99*sqrt(2)*(125* A*a^8*sgn(cos(d*x + c)) + 143*B*a^8*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2* c)^2 + 231*sqrt(2)*(65*A*a^8*sgn(cos(d*x + c)) + 69*B*a^8*sgn(cos(d*x + c) )))*tan(1/2*d*x + 1/2*c)^2 + 1155*sqrt(2)*(7*A*a^8*sgn(cos(d*x + c)) + 9*B *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))))*tan(1/2*d*x + 1/2*c)/((a*tan(1/2 *d*x + 1/2*c)^2 + a)^(11/2)*d)
Time = 18.11 (sec) , antiderivative size = 392, normalized size of antiderivative = 1.43 \[ \int \frac {(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac {11}{2}}(c+d x)} \, dx=\frac {\sqrt {a-\frac {a}{2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1}}\,\left (2\,{\sin \left (\frac {11\,c}{4}+\frac {11\,d\,x}{4}\right )}^2+\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,1{}\mathrm {i}-1\right )\,\left (\frac {A\,a^2\,\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,\left (-2\,{\sin \left (\frac {11\,c}{4}+\frac {11\,d\,x}{4}\right )}^2+\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )}{88\,d}+\frac {a^2\,\sin \left (\frac {9\,c}{2}+\frac {9\,d\,x}{2}\right )\,\left (5\,A+2\,B\right )\,\left (-2\,{\sin \left (\frac {11\,c}{4}+\frac {11\,d\,x}{4}\right )}^2+\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )}{72\,d}+\frac {a^2\,\sin \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )\,\left (13\,A+10\,B\right )\,\left (-2\,{\sin \left (\frac {11\,c}{4}+\frac {11\,d\,x}{4}\right )}^2+\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )}{56\,d}+\frac {a^2\,\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )\,\left (19\,A+20\,B\right )\,\left (-2\,{\sin \left (\frac {11\,c}{4}+\frac {11\,d\,x}{4}\right )}^2+\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )}{12\,d}+\frac {a^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (23\,A+26\,B\right )\,\left (-2\,{\sin \left (\frac {11\,c}{4}+\frac {11\,d\,x}{4}\right )}^2+\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )}{4\,d}+\frac {a^2\,\sin \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )\,\left (25\,A+24\,B\right )\,\left (-2\,{\sin \left (\frac {11\,c}{4}+\frac {11\,d\,x}{4}\right )}^2+\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )}{40\,d}\right )}{2\,\sqrt {-\frac {1}{2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1}}\,\left (2\,{\sin \left (\frac {c}{4}+\frac {d\,x}{4}\right )}^2-1\right )} \] Input:
int(((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^(5/2))/(1/cos(c + d*x))^(11 /2),x)
Output:
((a - a/(2*sin(c/2 + (d*x)/2)^2 - 1))^(1/2)*(sin((11*c)/2 + (11*d*x)/2)*1i + 2*sin((11*c)/4 + (11*d*x)/4)^2 - 1)*((A*a^2*sin((11*c)/2 + (11*d*x)/2)* (sin((11*c)/2 + (11*d*x)/2)*1i - 2*sin((11*c)/4 + (11*d*x)/4)^2 + 1))/(88* d) + (a^2*sin((9*c)/2 + (9*d*x)/2)*(5*A + 2*B)*(sin((11*c)/2 + (11*d*x)/2) *1i - 2*sin((11*c)/4 + (11*d*x)/4)^2 + 1))/(72*d) + (a^2*sin((7*c)/2 + (7* d*x)/2)*(13*A + 10*B)*(sin((11*c)/2 + (11*d*x)/2)*1i - 2*sin((11*c)/4 + (1 1*d*x)/4)^2 + 1))/(56*d) + (a^2*sin((3*c)/2 + (3*d*x)/2)*(19*A + 20*B)*(si n((11*c)/2 + (11*d*x)/2)*1i - 2*sin((11*c)/4 + (11*d*x)/4)^2 + 1))/(12*d) + (a^2*sin(c/2 + (d*x)/2)*(23*A + 26*B)*(sin((11*c)/2 + (11*d*x)/2)*1i - 2 *sin((11*c)/4 + (11*d*x)/4)^2 + 1))/(4*d) + (a^2*sin((5*c)/2 + (5*d*x)/2)* (25*A + 24*B)*(sin((11*c)/2 + (11*d*x)/2)*1i - 2*sin((11*c)/4 + (11*d*x)/4 )^2 + 1))/(40*d)))/(2*(-1/(2*sin(c/2 + (d*x)/2)^2 - 1))^(1/2)*(2*sin(c/4 + (d*x)/4)^2 - 1))
\[ \int \frac {(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac {11}{2}}(c+d x)} \, dx=\sqrt {a}\, a^{2} \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{6}}d x \right ) a +2 \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{5}}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{5}}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{4}}d x \right ) a +2 \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{4}}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{3}}d x \right ) b \right ) \] Input:
int((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(11/2),x)
Output:
sqrt(a)*a**2*(int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1))/sec(c + d*x) **6,x)*a + 2*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)**5, x)*b + int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1))/sec(c + d*x)**4,x)* a + 2*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)