Integrand size = 35, antiderivative size = 159 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+a \cos (c+d x)} \, dx=\frac {3 (5 A+7 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a d}-\frac {(3 A+5 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a d}-\frac {(3 A+5 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a d}+\frac {(5 A+7 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 a d}-\frac {(A+C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))} \]
3/5*(5*A+7*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(si n(1/2*d*x+1/2*c),2^(1/2))/a/d-1/3*(3*A+5*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/c os(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))/a/d+1/5*(5*A+7*C)* cos(d*x+c)^(3/2)*sin(d*x+c)/a/d-(A+C)*cos(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*c os(d*x+c))-1/3*(3*A+5*C)*sin(d*x+c)*cos(d*x+c)^(1/2)/a/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 8.26 (sec) , antiderivative size = 934, normalized size of antiderivative = 5.87 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+a \cos (c+d x)} \, dx=\frac {\cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\cos (c+d x)} \left (-\frac {2 (5 A+5 C+10 A \cos (c)+16 C \cos (c)) \csc (c)}{5 d}-\frac {4 C \cos (d x) \sin (c)}{3 d}+\frac {2 C \cos (2 d x) \sin (2 c)}{5 d}-\frac {2 \sec \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \left (A \sin \left (\frac {d x}{2}\right )+C \sin \left (\frac {d x}{2}\right )\right )}{d}-\frac {4 C \cos (c) \sin (d x)}{3 d}+\frac {2 C \cos (2 c) \sin (2 d x)}{5 d}\right )}{a+a \cos (c+d x)}+\frac {A \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec \left (\frac {c}{2}\right ) \sec (d x-\arctan (\cot (c))) \sqrt {1-\sin (d x-\arctan (\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \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 ^2(c)}}+\frac {5 C \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec \left (\frac {c}{2}\right ) \sec (d x-\arctan (\cot (c))) \sqrt {1-\sin (d x-\arctan (\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\arctan (\cot (c)))} \sqrt {1+\sin (d x-\arctan (\cot (c)))}}{3 d (a+a \cos (c+d x)) \sqrt {1+\cot ^2(c)}}-\frac {3 A \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1-\cos (d x+\arctan (\tan (c)))} \sqrt {1+\cos (d x+\arctan (\tan (c)))} \sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{2 d (a+a \cos (c+d x))}-\frac {21 C \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1-\cos (d x+\arctan (\tan (c)))} \sqrt {1+\cos (d x+\arctan (\tan (c)))} \sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{10 d (a+a \cos (c+d x))} \]
(Cos[c/2 + (d*x)/2]^2*Sqrt[Cos[c + d*x]]*((-2*(5*A + 5*C + 10*A*Cos[c] + 1 6*C*Cos[c])*Csc[c])/(5*d) - (4*C*Cos[d*x]*Sin[c])/(3*d) + (2*C*Cos[2*d*x]* Sin[2*c])/(5*d) - (2*Sec[c/2]*Sec[c/2 + (d*x)/2]*(A*Sin[(d*x)/2] + C*Sin[( d*x)/2]))/d - (4*C*Cos[c]*Sin[d*x])/(3*d) + (2*C*Cos[2*c]*Sin[2*d*x])/(5*d )))/(a + a*Cos[c + d*x]) + (A*Cos[c/2 + (d*x)/2]^2*Csc[c/2]*Hypergeometric PFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[d*x - Arc Tan[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]) + (5*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]*Se c[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]]]])/(3*d*(a + a*Cos[c + d*x])*Sqrt[1 + Cot[c]^2]) - (3*A*Co s[c/2 + (d*x)/2]^2*Csc[c/2]*Sec[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]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos [c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - (( Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]* Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(2*d*(a + a*Cos[c + d*x...
