Integrand size = 31, antiderivative size = 133 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\sec (c+d x)}} \, dx=\frac {2 (5 A+3 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 B \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 C \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 B \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}} \] Output:
2/5*(5*A+3*C)*cos(d*x+c)^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*sec(d *x+c)^(1/2)/d+2/3*B*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2) )*sec(d*x+c)^(1/2)/d+2/5*C*sin(d*x+c)/d/sec(d*x+c)^(3/2)+2/3*B*sin(d*x+c)/ d/sec(d*x+c)^(1/2)
Time = 0.83 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.71 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\sec (c+d x)}} \, dx=\frac {\sqrt {\sec (c+d x)} \left (6 (5 A+3 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+10 B \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+(5 B+3 C \cos (c+d x)) \sin (2 (c+d x))\right )}{15 d} \] Input:
Integrate[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/Sqrt[Sec[c + d*x]],x]
Output:
(Sqrt[Sec[c + d*x]]*(6*(5*A + 3*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/ 2, 2] + 10*B*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + (5*B + 3*C*Cos [c + d*x])*Sin[2*(c + d*x)]))/(15*d)
Time = 0.65 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.89, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.387, Rules used = {3042, 4709, 3042, 3502, 27, 3042, 3227, 3042, 3115, 3042, 3119, 3120}
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.
| \(\displaystyle \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\sec (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \cos (c+d x)+C \cos (c+d x)^2}{\sqrt {\sec (c+d x)}}dx\) |
| \(\Big \downarrow \) 4709 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)} \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right )dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A\right )dx\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2}{5} \int \frac {1}{2} \sqrt {\cos (c+d x)} (5 A+3 C+5 B \cos (c+d x))dx+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \int \sqrt {\cos (c+d x)} (5 A+3 C+5 B \cos (c+d x))dx+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (5 A+3 C+5 B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left ((5 A+3 C) \int \sqrt {\cos (c+d x)}dx+5 B \int \cos ^{\frac {3}{2}}(c+d x)dx\right )+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left ((5 A+3 C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+5 B \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx\right )+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left ((5 A+3 C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+5 B \left (\frac {1}{3} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left ((5 A+3 C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+5 B \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (5 B \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 (5 A+3 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {2 (5 A+3 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+5 B \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )\) |
Input:
Int[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/Sqrt[Sec[c + d*x]],x]
Output:
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*C*Cos[c + d*x]^(3/2)*Sin[c + d*x ])/(5*d) + ((2*(5*A + 3*C)*EllipticE[(c + d*x)/2, 2])/d + 5*B*((2*Elliptic F[(c + d*x)/2, 2])/(3*d) + (2*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d)))/5)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Simp[(c*Sec[a + b*x])^m*(c*Cos[a + b*x])^m Int[ActivateTrig[u]/(c*Cos[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSineIntegrandQ[u, x]
Leaf count of result is larger than twice the leaf count of optimal. \(307\) vs. \(2(116)=232\).
Time = 9.74 (sec) , antiderivative size = 308, normalized size of antiderivative = 2.32
| method | result | size |
| default | \(\frac {2 \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (24 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (-20 B -24 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (10 B +6 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+15 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-5 B \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}+9 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{15 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) | \(308\) |
| parts | \(\frac {2 A \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{\sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}-\frac {2 B \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{3 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}-\frac {2 C \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (-8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-3 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\right )}{5 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) | \(516\) |
Input:
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/sec(d*x+c)^(1/2),x,method=_RETURNVERBO SE)
Output:
2/15*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(24*C*cos(1/2 *d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+(-20*B-24*C)*sin(1/2*d*x+1/2*c)^4*cos(1/2 *d*x+1/2*c)+(10*B+6*C)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+15*A*(sin(1 /2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2* d*x+1/2*c),2^(1/2))-5*B*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2 *d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)+9*C*(sin(1/2*d*x+1/2*c)^ 2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^( 1/2)))/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/ 2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
Result contains complex when optimal does not.
Time = 0.09 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.18 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\sec (c+d x)}} \, dx=\frac {-5 i \, \sqrt {2} B {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} B {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 \, \sqrt {2} {\left (-5 i \, A - 3 i \, C\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 \, \sqrt {2} {\left (5 i \, A + 3 i \, C\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + \frac {2 \, {\left (3 \, C \cos \left (d x + c\right )^{2} + 5 \, B \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{15 \, d} \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/sec(d*x+c)^(1/2),x, algorithm="f ricas")
Output:
1/15*(-5*I*sqrt(2)*B*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*I*sqrt(2)*B*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*sqrt(2)*(-5*I*A - 3*I*C)*weierstrassZeta(-4, 0, weierstrassPInver se(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*sqrt(2)*(5*I*A + 3*I*C)*weie rstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c ))) + 2*(3*C*cos(d*x + c)^2 + 5*B*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/d
\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\sec (c+d x)}} \, dx=\int \frac {A + B \cos {\left (c + d x \right )} + C \cos ^{2}{\left (c + d x \right )}}{\sqrt {\sec {\left (c + d x \right )}}}\, dx \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)**2)/sec(d*x+c)**(1/2),x)
Output:
Integral((A + B*cos(c + d*x) + C*cos(c + d*x)**2)/sqrt(sec(c + d*x)), x)
\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\sec (c+d x)}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{\sqrt {\sec \left (d x + c\right )}} \,d x } \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/sec(d*x+c)^(1/2),x, algorithm="m axima")
Output:
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)/sqrt(sec(d*x + c)), x)
\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\sec (c+d x)}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{\sqrt {\sec \left (d x + c\right )}} \,d x } \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/sec(d*x+c)^(1/2),x, algorithm="g iac")
Output:
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)/sqrt(sec(d*x + c)), x)
Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\sec (c+d x)}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \] Input:
int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(1/cos(c + d*x))^(1/2),x)
Output:
int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(1/cos(c + d*x))^(1/2), x)
\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\sec (c+d x)}} \, dx=\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )}{\sec \left (d x +c \right )}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{2}}{\sec \left (d x +c \right )}d x \right ) c \] Input:
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/sec(d*x+c)^(1/2),x)
Output:
int(sqrt(sec(c + d*x))/sec(c + d*x),x)*a + int((sqrt(sec(c + d*x))*cos(c + d*x))/sec(c + d*x),x)*b + int((sqrt(sec(c + d*x))*cos(c + d*x)**2)/sec(c + d*x),x)*c