Integrand size = 43, antiderivative size = 209 \[ \int \cos ^4(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\sqrt {a} (35 A+40 B+48 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{64 d}+\frac {a (35 A+40 B+48 C) \sin (c+d x)}{64 d \sqrt {a+a \sec (c+d x)}}+\frac {a (35 A+40 B+48 C) \cos (c+d x) \sin (c+d x)}{96 d \sqrt {a+a \sec (c+d x)}}+\frac {a (A+8 B) \cos ^2(c+d x) \sin (c+d x)}{24 d \sqrt {a+a \sec (c+d x)}}+\frac {A \cos ^3(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{4 d} \]
1/64*(35*A+40*B+48*C)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))*a^ (1/2)/d+1/64*a*(35*A+40*B+48*C)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/96*a *(35*A+40*B+48*C)*cos(d*x+c)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/24*a*(A +8*B)*cos(d*x+c)^2*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/4*A*cos(d*x+c)^3* sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.17 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.43 \[ \int \cos ^4(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \left (C \operatorname {Hypergeometric2F1}\left (\frac {1}{2},3,\frac {3}{2},1-\sec (c+d x)\right )+B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},4,\frac {3}{2},1-\sec (c+d x)\right )+A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},5,\frac {3}{2},1-\sec (c+d x)\right )\right ) \sqrt {a (1+\sec (c+d x))} \tan \left (\frac {1}{2} (c+d x)\right )}{d} \]
(2*(C*Hypergeometric2F1[1/2, 3, 3/2, 1 - Sec[c + d*x]] + B*Hypergeometric2 F1[1/2, 4, 3/2, 1 - Sec[c + d*x]] + A*Hypergeometric2F1[1/2, 5, 3/2, 1 - S ec[c + d*x]])*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d*x)/2])/d
Time = 1.06 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.98, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.279, Rules used = {3042, 4574, 27, 3042, 4503, 3042, 4292, 3042, 4292, 3042, 4261, 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.
| \(\displaystyle \int \cos ^4(c+d x) \sqrt {a \sec (c+d x)+a} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {a \csc \left (c+d x+\frac {\pi }{2}\right )+a} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
| \(\Big \downarrow \) 4574 |
| \(\displaystyle \frac {\int \frac {1}{2} \cos ^3(c+d x) \sqrt {\sec (c+d x) a+a} (a (A+8 B)+a (5 A+8 C) \sec (c+d x))dx}{4 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \cos ^3(c+d x) \sqrt {\sec (c+d x) a+a} (a (A+8 B)+a (5 A+8 C) \sec (c+d x))dx}{8 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (a (A+8 B)+a (5 A+8 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx}{8 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\) |
| \(\Big \downarrow \) 4503 |
| \(\displaystyle \frac {\frac {1}{6} a (35 A+40 B+48 C) \int \cos ^2(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {a^2 (A+8 B) \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}}{8 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{6} a (35 A+40 B+48 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a^2 (A+8 B) \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}}{8 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\) |
| \(\Big \downarrow \) 4292 |
| \(\displaystyle \frac {\frac {1}{6} a (35 A+40 B+48 C) \left (\frac {3}{4} \int \cos (c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {a \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (A+8 B) \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}}{8 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{6} a (35 A+40 B+48 C) \left (\frac {3}{4} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (A+8 B) \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}}{8 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\) |
| \(\Big \downarrow \) 4292 |
| \(\displaystyle \frac {\frac {1}{6} a (35 A+40 B+48 C) \left (\frac {3}{4} \left (\frac {1}{2} \int \sqrt {\sec (c+d x) a+a}dx+\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (A+8 B) \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}}{8 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{6} a (35 A+40 B+48 C) \left (\frac {3}{4} \left (\frac {1}{2} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (A+8 B) \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}}{8 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\) |
| \(\Big \downarrow \) 4261 |
| \(\displaystyle \frac {\frac {1}{6} a (35 A+40 B+48 C) \left (\frac {3}{4} \left (\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {a \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {a \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (A+8 B) \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}}{8 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {a^2 (A+8 B) \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {1}{6} a (35 A+40 B+48 C) \left (\frac {3}{4} \left (\frac {\sqrt {a} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )}{8 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\) |
(A*Cos[c + d*x]^3*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(4*d) + ((a^2*(A + 8*B)*Cos[c + d*x]^2*Sin[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]]) + (a*(3 5*A + 40*B + 48*C)*((a*Cos[c + d*x]*Sin[c + d*x])/(2*d*Sqrt[a + a*Sec[c + d*x]]) + (3*((Sqrt[a]*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x ]]])/d + (a*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]])))/4))/6)/(8*a)
3.5.90.3.1 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_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[a*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*n*Sqrt[a + b*Csc[e + f*x]])), x] + Simp[a*((2*n + 1)/(2*b*d*n)) Int[Sqrt[a + b*Csc [e + f*x]]*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[n, -2^(-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[A*b^2*Co t[e + f*x]*((d*Csc[e + f*x])^n/(a*f*n*Sqrt[a + b*Csc[e + f*x]])), x] + Simp [(A*b*(2*n + 1) + 2*a*B*n)/(2*a*d*n) Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[ e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, A, B}, 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[((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. \(559\) vs. \(2(185)=370\).
