Integrand size = 43, antiderivative size = 212 \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=-\frac {4 a^2 (A-C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{d}+\frac {4 a^2 (2 A+3 B+2 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}-\frac {2 a^2 (5 A+3 B-C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 (4 A+3 B) \left (a^2+a^2 \cos (c+d x)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+a \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d} \]
2/3*A*(a+a*cos(d*x+c))^2*sec(d*x+c)^(3/2)*sin(d*x+c)/d-2/3*a^2*(5*A+3*B-C) *sin(d*x+c)/d/sec(d*x+c)^(1/2)+2/3*(4*A+3*B)*(a^2+a^2*cos(d*x+c))*sin(d*x+ c)*sec(d*x+c)^(1/2)/d-4*a^2*(A-C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x +1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^ (1/2)/d+4/3*a^2*(2*A+3*B+2*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2 *c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2 )/d
Time = 3.12 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.56 \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=\frac {a^2 \sec ^{\frac {3}{2}}(c+d x) \left (-24 (A-C) \cos ^{\frac {3}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+8 (2 A+3 B+2 C) \cos ^{\frac {3}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+2 (2 A+C+6 (2 A+B) \cos (c+d x)+C \cos (2 (c+d x))) \sin (c+d x)\right )}{6 d} \]
Integrate[(a + a*Cos[c + d*x])^2*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*S ec[c + d*x]^(5/2),x]
(a^2*Sec[c + d*x]^(3/2)*(-24*(A - C)*Cos[c + d*x]^(3/2)*EllipticE[(c + d*x )/2, 2] + 8*(2*A + 3*B + 2*C)*Cos[c + d*x]^(3/2)*EllipticF[(c + d*x)/2, 2] + 2*(2*A + C + 6*(2*A + B)*Cos[c + d*x] + C*Cos[2*(c + d*x)])*Sin[c + d*x ]))/(6*d)
Time = 1.43 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.97, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.395, Rules used = {3042, 4709, 3042, 3522, 27, 3042, 3454, 27, 3042, 3447, 3042, 3502, 3042, 3227, 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.
\(\displaystyle \int \sec ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sec (c+d x)^{5/2} (a \cos (c+d x)+a)^2 \left (A+B \cos (c+d x)+C \cos (c+d x)^2\right )dx\) |
\(\Big \downarrow \) 4709 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {(\cos (c+d x) a+a)^2 \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right )}{\cos ^{\frac {5}{2}}(c+d x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx\) |
\(\Big \downarrow \) 3522 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \int \frac {(\cos (c+d x) a+a)^2 (a (4 A+3 B)-3 a (A-C) \cos (c+d x))}{2 \cos ^{\frac {3}{2}}(c+d x)}dx}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {(\cos (c+d x) a+a)^2 (a (4 A+3 B)-3 a (A-C) \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)}dx}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (a (4 A+3 B)-3 a (A-C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3454 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \int \frac {3 (\cos (c+d x) a+a) \left (a^2 (3 A+3 B+C)-a^2 (5 A+3 B-C) \cos (c+d x)\right )}{2 \sqrt {\cos (c+d x)}}dx+\frac {2 (4 A+3 B) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {3 \int \frac {(\cos (c+d x) a+a) \left (a^2 (3 A+3 B+C)-a^2 (5 A+3 B-C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}}dx+\frac {2 (4 A+3 B) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {3 \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (a^2 (3 A+3 B+C)-a^2 (5 A+3 B-C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (4 A+3 B) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3447 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {3 \int \frac {-\left ((5 A+3 B-C) \cos ^2(c+d x) a^3\right )+(3 A+3 B+C) a^3+\left (a^3 (3 A+3 B+C)-a^3 (5 A+3 B-C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx+\frac {2 (4 A+3 B) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {3 \int \frac {-\left ((5 A+3 B-C) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^3\right )+(3 A+3 B+C) a^3+\left (a^3 (3 A+3 B+C)-a^3 (5 A+3 B-C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (4 A+3 B) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3502 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {3 \left (\frac {2}{3} \int \frac {a^3 (2 A+3 B+2 C)-3 a^3 (A-C) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx-\frac {2 a^3 (5 A+3 B-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 (4 A+3 B) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {3 \left (\frac {2}{3} \int \frac {a^3 (2 A+3 B+2 C)-3 a^3 (A-C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 a^3 (5 A+3 B-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 (4 A+3 B) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {3 \left (\frac {2}{3} \left (a^3 (2 A+3 B+2 C) \int \frac {1}{\sqrt {\cos (c+d x)}}dx-3 a^3 (A-C) \int \sqrt {\cos (c+d x)}dx\right )-\frac {2 a^3 (5 A+3 B-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 (4 A+3 B) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {3 \left (\frac {2}{3} \left (a^3 (2 A+3 B+2 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-3 a^3 (A-C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )-\frac {2 a^3 (5 A+3 B-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 (4 A+3 B) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {3 \left (\frac {2}{3} \left (a^3 (2 A+3 B+2 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 a^3 (A-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-\frac {2 a^3 (5 A+3 B-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 (4 A+3 B) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {3 \left (\frac {2}{3} \left (\frac {2 a^3 (2 A+3 B+2 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}-\frac {6 a^3 (A-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-\frac {2 a^3 (5 A+3 B-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 (4 A+3 B) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*A*(a + a*Cos[c + d*x])^2*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + ((2*(4*A + 3*B)*(a^3 + a^3*Cos[c + d*x]) *Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]) + 3*((2*((-6*a^3*(A - C)*EllipticE[( c + d*x)/2, 2])/d + (2*a^3*(2*A + 3*B + 2*C)*EllipticF[(c + d*x)/2, 2])/d) )/3 - (2*a^3*(5*A + 3*B - C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d)))/(3*a ))
