Integrand size = 35, antiderivative size = 230 \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \, dx=\frac {\sqrt {2} (A-B) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d}+\frac {2 A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}-\frac {2 (A-7 B) \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 (31 A-7 B) \sin (c+d x)}{105 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}-\frac {2 (43 A-91 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}} \]
(A-B)*arctanh(1/2*sin(d*x+c)*a^(1/2)*sec(d*x+c)^(1/2)*2^(1/2)/(a+a*sec(d*x +c))^(1/2))*2^(1/2)/d/a^(1/2)+2/7*A*sin(d*x+c)/d/sec(d*x+c)^(5/2)/(a+a*sec (d*x+c))^(1/2)-2/35*(A-7*B)*sin(d*x+c)/d/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c)) ^(1/2)+2/105*(31*A-7*B)*sin(d*x+c)/d/sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/ 2)-2/105*(43*A-91*B)*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^(1/2)
Time = 1.06 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.66 \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \, dx=\frac {-\frac {2 \left (-15 A+3 (A-7 B) \sec (c+d x)+(-31 A+7 B) \sec ^2(c+d x)+(43 A-91 B) \sec ^3(c+d x)\right ) \sin (c+d x)}{\sec ^{\frac {5}{2}}(c+d x)}-\frac {105 \sqrt {2} (A-B) \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \tan (c+d x)}{\sqrt {1-\sec (c+d x)}}}{105 d \sqrt {a (1+\sec (c+d x))}} \]
((-2*(-15*A + 3*(A - 7*B)*Sec[c + d*x] + (-31*A + 7*B)*Sec[c + d*x]^2 + (4 3*A - 91*B)*Sec[c + d*x]^3)*Sin[c + d*x])/Sec[c + d*x]^(5/2) - (105*Sqrt[2 ]*(A - B)*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Tan[ c + d*x])/Sqrt[1 - Sec[c + d*x]])/(105*d*Sqrt[a*(1 + Sec[c + d*x])])
Time = 1.43 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.13, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4510, 27, 3042, 4510, 27, 3042, 4510, 27, 3042, 4501, 3042, 4295, 219}
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.
| \(\displaystyle \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{7/2} \sqrt {a \csc \left (c+d x+\frac {\pi }{2}\right )+a}}dx\) |
| \(\Big \downarrow \) 4510 |
| \(\displaystyle \frac {2 \int -\frac {a (A-7 B)-6 a A \sec (c+d x)}{2 \sec ^{\frac {5}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}}dx}{7 a}+\frac {2 A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a (A-7 B)-6 a A \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}}dx}{7 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a (A-7 B)-6 a A \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{7 a}\) |
| \(\Big \downarrow \) 4510 |
| \(\displaystyle \frac {2 A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 \int -\frac {a^2 (31 A-7 B)-4 a^2 (A-7 B) \sec (c+d x)}{2 \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}}dx}{5 a}+\frac {2 a (A-7 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}}{7 a}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a (A-7 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^2 (31 A-7 B)-4 a^2 (A-7 B) \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}}dx}{5 a}}{7 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a (A-7 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^2 (31 A-7 B)-4 a^2 (A-7 B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{5 a}}{7 a}\) |
| \(\Big \downarrow \) 4510 |
| \(\displaystyle \frac {2 A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a (A-7 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 \int -\frac {a^3 (43 A-91 B)-2 a^3 (31 A-7 B) \sec (c+d x)}{2 \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{3 a}+\frac {2 a^2 (31 A-7 B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}}{5 a}}{7 a}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a (A-7 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^2 (31 A-7 B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^3 (43 A-91 B)-2 a^3 (31 A-7 B) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{3 a}}{5 a}}{7 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a (A-7 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^2 (31 A-7 B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^3 (43 A-91 B)-2 a^3 (31 A-7 B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}}{5 a}}{7 a}\) |
| \(\Big \downarrow \) 4501 |
| \(\displaystyle \frac {2 A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a (A-7 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^2 (31 A-7 B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (43 A-91 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-105 a^3 (A-B) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {\sec (c+d x) a+a}}dx}{3 a}}{5 a}}{7 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a (A-7 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^2 (31 A-7 B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (43 A-91 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-105 a^3 (A-B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}}{5 a}}{7 a}\) |
