Integrand size = 35, antiderivative size = 200 \[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=\frac {16 a^2 C \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {4 a^2 (3 A+C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}-\frac {2 a^2 (15 A-7 C) \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}-\frac {2 (5 A-C) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{5 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+a \cos (c+d x))^2 \sqrt {\sec (c+d x)} \sin (c+d x)}{d} \] Output:
16/5*a^2*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+4/3*a^2*(3*A+C)*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c, 2^(1/2))*sec(d*x+c)^(1/2)/d-2/15*a^2*(15*A-7*C)*sin(d*x+c)/d/sec(d*x+c)^(1 /2)-2/5*(5*A-C)*(a^2+a^2*cos(d*x+c))*sin(d*x+c)/d/sec(d*x+c)^(1/2)+2*A*(a+ a*cos(d*x+c))^2*sec(d*x+c)^(1/2)*sin(d*x+c)/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 4.56 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.88 \[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=\frac {a^2 \left (-96 i C+\frac {192 i C \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-e^{2 i (c+d x)}\right )}{\sqrt {1+e^{2 i (c+d x)}}}-\frac {80 i (3 A+C) e^{i (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-e^{2 i (c+d x)}\right )}{\sqrt {1+e^{2 i (c+d x)}}}+40 C \sin (c+d x)+3 C \sec (c+d x) \sin (3 (c+d x))+60 A \tan (c+d x)+3 C \tan (c+d x)\right )}{30 d \sqrt {\sec (c+d x)}} \] Input:
Integrate[(a + a*Cos[c + d*x])^2*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(3/2) ,x]
Output:
(a^2*((-96*I)*C + ((192*I)*C*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2*I)*( c + d*x))])/Sqrt[1 + E^((2*I)*(c + d*x))] - ((80*I)*(3*A + C)*E^(I*(c + d* x))*Hypergeometric2F1[1/4, 1/2, 5/4, -E^((2*I)*(c + d*x))])/Sqrt[1 + E^((2 *I)*(c + d*x))] + 40*C*Sin[c + d*x] + 3*C*Sec[c + d*x]*Sin[3*(c + d*x)] + 60*A*Tan[c + d*x] + 3*C*Tan[c + d*x]))/(30*d*Sqrt[Sec[c + d*x]])
Time = 1.36 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.96, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.486, Rules used = {3042, 4709, 3042, 3523, 27, 3042, 3455, 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.
\(\displaystyle \int \sec ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2 \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sec (c+d x)^{3/2} (a \cos (c+d x)+a)^2 \left (A+C \cos (c+d x)^2\right )dx\) |
\(\Big \downarrow \) 4709 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {(\cos (c+d x) a+a)^2 \left (C \cos ^2(c+d x)+A\right )}{\cos ^{\frac {3}{2}}(c+d x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+A\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx\) |
\(\Big \downarrow \) 3523 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \int \frac {(\cos (c+d x) a+a)^2 (4 a A-a (5 A-C) \cos (c+d x))}{2 \sqrt {\cos (c+d x)}}dx}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {(\cos (c+d x) a+a)^2 (4 a A-a (5 A-C) \cos (c+d x))}{\sqrt {\cos (c+d x)}}dx}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (4 a A-a (5 A-C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\right )\) |
\(\Big \downarrow \) 3455 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2}{5} \int \frac {(\cos (c+d x) a+a) \left (a^2 (15 A+C)-a^2 (15 A-7 C) \cos (c+d x)\right )}{2 \sqrt {\cos (c+d x)}}dx-\frac {2 (5 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^3 \cos (c+d x)+a^3\right )}{5 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{5} \int \frac {(\cos (c+d x) a+a) \left (a^2 (15 A+C)-a^2 (15 A-7 C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}}dx-\frac {2 (5 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^3 \cos (c+d x)+a^3\right )}{5 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{5} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (a^2 (15 A+C)-a^2 (15 A-7 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (5 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^3 \cos (c+d x)+a^3\right )}{5 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\right )\) |
\(\Big \downarrow \) 3447 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{5} \int \frac {-\left ((15 A-7 C) \cos ^2(c+d x) a^3\right )+(15 A+C) a^3+\left (a^3 (15 A+C)-a^3 (15 A-7 C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx-\frac {2 (5 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^3 \cos (c+d x)+a^3\right )}{5 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{5} \int \frac {-\left ((15 A-7 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^3\right )+(15 A+C) a^3+\left (a^3 (15 A+C)-a^3 (15 A-7 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (5 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^3 \cos (c+d x)+a^3\right )}{5 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\right )\) |
\(\Big \downarrow \) 3502 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{5} \left (\frac {2}{3} \int \frac {5 (3 A+C) a^3+12 C \cos (c+d x) a^3}{\sqrt {\cos (c+d x)}}dx-\frac {2 a^3 (15 A-7 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-\frac {2 (5 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^3 \cos (c+d x)+a^3\right )}{5 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{5} \left (\frac {2}{3} \int \frac {5 (3 A+C) a^3+12 C \sin \left (c+d x+\frac {\pi }{2}\right ) a^3}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 a^3 (15 A-7 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-\frac {2 (5 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^3 \cos (c+d x)+a^3\right )}{5 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\right )\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{5} \left (\frac {2}{3} \left (5 a^3 (3 A+C) \int \frac {1}{\sqrt {\cos (c+d x)}}dx+12 a^3 C \int \sqrt {\cos (c+d x)}dx\right )-\frac {2 a^3 (15 A-7 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-\frac {2 (5 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^3 \cos (c+d x)+a^3\right )}{5 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{5} \left (\frac {2}{3} \left (5 a^3 (3 A+C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+12 a^3 C \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )-\frac {2 a^3 (15 A-7 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-\frac {2 (5 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^3 \cos (c+d x)+a^3\right )}{5 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\right )\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{5} \left (\frac {2}{3} \left (5 a^3 (3 A+C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {24 a^3 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-\frac {2 a^3 (15 A-7 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-\frac {2 (5 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^3 \cos (c+d x)+a^3\right )}{5 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\right )\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{5} \left (\frac {2}{3} \left (\frac {10 a^3 (3 A+C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {24 a^3 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-\frac {2 a^3 (15 A-7 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-\frac {2 (5 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^3 \cos (c+d x)+a^3\right )}{5 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\right )\) |
