Integrand size = 35, antiderivative size = 180 \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\frac {a^{3/2} (14 A+11 B) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{8 d}+\frac {a^2 (14 A+11 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{8 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (6 A+7 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{12 d \sqrt {a+a \sec (c+d x)}}+\frac {a B \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{3 d} \]
1/8*a^(3/2)*(14*A+11*B)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2)) /d+1/8*a^2*(14*A+11*B)*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2 )+1/12*a^2*(6*A+7*B)*sec(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+ 1/3*a*B*sec(d*x+c)^(5/2)*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d
Time = 0.73 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.87 \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\frac {a^2 \left ((42 A+33 B) \arcsin \left (\sqrt {1-\sec (c+d x)}\right )+2 (6 A+11 B) \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x)+8 B \sqrt {1-\sec (c+d x)} \sec ^{\frac {5}{2}}(c+d x)+3 (14 A+11 B) \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))}\right ) \tan (c+d x)}{24 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \]
(a^2*((42*A + 33*B)*ArcSin[Sqrt[1 - Sec[c + d*x]]] + 2*(6*A + 11*B)*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^(3/2) + 8*B*Sqrt[1 - Sec[c + d*x]]*Sec[c + d *x]^(5/2) + 3*(14*A + 11*B)*Sqrt[-((-1 + Sec[c + d*x])*Sec[c + d*x])])*Tan [c + d*x])/(24*d*Sqrt[1 - Sec[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])
Time = 0.92 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.98, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4506, 27, 3042, 4504, 3042, 4290, 3042, 4288, 222}
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.
| \(\displaystyle \int \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2} (A+B \sec (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \frac {1}{3} \int \frac {1}{2} \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a} (3 a (2 A+B)+a (6 A+7 B) \sec (c+d x))dx+\frac {a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{6} \int \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a} (3 a (2 A+B)+a (6 A+7 B) \sec (c+d x))dx+\frac {a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{6} \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (3 a (2 A+B)+a (6 A+7 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\) |
| \(\Big \downarrow \) 4504 |
| \(\displaystyle \frac {1}{6} \left (\frac {3}{4} a (14 A+11 B) \int \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {a^2 (6 A+7 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{6} \left (\frac {3}{4} a (14 A+11 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a^2 (6 A+7 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\) |
| \(\Big \downarrow \) 4290 |
| \(\displaystyle \frac {1}{6} \left (\frac {3}{4} a (14 A+11 B) \left (\frac {1}{2} \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (6 A+7 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{6} \left (\frac {3}{4} a (14 A+11 B) \left (\frac {1}{2} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (6 A+7 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\) |
| \(\Big \downarrow \) 4288 |
| \(\displaystyle \frac {1}{6} \left (\frac {3}{4} a (14 A+11 B) \left (\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {a^2 (6 A+7 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\) |
| \(\Big \downarrow \) 222 |
| \(\displaystyle \frac {1}{6} \left (\frac {a^2 (6 A+7 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}+\frac {3}{4} a (14 A+11 B) \left (\frac {\sqrt {a} \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )\right )+\frac {a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\) |
(a*B*Sec[c + d*x]^(5/2)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(3*d) + ((a ^2*(6*A + 7*B)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(2*d*Sqrt[a + a*Sec[c + d* x]]) + (3*a*(14*A + 11*B)*((Sqrt[a]*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (a*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(d*Sqrt[a + a* Sec[c + d*x]])))/4)/6
3.3.32.3.1 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[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt [a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(a/(b*f))*Sqrt[a*(d/b)] Subst[Int[1/Sqrt[1 + x^2/a], x], x, b*(Cot[e + f*x]/Sqrt[a + b*Csc[e + f*x]])], x] /; FreeQ[{a , b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[a*(d/b), 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*b*d*Cot[e + f*x]*((d*Csc[e + f*x])^(n - 1)/( f*(2*n - 1)*Sqrt[a + b*Csc[e + f*x]])), x] + Simp[2*a*d*((n - 1)/(b*(2*n - 1))) Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[e + f*x])^(n - 1), x], x] /; Fre eQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[n, 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[-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]
Leaf count of result is larger than twice the leaf count of optimal. \(416\) vs. \(2(154)=308\).
