Integrand size = 43, antiderivative size = 165 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))} \, dx=\frac {(3 A-3 B+C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}+\frac {(5 A-3 B+3 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a d}+\frac {(5 A-3 B+3 C) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {(3 A-3 B+C) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {(A-B+C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))} \]
(3*A-3*B+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))/a/d+1/3*(5*A-3*B+3*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))/a/d+1/3*(5*A-3*B +3*C)*sin(d*x+c)/a/d/cos(d*x+c)^(3/2)-(A-B+C)*sin(d*x+c)/d/cos(d*x+c)^(3/2 )/(a+a*cos(d*x+c))-(3*A-3*B+C)*sin(d*x+c)/a/d/cos(d*x+c)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 8.37 (sec) , antiderivative size = 1332, normalized size of antiderivative = 8.07 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))} \, dx =\text {Too large to display} \]
Integrate[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(5/2)*(a + a*Cos[c + d*x])),x]
(Cos[c/2 + (d*x)/2]^2*Sqrt[Cos[c + d*x]]*(-(((2*A - 2*B + A*Cos[c] - B*Cos [c] + C*Cos[c])*Csc[c/2]*Sec[c/2]*Sec[c])/d) - (2*Sec[c/2]*Sec[c/2 + (d*x) /2]*(A*Sin[(d*x)/2] - B*Sin[(d*x)/2] + C*Sin[(d*x)/2]))/d + (4*A*Sec[c]*Se c[c + d*x]^2*Sin[d*x])/(3*d) + (4*Sec[c]*Sec[c + d*x]*(A*Sin[c] - 3*A*Sin[ d*x] + 3*B*Sin[d*x]))/(3*d)))/(a + a*Cos[c + d*x]) - (5*A*Cos[c/2 + (d*x)/ 2]^2*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c] ]]^2]*Sec[c/2]*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]] ]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + S in[d*x - ArcTan[Cot[c]]]])/(3*d*(a + a*Cos[c + d*x])*Sqrt[1 + Cot[c]^2]) + (B*Cos[c/2 + (d*x)/2]^2*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin [d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[ d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[C ot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(d*(a + a*Cos[c + d*x])*Sqr t[1 + Cot[c]^2]) - (C*Cos[c/2 + (d*x)/2]^2*Csc[c/2]*HypergeometricPFQ[{1/4 , 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[d*x - ArcTan[Cot[ c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]* Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(d*(a + a *Cos[c + d*x])*Sqrt[1 + Cot[c]^2]) - (3*A*Cos[c/2 + (d*x)/2]^2*Csc[c/2]*Se c[c/2]*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^ 2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]...
Time = 0.70 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.92, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {3042, 3520, 27, 3042, 3227, 3042, 3116, 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.
\(\displaystyle \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )}dx\) |
\(\Big \downarrow \) 3520 |
\(\displaystyle \frac {\int \frac {a (5 A-3 B+3 C)-a (3 A-3 B+C) \cos (c+d x)}{2 \cos ^{\frac {5}{2}}(c+d x)}dx}{a^2}-\frac {(A-B+C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {a (5 A-3 B+3 C)-a (3 A-3 B+C) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x)}dx}{2 a^2}-\frac {(A-B+C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {a (5 A-3 B+3 C)-a (3 A-3 B+C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx}{2 a^2}-\frac {(A-B+C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \frac {a (5 A-3 B+3 C) \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x)}dx-a (3 A-3 B+C) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)}dx}{2 a^2}-\frac {(A-B+C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {a (5 A-3 B+3 C) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx-a (3 A-3 B+C) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx}{2 a^2}-\frac {(A-B+C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}\) |
\(\Big \downarrow \) 3116 |
\(\displaystyle \frac {a (5 A-3 B+3 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )-a (3 A-3 B+C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\cos (c+d x)}dx\right )}{2 a^2}-\frac {(A-B+C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {a (5 A-3 B+3 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )-a (3 A-3 B+C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )}{2 a^2}-\frac {(A-B+C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {a (5 A-3 B+3 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )-a (3 A-3 B+C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )}{2 a^2}-\frac {(A-B+C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {a (5 A-3 B+3 C) \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )-a (3 A-3 B+C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )}{2 a^2}-\frac {(A-B+C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}\) |
-(((A - B + C)*Sin[c + d*x])/(d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x]))) + (a*(5*A - 3*B + 3*C)*((2*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*Sin[c + d *x])/(3*d*Cos[c + d*x]^(3/2))) - a*(3*A - 3*B + C)*((-2*EllipticE[(c + d*x )/2, 2])/d + (2*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]])))/(2*a^2)
3.5.61.3.1 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[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1)) I nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[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_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(a*A - b*B + a*C)*Cos[e + f*x]*(a + b* Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(f*(b*c - a*d)*(2*m + 1))), x ] + Simp[1/(b*(b*c - a*d)*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) + B*(b*c*m + a *d*(n + 1)) - C*(a*c*m + b*d*(n + 1)) + (d*(a*A - b*B)*(m + n + 2) + C*(b*c *(2*m + 1) - a*d*(m - n - 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c ^2 - d^2, 0] && LtQ[m, -2^(-1)]
Leaf count of result is larger than twice the leaf count of optimal. \(466\) vs. \(2(207)=414\).
