Integrand size = 43, antiderivative size = 169 \[ \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {(7 A-4 B+8 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 \sqrt {a} d}-\frac {\sqrt {2} (A-B+C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d}-\frac {(A-4 B) \sin (c+d x)}{4 d \sqrt {a+a \sec (c+d x)}}+\frac {A \cos (c+d x) \sin (c+d x)}{2 d \sqrt {a+a \sec (c+d x)}} \]
1/4*(7*A-4*B+8*C)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d/a^(1 /2)-(A-B+C)*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))* 2^(1/2)/d/a^(1/2)-1/4*(A-4*B)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/2*A*co s(d*x+c)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)
Time = 0.47 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.82 \[ \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\left ((7 A-4 B+8 C) \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right )-4 \sqrt {2} (A-B+C) \text {arctanh}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right )+\cos (c+d x) (-A+4 B+2 A \cos (c+d x)) \sqrt {1-\sec (c+d x)}\right ) \tan (c+d x)}{4 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \]
Integrate[(Cos[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sqrt[a + a*Sec[c + d*x]],x]
(((7*A - 4*B + 8*C)*ArcTanh[Sqrt[1 - Sec[c + d*x]]] - 4*Sqrt[2]*(A - B + C )*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]] + Cos[c + d*x]*(-A + 4*B + 2*A*C os[c + d*x])*Sqrt[1 - Sec[c + d*x]])*Tan[c + d*x])/(4*d*Sqrt[1 - Sec[c + d *x]]*Sqrt[a*(1 + Sec[c + d*x])])
Time = 1.00 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.07, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.302, Rules used = {3042, 4574, 27, 3042, 4510, 27, 3042, 4408, 3042, 4261, 216, 4282, 216}
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.
| \(\displaystyle \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a \sec (c+d x)+a}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^2 \sqrt {a \csc \left (c+d x+\frac {\pi }{2}\right )+a}}dx\) |
| \(\Big \downarrow \) 4574 |
| \(\displaystyle \frac {\int -\frac {\cos (c+d x) (a (A-4 B)-a (3 A+4 C) \sec (c+d x))}{2 \sqrt {\sec (c+d x) a+a}}dx}{2 a}+\frac {A \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {\cos (c+d x) (a (A-4 B)-a (3 A+4 C) \sec (c+d x))}{\sqrt {\sec (c+d x) a+a}}dx}{4 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a (A-4 B)-a (3 A+4 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a}\) |
| \(\Big \downarrow \) 4510 |
| \(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {\int -\frac {a^2 (7 A-4 B+8 C)-a^2 (A-4 B) \sec (c+d x)}{2 \sqrt {\sec (c+d x) a+a}}dx}{a}+\frac {a (A-4 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a (A-4 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^2 (7 A-4 B+8 C)-a^2 (A-4 B) \sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a (A-4 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^2 (7 A-4 B+8 C)-a^2 (A-4 B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 4408 |
| \(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a (A-4 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {a (7 A-4 B+8 C) \int \sqrt {\sec (c+d x) a+a}dx-8 a^2 (A-B+C) \int \frac {\sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a (A-4 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {a (7 A-4 B+8 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-8 a^2 (A-B+C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 4261 |
| \(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a (A-4 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {-8 a^2 (A-B+C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {2 a^2 (7 A-4 B+8 C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a (A-4 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^{3/2} (7 A-4 B+8 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-8 a^2 (A-B+C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 4282 |
| \(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a (A-4 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {16 a^2 (A-B+C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {2 a^{3/2} (7 A-4 B+8 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a (A-4 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^{3/2} (7 A-4 B+8 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {8 \sqrt {2} a^{3/2} (A-B+C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{2 a}}{4 a}\) |
(A*Cos[c + d*x]*Sin[c + d*x])/(2*d*Sqrt[a + a*Sec[c + d*x]]) - (-1/2*((2*a ^(3/2)*(7*A - 4*B + 8*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d - (8*Sqrt[2]*a^(3/2)*(A - B + C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/( Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/d)/a + (a*(A - 4*B)*Sin[c + d*x])/(d*S qrt[a + a*Sec[c + d*x]]))/(4*a)
3.6.16.3.1 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]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(b/d) Subst[Int[1/(a + x^2), x], x, b*(Cot[c + d*x]/Sqrt[a + b*Csc[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S ymbol] :> Simp[-2/f Subst[Int[1/(2*a + x^2), x], x, b*(Cot[e + f*x]/Sqrt[ a + b*Csc[e + f*x]])], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_ .) + (a_)], x_Symbol] :> Simp[c/a Int[Sqrt[a + b*Csc[e + f*x]], x], x] - Simp[(b*c - a*d)/a Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /; F reeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && 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[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]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), 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 - b*(A*(m + n + 1) + C*n)*Csc[e + f*x] , x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || EqQ[m + n + 1, 0])
Leaf count of result is larger than twice the leaf count of optimal. \(936\) vs. \(2(144)=288\).
