Integrand size = 43, antiderivative size = 169 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {(7 A-4 B+8 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{4 \sqrt {a} d}-\frac {\sqrt {2} (A-B+C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{\sqrt {a} d}-\frac {(A-4 B) \tan (c+d x)}{4 d \sqrt {a+a \cos (c+d x)}}+\frac {A \sec (c+d x) \tan (c+d x)}{2 d \sqrt {a+a \cos (c+d x)}} \]
1/4*(7*A-4*B+8*C)*arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/d/a^( 1/2)-(A-B+C)*arctanh(1/2*sin(d*x+c)*a^(1/2)*2^(1/2)/(a+a*cos(d*x+c))^(1/2) )*2^(1/2)/d/a^(1/2)-1/4*(A-4*B)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/2*A* sec(d*x+c)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)
Time = 0.84 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.70 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \left (-8 (A-B+C) \text {arctanh}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+\sqrt {2} (7 A-4 B+8 C) \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \sec (c+d x) (-A+4 B+2 A \sec (c+d x)) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{4 d \sqrt {a (1+\cos (c+d x))}} \]
Integrate[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^3)/Sqrt[a + a*Cos[c + d*x]],x]
(Cos[(c + d*x)/2]*(-8*(A - B + C)*ArcTanh[Sin[(c + d*x)/2]] + Sqrt[2]*(7*A - 4*B + 8*C)*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]] + 2*Sec[c + d*x]*(-A + 4*B + 2*A*Sec[c + d*x])*Sin[(c + d*x)/2]))/(4*d*Sqrt[a*(1 + Cos[c + d*x])])
Time = 1.13 (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, 3522, 27, 3042, 3463, 27, 3042, 3464, 3042, 3128, 219, 3252, 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.
| \(\displaystyle \int \frac {\sec ^3(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {a \cos (c+d x)+a}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \sqrt {a \sin \left (c+d x+\frac {\pi }{2}\right )+a}}dx\) |
| \(\Big \downarrow \) 3522 |
| \(\displaystyle \frac {\int -\frac {(a (A-4 B)-a (3 A+4 C) \cos (c+d x)) \sec ^2(c+d x)}{2 \sqrt {\cos (c+d x) a+a}}dx}{2 a}+\frac {A \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {(a (A-4 B)-a (3 A+4 C) \cos (c+d x)) \sec ^2(c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{4 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {a (A-4 B)-a (3 A+4 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a}\) |
| \(\Big \downarrow \) 3463 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {\int -\frac {\left (a^2 (7 A-4 B+8 C)-a^2 (A-4 B) \cos (c+d x)\right ) \sec (c+d x)}{2 \sqrt {\cos (c+d x) a+a}}dx}{a}+\frac {a (A-4 B) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{4 a}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a (A-4 B) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {\left (a^2 (7 A-4 B+8 C)-a^2 (A-4 B) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a (A-4 B) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {a^2 (7 A-4 B+8 C)-a^2 (A-4 B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 3464 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a (A-4 B) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {a (7 A-4 B+8 C) \int \sqrt {\cos (c+d x) a+a} \sec (c+d x)dx-8 a^2 (A-B+C) \int \frac {1}{\sqrt {\cos (c+d x) a+a}}dx}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a (A-4 B) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {a (7 A-4 B+8 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-8 a^2 (A-B+C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 3128 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a (A-4 B) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {16 a^2 (A-B+C) \int \frac {1}{2 a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}+a (7 A-4 B+8 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a (A-4 B) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {a (7 A-4 B+8 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {8 \sqrt {2} a^{3/2} (A-B+C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 3252 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a (A-4 B) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {-\frac {2 a^2 (7 A-4 B+8 C) \int \frac {1}{a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {8 \sqrt {2} a^{3/2} (A-B+C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a (A-4 B) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^{3/2} (7 A-4 B+8 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {8 \sqrt {2} a^{3/2} (A-B+C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{2 a}}{4 a}\) |
