Integrand size = 35, antiderivative size = 133 \[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\frac {a^{3/2} (12 A+7 B) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 d}+\frac {a^2 (4 A+5 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{4 d \sqrt {a+a \sec (c+d x)}}+\frac {a B \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{2 d} \]
1/4*a^(3/2)*(12*A+7*B)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/ d+1/4*a^2*(4*A+5*B)*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1 /2*a*B*sec(d*x+c)^(3/2)*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d
Time = 0.57 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.13 \[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\frac {a^2 \left (7 B \arcsin \left (\sqrt {1-\sec (c+d x)}\right )-12 A \arcsin \left (\sqrt {\sec (c+d x)}\right )+2 B \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x)+4 A \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))}+7 B \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))}\right ) \tan (c+d x)}{4 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \]
(a^2*(7*B*ArcSin[Sqrt[1 - Sec[c + d*x]]] - 12*A*ArcSin[Sqrt[Sec[c + d*x]]] + 2*B*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^(3/2) + 4*A*Sqrt[-((-1 + Sec[c + d*x])*Sec[c + d*x])] + 7*B*Sqrt[-((-1 + Sec[c + d*x])*Sec[c + d*x])])*Ta n[c + d*x])/(4*d*Sqrt[1 - Sec[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])
Time = 0.71 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.99, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {3042, 4506, 27, 3042, 4504, 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.
| \(\displaystyle \int \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2} (A+B \sec (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \frac {1}{2} \int \frac {1}{2} \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a} (a (4 A+B)+a (4 A+5 B) \sec (c+d x))dx+\frac {a B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a} (a (4 A+B)+a (4 A+5 B) \sec (c+d x))dx+\frac {a B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (a (4 A+B)+a (4 A+5 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {a B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\) |
| \(\Big \downarrow \) 4504 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} a (12 A+7 B) \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx+\frac {a^2 (4 A+5 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} a (12 A+7 B) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a^2 (4 A+5 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\) |
| \(\Big \downarrow \) 4288 |
| \(\displaystyle \frac {1}{4} \left (\frac {a^2 (4 A+5 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {a (12 A+7 B) \int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\) |
| \(\Big \downarrow \) 222 |
| \(\displaystyle \frac {1}{4} \left (\frac {a^{3/2} (12 A+7 B) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a^2 (4 A+5 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\) |
(a*B*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(2*d) + ((a ^(3/2)*(12*A + 7*B)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x] ]])/d + (a^2*(4*A + 5*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(d*Sqrt[a + a*Se c[c + d*x]]))/4
3.3.33.3.1 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_)]*(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. \(354\) vs. \(2(113)=226\).
Time = 8.10 (sec) , antiderivative size = 355, normalized size of antiderivative = 2.67
| method | result | size |
| default | \(\frac {a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sec \left (d x +c \right )^{\frac {5}{2}} \left (8 A \cos \left (d x +c \right )^{2} \sin \left (d x +c \right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}-12 A \cos \left (d x +c \right )^{3} \arctan \left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )-12 A \cos \left (d x +c \right )^{3} \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+14 B \cos \left (d x +c \right )^{2} \sin \left (d x +c \right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}-7 B \cos \left (d x +c \right )^{3} \arctan \left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )-7 B \cos \left (d x +c \right )^{3} \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+4 B \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\right )}{8 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\) | \(355\) |
| parts | \(\frac {A a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sqrt {\sec \left (d x +c \right )}\, \left (3 \cos \left (d x +c \right ) \arctan \left (\frac {\cos \left (d x +c \right )-\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )-3 \cos \left (d x +c \right ) \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+2 \sin \left (d x +c \right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\right )}{2 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}+\frac {B a \left (7 \cos \left (d x +c \right )^{2} \arctan \left (\frac {\cos \left (d x +c \right )-\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )-7 \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )^{2}+14 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \sin \left (d x +c \right ) \cos \left (d x +c \right )+4 \sin \left (d x +c \right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sec \left (d x +c \right )^{\frac {3}{2}}}{8 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\) | \(382\) |
1/8*a/d*(a*(1+sec(d*x+c)))^(1/2)*sec(d*x+c)^(5/2)/(cos(d*x+c)+1)/(-1/(cos( d*x+c)+1))^(1/2)*(8*A*cos(d*x+c)^2*sin(d*x+c)*(-1/(cos(d*x+c)+1))^(1/2)-12 *A*cos(d*x+c)^3*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))-12*A*cos(d*x+c)^3*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))+14*B*cos(d*x+c)^2*sin(d*x+c)* (-1/(cos(d*x+c)+1))^(1/2)-7*B*cos(d*x+c)^3*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))-7*B*cos(d*x+c)^3*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))+4*B *cos(d*x+c)*sin(d*x+c)*(-1/(cos(d*x+c)+1))^(1/2))
Time = 0.37 (sec) , antiderivative size = 410, normalized size of antiderivative = 3.08 \[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\left [\frac {{\left ({\left (12 \, A + 7 \, B\right )} a \cos \left (d x + c\right )^{2} + {\left (12 \, A + 7 \, B\right )} a \cos \left (d x + c\right )\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - \frac {4 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + \frac {4 \, {\left ({\left (4 \, A + 7 \, B\right )} a \cos \left (d x + c\right ) + 2 \, B a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{16 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}, \frac {{\left ({\left (12 \, A + 7 \, B\right )} a \cos \left (d x + c\right )^{2} + {\left (12 \, A + 7 \, B\right )} a \cos \left (d x + c\right )\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right ) + \frac {2 \, {\left ({\left (4 \, A + 7 \, B\right )} a \cos \left (d x + c\right ) + 2 \, B a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{8 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}\right ] \]
[1/16*(((12*A + 7*B)*a*cos(d*x + c)^2 + (12*A + 7*B)*a*cos(d*x + c))*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)/sqr t(cos(d*x + c)) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) + 4*((4*A + 7*B) *a*cos(d*x + c) + 2*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)^2 + d*cos(d*x + c)), 1/8*(((12*A + 7*B)*a*cos(d*x + c)^2 + (12*A + 7*B)*a*cos(d*x + c))*sqrt(-a)*arctan(2*sq rt(-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*((4*A + 7*B)*a*cos(d*x + c) + 2*B*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(c os(d*x + c)))/(d*cos(d*x + c)^2 + d*cos(d*x + c))]
Timed out. \[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 3389 vs. \(2 (113) = 226\).
Time = 0.64 (sec) , antiderivative size = 3389, normalized size of antiderivative = 25.48 \[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\text {Too large to display} \]
1/16*(4*(3*(a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2* sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - a*log (2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d *x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*s qrt(2)*sin(1/2*d*x + 1/2*c) + 2) - a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin( 1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d* x + 1/2*c) + 2))*cos(2*d*x + 2*c)^2 + 3*(a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin( 1/2*d*x + 1/2*c) + 2) - a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1 /2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt( 2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - a*log(2*co s(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2))*sin(2*d*x + 2*c)^2 + 4*sqrt( 2)*a*sin(3/2*d*x + 3/2*c) - 4*sqrt(2)*a*sin(1/2*d*x + 1/2*c) + 2*(2*sqrt(2 )*a*sin(3/2*d*x + 3/2*c) - 2*sqrt(2)*a*sin(1/2*d*x + 1/2*c) + 3*a*log(2*co s(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - 3*a*log(2*cos(1/2*d*x + 1/2 *c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*s...
\[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sqrt {\sec \left (d x + c\right )} \,d x } \]
Timed out. \[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\int \left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}} \,d x \]