Time = 2.29 (sec) , antiderivative size = 937, normalized size of antiderivative = 5.54
int(cos(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, method=_RETURNVERBOSE)
1/4/d/a*(-4*A*2^(1/2)*ln((cot(d*x+c)^2-2*cot(d*x+c)*csc(d*x+c)+csc(d*x+c)^ 2-1)^(1/2)-cot(d*x+c)+csc(d*x+c))*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d *x+c)+4*B*2^(1/2)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ln((cot(d*x+c)^2-2*co t(d*x+c)*csc(d*x+c)+csc(d*x+c)^2-1)^(1/2)-cot(d*x+c)+csc(d*x+c))*cos(d*x+c )-4*C*2^(1/2)*ln((cot(d*x+c)^2-2*cot(d*x+c)*csc(d*x+c)+csc(d*x+c)^2-1)^(1/ 2)-cot(d*x+c)+csc(d*x+c))*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)+2* A*cos(d*x+c)^2*sin(d*x+c)-4*A*2^(1/2)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*l n((cot(d*x+c)^2-2*cot(d*x+c)*csc(d*x+c)+csc(d*x+c)^2-1)^(1/2)-cot(d*x+c)+c sc(d*x+c))+7*A*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(sin(d*x+c)/(cos( d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)+4*B*2^(1/2)*ln((c ot(d*x+c)^2-2*cot(d*x+c)*csc(d*x+c)+csc(d*x+c)^2-1)^(1/2)-cot(d*x+c)+csc(d *x+c))*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-4*B*arctanh(sin(d*x+c)/(cos(d*x+ c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/ 2)*cos(d*x+c)-4*C*2^(1/2)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ln((cot(d*x+c )^2-2*cot(d*x+c)*csc(d*x+c)+csc(d*x+c)^2-1)^(1/2)-cot(d*x+c)+csc(d*x+c))+8 *C*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(- cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)-A*cos(d*x+c)*sin(d*x+c)+7*A*( -cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos( d*x+c)/(cos(d*x+c)+1))^(1/2))+4*B*cos(d*x+c)*sin(d*x+c)-4*B*(-cos(d*x+c)/( cos(d*x+c)+1))^(1/2)*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(co...
Time = 8.94 (sec) , antiderivative size = 521, normalized size of antiderivative = 3.08 \[ \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\left [\frac {4 \, \sqrt {2} {\left ({\left (A - B + C\right )} a \cos \left (d x + c\right ) + {\left (A - B + C\right )} a\right )} \sqrt {-\frac {1}{a}} \log \left (\frac {2 \, \sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \, \cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - {\left ({\left (7 \, A - 4 \, B + 8 \, C\right )} \cos \left (d x + c\right ) + 7 \, A - 4 \, B + 8 \, C\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (2 \, A \cos \left (d x + c\right )^{2} - {\left (A - 4 \, 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 )}{8 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}}, -\frac {{\left ({\left (7 \, A - 4 \, B + 8 \, C\right )} \cos \left (d x + c\right ) + 7 \, A - 4 \, B + 8 \, C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left (2 \, A \cos \left (d x + c\right )^{2} - {\left (A - 4 \, 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 ) - \frac {4 \, \sqrt {2} {\left ({\left (A - B + C\right )} a \cos \left (d x + c\right ) + {\left (A - B + C\right )} a\right )} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right )}{\sqrt {a}}}{4 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}}\right ] \]
integrate(cos(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1 /2),x, algorithm="fricas")
[1/8*(4*sqrt(2)*((A - B + C)*a*cos(d*x + c) + (A - B + C)*a)*sqrt(-1/a)*lo g((2*sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(-1/a)*cos(d*x + c)*sin(d*x + c) + 3*cos(d*x + c)^2 + 2*cos(d*x + c) - 1)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) - ((7*A - 4*B + 8*C)*cos(d*x + c) + 7*A - 4*B + 8*C) *sqrt(-a)*log((2*a*cos(d*x + c)^2 + 2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/c os(d*x + c))*cos(d*x + c)*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) + 2*(2*A*cos(d*x + c)^2 - (A - 4*B)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a*d*cos(d*x + c) + a*d), -1/4*(((7*A - 4*B + 8*C)*cos(d*x + c) + 7*A - 4*B + 8*C)*sqrt(a)*arctan(sqrt((a*cos(d *x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - (2*A*cos (d*x + c)^2 - (A - 4*B)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c) - 4*sqrt(2)*((A - B + C)*a*cos(d*x + c) + (A - B + C)*a)* arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt( a)*sin(d*x + c)))/sqrt(a))/(a*d*cos(d*x + c) + a*d)]