(A*Sec[c + d*x]*Tan[c + d*x])/(2*d*Sqrt[a + a*Cos[c + d*x]]) - (-1/2*((2*a ^(3/2)*(7*A - 4*B + 8*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d - (8*Sqrt[2]*a^(3/2)*(A - B + C)*ArcTanh[(Sqrt[a]*Sin[c + d*x]) /(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/d)/a + (a*(A - 4*B)*Tan[c + d*x])/(d *Sqrt[a + a*Cos[c + d*x]]))/(4*a)
3.5.7.3.1 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[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2/d Subst[Int[1/(2*a - x^2), x], x, b*(Cos[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + ( f_.)*(x_)]), x_Symbol] :> Simp[-2*(b/f) Subst[Int[1/(b*c + a*d - d*x^2), x], x, b*(Cos[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(b*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && Eq Q[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m + 1/2, 0])
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(A *b - a*B)/(b*c - a*d) Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Simp[(B*c - A*d)/(b*c - a*d) Int[Sqrt[a + b*Sin[e + f*x]]/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C - B*c*d + A*d^2))*Cos[e + f*x ]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(b*d*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^m* (c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + (c*C - B*d)*( a*c*m + b*d*(n + 1)) + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2* (n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ [m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0])
Leaf count of result is larger than twice the leaf count of optimal. \(1382\) vs. \(2(144)=288\).
Time = 14.66 (sec) , antiderivative size = 1383, normalized size of antiderivative = 8.18
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(1383\) |
| default | \(\text {Expression too large to display}\) | \(1768\) |
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+cos(d*x+c)*a)^(1/2),x, method=_RETURNVERBOSE)
-1/2*A*cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-4*a*(-8*2^(1/2) *ln(4*(a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+a)/cos(1/2*d*x+1/2*c))+7*ln( -4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)-2^(1/2)*(a *sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))+7*ln(4/(2*cos(1/2*d*x+1/2*c)+2^ (1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2 )*a^(1/2)+2*a)))*sin(1/2*d*x+1/2*c)^4+(-32*2^(1/2)*ln(4*(a^(1/2)*(a*sin(1/ 2*d*x+1/2*c)^2)^(1/2)+a)/cos(1/2*d*x+1/2*c))*a-4*2^(1/2)*(a*sin(1/2*d*x+1/ 2*c)^2)^(1/2)*a^(1/2)+28*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*a*co s(1/2*d*x+1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*a))*a+28 *ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)-2^(1/2 )*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))*a)*sin(1/2*d*x+1/2*c)^2+8*2 ^(1/2)*ln(4*(a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+a)/cos(1/2*d*x+1/2*c)) *a-2*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-7*ln(4/(2*cos(1/2*d*x+ 1/2*c)+2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1/2*c )^2)^(1/2)*a^(1/2)+2*a))*a-7*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(2^(1/2) *a*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a)) *a)/a^(3/2)/(2*cos(1/2*d*x+1/2*c)-2^(1/2))^2/(2*cos(1/2*d*x+1/2*c)+2^(1/2) )^2/sin(1/2*d*x+1/2*c)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d-1/2*B*cos(1/2*d*x+ 1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*a*(2^(1/2)*ln(2/(2*cos(1/2*d*x+1 /2*c)+2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1/2...
Leaf count of result is larger than twice the leaf count of optimal. 292 vs. \(2 (144) = 288\).