3.13.76.3.1 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[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_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 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^2)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[ e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a*d))), x] - Simp[b/(d*(n + 1)*(b*c + a*d)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp [a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n + 1) - B *(b*c*m - a*d*(n + 1)))*Sin[e + f*x], 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] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0 ])
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[((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])
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. \(800\) vs. \(2(242)=484\).
Time = 158.22 (sec) , antiderivative size = 801, normalized size of antiderivative = 3.78
method | result | size |
default | \(\text {Expression too large to display}\) | \(801\) |
parts | \(\text {Expression too large to display}\) | \(886\) |
int((a+a*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2),x, method=_RETURNVERBOSE)
-4/3*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2/(4*sin( 1/2*d*x+1/2*c)^4-4*sin(1/2*d*x+1/2*c)^2+1)/sin(1/2*d*x+1/2*c)^3*(-4*C*cos( 1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+12*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/ 2*c)^4-4*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)*E llipticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2-6*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))*sin(1/2*d*x+1/2*c)^2+6*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2* c)^4-6*B*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*Ell ipticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2+4*C*cos(1/2*d*x+1/ 2*c)*sin(1/2*d*x+1/2*c)^4-4*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c) ^2+6*C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)* (2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^2-7*sin(1/2*d*x+1/2*c) ^2*cos(1/2*d*x+1/2*c)*A+2*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3*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))-3*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+3*B*(2*sin(1/2*d*x+1 /2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(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)*C+2*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)*EllipticF(cos(1/2*d*x+1/2*c...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.08 \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=-\frac {2 \, {\left (i \, \sqrt {2} {\left (2 \, A + 3 \, B + 2 \, C\right )} a^{2} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - i \, \sqrt {2} {\left (2 \, A + 3 \, B + 2 \, C\right )} a^{2} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 i \, \sqrt {2} {\left (A - C\right )} a^{2} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 i \, \sqrt {2} {\left (A - C\right )} a^{2} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (C a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right ) + A a^{2}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{3 \, d \cos \left (d x + c\right )} \]
integrate((a+a*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(5 /2),x, algorithm="fricas")
-2/3*(I*sqrt(2)*(2*A + 3*B + 2*C)*a^2*cos(d*x + c)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - I*sqrt(2)*(2*A + 3*B + 2*C)*a^2*cos(d *x + c)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*I*sq rt(2)*(A - C)*a^2*cos(d*x + c)*weierstrassZeta(-4, 0, weierstrassPInverse( -4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*I*sqrt(2)*(A - C)*a^2*cos(d*x + c)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin (d*x + c))) - (C*a^2*cos(d*x + c)^2 + 3*(2*A + B)*a^2*cos(d*x + c) + A*a^2 )*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c))
Timed out. \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=\text {Timed out} \]
\[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {5}{2}} \,d x } \]
integrate((a+a*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(5 /2),x, algorithm="maxima")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)^2*s ec(d*x + c)^(5/2), x)
\[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {5}{2}} \,d x } \]
integrate((a+a*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(5 /2),x, algorithm="giac")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)^2*s ec(d*x + c)^(5/2), x)
Timed out. \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=\int {\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^2\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right ) \,d x \]