Time = 2.26 (sec) , antiderivative size = 560, normalized size of antiderivative = 2.68
| method | result | size |
| default | \(\frac {\left (48 A \cos \left (d x +c \right )^{4} \sin \left (d x +c \right )+56 A \cos \left (d x +c \right )^{3} \sin \left (d x +c \right )+64 B \sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}+105 A \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )+70 A \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )+120 B \,\operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )+80 B \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}+144 C \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )+96 C \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )+105 A \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+105 A \cos \left (d x +c \right ) \sin \left (d x +c \right )+120 B \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+120 B \cos \left (d x +c \right ) \sin \left (d x +c \right )+144 C \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+144 C \cos \left (d x +c \right ) \sin \left (d x +c \right )\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{192 d \left (\cos \left (d x +c \right )+1\right )}\) | \(560\) |
int(cos(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x, method=_RETURNVERBOSE)
1/192/d*(48*A*cos(d*x+c)^4*sin(d*x+c)+56*A*cos(d*x+c)^3*sin(d*x+c)+64*B*si n(d*x+c)*cos(d*x+c)^3+105*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)+70*A *cos(d*x+c)^2*sin(d*x+c)+120*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)+8 0*B*sin(d*x+c)*cos(d*x+c)^2+144*C*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arcta nh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c )+96*C*cos(d*x+c)^2*sin(d*x+c)+105*A*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ar ctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+105*A* cos(d*x+c)*sin(d*x+c)+120*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)/(cos(d*x+c)+1))^(1/2))+120*B*cos(d*x+c )*sin(d*x+c)+144*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))+144*C*cos(d*x+c)*sin(d*x +c))*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)
Time = 0.44 (sec) , antiderivative size = 404, normalized size of antiderivative = 1.93 \[ \int \cos ^4(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\left [\frac {3 \, {\left ({\left (35 \, A + 40 \, B + 48 \, C\right )} \cos \left (d x + c\right ) + 35 \, A + 40 \, B + 48 \, C\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (48 \, A \cos \left (d x + c\right )^{4} + 8 \, {\left (7 \, A + 8 \, B\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (35 \, A + 40 \, B + 48 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (35 \, A + 40 \, B + 48 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{384 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, -\frac {3 \, {\left ({\left (35 \, A + 40 \, B + 48 \, C\right )} \cos \left (d x + c\right ) + 35 \, A + 40 \, B + 48 \, C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left (48 \, A \cos \left (d x + c\right )^{4} + 8 \, {\left (7 \, A + 8 \, B\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (35 \, A + 40 \, B + 48 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (35 \, A + 40 \, B + 48 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{192 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \]
integrate(cos(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1 /2),x, algorithm="fricas")
[1/384*(3*((35*A + 40*B + 48*C)*cos(d*x + c) + 35*A + 40*B + 48*C)*sqrt(-a )*log((2*a*cos(d*x + c)^2 - 2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) + 2*(48*A*cos(d*x + c)^4 + 8*(7*A + 8*B)*cos(d*x + c)^3 + 2*(35*A + 40*B + 48*C)*cos(d*x + c)^2 + 3*(35*A + 40*B + 48*C)*cos(d*x + c))*sqrt((a*cos(d* x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c) + d), -1/192*(3*(( 35*A + 40*B + 48*C)*cos(d*x + c) + 35*A + 40*B + 48*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))) - (48*A*cos(d*x + c)^4 + 8*(7*A + 8*B)*cos(d*x + c)^3 + 2*(35*A + 40*B + 48 *C)*cos(d*x + c)^2 + 3*(35*A + 40*B + 48*C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c) + d)]
Timed out. \[ \int \cos ^4(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 9618 vs. \(2 (185) = 370\).
Time = 1.08 (sec) , antiderivative size = 9618, normalized size of antiderivative = 46.02 \[ \int \cos ^4(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \]
integrate(cos(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1 /2),x, algorithm="maxima")
1/768*(8*(4*(cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))^2 + sin( 2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))^2 + 2*cos(2/3*arctan2(sin (3*d*x + 3*c), cos(3*d*x + 3*c))) + 1)^(3/4)*(cos(3/2*arctan2(sin(2/3*arct an2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))), cos(2/3*arctan2(sin(3*d*x + 3*c) , cos(3*d*x + 3*c))) + 1))*sin(3*d*x + 3*c) - (cos(3*d*x + 3*c) - 1)*sin(3 /2*arctan2(sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))), cos(2/3*a rctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) + 1)))*sqrt(a) + 6*(cos(2/3*ar ctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))^2 + sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))^2 + 2*cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d* x + 3*c))) + 1)^(1/4)*((sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c) )) + 5*sin(1/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))))*cos(1/2*arcta n2(sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))), cos(2/3*arctan2(s in(3*d*x + 3*c), cos(3*d*x + 3*c))) + 1)) - (cos(2/3*arctan2(sin(3*d*x + 3 *c), cos(3*d*x + 3*c))) + 3*cos(1/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) - 4)*sin(1/2*arctan2(sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))), cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) + 1)))*sqrt(a ) + 15*sqrt(a)*(arctan2(-(cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3* c)))^2 + sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))^2 + 2*cos(2/ 3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) + 1)^(1/4)*(cos(1/2*arctan2 (sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))), cos(2/3*arctan2(...
\[ \int \cos ^4(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sqrt {a \sec \left (d x + c\right ) + a} \cos \left (d x + c\right )^{4} \,d x } \]
integrate(cos(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1 /2),x, algorithm="giac")
Timed out. \[ \int \cos ^4(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int {\cos \left (c+d\,x\right )}^4\,\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]