Time = 0.38 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.73 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {{\left ({\left (7 \, A - 4 \, B + 8 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (7 \, A - 4 \, B + 8 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) - 4 \, {\left ({\left (A - 4 \, B\right )} \cos \left (d x + c\right ) - 2 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right ) + \frac {8 \, \sqrt {2} {\left ({\left (A - B + C\right )} a \cos \left (d x + c\right )^{3} + {\left (A - B + C\right )} a \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} + \frac {2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{\sqrt {a}}}{16 \, {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\right )}} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c))^(1 /2),x, algorithm="fricas")
1/16*(((7*A - 4*B + 8*C)*cos(d*x + c)^3 + (7*A - 4*B + 8*C)*cos(d*x + c)^2 )*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*sqrt(a*cos(d*x + c) + a)*sqrt(a)*(cos(d*x + c) - 2)*sin(d*x + c) + 8*a)/(cos(d*x + c)^3 + c os(d*x + c)^2)) - 4*((A - 4*B)*cos(d*x + c) - 2*A)*sqrt(a*cos(d*x + c) + a )*sin(d*x + c) + 8*sqrt(2)*((A - B + C)*a*cos(d*x + c)^3 + (A - B + C)*a*c os(d*x + c)^2)*log(-(cos(d*x + c)^2 + 2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*s in(d*x + c)/sqrt(a) - 2*cos(d*x + c) - 3)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1))/sqrt(a))/(a*d*cos(d*x + c)^3 + a*d*cos(d*x + c)^2)
\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\int \frac {\left (A + B \cos {\left (c + d x \right )} + C \cos ^{2}{\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\sqrt {a \left (\cos {\left (c + d x \right )} + 1\right )}}\, dx \]
Integral((A + B*cos(c + d*x) + C*cos(c + d*x)**2)*sec(c + d*x)**3/sqrt(a*( cos(c + d*x) + 1)), x)
Leaf count of result is larger than twice the leaf count of optimal. 76209 vs. \(2 (144) = 288\).
Time = 2.81 (sec) , antiderivative size = 76209, normalized size of antiderivative = 450.94 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\text {Too large to display} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c))^(1 /2),x, algorithm="maxima")
-1/16*((4*sqrt(2)*cos(6*d*x + 6*c)^2*sin(3/2*d*x + 3/2*c) + 16*sqrt(2)*cos (5*d*x + 5*c)^2*sin(3/2*d*x + 3/2*c) + 36*sqrt(2)*cos(4*d*x + 4*c)^2*sin(3 /2*d*x + 3/2*c) + 64*sqrt(2)*cos(3*d*x + 3*c)^2*sin(3/2*d*x + 3/2*c) + 36* sqrt(2)*cos(2*d*x + 2*c)^2*sin(3/2*d*x + 3/2*c) + 4*sqrt(2)*sin(6*d*x + 6* c)^2*sin(3/2*d*x + 3/2*c) + 16*sqrt(2)*sin(5*d*x + 5*c)^2*sin(3/2*d*x + 3/ 2*c) + 36*sqrt(2)*sin(4*d*x + 4*c)^2*sin(3/2*d*x + 3/2*c) + 64*sqrt(2)*sin (3*d*x + 3*c)^2*sin(3/2*d*x + 3/2*c) + 36*sqrt(2)*sin(2*d*x + 2*c)^2*sin(3 /2*d*x + 3/2*c) - 8*(3*sqrt(2)*cos(3/2*d*x + 3/2*c)*sin(2*d*x + 2*c) - 3*s qrt(2)*cos(2*d*x + 2*c)*sin(3/2*d*x + 3/2*c) + 2*sqrt(2)*cos(3/2*d*x + 3/2 *c)*sin(d*x + c) + (sqrt(2)*sin(9/2*d*x + 9/2*c) + 