Input:
Int[(a + a*Cos[c + d*x])^2*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(3/2),x]
Output:
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*A*(a + a*Cos[c + d*x])^2*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]) + ((-2*(5*A - C)*Sqrt[Cos[c + d*x]]*(a^3 + a ^3*Cos[c + d*x])*Sin[c + d*x])/(5*d) + ((2*((24*a^3*C*EllipticE[(c + d*x)/ 2, 2])/d + (10*a^3*(3*A + C)*EllipticF[(c + d*x)/2, 2])/d))/3 - (2*a^3*(15 *A - 7*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d))/5)/a)
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)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1 ) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, 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_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C + 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*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(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, 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]
Time = 9.26 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.34
method | result | size |
default | \(\frac {4 a^{2} \left (12 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-32 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+15 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) A -15 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+13 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) C -5 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+12 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{15 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) | \(268\) |
parts | \(\text {Expression too large to display}\) | \(853\) |
Input:
int((a+a*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2),x,method=_RETUR NVERBOSE)
Output:
4/15*a^2*(12*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-32*C*cos(1/2*d*x+1/ 2*c)*sin(1/2*d*x+1/2*c)^4+15*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)*A-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)*EllipticF( cos(1/2*d*x+1/2*c),2^(1/2))+13*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)*C-5 *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))+12*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)))/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 = 0.10 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.88 \[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} {\left (3 \, A + C\right )} a^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} {\left (3 \, A + C\right )} a^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 12 i \, \sqrt {2} C a^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 12 i \, \sqrt {2} C a^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (3 \, C a^{2} \cos \left (d x + c\right )^{2} + 10 \, C a^{2} \cos \left (d x + c\right ) + 15 \, A a^{2}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{15 \, d} \] Input:
integrate((a+a*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2),x, algori thm="fricas")
Output:
-2/15*(5*I*sqrt(2)*(3*A + C)*a^2*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 5*I*sqrt(2)*(3*A + C)*a^2*weierstrassPInverse(-4, 0, co s(d*x + c) - I*sin(d*x + c)) - 12*I*sqrt(2)*C*a^2*weierstrassZeta(-4, 0, w eierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 12*I*sqrt(2)*C *a^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*si n(d*x + c))) - (3*C*a^2*cos(d*x + c)^2 + 10*C*a^2*cos(d*x + c) + 15*A*a^2) *sin(d*x + c)/sqrt(cos(d*x + c)))/d
Timed out. \[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=\text {Timed out} \] Input:
integrate((a+a*cos(d*x+c))**2*(A+C*cos(d*x+c)**2)*sec(d*x+c)**(3/2),x)
Output:
Timed out
\[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {3}{2}} \,d x } \] Input:
integrate((a+a*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2),x, algori thm="maxima")
Output:
integrate((C*cos(d*x + c)^2 + A)*(a*cos(d*x + c) + a)^2*sec(d*x + c)^(3/2) , x)
\[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {3}{2}} \,d x } \] Input:
integrate((a+a*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2),x, algori thm="giac")
Output:
integrate((C*cos(d*x + c)^2 + A)*(a*cos(d*x + c) + a)^2*sec(d*x + c)^(3/2) , x)
Timed out. \[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=\int \left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^2 \,d x \] Input:
int((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(3/2)*(a + a*cos(c + d*x))^2,x )
Output:
int((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(3/2)*(a + a*cos(c + d*x))^2, x)
\[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=a^{2} \left (2 \left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )d x \right ) a +\left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{4} \sec \left (d x +c \right )d x \right ) c +2 \left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )d x \right ) a +\left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )d x \right ) a \right ) \] Input:
int((a+a*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2),x)
Output:
a**2*(2*int(sqrt(sec(c + d*x))*cos(c + d*x)*sec(c + d*x),x)*a + int(sqrt(s ec(c + d*x))*cos(c + d*x)**4*sec(c + d*x),x)*c + 2*int(sqrt(sec(c + d*x))* cos(c + d*x)**3*sec(c + d*x),x)*c + int(sqrt(sec(c + d*x))*cos(c + d*x)**2 *sec(c + d*x),x)*a + int(sqrt(sec(c + d*x))*cos(c + d*x)**2*sec(c + d*x),x )*c + int(sqrt(sec(c + d*x))*sec(c + d*x),x)*a)