Time = 0.68 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.93, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3521, 27, 3042, 3227, 3042, 3115, 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.
| \(\displaystyle \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a \cos (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{a \sin \left (c+d x+\frac {\pi }{2}\right )+a}dx\) |
| \(\Big \downarrow \) 3521 |
| \(\displaystyle \frac {\int -\frac {1}{2} \cos ^{\frac {3}{2}}(c+d x) (a (3 A+5 C)-a (5 A+7 C) \cos (c+d x))dx}{a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \cos ^{\frac {3}{2}}(c+d x) (a (3 A+5 C)-a (5 A+7 C) \cos (c+d x))dx}{2 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a (3 A+5 C)-a (5 A+7 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{2 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle -\frac {a (3 A+5 C) \int \cos ^{\frac {3}{2}}(c+d x)dx-a (5 A+7 C) \int \cos ^{\frac {5}{2}}(c+d x)dx}{2 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {a (3 A+5 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx-a (5 A+7 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}dx}{2 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle -\frac {a (3 A+5 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-a (5 A+7 C) \left (\frac {3}{5} \int \sqrt {\cos (c+d x)}dx+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )}{2 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {a (3 A+5 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-a (5 A+7 C) \left (\frac {3}{5} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )}{2 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle -\frac {a (3 A+5 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-a (5 A+7 C) \left (\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )}{2 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle -\frac {a (3 A+5 C) \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-a (5 A+7 C) \left (\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )}{2 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\) |
-(((A + C)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x]))) - (a *(3*A + 5*C)*((2*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*Sqrt[Cos[c + d*x]]* Sin[c + d*x])/(3*d)) - a*(5*A + 7*C)*((6*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*d)))/(2*a^2)
3.2.52.3.1 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[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[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_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[a*(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)) I nt[(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)) - C*(a*c*m + b*d*(n + 1)) + (a*A*d*(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, 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)]
Time = 6.74 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.74
| method | result | size |
| default | \(\frac {\sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (15 A F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+45 A E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+25 C F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+63 C E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )-48 C \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+56 C \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (30 A +30 C \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-15 A -23 C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{15 a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\) | \(276\) |
1/15*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(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)*(15*A *EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+45*A*EllipticE(cos(1/2*d*x+1/2*c),2 ^(1/2))+25*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+63*C*EllipticE(cos(1/2* d*x+1/2*c),2^(1/2)))-48*C*sin(1/2*d*x+1/2*c)^8+56*C*sin(1/2*d*x+1/2*c)^6+( 30*A+30*C)*sin(1/2*d*x+1/2*c)^4+(-15*A-23*C)*sin(1/2*d*x+1/2*c)^2)/a/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)/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 = 263, normalized size of antiderivative = 1.65 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+a \cos (c+d x)} \, dx=\frac {2 \, {\left (6 \, C \cos \left (d x + c\right )^{2} - 4 \, C \cos \left (d x + c\right ) - 15 \, A - 25 \, C\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 5 \, {\left (\sqrt {2} {\left (-3 i \, A - 5 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-3 i \, A - 5 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 \, {\left (\sqrt {2} {\left (3 i \, A + 5 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (3 i \, A + 5 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 9 \, {\left (\sqrt {2} {\left (-5 i \, A - 7 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-5 i \, A - 7 i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 9 \, {\left (\sqrt {2} {\left (5 i \, A + 7 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (5 i \, A + 7 i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{30 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \]
1/30*(2*(6*C*cos(d*x + c)^2 - 4*C*cos(d*x + c) - 15*A - 25*C)*sqrt(cos(d*x + c))*sin(d*x + c) - 5*(sqrt(2)*(-3*I*A - 5*I*C)*cos(d*x + c) + sqrt(2)*( -3*I*A - 5*I*C))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 5*(sqrt(2)*(3*I*A + 5*I*C)*cos(d*x + c) + sqrt(2)*(3*I*A + 5*I*C))*weie rstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 9*(sqrt(2)*(-5*I*A - 7*I*C)*cos(d*x + c) + sqrt(2)*(-5*I*A - 7*I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 9*(sqrt(2)*(5 *I*A + 7*I*C)*cos(d*x + c) + sqrt(2)*(5*I*A + 7*I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/(a*d*cos(d* x + c) + a*d)
Timed out. \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+a \cos (c+d x)} \, dx=\text {Timed out} \]
\[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+a \cos (c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{\frac {3}{2}}}{a \cos \left (d x + c\right ) + a} \,d x } \]
\[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+a \cos (c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{\frac {3}{2}}}{a \cos \left (d x + c\right ) + a} \,d x } \]
Timed out. \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+a \cos (c+d x)} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^{3/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )}{a+a\,\cos \left (c+d\,x\right )} \,d x \]