| \(\Big \downarrow \) 4295 |
| \(\displaystyle \frac {2 A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a (A-7 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^2 (31 A-7 B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {210 a^3 (A-B) \int \frac {1}{2 a-\frac {a^2 \sin (c+d x) \tan (c+d x)}{\sec (c+d x) a+a}}d\left (-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {2 a^3 (43 A-91 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}}{3 a}}{5 a}}{7 a}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {2 A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a (A-7 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^2 (31 A-7 B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (43 A-91 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-\frac {105 \sqrt {2} a^{5/2} (A-B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{3 a}}{5 a}}{7 a}\) |
(2*A*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2)*Sqrt[a + a*Sec[c + d*x]]) - ((2 *a*(A - 7*B)*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x] ]) - ((2*a^2*(31*A - 7*B)*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]*Sqrt[a + a *Sec[c + d*x]]) - ((-105*Sqrt[2]*a^(5/2)*(A - B)*ArcTanh[(Sqrt[a]*Sqrt[Sec [c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/d + (2*a^3*( 43*A - 91*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]) )/(3*a))/(5*a))/(7*a)
3.3.54.3.1 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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*b*(d/(a*f)) Subst[Int[1/(2*b - d*x^2), x], x, b*(Cot[e + f*x]/(Sqrt[a + b*Csc[e + f*x]]*Sqrt[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_)*(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[(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[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 - A*b*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[n, 0]
Time = 3.81 (sec) , antiderivative size = 387, normalized size of antiderivative = 1.68
| method | result | size |
| default | \(\frac {\sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (105 A \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {2}\, \sec \left (d x +c \right )^{2}+30 A \sin \left (d x +c \right )-105 B \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {2}\, \sec \left (d x +c \right )^{2}+105 A \sqrt {2}\, \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sec \left (d x +c \right )^{3}-6 A \tan \left (d x +c \right )-105 B \sqrt {2}\, \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sec \left (d x +c \right )^{3}+42 B \tan \left (d x +c \right )+62 A \tan \left (d x +c \right ) \sec \left (d x +c \right )-14 B \tan \left (d x +c \right ) \sec \left (d x +c \right )-86 A \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}+182 B \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}\right )}{105 d a \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {7}{2}}}\) | \(387\) |
| parts | \(\frac {A \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (105 \sqrt {2}\, \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \sec \left (d x +c \right )^{2}+30 \sin \left (d x +c \right )+105 \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {2}\, \sec \left (d x +c \right )^{3}-6 \tan \left (d x +c \right )+62 \sec \left (d x +c \right ) \tan \left (d x +c \right )-86 \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}\right )}{105 d a \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {7}{2}}}-\frac {B \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (15 \sqrt {2}\, \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \sec \left (d x +c \right )+15 \sqrt {2}\, \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \sec \left (d x +c \right )^{2}-6 \sin \left (d x +c \right )+2 \tan \left (d x +c \right )-26 \sec \left (d x +c \right ) \tan \left (d x +c \right )\right )}{15 d a \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {5}{2}}}\) | \(408\) |