Time = 8.21 (sec) , antiderivative size = 417, normalized size of antiderivative = 2.32
| method | result | size |
| default | \(-\frac {a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sec \left (d x +c \right )^{\frac {7}{2}} \left (42 A \cos \left (d x +c \right )^{4} \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )-42 A \cos \left (d x +c \right )^{4} \arctan \left (\frac {\cos \left (d x +c \right )-\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )-84 A \sin \left (d x +c \right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )^{3}+33 B \cos \left (d x +c \right )^{4} \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )-33 B \cos \left (d x +c \right )^{4} \arctan \left (\frac {\cos \left (d x +c \right )-\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )-66 B \sin \left (d x +c \right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )^{3}-24 A \cos \left (d x +c \right )^{2} \sin \left (d x +c \right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}-44 B \cos \left (d x +c \right )^{2} \sin \left (d x +c \right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}-16 B \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\right )}{48 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\) | \(417\) |
| parts | \(\frac {A a \left (7 \cos \left (d x +c \right )^{2} \arctan \left (\frac {\cos \left (d x +c \right )-\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )-7 \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )^{2}+14 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \sin \left (d x +c \right ) \cos \left (d x +c \right )+4 \sin \left (d x +c \right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sec \left (d x +c \right )^{\frac {3}{2}}}{8 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}+\frac {B a \left (33 \cos \left (d x +c \right )^{3} \arctan \left (\frac {\cos \left (d x +c \right )-\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )-33 \cos \left (d x +c \right )^{3} \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+66 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}+44 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \sin \left (d x +c \right ) \cos \left (d x +c \right )+16 \sin \left (d x +c \right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sec \left (d x +c \right )^{\frac {5}{2}}}{48 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\) | \(444\) |
-1/48*a/d*(a*(1+sec(d*x+c)))^(1/2)*sec(d*x+c)^(7/2)/(cos(d*x+c)+1)/(-1/(co s(d*x+c)+1))^(1/2)*(42*A*cos(d*x+c)^4*arctan(1/2*(cos(d*x+c)+sin(d*x+c)+1) /(cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2))-42*A*cos(d*x+c)^4*arctan(1/2*(c os(d*x+c)-sin(d*x+c)+1)/(cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2))-84*A*sin (d*x+c)*(-1/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^3+33*B*cos(d*x+c)^4*arctan(1/ 2*(cos(d*x+c)+sin(d*x+c)+1)/(cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2))-33*B *cos(d*x+c)^4*arctan(1/2*(cos(d*x+c)-sin(d*x+c)+1)/(cos(d*x+c)+1)/(-1/(cos (d*x+c)+1))^(1/2))-66*B*sin(d*x+c)*(-1/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^3- 24*A*cos(d*x+c)^2*sin(d*x+c)*(-1/(cos(d*x+c)+1))^(1/2)-44*B*cos(d*x+c)^2*s in(d*x+c)*(-1/(cos(d*x+c)+1))^(1/2)-16*B*cos(d*x+c)*sin(d*x+c)*(-1/(cos(d* x+c)+1))^(1/2))
Time = 0.34 (sec) , antiderivative size = 458, normalized size of antiderivative = 2.54 \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\left [\frac {3 \, {\left ({\left (14 \, A + 11 \, B\right )} a \cos \left (d x + c\right )^{3} + {\left (14 \, A + 11 \, B\right )} a \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - \frac {4 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {a} \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 )}} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + \frac {4 \, {\left (3 \, {\left (14 \, A + 11 \, B\right )} a \cos \left (d x + c\right )^{2} + 2 \, {\left (6 \, A + 11 \, B\right )} a \cos \left (d x + c\right ) + 8 \, B a\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 )}}}{96 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}, \frac {3 \, {\left ({\left (14 \, A + 11 \, B\right )} a \cos \left (d x + c\right )^{3} + {\left (14 \, A + 11 \, B\right )} a \cos \left (d x + c\right )^{2}\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \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 )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right ) + \frac {2 \, {\left (3 \, {\left (14 \, A + 11 \, B\right )} a \cos \left (d x + c\right )^{2} + 2 \, {\left (6 \, A + 11 \, B\right )} a \cos \left (d x + c\right ) + 8 \, B a\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 )}}}{48 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}\right ] \]
[1/96*(3*((14*A + 11*B)*a*cos(d*x + c)^3 + (14*A + 11*B)*a*cos(d*x + c)^2) *sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*(cos(d*x + c)^2 - 2*cos(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) + 4*(3*(14 *A + 11*B)*a*cos(d*x + c)^2 + 2*(6*A + 11*B)*a*cos(d*x + c) + 8*B*a)*sqrt( (a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos (d*x + c)^3 + d*cos(d*x + c)^2), 1/48*(3*((14*A + 11*B)*a*cos(d*x + c)^3 + (14*A + 11*B)*a*cos(d*x + c)^2)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d* x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^ 2 - a*cos(d*x + c) - 2*a)) + 2*(3*(14*A + 11*B)*a*cos(d*x + c)^2 + 2*(6*A + 11*B)*a*cos(d*x + c) + 8*B*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*si n(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^3 + d*cos(d*x + c)^2)]
Timed out. \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 4606 vs. \(2 (154) = 308\).
Time = 0.69 (sec) , antiderivative size = 4606, normalized size of antiderivative = 25.59 \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\text {Too large to display} \]
-1/96*(6*(56*sqrt(2)*a*cos(7/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 2 4*sqrt(2)*a*cos(5/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))*s in(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 12*sqrt(2)*a *sin(3/2*d*x + 3/2*c) + 28*sqrt(2)*a*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 4*(3*sqrt(2)*a*sin(3/2*d*x + 3/2*c) + 7*sqrt(2)* a*sin(7/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 3*sqrt(2) *a*sin(5/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 7*sqrt(2 )*a*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*cos(8/3* arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 8*(3*sqrt(2)*a*sin( 3/2*d*x + 3/2*c) - 7*sqrt(2)*a*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3 /2*d*x + 3/2*c))))*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2 *c))) - 7*(a*cos(8/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^ 2 + 4*a*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + a *sin(8/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 4*a*sin( 8/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))*sin(4/3*arctan2(s in(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 4*a*sin(4/3*arctan2(sin(3/2* d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*(2*a*cos(4/3*arctan2(sin(3/2*d* x + 3/2*c), cos(3/2*d*x + 3/2*c))) + a)*cos(8/3*arctan2(sin(3/2*d*x + 3/2* c), cos(3/2*d*x + 3/2*c))) + 4*a*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), ...
\[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )^{\frac {3}{2}} \,d x } \]
Timed out. \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\int \left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2} \,d x \]