3*sqrt(2)*sin(7/2*d*x + 7/2*c) - 3*sqrt(2)*sin(5/2*d*x + 5/2*c) - sqrt(2)*sin(3/2*d*x + 3/2*c))*c os(6*d*x + 6*c) + 2*(sqrt(2)*sin(9/2*d*x + 9/2*c) + 3*sqrt(2)*sin(7/2*d*x + 7/2*c) - 3*sqrt(2)*sin(5/2*d*x + 5/2*c) - sqrt(2)*sin(3/2*d*x + 3/2*c))* cos(5*d*x + 5*c) - (3*sqrt(2)*sin(4*d*x + 4*c) + 4*sqrt(2)*sin(3*d*x + 3*c ) + 3*sqrt(2)*sin(2*d*x + 2*c) + 2*sqrt(2)*sin(d*x + c))*cos(9/2*d*x + 9/2 *c) + 3*(3*sqrt(2)*sin(7/2*d*x + 7/2*c) - 3*sqrt(2)*sin(5/2*d*x + 5/2*c) - sqrt(2)*sin(3/2*d*x + 3/2*c))*cos(4*d*x + 4*c) - 3*(4*sqrt(2)*sin(3*d*x + 3*c) + 3*sqrt(2)*sin(2*d*x + 2*c) + 2*sqrt(2)*sin(d*x + c))*cos(7/2*d*x + 7/2*c) - 4*(3*sqrt(2)*sin(5/2*d*x + 5/2*c) + sqrt(2)*sin(3/2*d*x + 3/2*c) )*cos(3*d*x + 3*c) + 3*(3*sqrt(2)*sin(2*d*x + 2*c) + 2*sqrt(2)*sin(d*x ...
Leaf count of result is larger than twice the leaf count of optimal. 303 vs. \(2 (144) = 288\).
Time = 0.42 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.79 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=-\frac {\frac {4 \, \sqrt {2} {\left (A \sqrt {a} - B \sqrt {a} + C \sqrt {a}\right )} \log \left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {4 \, \sqrt {2} {\left (A \sqrt {a} - B \sqrt {a} + C \sqrt {a}\right )} \log \left (-\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {{\left (7 \, A - 4 \, B + 8 \, C\right )} \log \left ({\left | \frac {1}{2} \, \sqrt {2} + \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{\sqrt {a} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {{\left (7 \, A - 4 \, B + 8 \, C\right )} \log \left ({\left | -\frac {1}{2} \, \sqrt {2} + \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{\sqrt {a} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {2 \, {\left (2 \, \sqrt {2} A \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 \, \sqrt {2} B \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + \sqrt {2} A \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, \sqrt {2} B \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{8 \, d} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c))^(1 /2),x, algorithm="giac")
-1/8*(4*sqrt(2)*(A*sqrt(a) - B*sqrt(a) + C*sqrt(a))*log(sin(1/2*d*x + 1/2* c) + 1)/(a*sgn(cos(1/2*d*x + 1/2*c))) - 4*sqrt(2)*(A*sqrt(a) - B*sqrt(a) + C*sqrt(a))*log(-sin(1/2*d*x + 1/2*c) + 1)/(a*sgn(cos(1/2*d*x + 1/2*c))) - (7*A - 4*B + 8*C)*log(abs(1/2*sqrt(2) + sin(1/2*d*x + 1/2*c)))/(sqrt(a)*s gn(cos(1/2*d*x + 1/2*c))) + (7*A - 4*B + 8*C)*log(abs(-1/2*sqrt(2) + sin(1 /2*d*x + 1/2*c)))/(sqrt(a)*sgn(cos(1/2*d*x + 1/2*c))) - 2*(2*sqrt(2)*A*sqr t(a)*sin(1/2*d*x + 1/2*c)^3 - 8*sqrt(2)*B*sqrt(a)*sin(1/2*d*x + 1/2*c)^3 + sqrt(2)*A*sqrt(a)*sin(1/2*d*x + 1/2*c) + 4*sqrt(2)*B*sqrt(a)*sin(1/2*d*x + 1/2*c))/((2*sin(1/2*d*x + 1/2*c)^2 - 1)^2*a*sgn(cos(1/2*d*x + 1/2*c))))/ d
Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{{\cos \left (c+d\,x\right )}^3\,\sqrt {a+a\,\cos \left (c+d\,x\right )}} \,d x \]