Time = 8.82 (sec) , antiderivative size = 467, normalized size of antiderivative = 2.83
method | result | size |
default | \(-\frac {\sqrt {-\left (-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\frac {\left (A -B +C \right ) \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}}+2 A \left (-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}}{6 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{2}\right )^{2}}+\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{3 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}}\right )+\frac {\left (2 B -2 A \right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \left (2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}\right )}{a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\) | \(467\) |
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+cos(d*x+c)*a),x,me thod=_RETURNVERBOSE)
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/a*((A-B+C)*(cos (1/2*d*x+1/2*c)*(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))-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)/cos(1/2*d*x+1/2*c)/(- 2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+2*A*(-1/6*cos(1/2*d*x+1 /2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/ 2*c)^2-1/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1) ^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos( 1/2*d*x+1/2*c),2^(1/2)))+(2*B-2*A)/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2 *c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d *x+1/2*c)^2*cos(1/2*d*x+1/2*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)*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 higher order function than in optimal. Order 9 vs. order 4.
Time = 0.12 (sec) , antiderivative size = 347, normalized size of antiderivative = 2.10 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))} \, dx=-\frac {2 \, {\left (3 \, {\left (3 \, A - 3 \, B + C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, A - 3 \, B\right )} \cos \left (d x + c\right ) - 2 \, A\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - {\left (\sqrt {2} {\left (-5 i \, A + 3 i \, B - 3 i \, C\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (-5 i \, A + 3 i \, B - 3 i \, C\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - {\left (\sqrt {2} {\left (5 i \, A - 3 i \, B + 3 i \, C\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (5 i \, A - 3 i \, B + 3 i \, C\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, {\left (\sqrt {2} {\left (-3 i \, A + 3 i \, B - i \, C\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (-3 i \, A + 3 i \, B - i \, C\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, {\left (\sqrt {2} {\left (3 i \, A - 3 i \, B + i \, C\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (3 i \, A - 3 i \, B + i \, C\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{6 \, {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\right )}} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c) ),x, algorithm="fricas")
-1/6*(2*(3*(3*A - 3*B + C)*cos(d*x + c)^2 + 2*(2*A - 3*B)*cos(d*x + c) - 2 *A)*sqrt(cos(d*x + c))*sin(d*x + c) - (sqrt(2)*(-5*I*A + 3*I*B - 3*I*C)*co s(d*x + c)^3 + sqrt(2)*(-5*I*A + 3*I*B - 3*I*C)*cos(d*x + c)^2)*weierstras sPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - (sqrt(2)*(5*I*A - 3*I*B + 3*I*C)*cos(d*x + c)^3 + sqrt(2)*(5*I*A - 3*I*B + 3*I*C)*cos(d*x + c)^2)* weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*(sqrt(2)*(-3 *I*A + 3*I*B - I*C)*cos(d*x + c)^3 + sqrt(2)*(-3*I*A + 3*I*B - I*C)*cos(d* x + c)^2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*(sqrt(2)*(3*I*A - 3*I*B + I*C)*cos(d*x + c)^3 + sqrt (2)*(3*I*A - 3*I*B + I*C)*cos(d*x + c)^2)*weierstrassZeta(-4, 0, weierstra ssPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/(a*d*cos(d*x + c)^3 + a *d*cos(d*x + c)^2)
Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))} \, dx=\text {Timed out} \]
\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )} \cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c) ),x, algorithm="maxima")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)/((a*cos(d*x + c) + a)*co s(d*x + c)^(5/2)), x)
\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )} \cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c) ),x, algorithm="giac")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)/((a*cos(d*x + c) + a)*co s(d*x + c)^(5/2)), x)
Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{{\cos \left (c+d\,x\right )}^{5/2}\,\left (a+a\,\cos \left (c+d\,x\right )\right )} \,d x \]