1/105/d/a*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)/sec(d*x+c)^(7/2)*(105*A* arctan(1/2*sin(d*x+c)*2^(1/2)/(cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2))*(- 1/(cos(d*x+c)+1))^(1/2)*2^(1/2)*sec(d*x+c)^2+30*A*sin(d*x+c)-105*B*arctan( 1/2*sin(d*x+c)*2^(1/2)/(cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2))*(-1/(cos( d*x+c)+1))^(1/2)*2^(1/2)*sec(d*x+c)^2+105*A*2^(1/2)*(-1/(cos(d*x+c)+1))^(1 /2)*arctan(1/2*sin(d*x+c)*2^(1/2)/(cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2) )*sec(d*x+c)^3-6*A*tan(d*x+c)-105*B*2^(1/2)*(-1/(cos(d*x+c)+1))^(1/2)*arct an(1/2*sin(d*x+c)*2^(1/2)/(cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2))*sec(d* x+c)^3+42*B*tan(d*x+c)+62*A*tan(d*x+c)*sec(d*x+c)-14*B*tan(d*x+c)*sec(d*x+ c)-86*A*tan(d*x+c)*sec(d*x+c)^2+182*B*tan(d*x+c)*sec(d*x+c)^2)
Time = 0.28 (sec) , antiderivative size = 422, normalized size of antiderivative = 1.83 \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \, dx=\left [-\frac {\frac {105 \, \sqrt {2} {\left ({\left (A - B\right )} a \cos \left (d x + c\right ) + {\left (A - B\right )} a\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} + \frac {2 \, \sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{\sqrt {a}} - \frac {4 \, {\left (15 \, A \cos \left (d x + c\right )^{4} - 3 \, {\left (A - 7 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (31 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{2} - {\left (43 \, A - 91 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{210 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}}, -\frac {105 \, \sqrt {2} {\left ({\left (A - B\right )} a \cos \left (d x + c\right ) + {\left (A - B\right )} a\right )} \sqrt {-\frac {1}{a}} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} \sqrt {\cos \left (d x + c\right )}}{\sin \left (d x + c\right )}\right ) - \frac {2 \, {\left (15 \, A \cos \left (d x + c\right )^{4} - 3 \, {\left (A - 7 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (31 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{2} - {\left (43 \, A - 91 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{105 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}}\right ] \]
[-1/210*(105*sqrt(2)*((A - B)*a*cos(d*x + c) + (A - B)*a)*log(-(cos(d*x + c)^2 + 2*sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c) )*sin(d*x + c)/sqrt(a) - 2*cos(d*x + c) - 3)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1))/sqrt(a) - 4*(15*A*cos(d*x + c)^4 - 3*(A - 7*B)*cos(d*x + c)^3 + (31*A - 7*B)*cos(d*x + c)^2 - (43*A - 91*B)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a*d*cos(d*x + c) + a*d), -1/105*(105*sqrt(2)*((A - B)*a*cos(d*x + c) + (A - B)*a)*sqrt(-1/ a)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(-1/a)*sqrt( cos(d*x + c))/sin(d*x + c)) - 2*(15*A*cos(d*x + c)^4 - 3*(A - 7*B)*cos(d*x + c)^3 + (31*A - 7*B)*cos(d*x + c)^2 - (43*A - 91*B)*cos(d*x + c))*sqrt(( a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a*d*co s(d*x + c) + a*d)]
Timed out. \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 805 vs. \(2 (195) = 390\).
Time = 0.59 (sec) , antiderivative size = 805, normalized size of antiderivative = 3.50 \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \, dx=\text {Too large to display} \]
-1/840*(sqrt(2)*(525*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*cos(4/7*arctan2(sin(7/2*d*x + 7/2*c), c os(7/2*d*x + 7/2*c)))*sin(7/2*d*x + 7/2*c) + 21*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) - 525*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))) + 17 5*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))) - 21*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))) - 420*log(cos(1/7*arctan2(sin(7/2*d*x + 7/2*c), cos (7/2*d*x + 7/2*c)))^2 + sin(1/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))^2 + 2*sin(1/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c ))) + 1) + 420*log(cos(1/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2 *c)))^2 + sin(1/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))^2 - 2*sin(1/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 1) - 30* sin(7/2*d*x + 7/2*c) + 21*sin(5/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d* x + 7/2*c))) - 175*sin(3/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2 *c))) + 525*sin(1/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))))* A/sqrt(a) - 14*sqrt(2)*(60*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*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) - 60*cos(5/2*d*x + 5/2*c)*s in(4/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 5*cos(5/2...
\[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \, dx=\int { \frac {B \sec \left (d x + c\right ) + A}{\sqrt {a \sec \left (d x + c\right ) + a} \sec \left (d x + c\right )^{\frac {7}{2}}} \,d x } \]
Timed out. \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}} \,d x \]