\[ \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \cos ^{2}{\left (c + d x \right )}}{\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )}}\, dx \]
Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*cos(c + d*x)**2/sqrt(a*( sec(c + d*x) + 1)), x)
\[ \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{2}}{\sqrt {a \sec \left (d x + c\right ) + a}} \,d x } \]
integrate(cos(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1 /2),x, algorithm="maxima")
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*cos(d*x + c)^2/sqrt(a*se c(d*x + c) + a), x)
Leaf count of result is larger than twice the leaf count of optimal. 628 vs. \(2 (144) = 288\).
Time = 2.03 (sec) , antiderivative size = 628, normalized size of antiderivative = 3.72 \[ \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\frac {4 \, \sqrt {2} {\left (A - B + C\right )} \log \left ({\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2}\right )}{\sqrt {-a} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} + \frac {{\left (7 \, A - 4 \, B + 8 \, C\right )} \log \left ({\left | {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - a {\left (2 \, \sqrt {2} + 3\right )} \right |}\right )}{\sqrt {-a} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {{\left (7 \, A - 4 \, B + 8 \, C\right )} \log \left ({\left | {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} + a {\left (2 \, \sqrt {2} - 3\right )} \right |}\right )}{\sqrt {-a} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} + \frac {4 \, \sqrt {2} {\left (17 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{6} A \sqrt {-a} - 12 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{6} B \sqrt {-a} - 57 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{4} A \sqrt {-a} a + 76 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{4} B \sqrt {-a} a + 19 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} A \sqrt {-a} a^{2} - 36 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} B \sqrt {-a} a^{2} - 3 \, A \sqrt {-a} a^{3} + 4 \, B \sqrt {-a} a^{3}\right )}}{{\left ({\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{4} - 6 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} a + a^{2}\right )}^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}}{8 \, d} \]
integrate(cos(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1 /2),x, algorithm="giac")
1/8*(4*sqrt(2)*(A - B + C)*log((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*ta n(1/2*d*x + 1/2*c)^2 + a))^2)/(sqrt(-a)*sgn(cos(d*x + c))) + (7*A - 4*B + 8*C)*log(abs((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c) ^2 + a))^2 - a*(2*sqrt(2) + 3)))/(sqrt(-a)*sgn(cos(d*x + c))) - (7*A - 4*B + 8*C)*log(abs((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2 *c)^2 + a))^2 + a*(2*sqrt(2) - 3)))/(sqrt(-a)*sgn(cos(d*x + c))) + 4*sqrt( 2)*(17*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a ))^6*A*sqrt(-a) - 12*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*B*sqrt(-a) - 57*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(- a*tan(1/2*d*x + 1/2*c)^2 + a))^4*A*sqrt(-a)*a + 76*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4*B*sqrt(-a)*a + 19*(sqrt(- a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*A*sqrt(-a )*a^2 - 36*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*B*sqrt(-a)*a^2 - 3*A*sqrt(-a)*a^3 + 4*B*sqrt(-a)*a^3)/(((sqrt(-a) *tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4 - 6*(sqrt(- a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*a + a^2)^ 2*sgn(cos(d*x + c))))/d
Timed out. \[ \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}} \,d x \]